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 OR ID ID OR = NV NV UT UT CA CA AZ AZ
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 F 1 = number of faces with 1 neighbour F 2 = number of faces with 2 neighbours F 3 = number of faces with 3 neighbours etc. so F = F 1 + F 2 + F 3 + F 4 + . . . • Counting the number of edges at each vertex and surrounding each face gives 2 E = 3 V (assuming special) = F 1 + 2 F 2 + 3 F 3 + 4 F 4 + . . .
Unavoidability • 6 x Euler’s formula is 6 V – 6 E + 6 F = 12 • Substituting F = F 1 + F 2 + F 3 + F 4 + . . . and 2 E = 3 V = F 1 + 2 F 2 + 3 F 3 + 4 F 4 + . . . gives 5F 1 + 4 F 2 + 3 F 3 + 2 F 4 + F 5 – F 7 – 2 F 8 – . . . = 12 • Since LHS must be positive, F 1 , F 2 , …, F 5 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 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
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 Dual graph Region’s number of neighbours Degree of vertex No neighbours Isolated vertex n mutually adjacent regions K n = the complete graph on n vertices 5 mutually adjacent regions K 5 is non-planar can’t exist in the plane V – E + F = 2 V – E + F = 2 ( V & F switched) Special (3 regions meet at Triangulation (every face is a every vertex) 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
Torus revisited • A torus can be formed by rolling rectangle ABCD into a cylinder, then joining the ends together P A B D C • 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 • The rectangle is the fundamental polygon • The edges labelled a are identified in the orientation indicated; similarly for b a a b b b b a • The expression aba -1 b -1 describes the torus • K 7 can be drawn on a torus, so χ (torus) ≥ 7
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 -1 b (almost the same as the torus except the final b does not have index –1)
Klein bottle properties • K 6 can be drawn on a Klein bottle a • So χ (KB) ≥ 6 • V = 6 E = 15 b b F = 9 • ε (KB) = 0 = ε (torus) a • But the Klein bottle is different because it is one-sided or non-orientable (to avoid self- intersection requires 4-dimensional space)
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. f e hexagon → g abd -1 ac -1 f triangle → c b -1 ee c pentagon → a fg -1 dgc -1 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 χ (cube) = 3 c b d b d c a a g f e g f e abb -1 c -1 d -1 dca -1 e -1 f -1 fegg -1
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