k-1-x - Webcourse

‫נושאים מתקדמים‬
‫באלגוריתמים מבוזרים‬
‫שמואל זקס‬
[email protected], www.cs.technion.ac.il/~zaks
A. Distributed algorithms
Example 1: synchony
9
8
7
6
5
4
3
2
1
84 days,
days, 12 professor
professor
Exercise: find a trade-off between no. of days
and no. of professors.
Example 2: leader election
x
4
x
5
x
8
?
x
6
9
Exercise: find a better algorithm to find the
maximum, prove correctness and analyze
performance.
Example 3: faults
Impossibility of consensus
The Byzantine Generals Problem
Example 4: snapshot
B. Self Stabilization
Example 5: self stabilization
clock synchronization
shared memory, synchronous
xx
8x7x
6
8x
7x
6
8 76
x6x7
8
x6 x7 8
6
Program at each processor p :
choose a neighbor x of p ;
if t(p) = t(x) then t(p):= t(p)+1;
Stabilizing if all start with the same
time,
however …
7
if t(p) = t(x) then t(p):= t(p)+1;
6
7
6
4
7
Is it stabilizing? no! (may even deadlock)
8
Program at each processor p :
choose a neighbor x of p ;
if t(p) = t(x) then t(p):= t(p)+1;
if t(p) < t(x) then t(p):= t(x);
Is it stabilizing?
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if t(p) = t(x) then t(p):= t(p)+1;
if t(p) < t(x) then t(p):= t(x);
xx
8x7x
6
8 76
xx
8 7 4
x7 x7 8
xx
7 7 8
Is it stabilizing? yes!
10
if t(p) = t(x) then t(p):= t(p)+1;
if t(p) < t(x) then t(p):= t(x);
xx
8x7x
6
8 76
xx
8 7 4
x7 x8 9
xx
7 8 9
Is it stabilizing? no!
11
Program at each processor p :
choose a fairly;
neighbor x of p
if t(p) = t(x) then t(p):= t(p)+1;
if t(p) < t(x) then t(p):= t(x);
Is it stabilizing?
yes!
Exercise: prove and analyze time until
stabilization.
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C. Networks (optical)
p1
lightpaths
Valid coloring
w( p1 )  w( p2 )
p2
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Saving a switch:
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Example 7: min ADMs
colors = 2
ADMs = 8
colors=3
ADMs = 7
Exercise: prove NP=hardness
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Example 8: min ADMs
w/grooming
colors=3
switches=10
g=2
colors=2
switches=9
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Example 9: approximation algorithms
N: # of lightpaths
ALG: #ADMs used by the algorithm
OPT: #ADMs used by an optimal solution
ALG  2N
N  OPT
ALG  2 x OPT
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Example 10: min regenerators
3. Regenerators
d=2
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set A = 3
set B = 1
d=2
g=1
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set A and set B = 4
d=2
g=1
20
set A or set B = 3
d=2
g=1
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Example 9: on-line algorithms
 Input arrives one at a time, and a decision is
made (and cannot be changed).
 In the minADM problem: lightpaths arrive
one at a time, and need to be colored.
 Competitive analysis
 An on-line algorithm A is c-competitive if
A(I)  c OPT(I)
for any input sequence I.
(A(I) and OPT(I) are #ADMs used by A and
by an optimal offline algorithm OPT.)
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Any algorithm ≥ 7/4 , even for a ring
Case a:
7/4=1.75
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Case b:
Case b1:
6/3 = 2
Case b2:
5/3 = 1.67
any algorithm ≥ 1.67
Exercise: prove any algorithm ≥ 1.75
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6.5 any algorithm for a path ≥ 3/2
k paths
k=12
x=6
k-1 spaces:
x between same color
k-1-x between different colors
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So far: any algorithm uses 2k ADMs
now – a short path at each gap of diffferent colo
(k-1-x such gaps
k=12, x=6, 12-1-6=5
Any algorithm uses at least one more ADM
for each (ALG uses exactly one)
So: any algorithm
≥ 2k + (k-1-x) ADMs
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So far: use ≥ 2k + (k-1-x) ADMs
now – two long paths at each of the k gap of same color
Any algorithm must use 2 ADMs for each
So: any algorithm ≥ 2k + (k-1-x) + 4x =
= 3k+3x-1 ADMs
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We showed: any algorithm uses ≥ 3k+3x-1 ADMs
2k ADMsOPT: the short paths ≤
2x ADMsfor the long paths
OPT  2k + 2x
any algorithm/OPT  3/2 – 1/(2k)
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Traffic grooming
 low capacity requests can be groomed into
high capacity wavelengths (colors).
 colors can be assigned such that at most g
lightpaths with the same color can share an
edge
 g is the grooming factor
#ADMs=12
#ADMs=9
Example 6: min ADM, no grooming
Example 7: min ADM, no grooming,
online
Example 8: min regenerators, no
grooming
Example 9: min locations for
regenerators, no grooming
Many parameters, many
optimizatoin problems
Many algorithmics issues:
Algorithms, complexity, On-line,
approximability, inapproximability,
etc.
Many open problems, of practical
importance and theoretical
interest