Every planar graph is 4-colourable a proof without computer

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1 Peter Dörre Department of Informatcs and Natural Scences Fachhochschule Südwestfalen (Unversty of Appled Scences) Frauenstuhlweg 31, D Iserlohn, Germany Emal: doerre(at)fh-swf.de Mathematcs Subject Classfcaton: 05C15 Abstract Colourng of planar graphs can be treated as a specal lst-colourng problem wth selected lsts for near-trangulatons. The new dea s to use sublsts of a common lst of four colours, to enforce a common colour n all lsts, and to admt on the boundng cycle at most one vertex wth a lst of at least two colours. By these condtons colours and lsts are excluded, whch have to be consdered n lstcolourng (leadng to the well-known result that planar graphs are 5-choosable). The essental applcaton s a proof of the 4-colour theorem. Keywords: planar graph; near-trangulaton; vertex colourng, chromatc number; 4-colourable, 5-choosable, lst-colourng wth selected lsts. 1 Introducton In response to Kempe s ncorrect proof of the 4-colour conjecture n 1879 [3], Heawood [2] publshed a proof of the 5-colour theorem for planar graphs, together wth a counter-example whch exposed an essental error n Kempe s approach. The 4-colour theorem was frst proved by Appel and Haken (see e.g. [1]) n 1976 wth the help of extensve computer calculatons. Robertson et al. [4] gave an mproved and ndependent verson of ths type of proof n The frst proof of the correspondng lst-colourng theorem for planar graphs every planar graph s 5-choosable was provded by Thomassen [5] n Ths theorem s best possble: e.g. Vogt [6] found planar graphs whch are not 4-choosable. 4CT_2008Dec23.doc page 1

2 Extensve use of computer programs as a proof-tool remans a source of controversy: such a type of proof s stll not unanmously accepted among mathematcans. Hence the search for old-fashoned proofs whch can be checked by hand should stll be consdered as a worthwhle enterprse, at least for a posteror verfcaton of a computer-based proof. Hopes that lst-colourng deas could be helpful to fnd such a proof have not been fulflled to date. In what follows such a proof by hand s presented. It does not yeld a colourng algorthm. The new dea s to use sublsts of a common lst of four colours, to enforce a common colour n all lsts, and to admt on the boundng cycle at most one vertex wth a lst of at least two colours. The strategy behnd ths dea s to avod counterexamples at low graph orders. 2 Theorems for the colourng of planar graphs Theorem 1. The chromatc number of a planar graph s not greater than four. The theorem s expressed n the vertex-colourng context wth the usual assumptons,.e. a coloured map n the plane or on a sphere s represented by ts dual (smple) graph G wth coloured vertces. When two vertces vk, vl G are connected by an edge vv k l (.e. the countres on the map have a common lne-shaped border), a (proper) colourng requres dstnct colours cv ( k ) and cv ( l ). Wthout loss of generalty the proof can be restrcted to (planar) near-trangulatons. A planar graph G s called a near-trangulaton f t s connected, wthout loops, and every nteror regon s (bounded by) a trangle. A regon s a trangle f t s ncdent wth exactly 3 edges. The exteror regon s bounded by the outer cycle. A (full) trangulaton s the specal case of a near-trangulaton, when also the nfnte exteror regon s bounded by a trangle (3-cycle). It follows from Euler s polyhedral formula that a planar graph wth n 3 vertces has at most 3n 6 edges, and the trangulaton s the edgemaxmal graph. Every planar graph H can be generated from a trangulaton G by removng edges and dsconnected vertces, therefore H G holds. As removal of edges reduces the number of restrctons for colourng, the chromatc number of H s not greater than the correspondng number of G. Instead of Theorem 1, t s convenent to prove the followng Theorem 2. Let G be a planar near-trangulaton wth outer cycle C = v1v2 vk 1vk v1, k 3. Assume that there s a common colour lst Lv ( ) = L0 = {1,2,3,4} of four colours for all vertces v G C, and a common colour α n all lsts of G. Assume further that there s a lst Lv ( ) { α} of at least three colours from L 0 for every vertex v C, wth the excepton of at most one arbtrarly selectable vertex w C, not adjacent to v 1, wth a lst of at least two colours, contanng α and not contanng another colour β Lv ( 2). Then there are always two dstnct colours α and β, such that two adjacent vertces v 1, v 2 C can be coloured wth cv ( 1) = α and cv ( 2) = β from ther lsts, and the colourng of v 1 and v 2 can be extended to a colourng of C and ts nteror G C. 4CT_2008Dec23.doc page 2

3 3 Proof of Theorem 2 We perform nducton wth respect to the number n = G of vertces. For n = k = 3 we have G = C = v1v2 v3v1, and the proof s trval. Let n 3 and the theorem be true for n vertces. Then consder n the nducton step a near-trangulaton G wth n + 1 vertces and k 3 vertces on the outer cycle C. Case 1. If C has a chord vv 1 j wth 3 j k 1 (see Fg. 1), the unon of vv 1 j wth each of the two components of G { v1, v j } yelds two nduced subgraphs G 1 and G 2 wth G1 G2 = v1vj. If there exsts a lst of only two colours, we may assume wthout loss of generalty that t belongs to a vertex on the outer cycle of G 1. If all vertces on cycles are enumerated n clockwse order, the outer cycles of G 1 and G 2 (each wth at least 3 vertces) are C1 = v1v2 vj v1 G1 and C2 = v1vj vk v1 G2, respectvely. If there s a lst of only two colours, we can always fnd a thrd colour β Lv ( 2) not contaned n ths lst, as Lv ( 2) 3. Otherwse, we choose β Lv ( 2) arbtrarly. Then we assgn cv ( 1) = α, cv ( 2) = β, and apply the nducton hypothess to C 1 and ts nteror. Next we fx the colours n G 1, such that cv ( 1) = α and a colour β Lv ( j ) s assgned to v j. As there s no vertex wth a lst of two colours n G 2 (hence there s no condton that cv ( j ) = β s not contaned n ths lst), we extend the colourng of v 1 and v l to a colourng of C 2 and ts nteror to obtan a colourng of G. v k v k 1 C 2 G 2 v 1 α v 2 β G 1 v j C 1 v k 1 v k P u 2 u l u 1 G v k v 1 C' v 2 α β Fg. 1: The outer cycle has a chord vv 1 j. Fg. 2: The case wthout a chord. Case 2. Now C has no chord (see Fg. 2). Let Nv ( k) = { v1, u1, u2,, ul, vk, l 1, be the set of neghbours of v k, enumerated n clockwse order. As the nteror of C s trangulated, vertces v 1 and vk 1 are connected by a path P = v1u1u2 ulvk 1. Then C = P ( C v k ) s a cycle (as C has no chord), and s the outer cycle of G = G v. k 4CT_2008Dec23.doc page 3

4 If G s a trangulaton, then vk 1 = v2, and a lst of two colours cannot exst. We begn wth settng cv ( 1) = α and cv ( k 1) = β, wth an arbtrary choce of β Lv ( k 1). Note that there s at least one colour γ Lv ( k ) \ { αβ, }. Next we defne new lsts of three colours L ( u ) = L( u )\{ γ } ( = 1,, l ), and set L ( v) = L( v) for the remanng vertces v G u1 u l {,, }. Now a lst wth at least 3 colours s avalable for every vertex v C, and a lst L ( v) = L0 s assgned to all vertces v n the nteror of C. We can therefore apply the nducton hypothess to C and ts nteror wth the new lsts L. In the resultng colourng of G, no neghbour of v k obtans the colour γ Lv ( k ). Hence we complete the colourng of G by assgnng t to v k. Now G s a near-trangulaton whch s not a trangulaton ( k 4 ). Frst assume that Lv ( k 1) = 3 and consder the colours n ths lst. Besdes α contaned n every lst, we always fnd always a second colour n the ntersecton Lv ( k 1) Lv ( 2), whch we denote by β and use for the colourng of v 2, together wth cv ( 1) = α. Then consder the colours n Lv ( k ) \ { α, β }: f there are two dstnct colours γ and δ, one of them can be removed from Lv ( k 1), such that the remanng lst stll contans three colours. Wthout loss of generalty let γ be ths colour,.e. Lv ( k 1) = { α, βδ, }. Next we contnue as n the case of a trangulaton by defnng new lsts L ( u ) = L( u ) \ { γ } ( = 1,, l ) and L ( v) = L( v) for the remanng vertces v G { u1,, u l }. Then we apply the nducton hypothess to C and ts nteror wth the new lsts L, and complete the colourng of G by assgnng γ to v k. Otherwse, f Lv ( k ) \ { α, β } s a lst of only one colour, say γ, then β Lv ( k ) holds, and we reserve β for v k. As β Lv ( k 1) holds, L ( vk 1) = L( vk 1)\{ β} can now be a lst of only two colours. Then there s at most one vertex w= vk 1, not adjacent to v 1, wth a lst L ( vk 1) of at least two colours, contanng α and another colour, other than β. Next we defne new lsts L ( u ) = L( u ) \ { β} ( = 1,, l ), and set L ( v) = L( v) for the remanng vertces v G { u1,, ul, vk. Then a lst wth at least 3 colours s avalable for every vertex v C { vk, and a lst L ( v) = L0 s assgned to all vertces v n the nteror of C. Agan we apply the nducton hypothess to C and ts nteror wth the new lsts L. In the resultng colourng of G, no neghbour of v k obtans the colour β Lv ( k ). Hence we complete the colourng of G by assgnng β to v k. Fnally assume that Lv ( k 1) = 2 holds from the begnnng. Then we always fnd a colour β n Lv ( 2) wth β Lv ( k 1), and hence assgn cv ( 1) = α and cv ( 2) = β. Now ether β Lv ( k ) or β Lv ( k ) holds. In the frst case, we reserve β for the colourng of v k, contnue wth defnng new lsts L and then perform the nducton step, as was descrbed n the last paragraph. Otherwse, β Lv ( k ) mples that there are two dstnct colours γ and δ 4CT_2008Dec23.doc page 4

5 n Lv ( k ) \ { α, β }. One of them, say γ, s not n Lv ( k 1), such that the new lst L ( vk 1) = L( vk 1)\{ γ } remans a lst of two colours, namely α and δ. Next we reserve γ for the colourng of v k, and contnue as n the prevous case wth Lv ( k 1) = 3 and two dstnct colours γ and δ n Lv ( k ) \ { α, β }. Acknowledgments Many deas employed n ths paper orgnate from the elegant and sophstcated constructon of Thomassen s proof that planar graphs are 5-choosable [5]. The author s ndebted to all who contrbuted to the progress of ths work by fndng errors n prevous versons. References [1] K. Appel and W. Haken: The soluton of the four-color-map problem. Sc. Amer. 237 (1977), [2] P. J. Heawood: Map-colour theorem. Quart. J. Pure Appl. Math. 24 (1890), [3] A. B. Kempe: On the geographcal problem of the four colors, Amer. J. Math. 2 (1879), [4] N. Robertson, D. P. Sanders, P. D. Seymour and R. Thomas: A new proof of the four colour theorem. Electron. Res. Announc. Amer. Math. Soc. 2 (1996), (electronc) [5] C. Thomassen: Every planar graph s 5-choosable. J. Combn. Theory Ser. B 62 (1994), [6] M. Vogt: Lst colourngs of planar graphs. Dscrete Math. 120 (1993), CT_2008Dec23.doc page 5