Colouring maps on impossible surfaces. Chris Wetherell Radford College / ANU Secondary College CMA Conference 2014

Transcription

1 Colouring maps on impossible surfaces Chris Wetherell Radford College / ANU Secondary College CMA Conference 2014

2 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)

3

4

5 χ(chessboard) = 2

6 χ(wobbly chessboard) = 2

7 χ(cube) = 3

8 χ(cube) = 3

9 χ(icosahedron) = 3

10 χ(tetrahedron) = 4

11 Nevada χ(usa) = 4

12 χ(usa) = 4 OR ID OR ID CA NV UT = CA NV UT AZ AZ

13 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

14 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

15 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)

16 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.

17 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.

18 Euler s formula basic step V = 2 E = 3 F = 3 So V E + F = 2

19 Euler s formula inductive step V is unchanged E increases by 1 F increases by 1 So V E + F is unchanged

20 Euler s formula inductive step V increases by 2 E increases by 3 F increases by 1 So V E + F is unchanged

21 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.

22 Specialised icosahedron 5 regions meet at every vertex Turning it into a special graph is equivalent to truncating the vertices

23 Let Unavoidability 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 Counting the number of edges at each vertex and surrounding each face gives 2E = 3V (assuming special) = F 1 + 2F 2 + 3F 3 + 4F

24 Unavoidability 6 x Euler s formula is 6V 6E + 6F = 12 Substituting F = F 1 + F 2 + F 3 + F and 2E = 3V = F 1 + 2F 2 + 3F 3 + 4F gives 5F 1 + 4F 2 + 3F 3 + 2F 4 + F 5 F 7 2F 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

25 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

26 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!

27 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.

28 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

29 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?

30 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

31 Dual properties Map Dual graph Region s number of neighbours Degree of vertex 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 Isolated vertex K n = the complete graph on n vertices K 5 is non-planar V E + F = 2 (V & F switched) Triangulation (every face is a triangle) Bipartite

32 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

33 A Torus revisited A torus can be formed by rolling rectangle ABCD into a cylinder, then joining the ends together B P 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)

34 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 b a The expression aba -1 b -1 describes the torus K 7 can be drawn on a torus, so χ(torus) 7

35 Torus revisited Does Euler s Formula still apply? V = 7 E = 21 F = 14 So V E + F = 0 By similar reasoning any map or graph b a drawn on any torus satisfies this equation Constant on RHS is the Euler characteristic, ε ε (sphere) = 2 and ε (torus) = 0 a b

36 An impossible surface finally! Consider aba -1 b (almost the same as the torus except the final b does not have index 1)

37 Klein bottle properties K 6 can be drawn on a Klein bottle a So χ(kb) 6 V = 6 E = 15 F = 9 b ε (KB) = 0 = ε (torus) a But the Klein bottle is different because it is one-sided or non-orientable (to avoid selfintersection requires 4-dimensional space) b

38 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 triangle pentagon abd -1 ac -1 f b -1 ee fg -1 dgc -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 others (or itself) when like edges are identified f c a g e g f c

39 χ(cube) = 3 c b d b d c a a g g f f e e abb -1 c -1 d -1 dca -1 e -1 f -1 fegg -1

40 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 -1 c -1 d -1 dca -1 e -1 f -1 fegg -1 = ac -1 ca -1 e -1 e = 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

41 Classification of closed surfaces (1860s) Theorem: Every closed surface is equivalent to exactly one of the following normal forms. Symbol Description Normal form Orientable ε aa S 0 Sphere Yes 2 1 S p Sphere with p handles a 1 b a b 1 1 a p b p a 1 p b 1 p Yes 2 2p N q Sphere with q cross-caps c 1 c 1 c 2 c 2 c c q q No 2 q Corollary: Closed surfaces with the same Euler characteristic and orientability are equivalent.

42 Heawood s Inequality (1890) Theorem: For any closed surface S, other than the sphere, χ( S) ε( S). Remarkably, by the Four Colour Theorem this inequality is also true for the sphere, but Heawood s method of proof fails in this case

43 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

44 Map Colour Theorem (1968) Theorem: (1) χ(klein bottle) = 6 (2) For every other surface S, other than the sphere, χ( S) = 7 where [x] is the integer part of x ε( S) Note that the formula yields the value χ = 7 when ε = 0, which is correct for the torus but not for the Klein bottle 2,

45 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:

46 References Appel, K. & Haken, W. & Koch, J., Every Planar Map is Four Colorable, Contemporary Mathematics; vol. 98, American Mathematical Society, USA, Barnette, D., Map Coloring, Polyhedra, and the Four-Color Problem, The Mathematical Association of America, USA, Ringel, G., Map Color Theorem, Springer-Verlag, New York, Various, Four color theorem, en.wikipedia.org/wiki/four_color_theorem (accessed August 2014). Image sources Africa map continent_coloring_pages_africa_map_template.gif Icosahedron canberracreatives.com.au/wp-content/uploads/2013/08/ icosehedron.png Klein bottle USA map world.mathigon.org/resources/graph_theory/colouring.png Soccer ball soccer-ball.jpg