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Olympiads in Informatics, 2016, Vol. 10, 87–98
© 2016 IOI, Vilnius University
DOI: 10.15388/ioi.2016.06
87
Understanding Unsolvable Problem
Jonathan Irvin GUNAWAN
Undergraduate Student
School of Computing, National University of Singapore
Computing 1, 13 Computing Drive, Singapore 117417
e-mail: [email protected]
Abstract. In recent IOIs, there are several problems that seem unsolvable, until we realise that
there is a special case to the problem that makes it tractable. In IOI 2014, the problem ‘Friend’
appears to be a standard NP-hard Maximum Independent Set problem. However, the graph is generated in a very special way, hence there is a way to solve the problem in polynomial time. There
were several contestants who didn’t identify the special case in this problem, and hence were stuck
at the problem. In this paper, we will study a well-known technique called reduction to show that
a problem we are currently tackling is intractable. In addition, we introduce techniques to identify
special cases such that contestants will be prepared to tackle these problems.
Keywords: special case, unsolvable, NP-hard.
1. Introduction
The problem ‘Friend’ in IOI 2014 required contestants to find a set of vertices with maximum total weight, such that no two vertices in the set are sharing a common edge. This
is a classical Weighted Maximum Independent Set problem. We can show that Weighted Maximum Independent Set problem is NP-hard by reduction from 3-SAT (Cormen
et al., 2009). Since the formulation of NP-completeness 4 decades ago, no one has been
able to propose a solution to any NP-hard problem in polynomial time. Clearly, it is not
expected that a high school student can solve the problem in 5 hours. None of the Indonesian IOI 2014 team solved this problem during the contest. After returning from the
competition, I asked the Indonesian team about this problem. None of the team members
were aware of the fact that Maximum Independent Set is an NP-hard problem, and thus
were stuck trying to solve a general Maximum Independent Set problem.
A similar problem also occurred in IOI 2008. The problem ‘Island’ required contestants to find a longest path in a graph with 1,000,000 vertices. The longest path problem
is a classic NP-hard problem which can be reduced from the Hamiltonian path problem.
If a contestant is not aware that the longest path problem is difficult to solve, the contestant may spend a lot of his/her time just to tackle the general longest path problem,
without realising that there is a special case to the given graph.
88
J.I. Gunawan
Generally, some contestants spend too much thinking time trying to solve something
that is believed to be unsolvable. If only they realise that their attempt is intractable,
they may try a different approach and find a special case of this problem. In section 2
of this paper, we will introduce a classic reduction technique often used in theoretical
computer science research. In the context of competitive programming, we may find out
that a problem which we are attempting is unlikely to be solvable. After realizing that
a problem is intractable, we are going to discuss how to proceed to solve the problem
in section 3. Finally, in section 4 we will take a look at some common special cases in
competitive programming that can be used to solve these kind of problems.
Fig. 1. The result of Indonesian team in IOI 2014, taken from http://stats.ioinformatics.org. The red squared column highlights the ‘Friend’ problem.
Fig. 2. IOI 2014 tasks statistics, taken from http://stats.ioinformatics.org. ‘Friend’
problem is the second least accepted problem in IOI 2014. It may be because some contestants (at least all the Indonesians) were stuck at trying to solve a general case of Maximum
Independent Set.
Fig. 3. The result of Indonesian team in IOI 2008, taken from http://stats.ioinformatics.org. The red squared column highlights the ‘Island’ problem.
89
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apply
amaximum
well-known
technique
called
reduction.
Suppose
we
most
common
issolution.
to
apply
aV
well-known
technique
called
Suppose
we
know
that
ry
edge,
at
one
of
its
endpoint
in
the
subset.
is
amost
graph
problem
of
finding
subset
of
such
that
for
every
edge,
most
graph
graph
problem
problem
of
of
finding
ais
athe
maximum
maximum
subset
subset
of
of
nodes
nodes
such
graph
that
that
problem
for
for
every
ofaanodes
edge,
finding
edge,
at
atendpoint
aan
most
maximum
one
one
ofsubset
its
its
endpoint
endpoint
of
nodes
isknow
is in
su
in
MAX-INDEPENDENT-SET
problem.
V,
E)
aINDEPENDENT-SET
VERTEX-COVER
of
the
graph
graph
G(V,
G(V,
E)
of
finding
aleast
maximum
subset
of
nodes
such
that
for
every
edge,
at
most
one
ofaisreduction.
its
is of
in
thewell.
subset.
Suppose
we
already
know
that
MIN-VERTEX-COVER
is
NP-hard
problem.
Therefore,
we
can
sh
the
subset.
Suppose
we
already
know
that
MIN-VERTEX-COVER
is
NP-hard
problem.
Therefore,
we
can
sh
the
subset.
Suppose
we
already
know
that
MIN-VERTEX-COVER
a
NP-hard
problem.
Therefore,
we
graph
G(V,
E)
ing
Intractability
of
a
Problem
through
Reduction
em
X
as
Therefore,
problem
Y
is
also
impossible
to
solve.
most
common
way
is
to
apply
well-known
technique
called
reduc
Proof.
Proof.
Let
us
us
assume
assume
that
that
VSuppose
Vis
−
−in
SSXthe
is
is
not
not
aaby
VERTEX-COVER.
VERTEX-COVER.
Ther
Ther
of
nodes
such
that
for
every
edge,
at
least
one
of
itsthat
endpoint
subset.
MAX-INDEPENDENT
of
graph
G(V,
E),
then
Vthat
−
SMIN-VERTEX-COVER
is
anot
VERTEX-COVER
of
the
Proof.
Let
us
assume
that is
aaat
VERTEX-COVER.
Therefore,
there
are
two
common
way
is
to
apply
athat
well-known
called
reduction.
we
know
that
problem
XA,
is
dmost
reduction.
Suppose
we
know
problem
Xis
isnot
impossible
to
solve,
and
we
also
know
that
we
can
solve
problem
X
using
problem
Y
as
aBb∈
impossible
to
solve,
and
we
also
know
that
we
can
solve
using
problem
Y
as
acan
black-bo
aENDENT-SET
maximum
subset
of
nodes
such
that
for
every
edge,
most
one
of
endpoint
is
in
the
subset.
we
already
know
MIN-VERTEX-COVER
is
atwo
NP-hard
problem.
the
thewe
subset.
subset.
Suppose
Suppose
we
we
already
already
know
know
that
MIN-VERTEX-COVER
the
subset.
is
Suppose
a−
aproblem
NP-hard
NP-hard
we
problem.
problem.
already
know
Therefore,
that
MIN-VERTEX
we
we
can
show
show
Proof.
Let
us
that
VSuppose
−
S
VERTEX-COVER.
Therefore,
there
are
B
/
Proof.
Let
us
assume
that
V
−
S technique
is
not
aLet
VERTEX-COVER.
there
are
two
vertices
A,
Let
us
assume
that
Visits
STherefore,
isTherefore,
not
aby
VERTEX-COVER.
Therefore,
that
are
attempting
isassume
unlikely
to
have
an
solution.
The
ppose
we
already
know
that
MIN-VERTEX-COVER
is
NP-hard
problem.
can
show
that
MAX-INDEPENDENT-SET
isProof.
also
aand
NP-hard
problem
by
reducing
MIN-VERTEX-COVER
problem
int
Proof.
Let
us
assume
that
VMIN-VERTEX-COVER
−
STherefore,
is
not
a1vertices
VERTEX-CO
that
MAX-INDEPENDENT-SET
is
also
NP-hard
problem
by
reducing
aaknow
MIN-VERTEX-COVER
problem
int
that
MAX-INDEPENDENT-SET
isaimmediate
also
aSaS
NP-hard
problem
by
reducing
awe
probl
impossible
to
solve,
and
we
also
that
we
can
solve
problem
Xend
by
1
et
us
assume
that
V
−
S
is
not
a
VERTEX-COVER.
Therefore,
there
are
two
vertices
A,
B
∈
/
V
−
S
Proof.
Proof.
Let
Let
us
us
assume
assume
that
that
V
V
−
−
is
is
not
not
a
a
VERTEX-COVER.
VERTEX-COVER.
Therefore,
Therefore,
there
there
are
are
two
two
vertic
vertic
there
are
two
vertices
A,
B
∈
/
V
−
S
is
not
a
VERTEX-COVER.
Therefore,
there
are
two
vertices
A,
B
∈
/
V
−
S
and
there
there
is
is
an
an
edge
edge
connecting
connecting
A
A
and
and
B.
B.
Since
Since
A,
A,
B
B
∈
/
∈
/
V
V
−
−
S,
S,
we
we
graph
problem
of
finding
a
maximum
subset
of
nodes
such
that
for
every
edge,
at
most
one
of
its
vertices and
there
is
an
edge
connecting and .
Since impossible
to
solve,
and
we
also
know
that
we
can
solve
problem
X
by
using
problem
Y
as
a
black-box
.
If
we
can
m
X
by
using
problem
Y
as
a
black-box
.
If
we
can
solve
problem
Y,
then
we
can
solve
problem
X
as
well.
Therefore,
problem
Y
is
also
impossib
ssume
that
V
−
S
is
not
a
VERTEX-COVER.
Therefore,
there
are
two
vertices
A,
B
∈
/
V
−
S
solve
problem
Y,
then
we
can
solve
problem
X
as
well.
Therefore,
problem
Y
is
also
impossible
to
so
ready
know
that
MIN-VERTEX-COVER
is
a
NP-hard
problem.
Therefore,
we
can
show
that
MAX-INDEPENDENT-SET
is
also
a
NP-hard
problem
by
reducing
a
MIN-VERTEX
that
that
MAX-INDEPENDENT-SET
MAX-INDEPENDENT-SET
is
is
also
also
a
a
NP-hard
NP-hard
problem
problem
that
by
by
MAX-INDEPENDENT-SET
reducing
reducing
a
a
MIN-VERTEX-COVER
MIN-VERTEX-COVER
is
also
a
problem
problem
NP-hard
into
into
proble
a
a
and
there
is
an
edge
connecting
A
and
B.
Since
A,
B
∈
/
V
−
S,
we
have
A,
B
∈
S.
As
A
and
B
are
co
We
will
first
prove
the
following
two
lemmas.
Proof.
Let
us
assume
that
V
−
S
is
not
a
VERTEX-COVER.
Therefore,
there
are
two
vertice
and
there
is
an
edge
connecting
A
and
B.
Since
A,
B
∈
/
V
−
S,
we
have
A,
B
∈
S.
As
A
and
B
are
c
and
is
an
edge
connecting
A
and
B.
A,A
Band
∈
/a VB.−Since
S, weA,
have
ow
that
the
problem
that
we
are
attempting
isB
unlikely
to
have
an
immediate
solution.
The
nown
technique
called
reduction.
Suppose
wethere
know
that
problem
XBis
isA,
DEPENDENT-SET
is∈
also
athere
NP-hard
problem
by
reducing
atwo
MIN-VERTEX-COVER
problem
into
−
is
not
VERTEX-COVER.
Therefore,
there
are
two
vertices
A,
B
/is
−
SSince
VV
−
SSMAX-INDEPENDENT-SET
is
not
aathat
VERTEX-COVER.
Therefore,
there
are
A,
B
//∈
VVA
−
S
MAX-INDEPENDENT-SET
problem.
and
there
anS.
edge
connecting
B
∈
/A
problem.
Therefore,
there
are
two
vertices
A,
∈
/be
V
−
S
MAX-INDEPENDENT-SET
problem.
stration.
Recall
NP-hard
problems
have
yet
solved
in
polynomial
solve
problem
Y,vertices
then
we
can
solve
problem
X
as
well.
Therefore,
pro
etCOVER.
isBMAX-INDEPENDENT-SET
an
edge
connecting
A
and
B.
Since
A,
B
∈
/to
−
S,
we
have
A,
∈
As
B
are
connected
two
vertices
A,
B
/and
Vimpossible
−
S
and
there
isproblem.
isto
an
an
edge
edge
connecting
connecting
A
A
and
and
B.
B.
Since
Since
A,
B
B
∈
/∈
∈
Vby
V
−
−
S,
S,
we
we
have
have
A,
A,
B
B ∈∈Therefore
S.
S. As
As A
AThis
and
and
A,
∈B.
S.
As
A
and
B
are
connected
and
Since
A,
B
∈
/and
Vwe
−
S,
we
have
A,
B
∈
S.
As
A
B
are
connected
by
by
an
an
edge,
edge,
we
we
note
note
that
that
S
S
not
not
an
an
INDEPENDENT-SET.
INDEPENDENT-SET.
Thi
,note
we
have .V
As and
and Since
are
connected
an
edge,
we
note
the
subset.
Suppose
we
already
that
MIN-VERTEX-COVER
isinto
aand
NP-hard
problem.
Therefore,
we
solve
problem
Y,
then
can
solve
problem
Xknow
as
well.
Therefore,
problem
Y
isis
also
impossible
to
solve.
ore,
problem
Y
isNP-hard
also
solve.
edge
connecting
A
B.
Since
A,
B
∈
/
V
−
S,
we
have
A,
∈
S.
As
A
and
B
are
connected
T-SET
is
also
a
problem
by
reducing
a
MIN-VERTEX-COVER
problem
a
MAX-INDEPENDENT-SET
problem.
MAX-INDEPENDENT-SET
problem.
MAX-INDEPENDENT-SET
problem.
1 B that
by
an
edge,
we
that
S
is
not
an
INDEPENDENT-SET.
This
is
a
contradiction.
V
−
there
is
an
edge
connecting
A
and
B.
A,
B
∈
/
V
−
S,
we
have
A,
B
∈
S.
As
A
and
B
by
an
edge,
we
note
that
S
is
not
an
INDEPENDENT-SET.
This
is
a
contradiction.
Therefore
V
by
an
edge,
we
note
S
is
not
an
INDEPENDENT-SET.
This
is
a
c
is
to
apply
a
well-known
technique
called
reduction.
Suppose
we
know
that
problem
X
is
at
we
can
solve
problem
X
by
using
problem
Y
as
a
black-box
.
If
we
can
NDENT-SET
problem.
ecting
Aand
and
B.
Since
A,
B
/This
VVINDEPENDENT-SET.
−
S,
we
have
A,
B
∈
S.
As
A
and
BVSare
are
connected
tifying
Intractability
of
aare
Problem
Reduction
ecting
and
B.
A,
∈
/∈
V
−
we
A,
B
S.
As
A
and
B
byaisINDEPENDENT-SET.
an
edge,
we
note
that
SVMIN-VERTEX-COVER
is−isisnot
an
INDEPENDENT
stge,
not
aA
VERTEX-COVER.
Therefore,
there
two
vertices
A,
B
∈
/−
−is
Sconnected
∈
/As
Vwe
S,
we
have
∈VB
S.
As
A
and
Bhave
are
connected
TEX-COVER
is
aSince
graph
problem
that
finding
aanot
minimum
subset
Lemma
If
SWe
⊆
is
INDEPENDENT-SET
of
graph
G(V,
E),
then
is
aa VERTEX-COVE
note
that
SA,
is
not
an
This
athrough
contradiction.
Therefore
VSis
S
isyethave
a to be
A−
B
are
connected
by
byB
an
an
edge,
edge,
we
we
note
note
that
that
S
S
isis∈
not
an
an
INDEPENDENT-SET.
This
This
aa −
contradiction.
contradiction.
Ther
The
that 1.
is
not
an
INDEPENDENT-SET.
This
is
contradiction.
Therefore is
contradiction.
Therefore
−
S
is
aan
an
INDEPENDENT-SET.
isS,
ainvolves
contradiction.
Therefore
V
areducing
VERTEX-COVER.
VERTEX-COVER.
that
MAX-INDEPENDENT-SET
is
also
NP-hard
problem
by
aproblems
prob
are
using
NP-hard
problems
for
illustration.
Recall
that
NP-hard
problems
yet
to
be
note
that
S
is
not
an
INDEPENDENT-SET.
This
is
a
contradiction.
Therefore
V
−
S
a
We
are
using
NP-hard
problems
for
illustration.
Recall
that
NP-hard
have
solved
T
problem.
1 This
VERTEX-COVER.
by
an
edge,
we
note
that
S
is
not
an
INDEPENDENT-SET.
is
a
contradiction.
There
VERTEX-COVER.
VERTEX-COVER.
we
also
know
can
solve
problem
using
problem
Y connected
asisaablack-box
.is
we can
em
X
as
well.
Therefore,
problem
Ythe
isThis
also
impossible
is
not
an
INDEPENDENT-SET.
This
is
contradiction.
Therefore
− SS for
isIfaaillustration.
SS one
is
an
INDEPENDENT-SET.
is
contradiction.
Therefore
VV −
We
will
first
prove
the
following
two
lemmas.
VERTEX-COVER.
and
B.not
Since
A,
Bathat
∈
/contradiction.
VSiswe
−
S,
we
have
A,
Bproblem.
∈XaaS.
As
and
B are
We
will
first
prove
the
following
two
lemmas.
NT-SET.
This
isWe
Therefore
Vby
−
SA
isto
asolve.
will
first
prove
following
two
lemmas.
st
of
its
endpoint
in
subset.
MAX-INDEPENDENT-SET
We
are
using
NP-hard
problems
Recall
that NP-hard
VERTEX-COVER.
graph
G(V,
E)
-COVER.
tion.
Therefore
V
−
is
athe
VERTEX-COVER.
VERTEX-COVER.
MAX-INDEPENDENT-SET
Wecan
are
using
NP-hard
problems
illustration.
Recall
that
NP-hard
problems
have
yet
to
be solved
in involves
polynomial
NP-hard
problems
have
yet
to
be
solved
in
polynomial
time
for
more
than
4lemmas.
decades.
MIN-VERTEX-COVER
issolve.
a problem
graph
problem
that
findin
ER.
time
for
more
than
4are
decades.
MIN-VERTEX-COVER
is
aimmediate
graph
that
involves
finding
min
We
will
first
prove
the
following
two
lemmas.
We
We
will
will
first
first
prove
prove
the
the
following
following
two
two
lemmas.
We
will
first
prove
the
following
two
lemmas.
VERTEX-COVER.
to
know
that
the
problem
that
weisfor
attempting
is unlikely
to
have
an
solution.
en
we
solve
problem
X
as
well.
Therefore,
problem
Y
is
also
impossible
to
Lemma
Lemma
2.
2.
If
If
S
S
⊆
⊆
V
V
is
is
a
a
VERTEX-COVER
VERTEX-COVER
of
ofThe
graph
graph
G(V,
G(V,
E),
E),a prob
then
the
t
prove
the
following
two
lemmas.
an
INDEPENDENT-SET.
This
a
contradiction.
Therefore
V
−
S
is
a
bset
of
nodes
such
that
for
every
edge,
at
most
one
of
its
endpoint
is
in
time
for
more
than
4
decades.
MIN-VERTEX-COVER
is
a graph
Lemma
2.
If
S
⊆
V
is
a
VERTEX-COVER
of
graph
G(V,
E),
then
V
−
S
is
an
INDEPENDENT-SE
Lemma
2.
If
S
⊆
V
is
a
VERTEX-COVER
of
graph
G(V,
E),
then
V
−
S
is
an
INDEPENDENT-S
Lemma
2.
If
S
⊆
V
is
a
VERTEX-COVER
of
graph
G(V,
E),
then
V of
−
time
for
more
than
4
decades.
MIN-VERTEX-COVER
is
a
graph
problem
that
involves
finding
a
minimum
subset
ph
problem
that
involves
finding
a
minimum
subset
of
nodes
such
that
for
every
edge,
at
least
one
of
its
endpoint
is
in
the
subset.
MAX-INDEP
Lemma
1.
If
S
⊆
V
is
an
INDEPENDENT-SET
of
graph
G(V,
E),
then
V
−
S
is
a
VERTEX-COVER
of
Lemma
2.
If
S
⊆
V
is
a
VERTEX-COVER
of
graph
G∈
of
nodes
such
that
for
every
edge,
at
least
one
of
its
endpoint
is
in
the
subset.
MAX-INDEPENDE
following
two
lemmas.
Lemma
1.
If
S
⊆
V
is
an
INDEPENDENT-SET
of
G(V,
E),
then
V
−
S
is
a
VERTEX-COVER
Lemma
2.
If is
a
VERTEX-COVER
graph
,
then an
INLemma
1.
If
S
⊆
V
is
an
INDEPENDENT-SET
graph
G(V,
E),
then
V
−
S
is
a
VERTEX-COV
nX-COVER
way
to
apply
aLet
well-known
technique
called
reduction.
Suppose
we
know
that
problem
X SS
isofisisthe
Proof.
usE),
assume
that
VS
is
not
aG(V,
VERTEX-COVER.
Therefore,
there
two
vertices
A,
B
2.
IfVan
Sis
⊆
V
is
a SS
VERTEX-COVER
graph
E),
then
Vthat
−
Sof
is
an
INDEPENDENT-SET
Lemma
Lemma
2.
2.
IfIfthe
SSof
⊆
⊆
VV−
is
isS
aaG(V,
VERTEX-COVER
VERTEX-COVER
of
graph
graph
G(V,
G(V,
E),
E),
then
then
VVare
−
−
an
anVINDEPEND
INDEPEND
stration.
Recall
that
NP-hard
problems
have
yet
to
be
solved
in
polynomial
S
isLemma
INDEPENDENT-SET
of
of
graph
G(V,
then
V
−
is
an
INDEPENDENT-SET
of
the
graph
graph
G(V,
E)
E)
at
MIN-VERTEX-COVER
a
NP-hard
problem.
Therefore,
we
can
show
of
nodes
such
for
every
edge,
at
least
one
of
its
endpoint
is
in
t
⊆
is
a
VERTEX-COVER
of
graph
G(V,
E),
then
V
−
S
is
an
INDEPENDENT-SET
of
the
Lemma
1.
If
S
⊆
V
is
an
INDEPENDENT-SET
of
graph
G(V,
E),
then
−
S
is
a
Lemma
1.
1.
Lemma
1.
If
If
⊆
⊆
V
V
is
is
an
an
INDEPENDENT-SET
INDEPENDENT-SET
of
of
graph
graph
G(V,
G(V,
E),
E),
If
then
then
S
⊆
V
V
−
−
is
S
S
an
is
is
INDEPENDENT-SET
a
a
VERTEX-COVER
VERTEX-COVER
of
of
of
the
the
gra
graph
G(V,
E)
Lemma
2.
If
S
⊆
V
is
a
VERTEX-COVER
of
graph
G(V,
E),
then
V
−
S
is
an
INDEPENDE
graph
G(V,
E)
V
graph
G(V,
E)
We
will
first
prove
the
following
two
lemmas.
1
of
nodes
such
that
for
every
edge,
at
least
one
of
its
endpoint
is
in
the
subset.
MAX-INDEPENDENT-SET
is
a
t-hard
is
in
the
subset.
MAX-INDEPENDENT-SET
is
a
S
⊆
V
is
an
INDEPENDENT-SET
of
graph
G(V,
E),
then
V
−
S
is
a
VERTEX-COVER
of
the
graph
problem
of
finding
a
maximum
subset
of
nodes
such
that
for
every
edge,
at
most
one
VERTEX-COVER
of
graph
G(V,
E),
then
V
−
S
is
an
INDEPENDENT-SET
of
the
VERTEX-COVER
of
graph
G(V,
E),
then
Vmaximum
−
Sproblems
isB.
an
INDEPENDENT-SET
ofhave
the
graph
G(V,
E)
DEPENDENT-SET
of
the
graph graph
G(V,
E)
graph
problem
of
finding
aproblem
of
nodes
for
every
at
most
oneBofare
itscoeV
graph
G(V,
E)
G(V,
E),
then
Vathere
−
S
isthat
anG(V,
INDEPENDENT-SET
ofby
the
graph
G(V,
E)
.subset
solve,
and
we
also
know
we
can
solve
X
using
problem
Ysuch
as
athat
black-box
.B
Ifedge,
weS.can
and
is
an
edge
connecting
A
and
Since
A,
B
∈
/
V
−
S,
we
A,
∈
As
A
and
V,
E)
problems
for
illustration.
Recall
that
NP-hard
have
yet
to
be
solved
in
polynomial
NDEPENDENT-SET
of
the
graph
graph
G(V,
E)
E)
TEX-COVER
is
graph
problem
that
involves
finding
a
minimum
subset
a
NP-hard
problem
by
reducing
a
MIN-VERTEX-COVER
problem
into
a
graph
of is
finding
maximum
subset
nodes
such
that
for
e
ngraph
INDEPENDENT-SET
ofaG(V,
graph
G(V,
E),
then
V −problem
SThe
is proof
agraph
VERTEX-COVER
of
the
graph
E)
graph
graph
G(V,
G(V,
E)
E)
G(V,
E)athe
graph
E)its
problem
of
finding
maximum
subset
ofis
nodes
such
that
for
every
edge,
atto
one
endpoint
is
for
every
edge,
at
most
one
endpoint
in
Proof.
Proof.
The
proof
is
actually
actually
similar
similar
to
to
the
the
previous
previous
lemma.
lemma.
Let
Let
us
usinassu
assu
the
subset.
we
already
know
that
MIN-VERTEX-COVER
is−
aof
NP-hard
problem.
The
the
subset.
Suppose
we
already
know
that
MIN-VERTEX-COVER
ismost
athat
NP-hard
problem.
Therefore,
X-COVER
ofMIN-VERTEX-COVER
graph
G(V,
E),
then
VSuppose
S
is
an
INDEPENDENT-SET
Y,decades.
then
can
solve
problem
X
as
well.
Therefore,
problem
Y
isof
also
impossible
solve.
Proof.
The
is
actually
similar
to
the
previous
lemma.
Let
us
assume
that
V
Sof
not
aVERTEX-COV
INDEPENDE
by
an
edge,
we
note
S
is
not
an
INDEPENDENT-SET.
This
is
athat contradiction.
Therefore
V
−
4at
is
graph
problem
that
involves
finding
aus
minimum
subset
Proof.
The
proof
isof
actually
similar
to
the
previous
lemma.
Let
assume
Sis its
is
not
avertices
INDEPEND
st
one
ofwe
its
endpoint
is
in
subset.
MAX-INDEPENDENT-SET
issimilar
aof
Proof.
The
proof
isSuppose
actually
to
the
previous
lemma.
Let
us
assume
tha
Lemma
1.proof
If the
S
⊆
Vthat
isa−
an
INDEPENDENT-SET
graph
G(V,
E),
then
VV
−
is
aprevious
Proof.
The
proof
is
actually
similar
to
the
previous
lemma.
Let
us
assume
is
the
subset.
we
already
know
that
MIN-VERTEX-COVER
is
Proof.
Let
us
assume
that
V
−
S
is
not
a
VERTEX-COVER.
Therefore,
there
are
two
vertices
A,
B
∈
/
V
−
Proof.
The
proof
is
actually
similar
to
the
lemma.
Proof.
Let
us
assume
that
V
−
S
is
not
a
VERTEX-COVER.
Therefore,
there
are
two
vertices
A,
B
∈
/
V
−
Proof.
Let
us
assume
that
V
−
S
is
not
a
VERTEX-COVER.
Therefore,
there
are
two
A,
B
proof
actually
similar
to
the
previous
lemma.
Let
usa assume
that
V
−
S
isproblem
not
aA,
INDEPENDENT-SET.
Proof.
Proof.
The
The
proof
proof
is
isus
actually
actually
similar
similar
to
to
the
the
previous
previous
lemma.
lemma.
Let
Let
us
us
assume
assume
that
V
V−
−
SSthere
isisis
not
not
aa−
IND
IND
the
subset.
Suppose
we
already
know
that
MIN-VERTEX-COVER
isTherefore,
alemma.
NP-hard
problem.
we
can
show
VER
isSprevious
aisis
NP-hard
problem.
Therefore,
we
can
show
the
Visthe
−
not
aus
INDEPENDENT-SET.
to
lemma.
Let
us
assume
that
VLet
−aA,
SA,
isassume
INDEPENDENT-SET.
Therefore,
Therefore,
there
there
are
two
two
vertices
vertices
A,
B
B
∈reducing
∈Vtwo
VaVisTherefore,
−
−
Sthat
SA
and
there
there
is
an
an
edge
edge
that
MAX-INDEPENDENT-SET
also
aare
NP-hard
by
aand
MIN-VERTEX-CO
that
MAX-INDEPENDENT-SET
isnot
also
aisthe
NP-hard
problem
by
reducing
MIN-VERTEX-COVER
pr∈
of
actually
similar
to
the
previous
lemma.
us
that
V
−
S
is
not
a
INDEPENDENT-SET.
Proof.
Let
assume
that
V
−
S
is
not
a
VERTEX-COVER.
Therefore,
are
two
Proof.
Proof.
Proof.
Let
Let
us
assume
assume
that
that
V
V
−
−
S
S
is
is
not
not
a
VERTEX-COVER.
VERTEX-COVER.
Therefore,
Let
us
assume
there
there
that
are
are
two
−
S
vertices
vertices
is
not
A,
A,
a
VERTEX-CO
B
B
∈
/
∈
/
V
V
−
S
S
Therefore,
there
are
two
vertices
B
∈
V
−
S
and
there
is
an
edge
connecting
A
and
B.
Since
A,
B
VERTEX-COVER.
for
every
edge,
at
least
one
of
its
endpoint
is
in
the
subset.
MAX-INDEPENDENT-SET
a
Proof.
The
proof
is
actually
similar
to
previous
Let
us
assume
that
V
−
S
is
not
a
INDE
Therefore,
there
are
two
vertices
B
∈
V
−
S
and
there
is
an
edge
connecting
and
B.
Since
A,
B
bset
of
nodes
such
that
for
every
edge,
at
most
one
of
its
endpoint
is
in
Therefore,
there
are
vertices
A,
B
∈
V
−
S
and
there
is
an
edge
conne
not
a
INDEPENDENT-SET.
Therefore,
there
two
vertices and
graph
G(V,
E)
that
Vprevious
−
San
is
not
aconnecting
VERTEX-COVER.
Therefore,
there
are
two
vertices
A,
B
∈
/NP-hard
VB
ysassume
similar
to
the
previous
lemma.
Let
us
assume
that
−B.
SA,
is
not
aA,
INDEPENDENT-SET.
that
MAX-INDEPENDENT-SET
isA,
also
aS.
problem
by
reduc
yye,
similar
to
lemma.
Let
assume
that
VV
−
S
is
not
INDEPENDENT-SET.
and
there
is
an
edge
connecting
Aatwo
and
B.
Since
A,
B
V
−
S,
we
have
A,
B
∈A,
S.
As
ASVan
and
Band
are
connec
Therefore,
there
are
two
vertices
A,
B
∈
Vedge,
−
S
and
ther
and
there
is
A
and
B.
Since
A,
B
∈
/a/V
V
we
have
B
∈
As
A
and
B
are
connec
ma.
Let
us
assume
that
Vedge
−
SB
is
not
aus
INDEPENDENT-SET.
and
there
is
an
edge
A
and
Since
B
∈
/S,
V
−
S,
we
have
B
∈−
S.
As
A
B
are
c
there
are
two
vertices
A,
∈
Vthere
−
STherefore,
and
there
an
edge
connecting
A
and
B.
Since
A,
∈
−
S,
not
aMAX-INDEPENDENT-SET
INDEPENDENT-SET.
Therefore,
Therefore,
there
are
are
two
vertices
vertices
A,
B
B
∈V
∈
V
−
−
S
S
and
and
there
there
is
is
an
an
edge
edge
connecting
connecting
A
A
and
and
B.
B.
Sin
Sin
isconnecting
also
NP-hard
problem
by
reducing
aand
MIN-VERTEX-COVER
problem
into
aB
reducing
athe
MIN-VERTEX-COVER
problem
into
ais
A
and
B.
Since
A,
∈
Vthere
−
S,
,that
B
∈
VS
−
S
and
there
is
an
edge
connecting
A
and
B.
Since
A,
B−
∈
V
−
S,
we
we
have
have
A,
A,
B
B
∈
/∈
/∈
S.
S.
As
As
A
AA,
and
B
B
are
are
connected
connected
by
by
an
edge,
we
we
note
note
MAX-INDEPENDENT-SET
problem.
MAX-INDEPENDENT-SET
problem.
ng
NP-hard
problems
for
illustration.
Recall
that
NP-hard
problems
have
yet
to
be
solved
in
polynomial
eat
are
vertices
A,
B
∈
VB
−
S
and
there
is
an
edge
connecting
A
and
B.
Since
A,
B
∈
V
−we
S,
tcting
VMIN-VERTEX-COVER
−two
is
not
aan
VERTEX-COVER.
there
are
two
vertices
B
∈
/A,
V
−
S
and
there
is
an
edge
connecting
A
and
B.
Since
A,
B
∈
/S,
V
−
S,
have
A,
B
∈
S.
As
and
and
there
there
is
is
an
edge
edge
connecting
connecting
A
A
and
and
B.
B.
Since
Since
A,
A,
B
B
∈
/
∈
/
and
V
−
−
there
S,
S,
we
we
is
have
an
have
edge
A,
B
B
connecting
∈
∈
S.
S.
As
As
A
A
A
and
and
and
B
B
B.
are
are
Since
connected
connected
A,
∈
/
we
have
A,
B
∈
/
S.
As
A
and
B
are
connected
by
an
edge,
we
note
that
S
is
not
a
VERTEX-COVER.
T
nding
a
maximum
subset
of
nodes
such
that
for
every
edge,
at
most
one
of
its
endpoint
is
in
Therefore,
are
two
vertices
S
there
is
an
edge
connecting
A
and
B.
Sinc
we
have
A,
B
∈
/
S.
As
B
are
connected
by
an
edge,
we
note
that
S
is
not
a
VERTEX-COVER.
is
edge
connecting and .
Since ,
have .
As and is
a
NP-hard
problem.
Therefore,
we
can
show
we
have
A,
B
∈
/
S.
As
A
and
B
are
connected
by
an
edge,
we
note
that
S
n
edge
connecting
A
and
B.
Since
A,
B
∈
/
V
−
S,
we
have
A,
B
∈
S.
As
A
and
B
are
connected
ertices
A,
B
∈
V
−
S
and
there
is
an
edge
connecting
A
and
B.
Since
A,
B
∈
V
−
S,
emmas.
MAX-INDEPENDENT-SET
problem.
ertices
A,
B
∈
V
−
S
and
there
is
an
edge
connecting
A
and
B.
Since
A,
B
∈
V
−
by
an
edge,
we
note
that
S
is
not
an
INDEPENDENT-SET.
This
is
a
contradiction.
Therefore
V
−
S
we
have
A,
B
∈
/
S.
As
A
and
B
are
connected
by
an
edg
o
the
previous
lemma.
Let
us
assume
that
V
−
S
is
not
a
INDEPENDENT-SET.
by
an
edge,
we
note
that
S
is
not
an
INDEPENDENT-SET.
This
is
a
contradiction.
Therefore
V
−
S
here
is
an
edge
connecting
Aconnected
and
B.
A,
B
∈
Vnot
−
S,of
by
an
edge,
we
note
that
S
is
not
an
INDEPENDENT-SET.
This
isnote
aINDEPENDENT-SET.
Therefore
VBi
Lemma
2.
If
S
⊆
Vby
isedge,
aSince
VERTEX-COVER
graph
G(V,
E),
then
V
−
Scontradiction.
is
an
INDEPENDENT-SE
B
∈
/
S.
As
A
and
B
are
by
an
edge,
we
note
that
S
is
not
a
VERTEX-COVER.
This
is
a
and
B.
Since
A,
B
∈
V
−
S,
we
we
have
have
A,
A,
B
B
∈
/
∈
/
S.
S.
As
As
A
A
and
and
B
B
are
are
connected
connected
by
by
an
an
edge,
edge,
we
we
note
that
that
S
S
is
is
not
not
a
a
VERTEX-C
VERTEXMAX-INDEPENDENT-SET
problem.
is
not
a
VERTEX-COVER.
This
is
a
eA,
connected
by
an
edge,
we
note
that
S
is
not
a
VERTEX-COVER.
This
is
a
contradiction.
contradiction.
Therefore
Therefore
V
V
−
−
S
S
is
is
an
an
INDEPENDENT-SET.
Proof.
Let
us
assume
that
V
−
S
is
a
VERTEX-COVER.
Therefore,
there
are
two
vertices
A,
than
4
decades.
MIN-VERTEX-COVER
is
a
graph
problem
that
involves
finding
a
minimum
subset
S.
As
A
and
B
are
connected
an
edge,
we
note
that
S
is
not
a
VERTEX-COVER.
This
is
a
ecting
A
and
B.
Since
A,
B
∈
/
V
−
S,
we
have
A,
B
∈
S.
As
A
and
B
are
connected
by
an
we
note
that
S
is
not
an
INDEPENDENT-SET.
This
is
a
contradiction
by
by
an
an
edge,
edge,
we
we
note
note
that
that
by
edge,
note
that
SS
is
not
an
an
INDEPENDENT-SET.
INDEPENDENT-SET.
This
This
isisedge,
aan
aTherefore
contradiction.
contradiction.
Sshow
isisis
not
Therefore
INDEPENDENT
V−
− SS isis aa
is
are
connected
by
an
edge,
we
note
that problem
iscontradiction.
notanaBVERTEX-COVER.
aTherefore
concontradiction.
Therefore
V
Sthat
isand
an
INDEPENDENT-SET.
we
already
know
that
MIN-VERTEX-COVER
isSSan
ais
NP-hard
problem.
we
can
we
have
A,VERTEX-COVER.
B
∈
/not
S.
As
A
areisisaTherefore
connected
by
an
we
that
S
isan
not
a VVERTEX-C
contradiction.
V−
−
S
is
INDEPENDENT-SET.
and
NP-hard
problem
aiswe
MIN-VERTEX-COVER
into
aSwe
contradiction.
V
−Therefore,
INDEPENDENT-SET.
eVERTEX-COVER.
note
that
S
not
an
INDEPENDENT-SET.
This
aVERTEX-COVER.
contradiction.
V
−
S
is
aINDEPENDENT-SE
and
B
are
connected
by
an
we
note
that
isB
not
aan
This
isnote
B
are
connected
by
an
edge,
note
not
VERTEX-COVER.
This
is
VERTEX-COVER.
Therefore
VaaThis
−
S
an
B
∈
V
−note
S
and
is
an
edge
connecting
A
and
B.
Since
S,
VERTEX-COVER.
edge,
we
that
Sby
isreducing
not
aTherefore
This
alemmas.
VERTEX-COVER.
graph
G(V,
E)
tion.
Therefore
Visthere
−
S
is
an
INDEPENDENT-SET.
This
is
aedge,
contradiction.
contradiction.
Therefore
Therefore
V
VA
−
−
SS
isis
an
INDEPENDENT-SET.
INDEPENDENT-SET.
an
INDEPENDENT-SET.
We
will
first
prove
the
following
twoA,A,
lemmas.
and
there
is
an
edge
connecting
and
B.
Since
B∈the
∈
/VV−−
S,
we
have
A,
B
∈
We
will
first
prove
the
following
two
that
for
every
edge,
at
least
one
of
its
endpoint
is
in
the
subset.
MAX-INDEPENDENT-SET
isS.a As A and B are
ENT-SET
of
graph
G(V,
E),
then
V
−
S
is
a
VERTEX-COVER
of
Therefore
V
−
S
an
INDEPENDENT-SET.
S
is
not
an
INDEPENDENT-SET.
This
is
a
contradiction.
Therefore
V
−
S
is
a
tradiction.
Therefore is
an
INDEPENDENT-SET.
VERTEX-COVER.
VERTEX-COVER.
VERTEX-COVER.
VERTEX-COVER.
NDENT-SET
is
also
a
NP-hard
problem
by
reducing
a
MIN-VERTEX-COVER
problem
into
a
contradiction.
Therefore
V
−
S
is
an
INDEPENDENT-SET.
Theorem
1.
1.
IfIf SS
⊆
⊆ofVThis
V
isis
aaisfollowing
MAX-INDEPENDENT-SET
MAX-INDEPENDENT-SET
of
of
graph
graph G(V
G(V
ER.
− SSWe
an INDEPENDENT-SET.
INDEPENDENT-SET.
We
first
prove
the
two
lemmas.
−
isis an
connected
by
edge,
we
note
is
aTheorem
VERTEX-COVER.
a G(V,
-SET.
Theorem
1.
IfVVfollowing
S
⊆note
Vthat
isthat
graph
E),
then
VV−−
Sof
isgraph
Theorem
1.
If
S
⊆
Vnodes
isaaSaMAX-INDEPENDENT-SET
MAX-INDEPENDENT-SET
of
graph
then
S
isa aMIN-VERTEX
MIN-VERTE
will
first
prove
the
two
lemmas.
Theorem
1.of
Ifwill
S of
⊆
V
is at
aE),
MAX-INDEPENDENT-SET
G(V, E),of
t
byan
an
edge,
we
Snot
is that
not
an
INDEPENDENT-SET.
This
isis
contradiction.
Therefore
m
of finding
aV
maximum
subset
of
such
for
every
edge,
most
one
of
its
endpoint
is
in
Lemma
2.
S2.
⊆
is
a1.
VERTEX-COVER
of
graph
G(V,
E),
then
V
−MIN-VERTEX-COVER
S
an
INDEPENDENT-SET
ofoV
Theorem
1.
Ifus
SG(V,
⊆
V
isE),
ais
Lemma
2.
IfIf
S
⊆
is
a1.
VERTEX-COVER
graph
G(V,
then
V
−
S
an
INDEPENDENT-SET
Lemma
If
SIf ⊆S
V⊆
is
VERTEX-COVER
graph
G(V,
E),
then
V
−
S
is
an
INDEPENDENT-SE
NT-SET
problem.
Proof.
The
proof
is
actually
similar
to
the
previous
lemma.
Let
assume
that
V
−
S
is
not
a
INDEPENDE
m
If
S
⊆
is
a
MAX-INDEPENDENT-SET
of
graph
G(V,
E),
then
V
−
S
is
a
Theorem
Theorem
If
If
S
S
⊆
⊆
V
V
is
is
a
a
MAX-INDEPENDENT-SET
MAX-INDEPENDENT-SET
of
of
graph
graph
G(V,
G(V,
E),
E),
then
then
V
V
−
−
S
S
is
is
a
MINMIN
hen
V
−
S
is
a
MIN-VERTEX-COVER
NDEPENDENT-SET
of
graph
G(V,
E),
then
V
−
S
is
a
MIN-VERTEX-COVER
of
of
the
the
graph
graph
G(V,
G(V,
E).
E).
Lemma
1.
If
S
⊆
V
is
an
INDEPENDENT-SET
of
graph
G(V,
E),
then
V
−
S
is
a
VERT
Theorem
1.
is
a
MAX-INDEPENDENT-SET
of
graph ,
then Lemma
1.
If
V
is
an
INDEPENDENT-SET
of
graph
G(V,
E),
then
V
−
S
is
a
VERTEX-C
nS 1.
INDEPENDENT-SET.
⊆
V
is
a
MAX-INDEPENDENT-SET
of
graph
G(V,
E),
then
V
−
S
is
a
MIN-VERTEX-COVER
Lemma
2.
If
S
⊆
V
is
a
VERTEX-COVER
of
graph
G(V,
E),
then
V
−
an
INDE
Lemma
Lemma
2.
2.
Lemma
2.
If
If
S
S
⊆
⊆
V
V
is
is
a
a
VERTEX-COVER
VERTEX-COVER
of
of
graph
graph
G(V,
G(V,
E),
E),
then
then
If
V
V
S
−
−
⊆
S
S
V
is
is
is
an
an
a
VERTEX-COVER
INDEPENDENT-SET
INDEPENDENT-SET
of
graph
of
of
the
the
G
of
the
graph
G(V,
E).
Theorem
1.
If
S
⊆
V
is
a
MAX-INDEPENDENT-SET
of
graph
G(V,
E),
then
S
is
a
MIN-V
of
the
graph
G(V,
E).
of
the
graph
G(V,
E).
VERTEX-COVER.
uppose
we
that
MIN-VERTEX-COVER
isVgraph
a−
NP-hard
problem.
Therefore,
canofAshow
SLemma
⊆graph
VMIN-VERTEX-COVER
isthen
aalready
VERTEX-COVER
of
graph
G(V,
E),
then
V1.
S
is
the
MAX-INDEPENDENT-SET
of
graph
G(V,
E),
then
MIN-VERTEX-COVER
Lemma
Ifand
⊆an
V .INDEPENDENT-SET
isis
an
graph
E)∈
MAX-INDEPENDENT-SET
of
graph
G(V,
E),
then
VV
−
SSS−of
isis
aaS
MIN-VERTEX-COVER
graph
G(V,
E)
the
graph
G(V,
E).
graph
G(V,
E)
of
G(V,
E),
then
VE)
−graph
SG(V,
is
aG(V,
MIN-VERTEX-COVER
G(V,
lemmas.
Therefore,
there
are
two
vertices
A,
Bvertices
∈of
−
anINDEPENDENT-SET
edge
andofB.
Since
A, B
E).
S
aG(V,
of
the
graph
G(V,
E).
E).
1.
Ifgraph
S
⊆know
V of
is
an
INDEPENDENT-SET
G(V,
E),
then
V
−of
Sconnecting
is aweVERTEX-COVER
ofG(V,
the
G(V,
E),
VE)
−
S
is
athe
VERTEX-COVER
of
the
graph
E)
isgraph
MIN-VERTEXCOVER
of
the
graph graph
G(V,
E)
taph
aisE).
VERTEX-COVER.
Therefore,
there
are Vtwo
A,
B
∈
/ there
Vthat
−V
SV
V,
VERTEX-COVER
of
G(V,
E),
then
−Proof.
S
is
an
INDEPENDENT-SET
the
graph
G(V,
E)
graph
graph
G(V,
G(V,
E)
graph
G(V,
E)
of
the
graph
G(V,
E).
Proof.
We
We
know
know
that
−
−
S
S
is
is
a
a
vertex
vertex
cover
cover
by
by
lemma
lemma
1.
1.
The
The
on
o
DEPENDENT-SET
is
also
a
NP-hard
problem
by
reducing
a
MIN-VERTEX-COVER
problem
into
a
graph
G(V,
E)
DEPENDENT-SET
oflemmas.
graph
G(V,
E),
then
−
S
is aknow
MIN-VERTEX-COVER
ve
the following
twoWe
byby
Proof.
know
that
V
−
SVERTEX-COVER
is
athe
cover
lemma
1.
only
thing
for
us
pro
we
have
A,
Bis
∈
/actually
S.
As
AV
and
BVisthe
are
connected
by
an
edge,
we
note
that
S
is
not
aremains
VERTEX-COVER.
Proof.
We
know
that
S
avertex
vertex
cover
1.aThe
The
only
thing
that
remains
forby
usto
tothin
p
graph
G(V,
E)
Proof.
We
that
Vlemma
−
S
is
vertex
cover
by
lemma
1.cover
The
only
Lemma
2.
If
S
⊆
VA,
isB
a−∈
of
graph
G(V,
E),
then
V−
−
Sisthat
isa−
an
INDEPENDENT-S
B.
Since
A,
B
∈
/
V
−
S,
we
have
S.
As
A
and
B
are
connected
Proof.
The
proof
is
actually
similar
to
previous
lemma.
Let
us
assume
that
V
−
S
is
not
a
INDEPENDENT-S
Proof.
We
know
that
V
S
vertex
lemm
Proof.
The
proof
similar
to
previous
lemma.
Let
us
assume
that
V
−
S
is
not
a
INDEPENDENT-S
Proof.
The
proof
is
actually
similar
to
the
previous
lemma.
Let
us
assume
that
V
S
is
not
a
INDEPEND
ENT-SET
of
graph
G(V,
E),
then
V
−
S
is
a
VERTEX-COVER
of
the
Proof.
We
know
that is
vertex
cover
by
lemma
1.
The
only
thing
that
remains
We
know
that
V
−
S
is
a
vertex
cover
by
lemma
1.
The
only
thing
that
remains
for
us
to
prove
is
its
Proof.
Proof.
We
We
know
know
that
that
V
V
−
−
SSINDEPENDENT-SET.
is
a−
anot
vertex
vertex
cover
cover
by
by
lemma
lemma
1.
1.−
The
The
only
only
thing
that
that
remains
remains
for
for
ng
that
for
us
to
prove
is
its
vertex
cover
lemma
1.
The
only
that
remains
us
tous
prove
is
its
V
Vlemma
−
−
S
S
is
islemma.
not
not
minimum
minimum
vertex
vertex
cover.
cover.
Then,
Then,
t
minimality.
minimality.
Suppose
Suppose
ENDENT-SET
problem.
Proof.
Let
us
assume
that
Visalemma.
Sforvertex
not
athe
VERTEX-COVER.
Therefore,
there
are
two
ver
Proof.
Let
us
assume
aisProof.
VERTEX-COVER.
Therefore,
there
are
two
vertices
Ais
w
that
Vremains
−
Sby
is
a vertex
cover
by
lemma
The
only
thing
that
remains
for
us
to
prove
is
its
Proof.
The
proof
is
actually
similar
to
previous
Let
usand
that
VA,
−B
SV∈Ther
is
no
Proof.
Proof.
The
The
proof
proof
isis
actually
actually
similar
similar
to
to
the
the
previous
previous
lemma.
Let
Let
us
The
assume
assume
proof
that
isminimum
actually
V
V
−
S
Snot
similar
not
not
aassume
athing
INDEPENDENT-SET.
to
INDEPENDENT-SET.
the
previous
lemma.
Vcontradiction.
−thing
S
isSV
not
minimum
cover.
Then,
there
is
another
vertex
cover
S
minimality.
Suppose
contradiction.
Therefore
Vfor
−
is
an
Proof.
We
know
that
−
S
is
vertex
cover
by
The
only
thing
that
remains
for
u−
VThe
−
S
is1.
not
minimum
vertex
cover.
Then,
there
isvertex
another
vertex
cover
VV−
−
minimality.
Suppose
INDEPENDENT-SET
V
−
S
is
not
cover.
Then,
there
minimality.
Suppose
graph
G(V,
E)
oof
actually
similar
to
the
previous
lemma.
Let
us
assume
that
V
−
S
is
athat
INDEPENDENT-SET.
−
STherefore,
is1.
vertex
cover
by
lemma
1.
The
only
thing
that
remains
for
us
to
prove
is
its
INDEPENDENT-SET.
This
is
avertices
Therefore
V
−
S
is
anot
Proof.
us
assume
V
−1.
Sis
is
athere
VERTEX-COVER.
−
S
aaTherefore,
vertex
cover
by
lemma
1.
only
thing
that
for
us
to
prove
its
Therefore,
there
are
two
vertices
A,
B
∈
VSS
−
S
and
there
an
edge
connecting
A
B.
Since
V
V
−
Sis
is
not
minimum
vertex
cov
minimality.
Suppose
 and
there
are
two
A,
B
∈
V
−
S
there
is
an
edge
connecting
A
B.
Since
A,
B−
∈
ma
The
only
thing
that
remains
us
to
prove
is
its
Therefore,
there
two
vertices
A,
B
∈
V
−
S
there
is
an
edge
connecting
A
B.
Since
A,
B
that
Let
and
vertices
and
−
V
isisis
an
INDEPENDENT-SET
of
graph
G(V,
E),
then
Vremains
−
Sand
is
ais
VERTEX-COVER
of
for
us
to
prove
isare
its
minimality.
Suppose −
is
not
minimum
vertex
cover.
Then,
V
−
S
is
not
minimum
vertex
cover.
Then,
there
is
another
vertex
cover
V
−
S
where
ty.
Suppose
emains
for
us
to
prove
is
its
V
V
−
−
S
is
is
not
not
minimum
minimum
vertex
vertex
cover.
cover.
Then,
Then,
there
is
is
another
another
vertex
vertex
cov
co
minimality.
minimality.
Suppose
Suppose
Proof.
Let
us
assume
that
V
−
S
is
not
a
VERTEX-COVER.
Therefore,
there
are
two
A,
B
∈
/
V
S
the
.
there
are
two
vertices
A,
B
∈
/
V
−
another
vertex
cover
V
−
S
where
minimum
vertex
cover.
Then,
there
is
another
vertex
cover
V
−
S
where
|V
|V
−
−
S
S
|
|
<
<
|V
|V
−
S|,
S|,
which
which
implies
implies
that
that
|S
|S
|
|
>
>
|S|.
|S|.
By
By
lemma
lemma
2,
2,
S
S
and
there
is
an
edge
connecting
A
and
B.
Since
A,
B
∈
/
V
−
S,
we
have
A,
B
∈
S.
As
A
a






and
there
is
an
edge
connecting
A
and
B.
Since
A,
B
∈
/
V
−
S,
we
have
A,
B
∈
S.
As
A
and
B






V
−minimum
S
is
not
minimum
vertex
cover.
Then,
there
is|S|.
another
vertex
cover
VA,
−
S
where
ppose
ySsimilar
to
the
previous
lemma.
Let
us
assume
that
V
−
S
not
aedge
INDEPENDENT-SET.
Therefore,
there
two
vertices
A,
B
V
−
S
and
there
is
an
edge
connecting
A
and
Therefore,
Therefore,
there
there
are
two
two
vertices
vertices
A,
A,
B
B
∈∈
V
Vare
−
S
Sedge
and
there
there
Therefore,
is
is−
an
an
edge
edge
there
connecting
connecting
are
two
A
A
and
and
B.
A,
Since
Since
B
∈set.
VA,
A,
−
B
B
∈∈−
and
VVis
−
−
ther
S,
S,
|V
−
SB
|are
<
|V
−
S|,
which
implies
that
|S
|connecting
>
By
lemma
2,
S
is
an
independent
Sini
V
−
is
minimum
cover.
Then,
there
isthat
another
vertex
cove
minimality.
Suppose
|V
−
S
|//B
<1.
|VThe
−
S|,
which
implies
that
|Sby
| isan
>
|S|.
By
lemma
2,
is
an
independent
set.
Therefore,
S
vertex
∈
|V
S−
|S
<
|V
−
S|,
which
implies
that
|S
|vertices
>
|S|.
By
lemma
2,
S
an
 and
re
are
two
vertices
A,
∈
V
S
there
is
an
A
and
B.
B
V
−
S,
is
not
vertex
cover.
Then,
there
is
another
vertex
cover
−
SSince
where
and
there
is
an
connecting
A
and
B.
Since
A,
B
/aTherefore,
VS|S|.
S,
we
S
is
not
minimum
vertex
cover.
Then,
there
is
another
cover
VV
−
S
we
have
A,
B
S.
As
A
and
B
are
connected
by
an
edge,
we
note
that
S
is
not
aisVVB.
VERTEX-COVER.
This
|V
|∈
<
|V
−
S|,
which
implies
|S
|aconnected
>
By
le
and
vertex
ertex
cover
by
lemma
only
thing
that
remains
for
us
to
prove
is
its
we
have
A,
∈
S.
As
A
B
are
connected
edge,
we
note
that
S
is
not
ais
VERTEX-COVER.
This
∈
−
an
S
 ∈
over.
Then,
there
is
another
cover
V
−
where
we
have
A,
B
∈
/S
S.
As
A
and
B
are
connected
by
edge,
we
note
that
S
is
not
aais
VERTEX-COVER.
there
is
another
vertex
cover not
where , where
which
implies
that  cover
 and
Theorem
1.
If
⊆
V
is
a
MAX-INDEPENDENT-SET
of
graph
G(V,
E),
then
−
S
is
a
MIN-VERTEX
<
|V
−
S|,
which
implies
that
|S
|
>
|S|.
By
lemma
2,
S
is
an
independent
set.
Therefore,
S
not
vertex
V
−
S
where
|V
|V
−
−
S
S
|
|
<
<
|V
|V
−
−
S|,
S|,
which
which
implies
implies
that
that
|S
|S
|
|
>
>
|S|.
|S|.
By
By
lemma
lemma
2,
2,
S
S
an
an
independent
independent
set.
set.
The
Th
and
there
is
an
edge
connecting
A
and
B.
Since
A,
B
∈
/
V
−
S,
we
have
A,
B
∈
S.
As
A
and
B
are
an
the
S,
we
have
A,
B
∈
S.
As
A
and
B
are
connected
dependent
set.
Therefore,
is
not
a
that
|S
|
>
|S|.
By
lemma
2,
S
is
an
independent
set.
Therefore,
S
is
not
a
maximum
maximum
independent
independent
set.
set.
This
This
is
is
a
a
contradiction.
contradiction.
Therefore
Therefore
V
V
−
−
st
prove
the
following
two
lemmas.
by
edge,
we
note
that
S
is
not
an
INDEPENDENT-SET.
This
is
contradiction.
Th
Proof.
The
proof
is
actually
similar
to
previous
lemma.
Let
us
assume
that
−
S
is
not
INDEPEND
by
an
edge,
we
note
that
S
is
not
an
INDEPENDENT-SET.
This
is
a
contradiction.
Therefore



−
S|,
which
implies
that
|S
|we
>
|S|.
By
lemma
Sby
isan
an
independent
set.
Therefore,
is
not
ertices
A,A
Band
V
S
and
there
isB
an
edge
Aedge,
and
A,
Vnot
−
S,
have
A,
B
∈
/connecting
S.
As
A
and
BS
are
connected
by
an
edge,
we
note
that
S Viscover.
not
aisTher
VER
we
we
have
have
A,
A,
B
BB
/−
∈
/cover.
S.
S.
As
As
A
A
and
and
B−
are
are
connected
connected
by
an
edge,
we
have
we
note
note
that
that
∈
/is
S.
S
S∈
is
As
is
not
not
A
and
amaximum
aSThis
VERTEX-COVER.
VERTEX-COVER.
Ban
are
This
by
This
isis
edg
aS
aS
maximum
independent
set.
This
iswe
aimplies
contradiction.
Therefore
VB
S
is
minimum
vertex
|V
−
S
<
|V
−
S|,
which
that
|S
|∈
>
|S|.
By
lemma
2,
S
is
independent
set.
maximum
independent
set.
This
is2,
avertices
contradiction.
Therefore
V
−
S
isthe
the
minimum
vertex
cover.
∈
are
t∈
aS.
VERTEX-COVER.
Therefore,
there
two
A,
B
/we
VSince
−
SB
maximum
independent
set.
This
a −
contradiction.
Therefore
−
S an
the
that
∈
/by
As
are
connected
by
an
edge,
note
that
isB.
not
aA,
VERTEX-COVER.
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isaconnected
aminimum
implies
that
|S
|
>
|S|.
By
lemma
2,
S
is
an
independent
set.
Therefore,
S
is
not
a
by
an
edge,
we
note
S
is
an
INDEPENDENT-SET.
This
implies
that
|S
|
>
|S|.
2,
S
is
an
independent
set.
Therefore,
S
is
not
a
contradiction.
Therefore
V
−
S
is
an
INDEPENDENT-SET.
maximum
independent
set.
is
a
contradiction.
Ther
minimum
vertex
Then,
there
is
another
vertex
cover
V
−
S
where
.
By
lemma
2, is
an
independent
set.
Therefore, is
not
a
incontradiction.
Therefore
V
S
is
an
INDEPENDENT-SET.
lemma
2,
S
is
an
independent
set.
Therefore,
S
is
not
a
contradiction.
Therefore
V
−
S
is
an
INDEPENDENT-SET.
of
the
graph
G(V,
E).
m
independent
set.
This
is
a
contradiction.
Therefore
V
−
S
is
the
minimum
vertex
cover.
nt
set.
Therefore,
S
is
not
a
maximum
maximum
independent
independent
set.
set.
This
This
is
is
a
a
contradiction.
contradiction.
Therefore
Therefore
V
V
−
−
S
S
is
is
the
the
minimum
vertex
vertex
co
co
an
edge,
we
note
that
S
is
not
an
INDEPENDENT-SET.
This
is
a
contradiction.
Therefore
V
−
S
is
a
T.
This
is
a
contradiction.
Therefore
V
−
S
is
a
minimum
vertex
cover.
a contradiction.
Therefore
Vedge,
−
SSSwe
is
the
minimum
vertex
cover.
VERTEX-COVER.
Therefore,
there
are
two
vertices
A,
Band
∈
−
Sminimum
and
there
isTherefore
ancover.
edge
A
and B. vertex
Since A,
VERTEX-COVER.
VER
graph
G(V,
E),
then
V
SisisVTherefore
is
an
INDEPENDENT-SET
of
the
This
amaximum
contradiction.
V
SE),
the
vertex
and
Bof
are
connected
by
an
note
that
SThis
is
not
athen
VERTEX-COVER.
This
avertex
Therefore
−
S
isabove
an−two
INDEPENDENT-SET.
contradiction.
contradiction.
Therefore
Therefore
Vcontradiction.
V
−
−
an
an
INDEPENDENT-SET.
INDEPENDENT-SET.
contradiction.
S
an
me
that
Vis
−
isBy
not
aINDEPENDENT-SET.
VERTEX-COVER.
Therefore,
are
vertices
A, conclude
B
∈
/V−
V−
Sis
independent
set.
isVisVathere
contradiction.
Therefore
Visconnecting
S−
isthat
the
minimum
covB
B.
Since
B
∈
/isS
Vis
−
S,
we
have
A,
S.
As
A−
B
are
connected
isset.
−
From
From
the
the
above
theorem,
we
we
conclude
that
athe
aINDEPENDENT-S
MIN-VERTEX-CO
MIN-VERTEX-CO
SThis
⊆
an
INDEPENDENT-SET
of
graph
G(V,
V
Stheorem,
athe
VERTEX-COVER
ofeasily
Therefore
V
−dependent
is
anminimum
.endent
This
contradiction.
Therefore
−∈
S
is
the
minimum
vertex
cover.
set.
This
aVB
contradiction.
Therefore is
is
minimum
cover.
VERTEX-COVER.
aaA,
contradiction.
Therefore
−
S
is
the
minimum
vertex
cover.
hat
|SVis
|V
>
|S|.
lemma
2,
SS.
isis
an
independent
set.
Therefore,
not
a note
herefore
−
SS
the
vertex
cover.
From
the
above
theorem,
we
conclude
that
a
MIN-VERTEX-COVER
can
be
constructed
if
m
vertex
cover.
From
the
above
theorem,
we
conclude
that
a
MIN-VERTEX-COVER
can
be
easily
constructed
ifwc
VERTEX-COVER.
From
the
above
theorem,
we
conclude
that
a
MIN-VERTEX-COVER
we
have
A,
B
∈
/
As
A
and
B
are
connected
by
an
edge,
we
that
S
is
not
a
VERTEX-COVER
Theorem
1.We
Sknow
⊆
ishave
aBVcontradiction.
MAX-INDEPENDENT-SET
of
graph
G(V,
E),
then
−
SifVconclude
iswe
acan
MIN-VERTEX-COV
From
the
above
theorem,
we
ausconstru
MIN-V
−
Sabove
is
an theorem,
INDEPENDENT-SET.
Theorem
1.
IfIfwe
S
⊆
VVSisthat
is
MAX-INDEPENDENT-SET
graph
E),
then
VVthen
−
S
is
aremains
MIN-VERTEX-COV
Theorem
1.
Ifwe
⊆
is−
MAX-INDEPENDENT-SET
of
graph
G(V,
E),
−
S
isbe
athat
MIN-VERTE
ea)an
connecting
AMIN-VERTEX-COVER
and
B.
Since
A,
∈
/V
VSais
−
S,
we
have
A,
B by
∈V of
S.
As
A
and
B
are
connected
INDEPENDENT-SET.
This
aa2.
Therefore
−that
S we
is
aG(V,
Proof.
V
S
is
a
vertex
cover
lemma
1.
The
only
thing
that
for
to
pro
the
conclude
that
a
MIN-VERTEX-COVER
can
be
easily
constructed
have
a
From
From
the
the
above
above
theorem,
theorem,
we
we
conclude
conclude
that
a
a
MIN-VERTEX-COVER
MIN-VERTEX-COVER
can
be
easily
easily
constru
be
easily
constructed
if
a
onclude
that
a
can
be
easily
constructed
if
have
a
MAX-INDEPENDENT-SET.
Lemma
If
⊆
V
is
a
VERTEX-COVER
of
graph
G(V,
E),
then
V
−
S
is
an
INDEPEN
Lemma
2.
If
S
⊆
a
VERTEX-COVER
of
graph
G(V,
E),
then
V
−
S
is
an
INDEPENDENT
contradiction.
Therefore
V
−
S
is
the
minimum
vertex
cover.
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theorem,
weIfthat
that
aabove
MIN-VERTEX-COVER
can
beV easily
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a E),
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MAX-INDEPENDENT-SET
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SScan
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V
V
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MAX-INDEPENDENT-SET
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G(V,
E),
E),
IfaififVERTEX-COVER
Swe
then
then
⊆ have
V VVisof
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−
a graph
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MAX-INDEPENDENT-SET
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MIN-VERTEX-COVER
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−
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an
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fm,
⊆
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MAX-INDEPENDENT-SET
G(V,
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m,
we
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ahave
MIN-VERTEX-COVER
can
be
easily
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2.
Ifgraph
Sthat
⊆then
V
we
conclude
that
a⊆
MIN-VERTEX-COVER
be
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of
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the
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be
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ifofLemma
we
aof
of
graph
G(V,
E).
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SVis2.
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ishave
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S Sof
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graph
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previous
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us
assume
that
V
−
not
a
INDEPENDENT-SET.
NDENT-SET.
MAX-INDEPENDENT-SET
of
graph
then
V
−
S
is
a
MIN-VERTEX-COVER
the
G(V,
E).
of
of
the
the
graph
graph
G(V,
G(V,
E).
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of
the
graph
We
say
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problem
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assume
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S
Therefore,
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G(V,
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In
the
the
beginning
beginning
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of
this
this
section,
section,
we
we
assume
assume
we
know
know
that
that
MIN-V
MIN-V
1.
S
⊆
V
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MAX-INDEPENDENT-SET
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then
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−
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MIN-VERTE
V edge
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and
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and
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V1.
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Proof.
We
know
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is−
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vertex
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for
us
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VV
−
S
is
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vertex
cover
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The
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remains
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Proof.
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know
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V
S
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a
vertex
cover
by
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1.
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only
thing
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beginning
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section,
we
assume
we
know
that
MIN-VERTEX-COVER
is
aisVaNP-hard
probl
Inthe
the
beginning
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this
section,
we
assume
we
know
that
MIN-VERTEX-COVER
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NP-hard
prob
VER
graph
G(V,
E),
then
INDEPENDENT-SET
of
the
In
the
beginning
of
this
section,
we
assume
we
know
that
MIN-VERTE
n
connecting
A
and
B.
Since
A,
B
∈
/
V
−
S,
we
have
A,
∈
S.
As
A
and
B
are
connected
Proof.
The
proof
is
actually
similar
to
the
previous
lemma.
Let
us
assume
that
−
S
not
aits
IN
In
the
beginning
of
this
section,
we
assume
we
know
The
proof
is
actually
similar
to
the
previous
lemma.
Let
us
assume
that
V
−
S
not
a
INDEPE
Proof.
We
know
that
V
−
S
is
a
vertex
cover
by
lemma
1.
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thing
that
rema
Proof.
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We
We
know
know
that
that
V
V
−
−
S
S
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a
vertex
vertex
cover
cover
by
by
lemma
lemma
1.
1.
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The
We
only
only
know
thing
thing
that
that
that
V
−
remains
remains
S
is
a
vertex
for
for
us
us
cover
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to
prove
prove
by
lemm
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is
maximum
independent
set.
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is
a
contradiction.
Therefore
V
−
S
is
the
minimum
vertex
cover.
e
beginning
of
this
section,
we
assume
we
know
that
MIN-VERTEX-COVER
is
a
NP-hard
problem.
It
is
In
In
the
the
beginning
beginning
of
of
this
this
section,
section,
we
we
assume
assume
we
we
know
know
that
that
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MIN-VERTEX-COVER
is
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X-COVER
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is
we
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problem.
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is
good
good
to
to
know
know
as
as
many
many
NP-hard
NP-hard
problem
problem
as
as
possible.
possible.
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This
is
is
necessary
necessary

of
the
graph
G(V,
E).
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that
V
−
S
is
a
vertex
cover
by
lemma
1.
The
only
thing
that
remains
for
us
to
prove
is
its
nected
by
an
edge,
we
note
that
S
is
not
a
VERTEX-COVER.
This
is
a
Proof.
The
proof
isThis
actually
similar
to
the
Let
us
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this
section,
we
assume
weS
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isisknow
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NP-hard
problem.
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is lemma.
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minimality.
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not
vertex
cover.
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there
is
vertex
V
−
S
wh
Suppose
V
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isminimum
not
minimum
vertex
cover.
Then,
there
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minimality.
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good
to
know
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NP-hard
problem
as
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isthere
necessary
so
that
ifprevious
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proble
is
aofminimality.
VERTEX-COVER
of
graph
G(V,
E),
then
V
−
Swe
isLet
an
INDEPENDENT-SET
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the
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the
beginning
of
this
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assume
we
that
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isnew
a−
NP-ha
good
know
as
many
NP-hard
problem
as
possible.
necessary
so
that
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we
encounter
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new
prob
good
to
know
as
NP-hard
problem
as
possible.
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necessary
so
tha
Proof.
The
proof
isto
actually
similar
to
the
previous
lemma.
us
assume
Vis
−
not
aproblem
INDEPENDENT-SET.
us
assume
that
V
−
S
is
not
athere
we
note
that
S
is
not
an
INDEPENDENT-SET.
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is
ais
contradiction.
Therefore
Vedge
−
Sthat
is
aifA
Therefore,
two
A,
B
∈
V
−
Swe
and
there
is
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connecting
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sV
section,
we
assume
we
know
that
MIN-VERTEX-COVER
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NP-hard
problem.
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is
Since
section,
we
assume
we
know
that
MIN-VERTEX-COVER
aaas
NP-hard
problem.
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is
good
to
know
as
many
NP-hard
problem
as
possible.
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Therefore,
two
vertices
A,
B
∈
Vmany
−
S
and
is
an
edge
connecting
and
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 where
w
MIN-VERTEX-COVER
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NP-hard
problem.
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is
−
S
aencounter
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lemma
1.
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only
thing
that
remains
for
us
to
prove
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its
V
−
Sany
is
not
minimum
cover.
Then,
there
another
minimality.
Suppose
V
Vto
−
−
S
S
is
isare
not
not
minimum
minimum
cover.
cover.
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Then,
there
there
is
another
VS
−
S
vertex
vertex
is
not
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minimum
cover
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−
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minimality.
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ais
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so
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possible.
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isanother
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we
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encounter
aver
ano
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we
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X,
we
can
can
use
use
of
of
the
the
NP-hard
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problems
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there
are
vertices
A,
B
∈
V
−
S
and
there
is
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edge
many
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problem
as
possible.
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so
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if
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X,
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|V
−
S
|
<
|V
−
S|,
which
implies
that
|S
|
>
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lemma
2,
S
is
an
independent
set.
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S
is
no
|V
−
S
|
<
|V
−
S|,
which
implies
|S
|
>
|S|.
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lemma
2,
S
is
an
independent
set.
Therefore,
S
is
|V
−
S
|
<
|V
−
S|,
which
implies
that
|S
|
|S|.
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lemma
2,
S
is
an
independent
set.
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can
use
any
of
the
NP-hard
problems
that
we
know,
reduce
it
to
problem
X,
and
thus
prove
that
X
is
also
good
to
know
as
many
NP-hard
problem
as
possible.
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is
necessary
so
that
if
we
encounter
a
ne
can
use
any
of
the
NP-hard
problems
that
we
know,
reduce
it
to
problem
X,
and
thus
prove
that
X
is
also
can
use
any
of
the
NP-hard
problems
that
we
know,
reduce
it
to
problem
X,
From
the
above
theorem,
we
conclude
that
a
MIN-VERTEX-COVER
can
be
easily
constructed
if
w
Therefore,
there
are
two
vertices
A,
B
∈
V
−
S
and
there
is
an
edge
connecting
A
and
B.
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A,
B
∈
V
−
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n
edge
connecting
A
and
B.
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A,
B
∈
V
−
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 note
VER.
we
have
A,
B
∈
/
S.
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and
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note
that
S
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VERTEX


hard
problem
as
possible.
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is
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so
that
if
we
encounter
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new
problem
X,
Proof.
We
know
that
V
−
S
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a
vertex
cover
by
lemma
1.
The
only
thing
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remains
for
us
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pS
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hard
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problems
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we
know,
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we
have
A,
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∈
/
S.
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A
and
B
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connected
by
an
edge,
we
that
S
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not
a
VERTEX-COV
we
assume
we
know
that
MIN-VERTEX-COVER
is
a
NP-hard
problem.
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is



his
is
necessary
so
that
if
we
encounter
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new
problem
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we
S
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vertex
cover.
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there
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another
vertex
cover
V
−
S
where
−
S
|
<
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S|,
which
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>
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lemma
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S
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an
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s
|V
|V
−
−
S
S
|
|
<
<
|V
|V
−
−
S|,
S|,
which
which
implies
implies
that
that
|S
|S
|
|
>
>
|S|.
|S|.
By
By
lemma
lemma
|V
−
S
2,
2,
|
S
<
S
|V
is
is
an
an
−
S|,
independent
independent
which
implies
set.
set.
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that
Therefore,
|S
|
>
S
S
|S|.
is
is
not
not
By
a
le
a
ny
of
the
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know,
reduce
it
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thus
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X
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also
NP-hard.
|V
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counter
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problem
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any
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NP-hard
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problems
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that
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we
know,
know,
reduce
reduce
it
it
to
to
problem
problem
X,
X,
and
and
thus
thus
prove
prove
that
that
previous
lemma.
Let
us
assume
that
V
−
S
is
not
a
INDEPENDENT-SET.
and
thus
prove
that
X
is
also
NP-hard.
ms
that
we
know,
reduce
it
to
problem
X,
and
thus
prove
that
X
is
also
NP-hard.
1contradiction.
1have
−
S|,
which
that
|S
| know,
>
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2,
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isedge,
an
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SNP-hard.
isYYX,
not
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A,
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∈
/problem
S.
As
and
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connected
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weisthat
note
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and
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Xare
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Therefore
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−
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maximum
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Therefore
VX
−
S
the
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maximum
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set.
is
aV
Therefore
Vis
−
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is
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minimum
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We
We
say
say
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problem
isset.
reducible
reducible
to
to
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problem
can
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11
1not
we
have
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/we
S.
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by
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note
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S
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aminimum
VERTEX-COVER.
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aV −
note
that
S
is∈
aand
VERTEX-COVER.
contradiction.
Therefore
−
SINDEPENDENT-SET.
is
an
INDEPENDENT-SET.
dehe
problems
that
we
know,
reduce
to
problem
X,
and
thus
prove
that
X
also
NP-hard.
−
S
is1−
not
minimum
vertex
cover.
Then,
there
another
vertex
cover
minimality.
Suppose
We
say
that
problem
X
is
reducible
to
problem
Yathat
problems
that
know,
reduce
ititare
to
problem
X,
and
thus
that
X
isisis
also
NP-hard.
We
say
problem
X
is
reducible
to
problem
Y
contradiction.
VThis
Sis
isaNP-hard.
an
lem
possible.
This
isthat
necessary
so
that
ifThis
we
encounter
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problem
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We
say
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problem
XV
is
reducible
to
problem
Yprove
uce
itasto
problem
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thus
prove
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X
is
also
90
J.I. Gunawan
From the above theorem, we conclude that a MIN-VERTEX-COVER can be easily
constructed if we have a MAX-INDEPENDENT-SET.
In the beginning of this section, we assume we know that MIN-VERTEX-COVER
is a NP-hard problem. It is good to know as many NP-hard problem as possible. This
is necessary so that if we encounter a new problem X , we can use any of the NP-hard
problems that we know, reduce it to problem X , and thus prove that X is also NPhard.
3. How to Proceed
Suppose we already know that a problem is unsolvable (i.e. any known algorithm will
not solve this problem in time). In competition, it is impossible to complain that “This
is unsolvable, can you eliminate this problem?” to the judges, since the judges believe
How
How to
to proceed
proceed
they have a solution. Such a request is absurd when there are already several contestants
have
that problem.
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in aproblem
major competition
(e.g. ACM International
any
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in
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any known
knownsolved
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omplain that
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proceed
Suppose we already know that a problem is unsolvable (i.e. any known algorithm will not solve this pro
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try
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isis close
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a major for
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However, approximation
approximation
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Approximation
proximation
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to create
create this
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the judges
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In
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the
reason
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y that
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the contestant’s
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solution isis indeed
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α-approximation. However,
However,
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time). Since
Since this
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solution
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a minimisation
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I found in textbook
this
in
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in detail.
detail.
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2-approximation
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problem,
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cuss this
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3.2
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Pruning
programming (especially IOI). One of the reason is because to create this kind of problem, the judges
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algorithm. Therefore,
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we
Hamiltonian Paths with constraints (ACM ICPC World Finals 2010 problem statement n.d.), which is k
be a NP-hard problem. While finding all possible paths is impossible, the solution for this problem is
the exponential algorithm we use to find all possible paths (ACM ICPC World Finals 2010 Solutions n.d
nding
nding Small
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some point we know that it is impossible to visit the rest of the unvisited points, then we can prune the p
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the
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ve we
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ortoLemma
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ffollowing
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We
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prove
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ample
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of
F
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)
into
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)
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SS
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is
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of
of
graph
graph
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E),
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then
then
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−
−
S
S
is
is
VERTEX-C
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disjoint
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and
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such
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elements
in
A
is
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to
the
sum
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all
elements
in
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in
A
is
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to
the
all
elements
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∞
whether
can
create
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(N
),the
into
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ofwant
all
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We
the
sequence
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and
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(N
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to
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toSSknow
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we
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attempting
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there
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have
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note
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INDEPENDENT-SET.
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iswe
equal
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not
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100.
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graph
graph
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ks
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problem.
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by
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he
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Finding
Small
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n=1
n=1
mple
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in
A
is
equal
to
the
sum
of
all
elements
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B
whether
we
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partition
of
F
(N
)
into
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and
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sum
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2
on
on
way
way
is
to
to
apply
apply
a
a
well-known
well-known
technique
technique
called
called
reduction.
reduction.
Suppose
Suppose
we
we
know
know
that
that
problem
problem
X
X
is
is
n
edge,
we
note
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S
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is
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contradiction.
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V
−
S
is
a
This
looks
like
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classic
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problem.
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problem
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by
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ndVERTEX-COVER.
check
the
proof
still
holds
given
the
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constraint
to) two
the
problem.
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like
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PARTITION
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100]),
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can
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efore,
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itlooks
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find
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ansuch
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create
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into
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Let
us
us
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assume
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vertices
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ooN
solve,
solve,
and
we
we
also
also
know
know
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that
we
we
can
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problem
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by
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problem
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ato
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Suppose
there
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problem
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large
N
that
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expo
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value
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it
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TEX-COVER.
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like
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classic
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problem.
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problem
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SUM.
for
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large
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,
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be
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sense
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the
in
A
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toLemma
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em.
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able
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at
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able
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f
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−
1)
+
f
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−
2),
n
>
2
Lemma
2.
2.
Lemma
2.
If
If
S
S
⊆
⊆
V
V
is
is
a
a
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of
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graph
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S
E),
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⊆
then
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then
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a
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of
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of
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problem.
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problem
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by
reduction
from
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and
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there
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an
an
edge
connecting
connecting
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and
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Since
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−
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we
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m
m
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and for
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However,
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Sometimes
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le
ofthe
athen
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aefficient
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ned
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we
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more
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Alarge
and
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an
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inFan
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efficient
However,
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th
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mma
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graph
graph
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graph
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SUM.
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,
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that
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This
like
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classic
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problem.
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NP-hard
by
reduction
from
SUBSETN
F
=
{f
(n)}
and
we
define
F
(N
to
the
first
N
terms
of
F
.
We
want
to
know
by
by
an
edge,
edge,
we
we
note
note
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that
S
S
is
is
not
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an
an
INDEPENDENT-SET.
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This
is
is
a
a
contradiction.
contradiction.
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Therefo
is
defined
in
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defined
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inmore
the
th
aforementioned
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should
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formula
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way,
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that
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issense
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en
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Nsense
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aforementioned
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it
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ancc
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artition
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VERTEX-COVER.
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sing
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for
illustration.
illustration.
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NP-hard
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problems
have
have
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yet
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to
be
be
solved
solved
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in
polynomial
polynomial
all
the
elements
inside
the
array
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small
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100]),
this
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aforementioned
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we
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inspect
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recurrence
formula
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o
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lemma.
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proof
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Let
us
us
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assume
actually
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similar
V
V
−
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in
uld
inspect
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partitioned
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recurrence.
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m.
efficient
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this
isisisnwhere
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fe SUBSET-SUM.
all
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than
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Programming,
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thoug
Any
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with
length
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c
1,
n
=
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∨
=
2Therefore,
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part
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ere
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isthe
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toin
Nathere
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is
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us
assume
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V
−
S
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not
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INDEPENDENT-SET.
=
{f
(n)}
,
and
we
define
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first
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terms
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F
.
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want
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know
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3.
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consecutive
subsequence
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length
multiples
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Therefore,
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there
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two
vertices
vertices
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A,
B
B
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∈
V
V
−
−
S
S
and
and
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an
are
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A
A
B
and
and
∈
V
B.
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−
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Since
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and
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B
B
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∈
is
V
Van
−
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Fedg
is
n=1
However,
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sequence
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Lemma
2.
2.
If
If
⊆
⊆
V
V
is
is
a
a
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VERTEX-COVER
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graph
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V
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S
S
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f
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ch
h
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MAX-INDEPENDENT-SET
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refore,
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length
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consecutive
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with
multiples
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=
aforementioned
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we
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formula
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closely.
ed
recurrence.
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we
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inspect
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recurrence
more
graph
graph
G(V,
G(V,
E)
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of
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cem
PARTITION
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atF
most
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its
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note
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with
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multiples
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Proof.
ave
B
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/
S.
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A
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that
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is
a
l
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contradiction.
contradiction.
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Therefore
V
V
−
−
S
S
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an
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contradiction.
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−
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INDEPENDENT-SET.
∞
of
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+{f1),
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Proof.
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any
consecutive
subsequence
of
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with
length
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three,
which
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(n)}
we
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(N
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to
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first
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terms
of
F
.
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want
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Pick
any
consecutive
with
length
multiples
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which
n=1
ge
value
of
N
,
it
is
unlikely
to
be
able
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find
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algorithm
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A
and
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in
an
Any
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partitioned
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3.
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

Suppose
Suppose
we
we
already
already
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know
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MIN-VERTEX-COVER
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is
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NP-hard
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problem.
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Therefore,
we
we
can
can
show
show
ny consecutive
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F(a+
with
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Proof.
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The
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isis1),
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actually
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similar
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the
previous
previous
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lemma.
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us
uspartition
assume
assume
that
that
Vinto
Vinstance,
−
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isis not
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aa INDEPE
INDEP
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fby (a
1),
f{f
+
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for
some
aabe
<
We
can
F
into
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by
F
=
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(a),
fproof
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(a
2),
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for
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Ffor
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+
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for
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will
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and
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such
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elements
N/2
b−
a(a
−
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−
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any
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A
Theorem
Theorem
1.
1.
1.
If
If
S
S
⊆
⊆
V
V
is
is
a
MAX-INDEPENDENT-SET
MAX-INDEPENDENT-SET
of
of
graph
graph
If
S
⊆
G(V,
G(V,
V
a
MAX-INDEPENDENT-SET
then
then
V
V
−
a
a
MIN-VERTEX-COVER
MIN-VERTEX-COVER
of
graph
G(V
is
sequence
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defined
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very
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sense
that
F
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defined
using
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of
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also
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NP-hard
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by
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reducing
a
a
MIN-VERTEX-COVER
MIN-VERTEX-COVER
problem
problem
into
into
a
a
algorithm
for
solving
a
problem
like
SUBSET-SUM.
While
O(2
)
is
still
exponential
to
Therefore,
Therefore,
there
there
are
are
two
two
vertices
vertices
A,
A,
B
B
∈
∈
V
V
−
−
S
S
and
and
there
there
is
is
an
edge
edge
connecting
connecting
A
and
and
B.
B.
Since
Since
{f
(a),
f
+
1),
f
(a
+
2),
...f
(b)}
for
some
a
<
b.
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rtition
F can
into
ARTITION
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from
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{f
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tion into
 −2 2
elements
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
a
−
b
−
a
−
2so
−
a
b
−
a
−
2
N
N
3
3
{f
(a),
fIfPick
(a
+
1),
f
(a
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for
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a
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b.
We
can
partition
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into
orem
1. graph
Swe
⊆thus
Vproblem.
is
a
MAX-INDEPENDENT-SET
of
graph
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E),
then
V
−
S
is
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of
of
the
the
graph
G(V,
G(V,
E).
E).
of
the
graph
G(V,
E).
any
consecutive
subsequence
of
F
with
length
multiples
of
three,
which
we
will
denote
by
F
=
Proof.
Therefore,
should
inspect
the
recurrence
formula
more
closely.
h
in
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a
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using
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we
we have
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B
Bable
∈
/∈
/Fany
S.
S.
As
A
A=and
and
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B(a
are
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an
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note
note
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isis1∪three,
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Alength
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k{f
}}we
∪
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+that
11+
0+
≤
}
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{f
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valueconsecutive
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unlikely
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Pick
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which
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wil
Proof.
A
=
(a
+
3k),
0
≤
k
≤
}
{f
(a
1
+
3k),
0
≤
A
=
{f
(a
3k),
0
≤
k
≤
}
∪
{f
(a
3k),
0
≤
k
≤
any
of
with
multiples
of
three,
we
will
denote
by
F
=
b
−
a
−
2
b
−

b
−
a
−
2
3
3
b
−
a
−
2
3
3
he
graph
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{f1−(a),
fisapproach.
(a
+E).
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some
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<sense
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We
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into3 the } ∪ {f
this
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a
2 3k),
b−
ain
−
2−
3− (a
A
=
{f
(a
+
0
≤
k
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contradiction.
contradiction.
Therefore
Therefore
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V
−
S
S
is
is
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INDEPENDENT-SET.
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a
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+
0
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}
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sequence
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a
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that
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b
a
−
2
b
−
a
−
2
e Proof.
subsequence
of(b)}
F 1that
with
of
three
can
be
partitioned
two
{f
(a),
f0(a
(a
+
1),
f0(a
+
2),
(b)}
for
athat
<1into
b.
partition
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∪
{f
(a
+
+ 3k),
≤
kisproblem
≤
}...f
1),
f (a +} We
2),
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for
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a
<
We
can
partition
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into
Proof.
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know
know
that
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V{f
−
S
S+multiples
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a vertex
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by
lemma
lemma
know
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only
only
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thing
thing
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3
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Proof.
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We
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that
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V
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−
−
S
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is
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minimality.
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b
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a
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From
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MIN-VERTEX-COVER
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easily
constructed
if
we
have
a
an
anMAX-INDEPENDENT-SET.
edge
edge
connecting
connecting
and
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B
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∈
/3k)
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−
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we
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Bf∈
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and
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are
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and
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have
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and will
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We
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−
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|V
−
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Jonathan
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V
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−
S
S
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a
Jonathan
Irvin
GUNAWAN
Jonathan
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b −an
a−
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f
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f
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ubsequence
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this
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From
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n the
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good
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If
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V
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3
11We
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using
X
problem
by
Y
problem
as
a
black-box
Y
as
a
black-box
.
If
we
can
.
by
the
construction
of
then function.
by
construction
of
the
bysequence
the
construction
of time
the
function.
∞ function.
1,
1∨
=
2first
1,
n
=define
1then ∨solve
n F=
2n
) SUBSET-SUM
algorithm
running
for
various
input
sizes.
This
the
We
consider
the
=
{f
, and
weX
(N
)= well.
to
bebe
the
ofimpossible
F
We impossible
want
to know
NP-hard
with
large
Ncan
that
is
impossible
to
exponentially
(e.g.NY
Nterms
>
50).
Theorem
2.FIf =
is(n)}
divisible
by
three,
can
partitioned
two
multisets
ormula
n=1
fwe
(n)
fthe
(n)function.
=
solveproblem
problem
Y,
problem
then
Y,
then
solve
we
can
problem
solve
problem
as
well.
X
Therefore,
as
Therefore,
problem
problem
isinto
also
is .also
to solve.
to solve
Theorem
2.
If
N
is
divisible
by
three,
then
F
(N
)
can
be
partitioned
into Y
two
multisets
of equal
sum.
nstruction
of solve
by
the
construction
of
the
function.
by
the
construction
of
the
function.
1,
n
=
1
∨
n
=
2
3.3
Finding
Small
Constraints
f
N
on
a
MacBook
Pro
(Retina,
13-inch,
Early
2015)
f
(n
−
1)
+
f
(n
−
2),
n
>
2
fpartition
(n
−N
1)
+
fSUBSET-SUM
(n
2),
n >algorithm
2then
hether
can
asum.
of
Fthat
(N−function.
)may
into
two
disjoint
A
suchproblem
that
the
sum
allpartitioned
elements
ofcreate
equal
also
lookwe
small
constraints
For
acan
SUBSET-SUM
N
N/2
Theorem
2.example,
IfF
N
is)) can
divisible
by
then
F
(N
) canof
be
int
by
the
of
the
Theorem
2.
IfIf
by
three,
(N
be
partitioned
into
two
multisets
of
equal
f for
(n)
=
Theorem
2.
N
is
divisible
byhelp.
three,
then
Fmultisets
(N
beand
partitioned
into
two
multisets
of
equal sum.
sum.
omparison
ofother
O(2
)construction
and
O(2
) divisible
running
time
forBthree,
various
input
sizes.
This
 is
by
construction
of
the
function.
fthe
(n
−the
1)
+
felements
(n
−
2),
n
>
2
by
the
construction
of
the
function.
by
construction
of
the
function.
by
the
construction
of
the
function.
nrd
A
is
equal
to
the
sum
of
all
in
B
problem,
and
will
not
be
solvable
with
N
=
100.
When
we
are
given
the
constraint
that
∞
Proof.
If
N
is
divisible
by
three,
then
F
(N
)
is
a
prefix
of
F
with
length
multiples
of
three.
By
lemm
2.
If
N
is
divisible
by
three,
then
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum.
1,
n
=
1
∨
n
=
2
∞
Theorem
2.
If
N
is
divisible
by
three,
then
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
by
the
construction
of
the
function.
Theorem
2.
Ifthen N
isthe
divisible
by
three,
F>
(N
) problems
can
partitioned
into
two mult
is run
100
times
for
each
value
N, and
on
a(N
MacBook
13-inch,
Early
2015)
nsider
the
F
=
we
define
FPro
(N
)(Retina,
the
first
terms
of
F
.be
We
want
totoknow
We=sequence
are
using
We
are
NP-hard
using
problems
NP-hard
problems
for
illustration.
for
Recall
Recall
NP-hard
that
problems
NP-hard
have
yet
to
have
be of
yet
solved
be
in polynomia
solved
in p
quence
F
{fProof.
(n)}
,N
and
we
define
F three,
) to
to
beillustration.
first
Nabe
terms
ofthen
FN
We
want
tomultiples
know
rd
problem
with
large
that
isof
impossible
solve
exponentially
(e.g.
50).
If {f
is=(n)}
divisible
by
to
isthat
prefix
of .N
with
length
n=1
fN
(n)
2 length
Proof.
Ifof
N(N
is))F
divisible
bybe
three,
then
Finto
(N multiples
)multiples
is amultisets
prefix
F equal
with
length
Theorem
2. is
Ifdivisible
N
isn=1
divisible
by
three,
can
partitioned
two
of
sum.
Proof.
IfIf
by
three,
then
F
is
aa )prefix
of
F
of
three.
By
Proof.
N
is[0,
divisible
by
three,
then
Fthen
(N
is(N
prefix
of
F with
with
length
of of
three.
By lemma
lemm
∞
the
array
are
small
(e.g.
100]),
this
problem
can
be
solved
using
O(N
X
)to
Dynamic
can
be
partitioned
into
two
multisets
equal
sum..
f
(n
−
1)
+
f
(n
−
2),
n
>
2
eer
F
=
{f
(n)}
,
and
we
define
F
(N
)
to
be
the
first
N
terms
of
F
.
We
want
know
polynomial
time
using
the
reduction
proof
of
the
original
problem
(withwe
can
create
a
partition
of
F
(N
)
into
two
disjoint
multisets
A
and
B
such
that
the
sum
of
all
elements
three.
By
lemma
3, can
be
partitioned
into
two
multisets
of
equal
sum.
time
for
more
time
for
than
more
4
decades.
than
4
decades.
MIN-VERTEX-COVER
MIN-VERTEX-COVER
is
a
graph
is
problem
a
graph
that
problem
involves
that
finding
involves
a
finding
minimum
a
minimu
subse
ate
a
partition
of
F
(N
)
into
two
disjoint
multisets
A
and
B
such
that
the
sum
of
all
elements
ook
for
other
small
constraints
that
may
help.
For
example,
a
SUBSET-SUM
problem
Theorem
If
N
divisible
bypartitioned
three,
then
(N
)be
can
be
partitioned
into
of equa
n=1
Theorem
2.can
IfF
N
is))F
by
three,
then
F
(N
)two
besum.
partitioned
Theorem
2.
N
is)divisible
divisible
by
three,
Fsum..
(N
can
be
partitioned
into
two
multisets
of three.
equal
sum
Theorem
2.
IfIfN
N
is
by
three,
then
(N
can
partitioned
into
two
multisets
of
equal
sum
Nfunction.
isthe
divisible
by
three,
then
F2.
(N
istwo
aisIfproblem.
prefix
of
multiples
of
By
lemma
3,can
Fmultisets
(N
)of
Ifelements
is
then
F
(N
)problem
isdivisible
a prefix
of
length
Byin
Theorem
2. be
IfProof.
N
isthe
divisible
bydivisible
three,
then
(N
)then
be
partitioned
into
two
multisets
of
equal
can
beFF
into
two
ofprefix
equal
sum..
Proof.
Nby
isthree,
divisible
bylength
three,
then
Fmultisets
(N
)three.
isF
awith
of
F multiples
with
length
multiples
can
be
partitioned
into
multisets
of
equal
can
be
partitioned
into
two
multisets
ofwith
equal
sum..
looks
like
classic
PARTITION
PARTITION
isthe
NP-hard
by
reduction
from
SUBSETX
isThis
upper
bound
of
inside
the
array.
Even
though
the
running
time
of
the
then
Fnodes
(N
)will
can
partitioned
into
two
multisets
of
equal
sum.
partition
ofthe
Fcan
(N
)aholds
into
two
disjoint
multisets
Athen
and
Bequal
such
that
the
sum
of
allin
elements
ther
the
proof
still
given
the
additional
constraint
to
the
problem.
s
equal
to
sum
of
all
elements
in
B
of
of
such
nodes
that
such
for
every
that
for
edge,
every
at
least
edge,
one
at
least
of
its
one
endpoint
of
its
endpoint
is
in
subset.
is
the
MAX-INDEPENDENT-SET
subset.
MAX-INDEPENDENT
is a
e
sum
of
all
elements
in
B
oblem,
and
not
be
solvable
with
N
=
100.
When
we
are
given
the
constraint
that
Proof.
If
N
is
divisible
by
three,
F
(N
)
is
a
prefix
of
F
with
length
multiples
of
three.
By
lemma
∞
then
F
(N
)
be
partitioned
into
two
multisets
of
sum.
3.the
If
N
≡
1 be
(mod
3),
F
(N
))to
bebeof
partitioned
into
two
multisets
of(withequal
sum.
er
the
F
=
{f
(n)}
,can
and
we
define
Fusing
(N
to
be
the
first
Nofconstraint.
terms
of
FAnother
. We
want
to
know
artitioned
intoofTheorem
two
ofalgorithm
equal
sum..
asequence
problem
toofmultisets
be
solvable
in
polynomial
time
the
reduction
proof
ofalgorithm
the
original
problem
nto
two
multisets
sum.
Theorem
3.
If , three,
cannot
cannot
partitioned
into
two
multisets
of
can
be
partitioned
into
two
multisets
of
equal
sum..
into
two
multisets
equal
sum..
n=1
UM.
Therefore,
aequal
large
value
of
N
,ispartitioned
it
is
unlikely
be
able
to
an
that
finds
A
Bfunction.
inBy
anlengt
lauses
tothe
the
size
the
input,
still
fast
enough
for
the
given
sible
by
three,
then
F
(N
)100]),
can
be
partitioned
into
two
multisets
equal
sum.
Proof.
If
N
isthe
divisible
by
then
F
(N
)partitioned
isO(N
afind
prefix
of
F
with
length
multiples
of
three.
By
2
by
the
construction
the
Proof.
Ifnodes
N
is
divisible
by
three,
then
F
(N
)of
isbe
aone
prefix
of
F
with
Proof.
N
is
divisible
by
three,
then
F
(N
is
aF
prefix
of
F
with
length
multiples
ofand
three.
By
lemm
Proof.
IfIf
N
is
divisible
by
three,
then
F
(N
))of
is
a≡
prefix
of
F
with
length
multiples
of
three.
lemm
by
the
construction
of
the
function.
by
construction
the
function.
by
construction
of
the
function.
of
all
elements
infor
B
graph
problem
graph
of
problem
finding
of
a
finding
maximum
a
maximum
subset
of
subset
such
nodes
that
such
for
every
that
for
edge,
every
at
most
edge,
at
most
of
its
one
endpoint
of
its
endp
is
in
Proof.
If
N
is
divisible
by
three,
then
F
(N
)
is
a
prefix
of
with
length
multiples
three.
By
lemma
3,
F
(N
array
are
small
(e.g.
[0,
this
problem
can
be
solved
using
X
)
Dynamic
Theorem
3.
If
N
1
(mod
3),
F
(N
)
cannot
partitioned
into
two
m
can
be
partitioned
into
two
multisets
of
equal
sum..
Theorem
3.
If
N
≡
1
(mod
3),
F
(N
)
cannot
be
into
two
multisets
of
equal
sum.
the
construction
of
the
function.
Theorem
3.
If
N
≡
1
(mod
3),
F
(N
)
cannot
be
partitioned
into
two
multisets
of
equal
sum.
efficient
aHowever,
partition
ofofFN
(N
into
two
disjoint
multisets
A and
Bexample,
the sum
all
elements
equal
sum.
Fcan
(N))create
isway.
a prefix
prefix
ofand
F check
with
length
multiples
ofstill
three.
By
lemma
3,
Fsuch
(N
)that
itional
constraint),
whether
the
proof
holds
given
the
additional
constraint
toNofFthe
problem.
this
sequence
is
defined
in
athat
very
special
way,
in
the
sense
that
is
defined
using
the
not
F (N
nique
is
when
the
value
is)inside
too
large
(e.g.
N
<
40).
For
while
O(2
)elements.
F
(N
is
a≡
of
F
with
length
multiples
of
three.
By
lemma
3,
)contains
can
be
partitioned
into
two
multisets
of
equal
sum..
n.
can
be
partitioned
into
two
multisets
of
equal
sum..
can
be
partitioned
into
two
multisets
of
equal
sum..
can
be
partitioned
into
two
multisets
of
equal
sum..
Proof.
Suppose
F
(N
)
=
F
(N
)
−
f
(1).
Note
that
F
(N
)
N
−
1
Since
N
≡
1
(m
3.
If
N
1
(mod
3),
F
(N
)
cannot
be
partitioned
into
two
multisets
of
equal
sum.
l
case
in
a
NP-hard
problem.
Suppose
we
have
a
function
with
the
th
multiples
of
three.
By
lemma
3,
F
(N
)
Theorem
3.
If
N
≡
1
(mod
3),
F
(N
)
cannot
be
partitioned
into
two
multisets
of
equal
sum.
his
looks
like
a
classic
PARTITION
problem.
PARTITION
problem
is
NP-hard
by
reduction
from
SUBSETthe
subset.
the
Suppose
subset.
we
Suppose
already
we
know
already
that
know
MIN-VERTEX-COVER
MIN-VERTEX-COVER
is
a
NP-hard
is
a
NP-hard
problem.
problem.
Therefore,
Therefore,
we
can
show
we
can
be
partitioned
into
two
multisets
of
equal
sum..
a
classic
PARTITION
problem.
PARTITION
problem
is
NP-hard
by
reduction
from
SUBSEThe
upper
bound
of
the
elements
the
array.
Even
though
the
running
time
of
the
Theorem
3.
If
N
≡
1
(mod
3),
F
(N
)
cannot
be
partitioned
into
two
multisets
of
e
yis
three,
then
F
(Nall
is
a3.
prefix
of
F1 )can,
with
length
multiples
ofby
three.
By
lemma
3,
FNMiddle
(N
)1of
ual
to the
sum
of
elements
in
B
of
equal
sum..
 three,
Fdivisible
Theorem
2.
Ifthen
Nsum.
is
by
three
N/2
≡
=
Theorem
2.
If
N
istwo
divisible
by
three,
(N
) can
be
part
(N
Theorem
2.
If
N
is)divisible
divisible
by
three,
then
F(N
(N
)fis
can
be
partitioned
into
two
multisets
equal
sum.
Theorem
2.
If
N
divisible
then
F
(N
)two
can
be
partitioned
into
two
multise
Proof.
Suppose
F
(N
)
F
(N
)
−
f
(1).
Note
that
F
(N
)
contains
N
−
forementioned
recurrence.
Therefore,
we
should
inspect
recurrence
formula
more
closely.
Theorem
If
N
(mod
3),
F
(N
)
cannot
be
partitioned
into
multisets
of
equal
sum.
the
Proof.
Suppose
F
(N
=
F
(N
)
−
(1).
Note
that
F
(N
)
contains
−
elements.
Since
N
≡
1
(m
unlikely
to
run
in
one
second,
we
for
instance,
use
a
O(2
)
Meet
In
The
Proof.
Suppose
F
(N
)
=
F
(N
)
−
f
(1).
Note
that
F
)
contains
N
−
1
elements.
Since
N
≡
1
(m
heorem
2.
If
N
is
by
three,
then
F
)
can
be
partitioned
into
multisets
of
equal
f
equal
sum..
Proof.
Suppose .
Note
that contains elements.
n.
haveN
−
1
≡
0
(mod
3).
By
lemma
3,
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum.
The
sic
PARTITION
problem.
PARTITION
problem
is
NP-hard
by
reduction
from
SUBSETTherefore,
for
aoflarge
of3.NIfis,case
itstill
is≡
unlikely
to
be
able
to find
an
algorithm
that
finds
Awith
and the
Bofin
an sum.
that
that
MAX-INDEPENDENT-SET
is
also
aNP-hard
NP-hard
is
also
aproblem.
NP-hard
problem
by
problem
reducing
by
reducing
ahave
MIN-VERTEX-COVER
aB
MIN-VERTEX-COVER
problem
into
probla
or
asize
large
value
N
,sum..
itvalue
is
unlikely
be
able
to
find
an
algorithm
that
finds
and
in multisets
an
he
ofMAX-INDEPENDENT-SET
the
input,
the
algorithm
fast
enough
the
given
constraint.
Another
one
wo
multisets
equal
(mod
 3),into
Theorem
N
1two
(mod
3),
cannot
be
partitioned
into
two
equal
give
we
example
of
a1fSuppose
special
in
we
a)N
function
 N
hree,
then
Fof
)Fcan
partitioned
multisets
sum.
Theorem
If)F
N
≡
13).
FA
(N
cannot
be
partitioned
into
two
m
Theorem
3.
N
≡3),
13).
(mod
3),
F)F(N
(N
)F
cannot
be
partitioned
into
two
multisets
ofbe
equal
sum.
Theorem
3.
IfIfWhile
N
≡
1to
3),
F
)for
cannot
be
partitioned
two
multisets
of
equal
sum.
3.
rtitioned
into
two
multisets
of
equal
sum.
of
uppose
Flike
(N
)(N
=
(N
)be
−elements
Note
that
(N
contains
−
1be
elements.
Since
≡
1)F)can
(mod
3),
we
Proof.
F
(N
)have =
Fstill
(N
)sum.
−
(1).
Note
that
FA
(N
)elements
contains
−
1sum.
elements.
Since
N
≡
Theorem
3.
If
N
≡
(mod
F
(N
)aF
cannot
be
partitioned
into
two
multisets
equal
haveN
−
1F
≡
(mod
By
lemma
3,
Ftwo
(N
be
partitioned
into
tw
Proof.
Suppose
(N
)f(N
=0)equal
(N
)Suppose
−
(1).
Note
that
Fof
(N
contains
N
−
1sum.
element
N/2
equal
problem
SUBSET-SUM.
O(2
is
exponential
to
N
,fpartitioned
it
isof
much
faster
than
haveN
−
11sequence
≡
0(1).
(mod
3).
By
lemma
3,
F
(N
can
into
multisets
of
equal
The
Since ,into
we
.(mod
By
lemma
3, is(N
can
haveN
−all
≡
0not
(mod
By
lemma
3,
(N
)N
can
be
partitioned
into
two
multisets
of
sum.
The
Nin
sum
of
in
F
(N
)
is
even.
However,
the
sum
all
in
)
=
F
(N
)
+
f
(1),
wh
arge
value
of
N
,
it
is
unlikely
to
be
able
to
find
an
algorithm
that
and
B
an
))ever,
cannot
be
partitioned
into
two
multisets
of
equal
too
finds
Proof.
If
N
divisible
by
three,
nt
way.
However,
this
is
defined
in
a
very
special
way,
in
the
sense
that
F
is
defined
using
the
MAX-INDEPENDENT-SET
MAX-INDEPENDENT-SET
problem.
problem.
this
sequence
is
defined
in
a
very
special
way,
in
the
sense
that
F
is
defined
using
the
is
when
the
value
of
N
is
large
(e.g.
N
<
40).
For
example,
while
O(2
)
Proof.
If
N
is
divisible
by
three,
then
F
(N
)
is
a
prefix
of
F(m
w
Proof.
If
N
is
divisible
by
three,
then
F
(N
)
is
a
prefix
of
F
with
length
multiples
of
three.
By
lemma
3,
F
(N
)then
Proof.
If
N
is
divisible
by
three,
then
F
(N
)
is
a
prefix
of
F
with
length
multiples
of
Any
consecutive
subsequence
of
F
with
length
multiples
of
three
can
be
partitioned
into
two
multisets
emma
3.
Proof.
Suppose
F
)
=
F
(N
)
−
f
(1).
Note
that
F
(N
)
contains
N
−
1
elements.
Since
N
≡
1
(N
oof.
If
N
is
divisible
by
three,
then
F
(N
)
is
a
prefix
of
F
with
length
multiples
of
three.
By
lemma
3,
F
(N
)
cannot
be
partitioned
into
two
multisets
of
equal
sum.

oks
like
a
classic
PARTITION
problem.
PARTITION
problem
is
NP-hard
by
reduction
from
SUBSEThree,
then
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum.
ormula


N




1of≡N
03),
(mod
3).
By
lemma
3,
F
(N
)F
can
partitioned
into
two
of
equal
sum.
Therefore,
multisets
of
equal
sum.
haveN
1very
≡
(mod
3).
By
lemma
3,equal
FTherefore,
(N
)multisets
can
be
into
two
equal
sum
of
all
elements
F
)F
is
even.
However,
the
sum
of
all
elements
haveN
−multiples
1))way,
≡
0using
(mod
3).
By
lemma
3,defined
(N
can
be
partitioned
into
multisets
partitioned
into
two
multisets
of
equal
sum.
the
sum
of
all
elements
in  twice
(N
partitioned
that
can
solve
is
the
range
of
N
aaformula
O(2
)partitioned
solution.
sum
of
all
elements
in
)be
is
However,
the
of
all
elements
in
(N
=
F
(N
)of
+
ff(1),
wh
 in
 the
0
sum
of
all
elements
in
F
(N
is
However,
the
sum
of
all
elements
in
F
(N
)multisets
=
F
(N
)contains
+Since
(1),
wh
we
(N
by
the
construction
the
function.
nSuppose
=
1−
n
=
2special
mod
F
(N
cannot
be
partitioned
two
multisets
of
sum.
because
F
(N
)=
is
even
while
fsum
(1)
isinto
odd.
Therefore,
there
is
no
to
partition
F
(N
).
Proof.
Suppose
F
(N
=
F
(N
)(1).
−
(1).
Note
that
F
(N
))1can
contains
−
1)equal
elements.
N
≡−
this
sequence
defined
in
a∨
in
the
sense
that
Fsum
is
using
the
of
N/2
f
be
partitioned
into
two
multisets
mentioned
recurrence.
Therefore,
should
inspect
the
recurrence
formula
more
closely.
then
F
(N
)itaSuppose
is
ais
prefix
of
F
with
length
three.
By
lemma
3,
F
(N
)way
Proof.
Suppose
F
(N
)F
=
F
)−
−
felements.
(1).
Note
that
FN
)1two
Proof.
Suppose
F
(N
)of
=
F
(N
)even.
−
Note
that
(N
)(N
contains
N
−
1F
elements.
Since
N
≡ The
(m
Proof.
F
(N
)into
=
(N
)even.
−
ffof
(1).
Note
that
F
(N
)into
contains
N
−
1N
elements.
Since
N
≡
11N
(m
urrence.
Therefore,
we
should
inspect
the
recurrence
more
closely.
ikely
to
run
in)multiples
one
second,
we
can,
for
instance,
use
O(2
Meet
In
The
with
length
three.
By
lemma
3,
can
be
two
multisets
of
sum..
can
be
partitioned
into
two
equal
sum..
can
be
partitioned
two
of
equal
sum..
frefore,
equal
sum.
Proof.
F
(N
)all
F
(N
)unlikely
−
fF
(1).
Note
that
F≡
))an
contains
N
Since
≡
(mod
w
haveN
−
1of
≡
0multisets
(mod
3).
By
lemma
3,
F
(N
)(N
can
be
partitioned
into
two
of(N
equal
 into
 Middle
be
partitioned
two
multisets
of
equal
sum..
sum.3),
for
large
value
of
N
,elements
it
is
to
be
able
to
find
algorithm
that
finds
Amultisets
and
B
in
an
elements
 multisets
of

ln).
elements
in
F
)
is
even.
However,
the
sum
of
all
in
F
(N
)
=
F
(N
)
+
f
(1),
which
is
odd
1).
Note
that
F
(N
)
contains
N
−
1
elements.
Since
N
1
(mod
3),
we
sum
in
F
(N
)
is
even.
However,
the
sum
of
all
elements
in
)
=
F
(N
)
+
f
(p
is
even.
However,
the
sum
of
all
elements
in ,
which
is
because
F
(N
)
is
even
while
f
(1)
is
odd.
Therefore,
there
is
no
way
to
(N
sum
of
all
elements
in
F
(N
)
is
even.
However,
the
sum
of
all
elements
in
F
(N
)
=

because
F
(N
)
is
even
while
f
(1)
is
odd.
Therefore,
there
is
no
way
to
partition
F
(N
).
by
the
construction
of
the
function.


because
F
(N
)
is
even
while
f
(1)
is
odd.
Therefore,
there
is
no
way
to
partition
F
(N
).
more
N/2
Note
that
F
(N
)
contains
N
−
1
elements.
Since
N
≡
1
(mod
3),
we
haveN
−
1
≡
0
(mod
3).
By
lemma
3,
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum
ce.
Therefore,
we
should
inspect
the
recurrence
formula
closely.
hen
F
(N
)
is
a
prefix
of
F
with
length
multiples
of
three.
By
lemma
3,
F
(N
)



sets
of
equal
sum..
haveN
−
1
≡
0
(mod
3).
By
lemma
3,
F
(N
)
can
be
partitioned
into
tw
haveN
−
1
≡
0
(mod
3).
By
lemma
3,
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum.
Th
We
will
first
We
prove
will
first
the
prove
following
the
following
two
lemmas.
two
lemmas.
haveN
−
1
≡
0
(mod
3).
By
lemma
3,
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum.
Th
em
SUBSET-SUM.
O(2
) (N
is
still
tobe
N∨partitioned
, nitthe
is
faster
than
n
−
1)
+
f (n
−
2),
nisWhile
>
213).is
−F(N
1Theorem
elements.
Since
N
≡
haveN
−be
1this
≡
0sequence
By
lemma
3,
Fexponential
(N
)(1)
can
be
partitioned
into
two
of
equal
sum.
Therefore,
th
(mod
sum
all
elements
F
)while is
even.
However,
the
sum
of
all
elements
in to
F
(N
)two
=≡
F
) +). f3),
(1),F (N
wh
1,two
n
=
1to
=
2much
like
N
ay.
However,
defined
inwith
very
special
way,
in
sense
that
F
is
defined
using
the
Ifof
N
divisible
by
three,
then
F
(N
))multiples
can
two
multisets
equal
sum.
while
Any
subsequence
of
Fwe
length
of
three
can
partitioned
into
multisets
ma
3.
=
(N
)is)−can
fconsecutive
(1).
Note
that
Finto
(N
)≡
contains
N
−
1
elements.
Since
N
≡
1into
(mod
3),
we
Theorem
4.
If
2is
(mod
3).
F
(N
can
be
partitioned
into
two
multisets
of
equal
sum.
subsequence
of
F
with
length
of
can
be
partitioned
into
two
multisets
F
) Pick
even
while
f1(mod
is=
odd.
Therefore,
there
is
no
way
partition
F≡
(N
).
,nsecutive
(N
partitioned
multisets
of
equal
sum.
Therefore,
the
odd
because in
is3),
even
length
is
odd.
Therefore,
there
is
no
way
partition because
Fsubsequence
(N
)(N
even
f≡
is
odd.
there
ismultisets
no
way
to
partition
F
(N
Theorem
3.
Ifin
N
1)(N
(mod
because
Faof
(N
)be
is
even
while
fTherefore,
(1)
is
Therefore,
there
is
no
way
partition
F
Theorem
3.
If
N
1be
(mod
3),
Fwe
(N
)of
cannot
be
partitioned
i(
F

Theorem
3.
If2.
N
≡
(mod
3),
F
)F
cannot
partitioned
into
two
multisets
of
equal
sum.
any
consecutive
with
multiples
of
three,
which
will
denote
by
F
=
roof.
odd.
to
multiples
multisets
N
Theorem
3.
If
N
1three
(mod
3),
F
(N
)the
cannot
be
partitioned
into
two
of
equ
f(1)
(n)
heorem
3.
Ifsolve
N
≡
1
(mod
3),
F
(N
)
cannot
be
partitioned
into
two
multisets
of
equal
sum.
,ets
F of
(N
)can
can
be
partitioned
into
two
multisets
of
equal
sum.
Therefore,
the
3.4
Finding
Special
Cases
sum
all
elements
in
F
(N
)
is
even.
However,
sum
of
all
elements
F
(N
=
F
(N
)
+
f
equal
sum..
range
 (N
 of
sum
of
all
elements
in
F
(N
)
is
even.
However,
the
sum
of
all
elements
sum
of
all
elements
in
(N
)
is
even.
However,
the
sum
of
all
elements
in
F
(N
)
=
F
(N
)
+
f
(1),
w
sum
of
all
elements
in
F
(N
)
is
even.
However,
the
sum
of
all
elements
in
F
)
=
F
(N
)
+
f
(1),
w
that
it
is
twice
the
of
N
using
a
O(2
)
solution.
wo
multisets
of
equal
sum.
Therefore,
the
sum
of
all
elements
in
F
(N
)
is
even.
However,
the
sum
of
all
elements
in
F
(N
)
=
F
(N
)
+
f
(1),
which
is
od

Theorem
4.
If
≡know
2then
(mod
3).
F
(N
)−
can
be
partitioned
intomultisets
twoofmult
because
F
(N
)Sinto
is
while
ff3).
(1)
is (N
odd.
Therefore,
there
no
way
tothen
partition
F is
(N
).sum.
F
Theorem
4.
If
N
≡
22N
(mod
F
(N
)) +
can
be
partitioned
into
two
multisets
of
equal
Theorem
2.
If+
N
is
divisible
by
three,
Fis
(N
)closely.
can
be
partitioned
into
two
o(
Theorem
4.
If
N
≡
(mod
3).
(N
can
be
partitioned
into
two
multisets
of
equal
sum.
recurrence.
Therefore,
we
should
inspect
the
recurrence
formula
more
 E),
tive
subsequence
of
F
with
length
multiples
of
three
can
be
partitioned
into
two
multisets
feven
(n
−
1)
(n
−
n
>
2Nto
we
define
Fsum
)of
be
the
first
terms
of
F
We
want
ual
sum.
By
lemma
3,
Ff(N
(N
)to
can
be
partitioned
into
two
multisets
of
equal
sum.
Therefore,
the
.of
However,
the
all
elements
in
F
(N
)
=
F
)
f
(1),
which
is
odd
F
(N
)
cannot
be
partitioned
two
multisets
of
equal
sum.
Lemma
1.
Lemma
1.
If
S
⊆
V
If
is
an
⊆
INDEPENDENT-SET
V
is
an
INDEPENDENT-SET
of
graph
of
G(V,
graph
G(V,
then
E),
V
S
is
V
a
−
VERTEX-COVER
S
a
VERTEX-COV
the
dand
into
two
multisets
equal
sum.
 .2),

foned
(a),
f
(a
+
1),
(a
+
2),
...f
(b)}
for
some
a
<
b.
We
can
partition
F
into



 (N
sum
elements
in
F
(N
)4.
=
F
) B.
+
fSince
(1),
which
isof
odd
 (N
because
F
(N
is(1).
even
while
f3).
(1)
is
odd.
Therefore,
is
no
way
to
partition
F=
(NFF
).
If=2N
by
then
F
(N
)Therefore,
is
a≡
prefix
ofcan
F
with
length
multiples
of
three.
By
lemma
3,
is
 there
 Fway
F
(N
)contains
isSuppose
even
while
fiselements.
(1)
odd.
Therefore,
there
no
to
because
FF
(N
is
even
while
(1)
isto
odd.
Therefore,
there
no
way
to
partition
F
(N
).
because
F
(N
)f)which
is
even
while
fffbecause
(1)
is
odd.
Therefore,
there
way
to
partition
F
).
divisible
Proof.
(N
now
Proof.
Assign
(1)
A
and
(2)
f−isbe
(1)
=
f−
we
are
partition
(N
4.
If
N
(mod
3).
(N
))three,
can
be
partitioned
into
two
multisets
sum.
sHowever,
inProof.
F
(N
)≡the
F
(N
)Theorem
+
fno
(1),
is
odd
4.
If
N
≡
2Note
(mod
F
(N
)(N
partitioned
into
multisets
equal
sum.
Proof.
Suppose
F
(N
)multisets
(N
)F
−
f) )(=
IOI,
because
F
(N
)of
is
while
f)to
(1)
is
odd.
there
to
partition
F
(N
).into
)3).
Theorem
Ifthat
N
2=
FN
(N
)equal
can
be
partitioned
two
of
equ
(N
)no
=
F
(N
)Since
−
(1).
Note
that
(N
) con
Suppose
F
(N
)all
=
F
(N
)−
fall
F
(N
)paper.
N
1is
N≡≡
1of
(mod
3),
we
Proof.
Suppose
Fin
(N
)(N
F
(1).
Note
that
Ftwo
(N
)ftrying
contains
N
− is13),
elements.
ble
approach
in
and
thus
is−
main
focus
of
this
To
solve
IOI
2008
Island
nto
disjoint
multisets
A
and
B
such
that
the
sum
of
all
elements
oof.
Suppose
Fthere
(N
){f
=even
F
(N
fthe
(1).
Note
that
FFAssign
contains
−F
1(2),
elements.
Since
N
1to
(mod
we
(N
)two
is
even.
However,
the
sum
of
elements
(N
))(mod
=
F
(N
)fno
+
fway
(1),
which
is
odd
FProof.
(N
)Therefore,
cannot
be
partitioned
into
two
multisets
of
equal
sum.
dd.
is
way
to
partition
F
(N
).
∞
graph
G(V,
graph
E)
G(V,
E)
are no
 we
denote
Proof.
f
(1)
to
A
and
f
(2)
to
B.
Since
f
(1)
=
f
(2),
Theorem
4.
If
N
≡
2
(mod
3).
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum.
Any
consecutive
subsequence
of
F
with
length
multiples
of
three
can
be
partitioned
into
two
multisets
.onsecutive
r
the
sequence
F
=
(n)}
,
and
we
define
to
be
the
first
N
terms
of
F
.
We
want
to
know

Proof.
Assign
f
(1)
to
A
and
f
(2)
to
B.
Since
f
(1)
=
f
(2),
we
are
now
trying
to
partition
F
(N
)
=
Theorem
4.
If . can
be
partitioned
multisets
of
equal
Proof.
If
N
is
divisible
by
three,
then
F
(N
)
is
a
prefix
of
F
with
length
multiples
of
thre
dd.
Therefore,
there
is
no
way
to
partition
F
(N
).

Proof.
Assign
f
(1)
to
A
and
f
(2)
to
B.
Since
f
(1)
=
f
(2),
we
are
now
trying
to
partition
F
(N
)
=
can
be
partitioned
into
two
multisets
of
equal
sum..


Pick
any
consecutive
subsequence
of
F
with
length
multiples
of
three,
which
we
will
by
F
=
n=1
subsequence
of
F
with
length
multiples
of
three,
which
we
denote
by
Fthe
bfind
−
aa2−
23).
blemma
−
a 1obtain
−
2=
f(mod
(1)
−3).
fapproach.
(2).
F
(N
)We
will
have
N
−
elements.
Since
N
2will
(mod
3),
we
N
−
≡ 0By
(mod
3).
partition
F−
).
haveN
−
≡3,
0function.
3).
lemma
−
f (1).
Note
F
(N
) Since
contains
N
−
12≡
elements.
Since
Nlemma
≡
1be
(mod
3),
we
by
the
construction
of
1
ontains
10that
elements.
≡
(mod
3),
we
haveN
−
1≡
≡
0partitioned
(mod
By
F(mod
(N
) 2equal
can
be
partition
−1N
1≡
≡
(mod
3).
By
3,1−
F
(N
)can
can
be
partitioned
into
two
multisets
equal
sum.
Therefore,
the mu
3.4
Finding
Special
Cases
haveN
0to
(mod
By
3,
F≡the
(N
)3).
can
be
partitioned
into
two
multisets
of3
we
need
to
use
this
must
special
constraint
in
problem
such
BhaveN
veN
−
0(N
By
lemma
3,
F
(N
)(mod
be
partitioned
into
two
multisets
of
equal
sum.
Therefore,
the
hile
fTheorem
(1)
is
odd.
Therefore,
there
is
no
way
partition
F
(N
).
Theorem
4.
Ifpartitioned
N
≡
3).
F
(N
)A
can
into
two
multisets
of
sum.
be
Theorem
4.
If
N
2B
(mod
3).
F
(N
)of
be
partitioned
into
two
Theorem
4.
Iflemma
NN
≡
(mod
3).
F
(N
can
be
partitioned
into
two
multisets
of
equal
sum.
Theorem
4.
If
N
≡
22≤
(mod
3).
F
(N
)})we
can
partitioned
into
two
multisets
of
equal
sum.
A
=
{f
(a
+
3k),
0
k
≤
∪
{f
(a
+
1
+
3k),
0
≤
k
≤
}
can

ssign
f
(1)
to
A
and
f
(2)
to
B.
Since
f
(1)
=
(2),
are
now
trying
to
partition
F
(N
)
=
F
(N
)
−

Proof.
Assign
f
(1)
to
A
and
f
(2)
to
B.
Since
f
(1)
=
f
(2),
we
are
now
trying
to
partition
F
4.
If
N
≡
2
(mod
3).
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum.
f
(1)
−
f
(2).
F
(N
)
will
have
N
−
2
elements.
Since
N
≡
2
(mod
3),
we
Proof.
Assign
f
(1)
to
A
and
f
(2)
to
B.
Since
f
(1)
=
f
(2),
we
are
now
trying
to
um.
can
create
a
partition
of
F
(N
)
into
two
disjoint
multisets
and
such
that
the
sum
of
all
elements
sum.


f
(1)
−
f
(2).
F
(N
)
will
have
N
−
2
elements.
Since
N
≡
2
(mod
3),
we
obtain
N
−
2
≡
0
(mod
3).
can
be
into
two
multisets
of
equal
sum..


f
(1)
−
f
(2).
F
(N
)
will
have
N
−
2
elements.
Since
N
≡
2
(mod
3),
we
obtain
N
−
2
≡
0
(mod
3).

utive
subsequence
FF
length
multiples
of
we
will
denote
by
F
with
even.
,aum
(a
+
1),
f(b)}
for
some
aHowever,
<
b.
We
can
partition
F
into
2),
...f
some
acontains
<
b.
We
can
partition
Fthree,
into
a which
3,+for
F2),
(N
) (N
can
be
partitioned
into
multisets
of
equal
which
our
theorem.
3two
3of
−
(1).
Note
Fof
(N
)(b)}
N
−
1
elements.
Since
N
≡
1elements
(mod
3),
we
)ff+
can
be
partitioned
into
two
multisets
of
equal
sum.
sum
of
all
elements
inwhich
Fthe
)(N
is (N
ma
3,
F
(N
)(athat
can
be
partitioned
into
two
multisets
of
equal
Therefore,
the
Proof.
Proof.
Let
us
assume
Let
us
that
assume
V
−
that
S
is
not
−
S
athe
VERTEX-COVER.
is
not
VERTEX-COVER.
Therefore,
Therefore,
there
are
two
are
two
A,
vertices
B
∈
/odd
V)even.
A,
−
BF
S
oned
two
multisets
of
equal
sum.
Therefore,
the
sum
of
in
F(N
(N
)implies
is
sum
of
ofinto
all
elements
in...f
(N
)fis
is
even.
the
sum
of
all
elements
in
F(N
)=
=
F (N
(N
)+
+fHowever,
fvertices
(1),
sum
all
elements
in
F
(N
)sum.
isall
However,
the
sum
all
elements
in(N
Fis
=
ows
problem
to
be
solvable
in
polynomial
time.
We
can
check
whether
additional
Proof.
Assign
(1)
and
fN
(2)
to
B.
Since
feven.
(1)
=
fsum,
(2),
we
are
now
trying
to
partition
F
)all
=
be
m
of
all
elements
F
even.
However,
sum
of
all
elements
in
F
F
)there
(1),
which
odd
)2).
can
partitioned
into
two
multisets
of
equal
of
Theorem
3.
If
N
≡
1−
(mod
3),
FA
(N
)V
cannot
be
partitioned
into
two
multisets
of
equal
sum.
sum.
in
F
Fthe
(N
)for
will
have
N
−
2)We
elements.
Since
≡
2N
(mod
3),
we
obtain
Ninto
−which
2an
0multisets
(mod
3).
By
lemma
ultisets
of
equal
sum.
pproach
IOI,
and
thus
is
the
main
focus
of
this
paper.
To
solve
IOI
2008
Island
fbe
(1)
fAssign
(2).
Fto
(N
)two
will
have
−equal
2to
elements.
Since
N
≡
2)≡
(mod
3),
we
obtain
N
−iswhich
2obtain
≡
0F
(mo
Theorem
2.
If
N
is
divisible
by
three,
then
(N
)
3,
F
(N
)
can
be
partitioned
two
of
equal
sum,
f
(1)
−
f
(2).
(N
)
have
N
−
2
elements.
Since
N
≡
2
(mod
3),
we
N
al
to
the
sum
of
all
elements
in
B
 in
3,
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum,
which
implies
our
theorem.
3,
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum,
implies
theorem.
our
will
 impl


b
−
a
−
2
),
...f
(b)}
some
a
<
b.
can
partition
F
into


mod
3).
F
(N
)
can
partitioned
into
multisets
of
equal
sum.
Proof.
f
(1)
to
A
and
f
(2)
B.
Since
f
(1)
=
f
(2),
we
are
now
trying
to
partition
ma
3,
F
(N
)
can
be
partitioned
into
two
multisets
of
sum.
Therefore,
the


because
F
(N
)
is
even
while
f
(1)
is
ven.
However,
sum
of
all
in
F
(N
)Since
F
(N
)B
+
fno
(1),
which
is
odd
Assign
fno
to
A
and
to
Since
fby
(1)
=
f≡
(2),
weF
are
Proof.
Assign
f)a2elements
(1)
A0with
and
f≡is
(2)
to
Since
(1)
=
(2),
we
are
now
trying
to
partition
F
(N
)no−
and
is
an
there
edge
isF
connecting
an
connecting
A
and
B.
A
Since
and
A,
/(1)
VF
A,
−
B
S,
∈
/≡
we
V
have
−
S,
A,
we
B
∈
S.
A,
As
Bnotwo
A
∈
S.
and
As
B
A
are
connected
B
are
cs
Proof.
Assign
f(N
(1)
to
A
and
f≤
(2)
to
B.
Since
ffway
(1)
ffpartition
(2),
we
are
now
trying
to
(N
all
elements
in
Fthe
(N
)problem
=
(N
+
(1),
odd
because
(N
)=
is
even
while
fB.
(1)
is
odd.
Therefore,
there
is(N
Proof.
Assign f23k),
to and (mod
to . 3),
Since ,(2)
we
are
now
trying
to
because
Fthere
(N
iseven
even
while
fisedge
(1)
is
odd.
Therefore,
there
to
F
(N
).
because
F
(N
)Proof.
even
while
f∈
(1)
is
odd.
Therefore,
there
is
way
to
partition
F
kcause
any
consecutive
subsequence
length
multiples
of
three,
which
will
F
=
Proof.
Assign
f3,
to
A
and
ftoF
(2)
to
B.
fB.
(1)
=
f(N
(2),
we
are
now
trying
to
partition
Fpartition
(N
=
F
(N
))n
f)is
(1)
−
f(1)
(2).
F
)isof
will
N
−
2=
elements.
Since
N
2whether
(mod
3),
we
obtain
N
−
2theorem.
0)and
(mod
3).
b which
−
a
−
2sum,
bwe
−
ahave
−
2denote
blem.
PARTITION
NP-hard
by
reduction
from
SUBSET is
}.
B
=
{f
(a
+
+
≤
k
F
(N
)and
while
f
(1)
odd.
Therefore,
there
is
way
to
partition
F
(N
).
b
−
−
b
−
a
−
2
Theorem
3.
If
N
1
F
)
cannot
be
partitioned
into
multisets
of
equal
two
 have
Therefore,
solving
this
problem
is
reduced
to
checking
N
≡
1
(mod
3).
We
can
solve
this
can
be
partitioned
into
multisets
of
equal
which
implies
our
theorem.
B.
Since
f
(1)
=
f
(2),
we
are
now
trying
to
partition
F
(N
)
=
F
(N
)
−
eed
to
use
this
approach.
We
must
find
a
special
constraint
in
the
problem
such
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum,
which
implies
our
 k
F
)San
can
be
partitioned
into
two
multisets
equal
sum,
which
implies
our
theo
 are
 (a
 +
Aby
{f
(a
0note
≤
≤
}+
∪
{f
+
1f(N
+
3k),
0This
≤
≤
}contradiction.
3felements.
A
{f
(a
+
3k),
0we
≤
≤
}<
∪(N
{f
(a
+
1−
3k),
≤
}−
3k),
B.
Since
(1)
=
(2),
we
now
trying
tofNote
partition
(N
)(1),
F
)≡
−
F
(N
)an
=
F
(N
)f)partition
−
felements.
fpartition
(2).
F0F
(N
)−
will
have
N
−
2a1−
elements.
Since
N
≡
22≡
(mod
3),
we
sum
of
all
elements
in
F
(N
)−
=
F
(N
)≤
+
which
isk22Ifis
odd
Proof.
Suppose
F
)will
=be
F
(N
)N
−
(1).
contains
elements.
Since
≡
12whether
(mod
3),
we
there
f
)ven.
is=
odd.
Therefore,
is
no
way
to
F
).
(2).
F
)equal
will
have
N
2of
elements.
N
23).
(mod
3),
f=
(1)
−
(2).
F
(N
will
have
N(1)
−
2that
Since
N
(mod
3),
we
obtain
N(mod
−then
≡
(mod
3).
by
an
edge,
an
edge,
we
that
that
S3,
not
is
INDEPENDENT-SET.
not
an
INDEPENDENT-SET.
This
contradiction.
is
Therefore
Therefore
−
Slemm
is
Vowa1
partition . into
will
reduced
Since
fthe
(1)
−
ff(N
(2).
F
(N
))kis
will
have
N
2(N
elements.
Since
N
≡
(mod
3),
obtain
−
≡
00V
(mod
3).
no
way
to
partition
Fknote
(N
).
1),
f(2)
(a
2),
...f
(b)}
some
a
b.
We
can
F
now
trying
to
partition
F
(N
=
F
(N
Proof.
N
is
divisible
by
three,
F
(N
)N
is
a(N
b2(N
−
2)3),
bto
a=
−
2(N
fHowever,
−f+
fto
(2).
F
)fafor
have
−
2−
Since
≡
2in
(mod
3),
we
obtain
N
−
2Since
≡
0N
By
whether
Therefore,
solving
this
problem
to
checking
≡
3,
F
(N
)−
can
partitioned
into
two
multisets
of
sum,
implies
our
theorem.
s+
unlikely
to
be
able
to
find
that
finds
A
and
B
an
Therefore,
solving
this
problem
is
reduced
checking
N
≡
1a−
(mod
3).
We
can
solve
this
3(1)
3elements.
Therefore,
solving
this
problem
is−
reduced
to
checking
whether
N
≡)we
1sum.
(mod
3).
We
can
solve
this
3more
3have and
f(1)
to
B.
Since
(1)
=
falgorithm
(2),
we
are
now
trying
toN
partition
F
(N
)which
=
Fequal
(N
We
will
provide
examples
of
special
cases
in
the
following
section.
lements.
Since
N
≡
(mod
we
obtain
N
−
2
≡
0
(mod
3).
By
lemma
he
problem
be
solvable
in
polynomial
time.
We
can
check
whether
an
additional
Theorem
4.
If
N
≡
2
(mod
3).
F


Theorem
4.
If
N
≡
2
(mod
3).
F
(N
)
can
be
partitioned
into
oks
like
a
classic
PARTITION
problem.
PARTITION
problem
is
NP-hard
by
reduction
from
SUBSETTheorem
4.
If
2
(mod
3).
F
(N
)
can
be
partitioned
into
two
multisets
of




f
(a
+
3k),
0
≤
k
≤
}
∪
{f
(a
+
1
+
3k),
0
≤
k
≤
}
Theorem
4.
If
N
≡
2
(mod
3).
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
b−a−2
heorem
4.
If
N
≡
2
(mod
3).
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum.
Proof.
Suppose
(N
)
=
F
(N
)
−
f
(1).
Note
that
F
(N
)
contains
N
−
1
elements.
Sin
ements.
Since
N
≡
2
(mod
3),
we
obtain
N
−
2
≡
0
(mod
3).
lemma
(mod
3).
By
lemma
3,
F
can
be
partitioned
into
sum,
which
implies
is
odd.
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there
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).
haveN
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3,
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VERTEX-COVER.
VERTEX-COVER.
3,
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Therefore,
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We
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more
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of
special
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in1 the
following
section.
Therefore,
solving
this
problem
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N3).≡of
1 equal
(mod
3).o
3 into
fined
in
awill
very
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that
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is
using
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more
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of
special
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in
following
section.
We
will
more
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the
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multisets
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Therefore,
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3).
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)
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of
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tioned
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theorem.
multisets
of
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which
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our
theorem.
sum
of
all
elements
in
(N
)
even.
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sum
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Atheorem.
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{f
(arecurrence
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0−≤
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kB.
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}section.
rovide
more
of
in provide
the
section.
plies
will
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examples
of
special
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innow
the
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Proof.
Assign
fF(mod
(1)
to
AF
and
to
We
will
more
examples
of
special
cases
in
the
following
section.
Proof.
Assign
fG(V,
(1)
to
A
and
f=
(2)
to
Since
fwhether
(1)
=
fof
(2),
inspect
the
formula
more
closely.
Proof.
Assign
f0(1)
to3k),
A
and
(2)
B.
Since
fspecial
(1)
=
f+
(2),
we
now
trying
to
partition
(N
)We
=
F(N
(N)solve
)f−(2)
−
Proof.
Assign
f∪problem
(1)
to
A
fwe
(2)
to
fF
(1)
f1INDEPENDENT-SET
(2),
we
are
trying
to
oof.
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(1)
to
A
and
fsolving
(2)
to
B.
Since
f3(a
(1)
=
fcases
(2),
are
trying
to
partition
F
(N
) .the
=now
ned
into
two
ofTherefore,
equal
sum,
which
implies
our
theorem.
solving
this
isway,
reduced
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N
≡
1B.
3).
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can
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F
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can
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equal
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Therefore,
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problem
is
reduced
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N
≡p
Therefore,
solving
this
problem
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reduced
to
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N
≡
(mod
3).
can
solve
thi
Lemma
2.
Lemma
2.
If
S
⊆
V
If
is
S
a
⊆
VERTEX-COVER
V
is
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of
graph
of
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graph
E),
then
E),
V
−
then
S
is
V
an
−
S
is
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INDEPENDENT-SE
the
Therefore,
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3
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ay.
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defined
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F
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)
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be
Therefore,
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reduced
to
checking
whether
N
≡
1
(mod
3).
We
can
solve
this
in
O(1).
f
(a
+
3k)
+
f
+
1
+
3k)
=
f
(a
+
2
+
3k)
We
will
provide
more
examples
special
in
the
following
section.
Therefore,
this
problem
is
reduced
to
checking
whether sum
of
all
elements
in
F
(N
)
is
even.
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the
sum
of
all
elements
in
F
(N
)
=
F
(N


because
F
(N
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while
f
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).
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educed
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f
(1)
−
f
(2).
F
(N
)
will
have
N
−
2
e
2)
to
B.
Since
f
(1)
=
f
(2),
we
are
now
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partition
F
(N
)
=
(N
)
−
4
Some
example
special
cases
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),
we
are
now
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partition
F
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F
(N
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b−a−2
f
(1)
−
f
(2).
F
(N
)
will
have
−
2
elements.
Since
N
≡
2
(m
foned
(1)
−
f
(2).
F
(N
)
will
have
N
−
2
elements.
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N
≡
2
(mod
3),
we
obtain
N
−
2
≡
0
(mod
3).
By
lemma
b−a−2
f
(1)
−
f
(2).
(N
)
will
have
N
−
2
elements.
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N
≡
(mod
3),
we
obtain
N
−
b
−
a
−
2
1)
−
f
(2).
F
(N
)
will
have
N
−
2
elements.
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N
≡
2
(mod
3),
we
obtain
N
−
2
≡
0
(mod
3).
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lemma
educed
to
checking
whether
N
≡
1
(mod
3).
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can
solve
this
in
O(1).
B
will
have
the
same
sum,
as
for
every
0
≤
k
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We
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special
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the
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the
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We
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the
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graph
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G(V,
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recurrence.
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length
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three
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1
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We
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3partitioned
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Therefore,
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4
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Any
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can
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3k)
+
f
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1
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3k)
f
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2
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two
multisets
offour
equal
sum,
implies
our
Proof.
Proof.
The
proof
is
actually
is
similar
actually
to
similar
the
previous
to
the
lemma.
previous
Let
lemma.
us
assume
Let
us
that
assume
V
−
S
that
not
V
−
a
S
INDEPENDENT-SET
is
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a
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3k)
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2
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which
theorem.
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≡ occur
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We
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We
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Therefore,
solving
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Therefore,
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problem
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whet
Therefore,
solving
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problem
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≡
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will
have
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Therefore,
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this
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checking
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flength
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3k)
two
multisets
of
equal
sum,
which
implies
our
theorem.
Assign
f (1)
A
and
ftwo
(2)
tofB
B.
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fdenote
(1)
= Sfby(2),
we
are
now
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partition
Feven.
(N
)O(1).
=
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Therefore,
Therefore,
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there
and
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and
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and
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VA,
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Some
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FinN
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special
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competitive
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We
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speth
m
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checking
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me
example
of
special
cases
hether
N
≡
1
3).
We
can
solve
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in
O(1).
We
will
more
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following
We
will
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examples
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We
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e
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examples
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special
in
the
following
section.
Proof.
Assign
f
(1)
to
A
and
f
(2)
to
B.
Since
f
(1)
=
f
(2),
we
are
now
trying
to

f
(1)
−
f
(2).
F
(N
)
will
have
N
−
2
elements.
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N
≡
2
(mod
we
obtain
N
−
2
≡
0
(mod
3).
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lemma
me
cases
wecases
have
A,
we
BSome
have
∈
/of
S.
A,
B
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/+some
S.
and
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and
connected
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edge,
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an
we
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note
that
weF
note
S occur
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that
aS VERTEX-COVER.
iswhile
not fa(1)
VERTEX-COVER.
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4.1
graphs
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4.1
Planar
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4.1
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Planar
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f
(1)
−
f
(2).
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)
will
have
N
−
2
elements.
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N
≡
2
(mod
3),
we
obtain
N
−
2
≡
3,
F
(N
)
can
be
partitioned
into
two
multisets
of
equal
sum,
which
implies
our
theorem.
contradiction.
contradiction.
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Therefore
V
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S
is
V
an
−
INDEPENDENT-SET.
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es
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O(V
) 1 ,in
a100,
planar
graph.
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by
the
construction
of
the
function.
2to for
Proof.
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fassume
(1)
to
A
and
fmillion
(2)
B.
Since
fby
(1)
=
fdone
(2),
w
holds
V
≥ a3
using
the
Eu
graph,
even
though
the
constraint
states
that
V
≤
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graph,
000,
even
E
≤
though
100,
000
the
,
constraint
the
standard
states
DFS
that
solution
V
≤
DFS
solution
still
runs
under
one
second
(in
competitive
programming,
we
assume
that
under
one
second
(in
competitive
programming,
we
assume
that
1
million
operations
ca
we
assume
that
1
million
operations
can
be
done
in
1
second
fprogramming,
connected
components
using
DFS
in
a
general
graph
takes
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+
E),
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only
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)
in
under
one
second
(in
competitive
programming,
we
that
1
operations
can
be
in
Theorem
2.
If
N
is
divisible
by
three,
then
F
(N
)
can
be
partitioned
into
two
multise
(Halim components
& Halim 2013)).
the number of connected
using DFSTherefore,
in a general
graph
takes
O(V +
E),
but it
takes whether
O(V ) inN
a
 this
solving
problem
isthat
reduced
toonly
checking
f
(1)
−
f
(2).
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(N
)
will
have
N
−
2
elements.
Since
N
≡
2
(mo
has
O(E)
in
its
running
tim
1 million
operations
can
be
done
in
1
second
(Halim
and
Halim
2013)).
under
one
second
(in
competitive
programming,
we
under
assume
one
that
second
1
million
(in
competitive
operations
programming,
can
be
done
in
(Halim
&
Halim
2013)).
Theorem
2. Ifproblem
N&isHalim
divisible
then
F (N
)number
can
be partitioned
into
two
multisets
of
equal sum.in a planarw
Therefore,
if (Halim
the
requires
usthree,
to compute
the
of
connected
components
in
a
planar
planar graph.
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if 2013)).
thebyproblem
requires
us
to
compute
the
number
of
connected
components
three, then
We will
more
special
the
sectio
Proof.
N is000,
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(N ) isHalim
a2of
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of
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)2can
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partitioned
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multisets
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wh
Bellman-Ford
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(Halim
&states
Halim
2013)).
2013)).
hough
the constraint
that
V ≤If100,
E≤
000
, the
DFS
solution
still
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graph,
even
though
the constraint
states
that
V 100,
≤ 100,
000,
E(Halim
≤ 100,&000
, the
standard
DFS
solution still runs
4.1.2
Maximum
Clique
problem
Proof. If N is4.1.2.
divisible
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three,
thenProblem
F (N ) into
is a two
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Maximum
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Maximum
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4.1.2
Maximum
problem
under
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assume
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million
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in 1 second
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isinreduced
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can be partitioned
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p
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spe
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(mod
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(N
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will
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The
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uires us to findThe
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2013)).
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vertices
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be
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of
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et
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2001).
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Jonathan
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Jonathan
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Proof.
Assign
fcolour
(1)
toawe
Aplanar
and
fgraph,
(2)
tonot
B.
Since
(1)
=
f (2),
wethan
are
now
trying
to pa
orem,
anyatcolours
planar
graph
can be
coloured
with
at
most
four
. Since
most four
colours
are
required
does
exist
most
four
are required
tofour
colour
a to
planar
graph,
there
does
exist
athat
clique
with
more
vertices.
H
on
graph
is NP-hard.
with
more
than
vertices.
Hence,
can
solve
the
problem
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whether
there
Planar
graph
is athere
graph
can
begeneral
drawn
on
a four
flat
surface
with
 Since f (1) = f (2), we are now trying to partition F  (N ) = F (N ) −
Proof.
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f
(1)
to
A
and
f
(2)
to
B.
f
(1)
−
f
(2).
F
(N
)
will
have
N
−
2
elements.
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N
≡
2
(mod
3),
we
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N
−w2
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required
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planar
graph,
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does
not
exist
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vertices.
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wethe
can
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clique
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clique
with
four
vertices.
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there
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of
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graph
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fourthe
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the
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4.1 this
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ontral
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lem
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European
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this
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planar
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mpiad
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aindirected
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cted
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athat
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even though the constraint states that V ≤ 100, 000,
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000
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etin
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are
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plan
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4.1.2
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2 the Maximum
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onsome
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are
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lines.
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ines.
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visitable
vertices
lying
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eve
et
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to
the
graph
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examples.
(Halim
&
Halim
2013)).
2
2 lying on the ‘right’ line from every vertex lying
to vertex
print
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andus for
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wevisitable
shall
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and
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requires
to find
the maximum
set o
‘right’
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simplicity).
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examples.
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shall
call
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‘right’connected
vertices forbysimplicity).
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on the
(which
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shall
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vertices
and
ity).
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figure
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4.1.1general
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co
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However,
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may
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E
≤
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−
holds
for the
V ≥
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on general graph is NP-hard. However, it is easy to solve this6)
given the
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the number of connected
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vertex v is visitable from vertex u if thereJonathan
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Irvin
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even though the constraint states that V ≤ 100, 000, E ≤ 100, 000 , the stand
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≤ 100, 000, programming,
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Halim 2013)).we assume that 1 million operations can be done in 1 second
(Halim & Halim 2013)).
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Fig. 5. An instance of problem ‘Traffic’. In this example, the expected output is {4, 4, 0, 2},
since the top ‘left’ vertex can visit 4 ‘right’ vertices, the second top ‘left’ vertex can2 visit 4
of O(V ). It is very difficult to find a
‘right’ vertices, the third top ‘left’ vertex cannot visit any ‘right’ vertex, and the bottom ‘left’
important constraint on the problem.
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not visitable from any ‘left’ vertex. With this property, there is a O(V log V ) solution
solution (CEOI 2011 Solu
ht’
anar
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ensures that
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foraccording
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anar
scending
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order
ensures
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planar properties
ensures
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every that for every
(CEOI 2011 Solutions, n.d.).
tex,
contiguous
sequence
sequence,
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assuming
visitable
that
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contiguous
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ight’
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property,
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(CEOI
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Solutions
n.d.).
CEOI
2011 Solutions
n.d.).
4.2. Bipartite Graph
Bipartite graph is a graph in which th
edges connect a vertex from U and a
the problem statement does not explic
Bipartite graph is a graph in which the vertices can be partitioned into two disjoint sets used as an introduction for this pape
and such
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have
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onnect
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.
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. Some problems have a bipartite graph as an input although
graph
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blem
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aximum
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Cover
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Cover
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iscussed
MAX-INDEPENDENT-SET
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of the
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arlier section,
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number ofand
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1976),
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Bipartite
A directed acyclic graph is a directed g
that
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hat are
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4.2.1
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e discussed in an earlier section, both of the MIN-VERTEX-COVER and MAX-INDEPENDENT-SET problems
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tr.2.1
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em, the size of the minimum vertex cover in As
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we discussed
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hing
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the
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of the
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MAX-INDEPENDENT-SET
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nh an
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VER
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ber
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Therefore,
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are solvable
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owever,
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algorithm
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algorithm
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much
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and
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yhe&size
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in the
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equal
tominus
the independent
number
of the
vertices
the
sizeisofequal
the
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algorithm
is much
simpler
toTherefore,
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using
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ertices
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problems
rtices
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ching.
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4.3
Directed
Acyclic Graph
4.3. of
Directed
Acyclic
Graph
ected acyclic graph is a directed graph that does not contain any cycle. Similar to planar and bipartite graphs,
4.3acyclic
Directed Acyclic
are several graphAproblems
that aregraph
muchiseasier
to solve
if the
is a contain
directed
directed acyclic
a directed
graph
thatgraph
does not
any cycle.graph.
Similar to
4.3 Directed
Acyclic Graph
4.3 Directed
Graph
c Graph
Directed Acyclic
Graph Acyclic
G
planar and bipartite graphs, A
there
are several
thatgraph
are much
easiernot
to contain any c
directed
acyclicgraph
graph problems
is a directed
that does
Minimum
Path
Cover
A
acyclic
graph
is4.3.1
anot
graph
that
does
not graphs,
contain
cycle.
to
planar
solve
if the
graph
isSimilar
a directed
directed
acyclic
graph.
cnygraph
a directed
graph
that
does
contain
anyare
cycle.
Similar
to
planarany
andthat
bipartite
graphs,
cycle.
Similar
planar
and
bipartite
graphs,
aph
that isdoes
notto
contain
any
cycle.
tothere
planar
and
bipartite
several
graph
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much
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to and
solvebipartite
if the gr
g
there
are
several
graph
problems
thatif are
much
easier
to solveacyclic
if the graph
is a directed acyclic graph.
graph
problems
that
are
much
easier
to
solve
the
graph
is
a
directed
graph.
he
graph
is
a
directed
acyclic
graph.
re
much
easier
to
solve
if
the
graph
is
a
directed
acyclic
graph.
-PATH-COVER is a problem that requries us to find the minimum number of vertex-disjoint paths needed
4.3.1. Minimum Path Cover
4.3.1
Minimum
Path Cov
over all of the vertices in a graph. By a simple reduction from Hamiltonian Path, this
problem
is NP-hard.
MIN-PATH-COVER
is a Path
problem
that requires
us Path
to findCover
the minimum number of
4.3.1
Minimum
Cover
4.3.1
Minimum
Cover
4.3.1
Minimum
Path
Cover
aph has a Hamiltonian Path if and only if we MIN-PATH-COVER
only need one path to
all ofthat
the vertices.
However,
is cover
a problem
requries
us
to refind this
the minimu
vertex-disjoint paths needed to cover all of the vertices
in a graph.
By a simple
em
can
be
solved
in
polynomial
time
for
a
directed
acyclic
graph.
For
a
graph
G
=
(V,
E),
we
create
a new from
VER
is
a
problem
that
requries
us
to
find
the
minimum
number
of
vertex-disjoint
paths
needed
is
anumber
problem
that
requries
to isfind
the minimum
number
of vertex-disjoint
pathsHn
imum number
of duction
vertex-disjoint
paths
needed
requries
usMIN-PATH-COVER
to find
the minimum
of Path,
vertex-disjoint
needed
to cover
all us
ofpaths
the
vertices
in Aa graph
graph.
ByaaHamiltonian
simple
reduction
from
Hamiltonian
this
problem
NP-hard.
has


hhe
G
(Vout
∪
,all
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),
where
Voutifis=
= VA
and
E
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{(u,
v)
∈
Vout
×
Vthe
:ifvertices.
(u,
v)isonly
∈NP-hard.
E}.
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it can
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ahas
simple
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from
Hamiltonian
this
problem
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vertices
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graph.
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simple
reduction
from
Hamiltonian
Path,
problem
m
Hamiltonian
problem
NP-hard.
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a=simple
reduction
from
Hamiltonian
Path,
this
problem
is
NP-hard.
graph
a to
Hamiltonian
Path
and
ifPath,
we
only
need
Path
ifBy
and
only
we
need
one
path
cover
allthis
of
However,
this

n
bytoKonig’s
Theorem
that
G
has
a
matching
of
size
m
if
and
only
if
there
exist
|V
|
−
m
vertex-disjoint
graph
has
avertices.
Hamiltonian
Path
if
and
only
if
we
only
need
one
path
to
cover
all
of
the
vertices.
Howeve
amiltonian
Path
ifproblem
and
only
if
we
only
need
one
path
to
cover
all
of
the
vertices.
However,
this
ath
all
of
the
However,
this
only
if cover
weAonly
need
one
path
to
cover
all
of
the
vertices.
However,
this
problem
can
be
solved
in
polynomial
time
for
a
directed
acyclic
graph.
can be solved in polynomial time for a directed acyclic graph. For a graph  For
ssolved
that
cover
all ofacyclic
vertices
indirected
G.
for , awe
graph.
aout
graph
=
E), we
create
ainnew
problem
can
begraph.
solved
in
polynomial
time
for
a we
directed
For
a=graph
(V,
create
aph.
apolynomial
graph
Gthe
=time
(V,
E),
we
create
anew
new
me
forFor
aindirected
For
a graph
Ggraph =
(V,G
E),
create
aG,new
graph
=
(V
∪ Vacyclic
E (V,
), ,graph.
where
Vout
Vin
=GV= and
E  =we{(u,
v) ∈
create
aacyclic
where and E),
in
out
in
out


graph G = (V



∪inV∈=
E and
),it
V{(u,
V∈
=Then
V×and
E
=be
{(u,
v)by
∈Konig’s
Vhas
×
V
(u, v)
E}.mThen
it
∪V
EV
),
V=
=
Vv)
v)
VitTheorem
: (u,
v)
∈ E}.
Then
itmatching
can
∈
V,out
× where
V
v){(u,
∈ E}.
Then
can
be
.out
can
be
shown
Theorem
that =
Vin
=
and
V,Vout
×where
VEin =
: (u,
v)=
∈by
E}.
Then
it
can
shown
Konig’s
that
G
of∈size
if and
out
in
out
inV
out a
in :be
out
in
in :E(u,
ttv)
in

 Konig’s
by
Theorem
that
athat
of of
size
|V | − m vertex-d
has
matching
size has
and
only
there
exist vertex-disjoint
vertex-disjoint
’s
that
G
has
matching
of
sizeGexist
m
ifif and
only
ifif there
exist
|Vif |and
− monly
and
only ifshown
there
exist
−
monly
vertex-disjoint
4.4
Miscellaneous
s aTheorem
matching
of size
m|Vifa| and
ifof
there
|V
| matching
−
m
vertex-disjoint
paths
cover
all
themvertices
in
G.if there existpaths
that
cover all
all of
of the
the vertices in that
in G. .
vertices
in cover
G.
.all of the paths
4.4.1 Special case of CNF-SAT problem
4.4
Miscellaneous
will
use Google
Code Jam 4.4
2008 Round
1A ‘Milkshake’
for this example. This problem requires the
Miscellaneous
4.4problem
Miscellaneous
ous
4.4
Miscellaneous
4.4. Miscellaneous
4.4.1
Special
case with
of CNF-SAT
estant to find a solution with the minimum number of true variables that satisfy
a CNF-SAT
problem
up
4.4.1
Special
case
of
CNF-SAT
problem
4.4.1
Special
case
of
CNF-SAT
problem
SAT
problem
Special
case of CNF-SAT problem
We will
use Google Code Jam 2008 Round 1A ‘Milkshake’ problem fo
4.4.1. Special Case of CNF-SAT
Problem
gle
Code
Jam
2008
Round
1A
‘Milkshake’
problem
for
example.
Thiswith
problem
requires
We
will
use
Google
Code
Jam
2008
Round
1Athis
problem
for
example.
This of
problem
requir
m
for1A
this
example.
requires
und
‘Milkshake’
problem
this
example.
This
requires
the
contestant
to‘Milkshake’
find
a‘Milkshake’
solution
thethis
minimum
number
true variable
than
Irvin
GUNAWAN
WeThis
will problem
use for
Google
Code the
Jam
2008problem
Round
1A
problem
for
this the
example.
contestant
to
find
a problem
solution
with
the
minimum
number
of atrue
variables
that satisfy
a CNF-SAT
problem w
dinimum
a solution
withofThis
the
minimum
number
ofthetrue
variables
satisfy
CNF-SAT
withnumber
up
ables
that
satisfy
a true
CNF-SAT
up
number
variables
that satisfy
a CNF-SAT
problem
with
upwith
problem
requires
contestant
tothat
find
a solution
theproblem
minimum
of
true variables that satisfy a Jonathan
CNF-SATIrvin
problem
with up to 2,000 variables (Google
GUNAWAN
Code
2008 Round 1A, ‘Milkshake’ problem, n.d.). The CNF-SAT is a satisfiability
IrvinJam
GUNAWAN
UNAWANJonathan
problem given in a conjunctive normal form (i.e. conjunction of disjunction of literals)
which was proven to be NP-hard (Cook, 1971). Therefore, it is unlikely that there is an
algorithm to solve a CNF-SAT problem with 2,000 variables in less than 8 minutes3.
However, there is a special property in this problem, in which at most one unnegated
literal exists in each clause. Therefore, all clauses can be converted into Horn clauses.
With this property, a linear time algorithm exists. (Google Code Jam 2008 Round 1A,
‘Milkshake’ solution, n.d.).
3
In Google Code Jam, contestants are given 8 minutes to produce the output upon downloading the input.
97
5. Conclusion
In conclusion, we can use a well-known reduction technique to prove that a problem that
we are currently attempting to solve is impossible (or at least it is very hard such that no
people has been able to solve it for more than 40 years). In competitive programming
(including IOI), understanding this technique is essential so that we will not be stuck at
trying to solve an impossible problem, thus prompting us to find another way to solve
the problem. To prove that a problem is NP-hard, it is good to know as many NP-hard
problems as possible, so that we can reduce from any one of the problems that we know
to the new problem. Some of the classic NP-hard problems include 3-SAT, Vertex Cover,
Independent Set, and Subset Sum. After realizing that the problem is NP-hard, we must
be able to find the special case that makes the problem solvable. We must be able to find
a special property that breaks the reduction proof. Having a lot of practice on these kind
of problems will help us to familiarize with the possibilities of a special case. Some of
the common special cases include planar, bipartite, and directed acyclic graph.
References
ACM ICPC World Finals 2010 Problem Statement (n.d.). Available via internet:
http://icpc.baylor.edu/download/worldfinals/problems/2010WorldFinalProblemSet.pdf
ACM ICPC World Finals 2010 Solutions (n.d.). Available via internet:
http://www.csc.kth.se/~austrin/icpc/finals2010solutions.pdf
Bondy, J.A., Murty, U. S. R. (1976). Graph Theory with Applications, Vol. 290, London: Macmillan.
CEOI 2011 Solutions (n.d.). Available via internet:
http://ceoi.inf.elte.hu/probarch/11/ceoi2011booklet.pdf
CEOI 2011. ‘Traffic’ Problem (n.d.), available via internet:
http://ceoi.inf.elte.hu/probarch/11/trazad.pdf
Cook, S.A. (1971). The complexity of theorem-proving procedures. In: Proceedings of the Third Annual ACM
Symposium on Theory of Computing. ACM, 151–158.
Cormen, T.H., Leiserson, C.E., Rivest, R.L., Stein, C. (2009). Introduction to Algorithms (3rd ed.). MIT Press.
Forišek, M. (2015). International olympiad in informatics 2015 syllabus. Available via internet:
https://people.ksp.sk/~misof/ioi-syllabus/ioi-syllabus.pdf
Gonthier, G. (2005). A computer-checked proof of the four colour theorem. Available via internet:
http://research.microsoft.com/en-us/people/gonthier/4colproof.pdf
Google Code Jam 2008 Round 1A, ‘Milkshake’ Problem (n.d.). Available via internet:
https://code.google.com/codejam/contest/32016/dashboard#s=p1
Google Code Jam 2008 Round 1A, ‘Milkshake’ solution (n.d.). Available via internet:
https://code.google.com/codejam/contest/32016/dashboard#s=a&a=1
Halim, S., Halim, F. (2013). Competitve Programming 3. lulu.com
IOI 2008, ‘Island’ Problem (n.d.). Available via internet:
http://www.ioinformatics.org/locations/ioi08/contest/day1/islands.pdf
IOI 2014, ‘Friend’ Problem (n.d.). Available via internet:
http://www.ioinformatics.org/locations/ioi14/contest/day2/friend/friend.pdf
West, D. B. et al. (2001). Introduction to Graph Theory, Vol. 2. Prentice hall Upper Saddle River.
98
J.I. Gunawan
J.I. Gunawan is an undergraduate student studying Computer Science
in National University of Singapore. He participated a lot of programming contests, including IOI 2012 and 2013 and ACM ICPC World
Finals 2014 and 2015. He is also the lead of the Scientific Committee
of Indonesian Computing Olympiad team (TOKI) training Indonesian
teams for IOI in 2015–2016.

here - International Olympiad in Informatics

Transcription

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