Chris Wetherell Radford College / ANU Secondary College CMA

Colouring maps on
impossible surfaces
Chris Wetherell
Radford College / ANU Secondary College
CMA Conference 2014
Colouring rules
• To colour a map means assigning a colour to
each region (country, state, county etc.) so that
adjacent regions are different
• Regions meeting at a single point can share
the same colour
• Regions are physical, not political (or similar)
• A map’s chromatic number χ is the minimum
number of colours required to colour it (in?)
• This kind of problem is an example of topology
(as opposed to geometry)
χ(chessboard) = 2
χ(wobbly chessboard) = 2
χ(cube) = 3
χ(cube) = 3
χ(icosahedron) = 3
χ(tetrahedron) = 4
χ(USA) = 4
Nevada
χ(USA) = 4
OR
NV
CA
ID
UT
AZ
=
ID
OR
NV
CA
AZ
UT
A very brief history of
the Four Colour Theorem
• 1852: Francis Guthrie conjectures that 4 colours
always suffice for any map
• Early observations: Obviously 4 regions can be
mutually adjacent so there are maps requiring at
least 4 colours – e.g. tetrahedron
• Early misconceptions: Obviously 5 regions can
never be mutually adjacent (this was well-known)
so therefore… you never need 5 colours?
• Flawed reasoning: Nevada and its neighbours
require 4 colours, but no 4 regions are mutually
adjacent
A very brief history of
the Four Colour Theorem
• Early progress: Turn the question into a graph theory
problem (more on this later)
• 1850s-1970s: Lots of clever people doing lots of clever
things, lots of false alarms, lots of banging heads on walls
• 1976: Appel and Haken (with Koch) ‘prove’ the Four
Colour Theorem
• Philosophical dilemma: Method of proof requires
exhaustive checking of millions of potential colourings
for thousands of carefully constructed maps/graphs by
computer program – does this count?
• Recent years: Improvements made to exhaustive
method, but still no humanly verifiable proof known
Appel and Haken’s method
• construct an unavoidable list of sub-maps, i.e.
any map you can draw must contain at least
one of the sub-maps on the list
– this was done by hand (1476 sub-maps)
• show that every sub-map on the unavoidable
list is reducible, i.e. it can’t appear in any map
requiring 5 colours (by induction)
– this was checked with approximately 1200
hours of computer calculations (with Koch)
The Five Colour Theorem*
*it’s like the Four Colour Theorem, only bigger
• Consider the list of sub-maps:
U = {region with 1 neighbour, region with 2
neighbours, …, region with 5 neighbours}
• U is unavoidable
Theorem: Every map must contain at least one
region with at most 5 neighbours.
• U is reducible
Theorem: The smallest maps requiring more
than 5 colours cannot contain any regions with
fewer than 6 neighbours.
Euler’s formula
• Let
V = number of vertices (where 3 or more
regions meet)
E = number of edges (pieces of boundary
between pairs of vertices)
F = number of faces/regions (including
the infinite one)
• Theorem: V – E + F = 2.
Euler’s formula – basic step
•
•
•
•
V=2
E=3
F=3
So V – E + F = 2
Euler’s formula – inductive step
•
•
•
•
V is unchanged
E increases by 1
F increases by 1
So V – E + F is unchanged
Euler’s formula – inductive step
•
•
•
•
V increases by 2
E increases by 3
F increases by 1
So V – E + F is unchanged
Special maps
• A map is special if exactly 3 faces (regions)
meet at each vertex
• Theorem: If n colours suffice for every special
map, then n colours suffice for every map.
‘Specialised’ icosahedron
• 5 regions meet at every vertex
• Turning it into a special graph is equivalent to
truncating the vertices
Unavoidability
• Let
F1 = number of faces with 1 neighbour
F2 = number of faces with 2 neighbours
F3 = number of faces with 3 neighbours etc.
so F = F1 + F2 + F3 + F4 + . . .
• Counting the number of edges at each vertex
and surrounding each face gives
2E = 3V (assuming special)
= F1 + 2F2 + 3F3 + 4F4 + . . .
Unavoidability
• 6 x Euler’s formula is 6V – 6E + 6F = 12
• Substituting F = F1 + F2 + F3 + F4 + . . . and
2E = 3V = F1 + 2F2 + 3F3 + 4F4 + . . . gives
5F1 + 4F2 + 3F3 + 2F4 + F5 – F7 – 2F8 – . . . = 12
• Since LHS must be positive, F1, F2, …, F5 can’t
all equal 0
• That is, there is at least one face with 1, 2, 3, 4
or 5 neighbours
Reducibility (for 5 colours)
• Hypothetically, let M be a special map with the
fewest regions for which 5 colours is not enough
• Pick a region with at most 5 neighbours and
remove some edges
• M can be coloured with 5 colours! Contradiction
Adapting the proof to 4 colours
• Five Colour Theorem: Relatively easy (there are a
few technicalities that have been glossed over…)
• Kempe chains: Adapts the reducibility argument
when only 4 colours are available – almost works!
• Four Colour Theorem: Appel and Haken adapt the
RHS = 12 result, via ‘discharging rules’, to find larger
unavoidable sets, eventually settling on 486 rules to
construct 1476 sub-maps which are shown reducible
by computer-implemented generalisations of the
Kempe chain idea – extremely difficult!
Appel and Haken’s sense of humour
• Regarding the discharging methods which replace
one irreducible sub-map with several others:
“The reader is to be forgiven for thinking that
anyone who can think of this as good news
enjoys going to the dentist.”
• Regarding probabilistic arguments for the existence
of several different reducible unavoidable sets:
“[There are] a large number of possible proofs of
the Four-Color Theorem as a reward for our
patience, a larger number of proofs of the FourColor Theorem than anyone really wants to see.
Actually, one proof of this type is probably one
more than many people really want to see.”
Other surfaces
• The chromatic number χ of a surface is the
minimum number of colours required to colour
every map on that surface
• The Four Colour Theorem can be stated as
χ(plane) = 4
• Based on the flattening idea for a tetrahedron,
cube, icosahedron etc., it can also be stated as
χ(sphere) = 4
Torus
• A torus (doughnut) can be formed by adding a
handle to a sphere
+
=
• 4 mutually adjacent regions exist on a sphere
• 5 exist on the torus, so χ(torus) ≥ 5
• Is this the maximum ever needed?
Dual of a map
• Represent each region by a node or vertex
• Join vertices by an edge if the corresponding
regions are adjacent
• The graph formed is
the dual of the map
• Colouring the map
is equivalent to
colouring the
vertices of the graph
Dual properties
Map
Region’s number of neighbours
No neighbours
n mutually adjacent regions
5 mutually adjacent regions
can’t exist in the plane
V–E+F=2
Special (3 regions meet at
every vertex)
χ=2
Dual graph
Degree of vertex
Isolated vertex
Kn = the complete graph on n
vertices
K5 is non-planar
V – E + F = 2 (V & F switched)
Triangulation (every face is a
triangle)
Bipartite
Dual of a polyhedron
• The dual of a tetrahedron is…
another tetrahedron
• The dual of a cube is…
an octahedron, and vice versa
• The dual of an icosahedron is…
a dodecahedron, and vice versa
• The dual of a pyramid is another pyramid
• The dual of a prism is a bipyramid, and vice versa
Torus revisited
• A torus can be formed by rolling rectangle ABCD
into a cylinder, then joining the ends together
A
B
D
C
• Note that
P
– corners A, B, C and D have been identified, i.e. all of
them represent the same point P on the torus
– similarly, edges AB = DC and BC = AD (note order)
Torus revisited
• The rectangle is the fundamental polygon
• The edges labelled a are identified in the
orientation indicated; similarly for b
a
b
b
b
a
a
• The expression aba-1b-1 describes the torus
• K7 can be drawn on a torus, so χ(torus) ≥ 7
b
Torus revisited
• Does Euler’s Formula still apply?
a
• V=7
E = 21
F = 14
b
b
• So V – E + F = 0
• By similar reasoning
any map or graph
a
drawn on any torus satisfies this equation
• Constant on RHS is the Euler characteristic, ε
• ε (sphere) = 2 and ε (torus) = 0
An ‘impossible’ surface – finally!
• Consider aba-1b (almost the same as the torus
except the final b does not have index –1)
Klein bottle properties
• K6 can be drawn on a Klein bottle
• So χ(KB) ≥ 6
• V=6
E = 15
F=9
• ε (KB) = 0 = ε (torus)
a
b
a
• But the Klein bottle is different because it is
one-sided or non-orientable (to avoid selfintersection requires 4-dimensional space)
b
Constructing other surfaces
• Consider any list of expressions involving the
symbols a, b, c, d, … twice each, some of which
a
e
may have index –1, e.g.
hexagon → abd-1ac-1f
triangle → b-1ee
pentagon → fg-1dgc-1
f
e
g
c
c
a
g
f
• All such lists define a closed surface (without a
boundary), and vice versa
• Each row is a 2D ‘patch’ which is ‘sewn’ to the
others (or itself) when like edges are identified
χ(cube) = 3
c
d
b
b
d
c
a
a
f
g
g
e
f
e
abb-1c-1d-1dca-1e-1f -1fegg-1
Simplifying patchwork expressions
• The simplest rule effectively ‘cancels’ a symbol and its
inverse in the expected way, e.g. for the net of a cube
cube = abb-1c-1d-1dca-1e-1f -1fegg-1
= ac-1ca-1e-1e
= aa-1
= sphere!
a
a
a
• In general, this is not like normal algebra because the
rules are based on ‘cutting’ and ‘re-sewing’ along edges
• More complicated rules include ‘handle normalisation’,
‘cross-cap normalisation’, ‘handle conversion’…
Classification of closed surfaces (1860s)
• Theorem: Every closed surface is equivalent to
exactly one of the following normal forms.
Symbol Description
S0
Sp
Nq
Sphere
Normal form
Orientable
ε
aa −1
Yes
2
Yes
2 – 2p
No
2–q
Sphere with
a1b1a1−1b1−1 ap bpap−1bp−1
p handles
Sphere with
c1c1c2c2  cq cq
q cross-caps
• Corollary: Closed surfaces with the same Euler
characteristic and orientability are equivalent.
Heawood’s Inequality (1890)
• Theorem: For any closed surface S, other than
the sphere,
χ (S ) ≤
7 + 49 − 24 ε (S)
2
.
• Remarkably, by the Four Colour Theorem this
inequality is also true for the sphere, but
Heawood’s method of proof fails in this case
Thread Problem
• So far we have asked:
– given a surface, how many colours do you need
to colour every map?
– given a surface, what’s the largest complete
graph that can be drawn on it?
• The Thread Problem instead asks:
– given a complete graph, what’s the simplest
surface it can be drawn on?
• This was studied by many mathematicians and its
eventual solution in 1968 – without controversial
computer methods – led to…
Map Colour Theorem (1968)
• Theorem: (1) χ(Klein bottle) = 6
(2) For every other surface S, other than the
sphere,
 7 + 49 − 24 ε (S) 
χ (S ) = 
,
2


where [x] is the integer part of x.
• Note that the formula yields the value χ = 7
when ε = 0, which is correct for the torus but
not for the Klein bottle
Important lessons
• Sometimes the apparently ‘easy’ case is the
most difficult
• All doughnuts are topologically equivalent, so
complain to the manager at your local Donut
King if they don’t stock ones like this:
References
• Appel, K. & Haken, W. & Koch, J., Every Planar Map is Four Colorable,
Contemporary Mathematics; vol. 98, American Mathematical Society,
USA, 1989.
• Barnette, D., Map Coloring, Polyhedra, and the Four-Color Problem, The
Mathematical Association of America, USA, 1983.
• Ringel, G., Map Color Theorem, Springer-Verlag, New York, 1974.
• Various, Four color theorem, en.wikipedia.org/wiki/Four_color_theorem
(accessed August 2014).
Image sources
• Africa map www.icoolpages.com/wp-content/uploads/2014/05/
continent_coloring_pages_africa_map_template.gif
• Icosahedron canberracreatives.com.au/wp-content/uploads/2013/08/
icosehedron.png
• Klein bottle http://www.drchristiansalas.org.uk/klein1.gif
• USA map world.mathigon.org/resources/Graph_Theory/colouring.png
• Soccer ball http://researchsalad.files.wordpress.com/2011/12/
soccer-ball.jpg