Colouring maps on impossible surfaces Chris Wetherell Radford - - PowerPoint PPT Presentation

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(chessboard) = 2

χ(wobbly chessboard) = 2 χ(wobbly chessboard) = 2

χ(cube) = 3 χ(cube) = 3

χ(cube) = 3 χ(cube) = 3

χ(icosahedron) = 3 χ(icosahedron) =

χ(tetrahedron) = 4 χ(tetrahedron) =

χ(USA) = 4

Nevada

χ(USA) = 4

NV ID OR CA AZ UT NV ID OR 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

• ne 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.

• 5 regions meet at every vertex
• Turning it into a special graph is equivalent to

truncating the vertices

‘Specialised’ icosahedron

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
• r 5 neighbours

• 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

Reducibility (for 5 colours)

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
• ne 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
• f 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 Four- Color 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

• 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?

Torus

+ =

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 Dual graph Region’s number of neighbours Degree of vertex No neighbours Isolated vertex n mutually adjacent regions Kn = the complete graph on n vertices 5 mutually adjacent regions can’t exist in the plane K5 is non-planar V – E + F = 2 V – E + F = 2 (V & F switched) Special (3 regions meet at every vertex) Triangulation (every face is a triangle) χ = 2 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

• A torus can be formed by rolling rectangle ABCD

into a cylinder, then joining the ends together

• Note that

– 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

B C A D P

b a a b

• The rectangle is the fundamental polygon
• The edges labelled a are identified in the
• rientation indicated; similarly for b
• The expression aba-1b-1 describes the torus
• K7 can be drawn on a torus, so χ(torus) ≥ 7

Torus revisited

a b b

• Does Euler’s Formula still apply?
• V = 7

E = 21 F = 14

• So V – E + F = 0
• By similar reasoning

any map or graph drawn on any torus satisfies this equation

• Constant on RHS is the Euler characteristic, ε
• ε (sphere) = 2 and ε (torus) = 0

Torus revisited

b a a b

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)
• But the Klein bottle is different because it is
• ne-sided or non-orientable (to avoid self-

intersection requires 4-dimensional space) b a a b

• Consider any list of expressions involving the

symbols a, b, c, d, … twice each, some of which may have index –1, e.g. abd-1ac-1f b-1ee fg-1dgc-1

• All such lists define a closed surface (without a

boundary), and vice versa

• Each row is a 2D ‘patch’ which is ‘sewn’ to the
• thers (or itself) when like edges are identified

e e a a c f g g c f

Constructing other surfaces

hexagon → triangle → pentagon →

χ(cube) = 3 χ(cube) = 3

a b d c e f e c g d b g f a

abb-1c-1d-1dca-1e-1f -1fegg-1

• 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!

• 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’…

Simplifying patchwork expressions

a a a

Classification of closed surfaces (1860s)

• Theorem: Every closed surface is equivalent to

exactly one of the following normal forms.

• Corollary: Closed surfaces with the same Euler

characteristic and orientability are equivalent.

Symbol Description Normal form Orientable ε

S0

Sphere Yes 2

Sp

Sphere with p handles Yes 2 – 2p

Nq

Sphere with q cross-caps No 2 – q

1 1 1 1 1 1 1 1 − − − − p p p p

b a b a b a b a 

q qc

c c c c c 

2 2 1 1 1 −

aa

• Theorem: For any closed surface S, other than

the sphere,

• Remarkably, by the Four Colour Theorem this

inequality is also true for the sphere, but Heawood’s method of proof fails in this case

. 2 ) ( 24 49 7 ) ( S S ε χ − + ≤

Heawood’s Inequality (1890)

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, 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

, S S         − + = 2 ) ( 24 49 7 ) ( ε χ

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).

• 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

Image sources