Walk-powers and homomorphism bound of planar graphs
WWalk-powers and homomorphism bound of planar graphs
Reza Naserasr a , Sagnik Sen b , Qiang Sun ∗ a ( a ) LRI, CNRS and Universit´e Paris Sud, F-91405 Orsay Cedex, France( b ) Indian Statistical Institute, Kolkata, India October 19, 2018
Abstract
As an extension of the Four-Color Theorem it is conjectured that every planar graphof odd-girth at least 2 k + 1 admits a homomorphism to P C k = ( Z k , { e , e , · · · , e k , J } )where e i ’s are standard basis and J is all 1 vector. Noting that P C k itself is of odd-girth2 k + 1, in this work we show that if the conjecture is true, then P C k is an optimal such agraph both with respect to number of vertices and number of edges. The result is obtainedusing the notion of walk-power of graphs and their clique numbers.An analogous result is proved for bipartite signed planar graphs of unbalanced-girth 2 k .The work is presented on a uniform frame work of planar consistent signed graphs. Given a graph G , a signature on G is an assignment of signs, + or − , to the edges. The set ofnegative edges is normally denoted by Σ and will normally be referred to as the signature. A re-signing is to change the signs of all edges incident to a given set of vertices or, equivalently, edgesof an edge-cut. Two signatures are said to be equivalent if one can be obtained from the otherby a re-signing. A graph G together with a set of signatures equivalent to Σ is called a signedgraph and is denoted by [ G, Σ] where Σ is any member of the class of equivalent signatures. Asigned cycle with an even (odd) number of negative edges is called balanced ( unbalanced ). Itis easily observed that the balance of a cycle is invariant of re-signing. The unbalanced-girth ofa signed graph is the shortest length of its unbalanced cycles. A consistent signed graph is asigned graph in which every balanced cycle is of even length and all unbalanced cycles are of asame parity. Thus there are two types of consistent signed graphs:i. when all unbalanced cycles are of odd length, it can be shown that this is the case if andonly if Σ ≡ E ( G ), such a signed graph will be called an odd signed graph ;ii. when all unbalanced cycles are of even length, which will be the case if and only if G isbipartite, such a signed graph thus will be referred to as a singed bipartite graph . ∗ Corresponding author, supported by CSC. E-mail address: [email protected](Q. Sun). a r X i v : . [ m a t h . C O ] J a n .2 Homomorphisms and bounds Given two signed graphs [ G, Σ] and [ H, Σ (cid:48) ] we say there is a homomorphism of [ G, Σ] to [ H, Σ (cid:48) ],denoted [ G, Σ] → [ H, Σ (cid:48) ], if there is a signature Σ of G equivalent to Σ and a mapping ϕ : V ( G ) → V ( H ) such that xy ∈ E ( G ) implies ϕ ( x ) ϕ ( y ) ∈ E ( H ) and xy ∈ Σ if and only if ϕ ( x ) ϕ ( y ) ∈ Σ (cid:48) . It is easy to prove that if [ G, Σ] → [ H, Σ (cid:48) ], then unbalanced-girth of [ G, Σ] isat least as the unbalanced-girth of [ H, Σ (cid:48) ]. Give a class C of (signed) graphs we say a (signed)graph B bounds C if every member of C admits a homomorphism to B . For more on this subjectwe refer to [NRS13]. Projective cube of dimension d , denoted PC d , is the Cayley graph ( Z d , { e , e , · · · e d , J } ) where e i ’s are the standard basis and J is the all 1 vector of relevant length ( d here). It is obtainedby identifying antipodal vertices of the hypercube of dimension d + 1 or, equivalently, by addingedges between pairs of antipodal vertices of hypercube of dimension d . We define singed projec-tive cube of dimension d , denoted SPC d , to be the singed graph obtained from PC d by assigning+ to each edge corresponding to an e i and − to edges corresponding to J .Projective cubes, also known as folded cubes, are well-studied graphs. We refer to [NRS13]and references there for some properties of signed projective cubes and for a proof of the followingtwo theorems: Theorem 1.1.
Signed projective cube of dimension d is a consistent signed graph and hasunbalanced-girth d + 1 . It follows that if a signed graph admits a homomorphism to a signed projective cube, it mustbe a consistent signed graph. Such a mapping then becomes equivalent to a packing problem asthe following theorem claims:
Theorem 1.2.
A consistent signed graph ( G, Σ) admits a homomorphism to SPC d if and onlyif the edges set of G can be partitioned into d + 1 disjoint sets each of which induces a signatureequivalent to Σ . The following conjecture, introduced in [N07] and [G05] (also see [NRS12]) is the focus ofthis work:
Conjecture 1.3.
Given d ≥ , every planar consistent signed graph of unbalanced-girth d + 1 admits a homomorphism to SPC d . The conjecture is formed of two parts: for even values of d (by considering the signature inwhich all edges are negative) it claims that every planar graph of odd-girth at least d + 1 admitsa homomorphism to PC d . For odd values of d it says that every planar signed bipartite graphof unbalanced-girth at least d + 1 admits a homomorphism to SPC d . Since PC is isomorphicto K , the very first case of this conjecture is the Four-Color Theorem. This conjecture, for odd values of d was introduced in [N07] in relation to a question of J. Neˇsetˇrilwho asked if there is a triangle-free graph to which every triangle-free planar graph admits ahomomorphism. This question was answered in a larger frame work by P. Ossona de Mendezand J. Neˇsetˇril which is stated based on the following notation.2iven a finite set H of connected graphs we use F orb h ( H ) to denote the class of all graphswhich do not admit a homomorphism from any member of H . Similarly, given a set M of graphswe use F orb m ( M ) to denote the class of all graphs that have no member of M as a minor. Theorem 1.4. [NO08] Given a finite set M of graphs and a finite set H of connected graphs,there is graph in F orb h ( H ) to which every graph in F orb h ( H ) ∩ F orb m ( M ) admits a homomor-phism. The bound that are build using known proofs of this theorem are super exponential. To find theoptimal bound in this theorem, in general, is a very difficult question. Indeed this question, inparticular, contains the Hadwiger’s conjecture simply by taking M = H = { K n } . Conjecture 1.3proposes a smaller bound for the case of M = { K , K , } and H = { C k − } . For k = 1, ( C being a loop), since K is a planar graph, it is the optimal answer by the Four-Color Theorem.For k = 2, it is proved in [N13] that P C (4), known as the Clebsch graph, is the optimal bound.Here we prove that any bound of odd-girth 2 k + 1 for planar graphs of odd-girth 2 k + 1 has tohave at least 2 k vertcies each of degree at least 2 k + 1. This would imply that if Conjecture 1.3holds, then P C (2 k ) is an optimal bound. We prove an analogue result for the case of planarsigned bipartite graphs, even though analogue of Theorem 1.4 for signed bipartite graphs is notproved yet. In this section we consider the first part of Conjecture 1.3. This case deals with odd signedgraphs in which case one can assume all the edges are negative. Thus homomorphism problemhere is simply a homomorphism of graphs.To prove our result, in fact we prove a stronger claim in the following sense. Given a graph G and a positive integer k we define the k -th walk-power of G , denoted by G ( k ) , to be a graphwhose vertex set is also V ( G ) with two vertcies x and y being adjacent if there is a walk oflength k connecting x and y in G . This graph would be loopless only if k is odd and G hasodd-girth at least k + 2, thus this will be the only case of interest for us in this work. If φ is ahomomorphism of G to H , then it can easily be checked that φ is also a homomorphism of G ( k ) to H ( k ) . Thus to prove our claim we will prove the following stronger result. Theorem 2.1.
There is a planar graph G of odd-girth k + 1 with ω ( G (2 k − ) ≥ k . To prove the theorem we will in fact construct an example of such a graph. This constructionis based on the following local construction.
Lemma 2.2.
Let G be a graph obtained from subdividing edges of K such that in a planarembedding of G each of the four faces is a cycle of length k + 1 . Then G (2 k − is isomorphic to K k .Proof. Let a, b, c and d be the original vertices of the K from which G is constructed. For x, y ∈ { a, b, c, d } let P xy be the subdivision of xy , and let t xy be the length of this path. For aninternal vertex w of P xy , let P xw (or P wx ) be the part of P xy connecting w to x , let t xw be thelength of it.We have 3 ab + t bc + t ca = t ab + t bd + t da = t ac + t cd + t da = t bc + t cd + t db = 2 k + 1 . (1)From equation 1 we have t xy = t wz for { x, y, w, z } = { a, b, c, d } , (2)that is to say the if all four faces are of a same length, then parallel edge of K are subdividedthe same number of times (the parity of the length of faces is not important for this claim, theeven case will be used later).Let u and v be a pair of vertices of G . If they are both vertices of a facial cycle of G , thenthere is a walk of length 2 k − k + 1.If there is no facial cycle of G containing both u and v , then they are internal vertices (aftersubdivision) of two distinct parallel edges of K , thus we may assume, without loss of generality,that u is a vertex of the path P ab and v is a vertex of the path P cd .Note that by equation 2 we have t au + t bu = t cv + t dv = t ab = t cd . (3)If t ab = t cd is even (odd respectively), then t au and t bu have the same parity (different paritiesrespectively) and t cv and t dv have the same parity (different parities respectively). Moreover,since t cd is even (odd respectively) and t ac + t cd + t da = 2 k + 1, t ac and t ad have different parities(same parity respectively).Now one of the paths connecting u, v , say P ua ∪ P ac ∪ P cv , is of length t au + t ac + t cv , andanother path, say P ub ∪ P bd ∪ P dv , is of length t bu + t bd + t dv . By (3) we have ( t bu + t bd + t dv ) +( t au + t ac + t cv ) = 2( t ab + t bd ), hence t bu + t bd + t dv and t au + t ac + t cv have a same parity.Furthermore, since P ab ∪ P ad ∪ P bd forms a facial cycle we have t ab + t ad + t bd = 2 k + 1, thus2( t ab + t bd ) = 4 k + 2 − t bc ≤ k .Hence we have min { ( t au + t ac + t cv ) , ( t bu + t bd + t dv ) } ≤ k . Similarly, we can show that min { ( t au + t ad + t dv ) , ( t bu + t bc + t cv ) } ≤ k .But note that min { ( t au + t ac + t cv ) , ( t bu + t bd + t dv ) } and min { ( t au + t ad + t dv ) , ( t bu + t bc + t cv ) } have different parities irrespective of the parity of t ab = t cd . Therefore, there is a walk of length2 k − u to v . Proof of Theorem 2.1 . Consider a K on four vertices a, b, c and d . Let G be a subdivisionof this K where edges ab and cd each are subdivided into 2 k − G is a subdivisionof K in which all the four faces are cycles of length 2 k + 1. Hence by Lemma 2.2 we have ω ( G (2 k − ) = | V ( G ) | = 4 k.
4n the following we build a sequence of graphs G i , i = 1 , , · · · , k −
1, such that each G i +1 , i ≤ k −
2, contains G i as a subgraph, G i +1 is planar and of odd-girth 2 k + 1 and such that ω ( G (2 k − i +1 ) > ω ( G (2 k − i ). At the final step we will have ω ( G (2 k − k − ) ≥ k . We start with the following partial construction. Suppose G i is built and let P = uv v · · · v r w be a maximal thread, that is, a path P connecting u and w such that all v j ’s ( j ∈ { , , ..., r } )are of degree 2 in G i but u and w are of degree at least 3. Furthermore, assume that P is eitherpart of a path of length 2 k − a and b or part of a path of length 2 k − c and d .Since P is a thread, if we add a new edge uw in G i , the resulting graph will still be planar.So we add such an edge and subdivide it r times to obtain the new thread P (cid:48) = uv (cid:48) v (cid:48) · · · v (cid:48) r w .Consider a planar drawing of the graph in which P and P (cid:48) form a facial cycle of length 2 r + 2.In the face P P (cid:48) connect v and v (cid:48) r by a new edge. Subdivide this new edge 2 k − r − k − r edges, we draw it in dot line), so that each of the facial cycles containingthe new thread is of length 2 k + 1. Figure 1: thread PDenote by G (cid:48) i the resulting graph. We first note that G (cid:48) i is also of odd-girth 2 k + 1. Nowsuppose a maximal clique W of G (2 k − i contains v j of the thread P . Then we claim that W ∪ v (cid:48) j is also a clique of G (cid:48) (2 k − i .To prove this let x be any vertex of W . If x is not in P , then consider a walk of length 2 k − v j to x . Each time this walk uses a part of P , replace it with the corresponding part from P (cid:48) and this would give a walk of length 2 k − x to v (cid:48) j .If x ∈ P , then, without loss of generality, assume that P is part of a path of length 2 k − a and b . Consider the subgraph induced by this path together with c , P (cid:48) and the v ...v (cid:48) r thread we added to build G (cid:48) i . This induced subgraph is a subdivision of K in which5ll the faces are cycles of length 2 k + 1. Thus, by Lemma 2.2 there is a walk of length 2 k − x and v (cid:48) j . Extending this argument we observe that if all vertices of P are in W ,then W ∪ { v (cid:48) , v (cid:48) , · · · , v (cid:48) r } is a clique in G (cid:48) (2 k − i .Now we describe our general construction. At first we have G on 4 k vertices and twomaximal threads. By Lemma 2.2 all the vertices of these two threads are parts of the uniqueclique of order 4 k in G (2 k − . We apply the previously mentioned construction on both threadsto build G which will have four maximal threads each of length 2 k − k − a, b or c, d , for examplethe v (cid:48) k − v -thread drawn in dot line of Figure 1 is not considered). There is a clique of order4 k + 2(2 k −
2) in G (2 k − , and there are four maximal threads of length 2 k −
2, each is a partof a path of length 2 k − a and b or c and d .Continuing this construction, in general, there is a clique W i of G (2 k − i (2 ≤ i ≤ k − k + (cid:80) i − j =1 j (2 k − j −
1) and there are 2 i maximal threads of length 2 k − i which each is a part of a path of length 2 k − a to b or c to d .Note that G i at each step is a planar graph of odd-girth 2 k + 1. The clique W k − of G (2 k − k − has order equal to4 k + k − (cid:88) j =1 j (2 k − j −
1) = 4 k + (2 k − k − (cid:88) j =1 j − k − (cid:88) j =1 j j − = 4 k + [(2 k − k − − − − k − ) − ( − k − k − ]= 4 k + ( k k − k − k − + 2) − (2 − k + k k − k − )= 2 k . This completes the proof.
Corollary 2.3.
Let B be a graph of odd-girth k + 1 to which every planar graph of odd-girth k + 1 admits a homomorphism. Then | V ( B ) | ≥ k . Furthermore, if B is minimal with thisproperty, then δ ( B ) ≥ k + 1 .Proof. Let G be a graph build in the previous theorem. Since G is of odd-girth 2 k + 1, bythe assumption, it maps to B . Since B is also of odd-girth 2 k + 1, both B (2 k − and G (2 k − are simple graphs and G (2 k − → B (2 k − . Hence K k ⊂ B (2 k − which, in particular implies | V ( B ) | ≥ k .To prove the lower bound on minimum degree, we first introduce the following graph: let P be a path of length 2 k on vertices x , x , · · · , x k +1 connected in this order. Let P (cid:48) be obtainedfrom P by subdividing each edge 2 k − x i is at distance 2 k − x i +1 . Let y i , y i , · · · y i k − be the new vertices subdividing x i x i +1 and connected in this order in P (cid:48) . We addshort cut edges so that the shortest odd walk between each x i and x j becomes of length 2 k − x y , y y , y y , · · · , y k k − x k +1 . Now given a vertex u , the graph P u is a graphwhich is formed from a disjoint copy of P (cid:48) by connecting u to all x i ’s. Note that the graph P u is of odd-girth 2 k + 1 and that in P (2 k − u vertices of P (i.e., x i ’s) induce a (2 k + 1)-clique.Now since B is minimal, there exists a planar graph G B of odd-girth 2 k + 1 whose mappingsto B are always onto. Let G ∗ B be a new graph obtained from G B by adding a P u for each vertex6 of G B . This new graph is also of odd-girth 2 k + 1, thus, by the choice of B , it maps to B .Let φ be such a mapping of G ∗ B to B . This mapping induces a mapping of G B to B . Thus eachvertex v of B is image of a vertex u of G B by the choice of G B . But in the mapping G ∗ B to B ,all x i ’s of P u must map to distinct vertices all of whom are neighbours of φ ( u ) = v .Note that since P C (2 k ) is a (2 k + 1)-regular graph on 2 k vertices, it would be an optimalbound if Conjecture 1.3 holds. The development of the notion of homomorphisms for signed graphs has began very recentlyand, therefore, it is not yet known if an analogue of Theorem 1.4 would hold for the class ofsigned bipartite graphs. While we believe that would be the case, here we prove that
SP C k isthe optimal bound for the signed bipartite case of Conjecture 1.3 if the conjecture holds.To start, we introduce an analogue notion of walk-power. Let [ G, Σ] be a signed bipartitegraph with (
X, Y ) being the partition of vertices. Given an even integer r ≥ G, Σ] r to be a graph on V ( G ) where a pair u, v of vertices are adjacent if the following conditions hold: • u and v are in the same part of G ; • there are u, v -paths P and P , each of length at most r , such that one has an odd numberof negative edges and the other has even number of them.Note that the second condition is independent of the choice of a representative signature. Fur-thermore [ G, Σ] r is a graph (not signed) with no connection from X to Y .We remark that these two conditions together are to say that: for any choice of an equivalentsignature if u and v are identified then there would be an unbalanced cycle of even length atmost r . That can be analogue to the definition of G r for odd values of r where odd-girth of G is at least r + 2, in the following sense: first of all G can be regarded as a consistent signedgraph [ G, E ( G )]; secondly for any choice of equivalent signature Σ of [ G, E ( G )] r if identifyingpair u, v of vertices results in yet an odd (signed) graph (analogue of 1) but of unbalanced girthat most r (analogue of 2), then u and v are adjacent in [ G, E ( G )] r . While [ G, E ( G )] r could bea proper subgraph of G r , the claim and proof of Theorem 2.1 can be revised with this modifieddefinition.With the previous remark following lemma is easy to verify. Lemma 3.1.
Let [ G, Σ] and [ H, Π] be two signed bipartite graphs and let φ be a homomorphismof [ G, Σ] to [ H, Π] . Then for any positive even integer r , φ is also a homomorphism of [ G, Σ] r to [ H, Π] r . Thus if both graphs are of unbalanced girth at least r + 2, then [ G, Σ] r and [ H, Π] r areboth loopless, and, therefore, mapping φ would imply ω ([ G, Σ] r ) ≤ ω ([ H, Π] r ). Furthermore,assuming that G and H are both connected, since φ is also a mapping of G to H , it wouldpreserve bipartition. Thus in what follows we will built a signed bipartite planar graph [ G, Σ]of unbalanced girth 2 k such that each part of G contains a clique of size 2 k − in [ G, Σ] k − .To this end we start with the following lemma which is the signed bipartite analogue ofLemma 2.2. 7 emma 3.2. Let [ G, Σ] be a planar signed graph which is obtained from assigning a signatureto a subdivision of K in such a way that each of the four facial cycles is an unbalanced cycleof length k . Then [ G, Σ] (2 k − is isomorphic to disjoint copies of K (2 k − each induced on onepart of the bipartite graph G .Proof. We consider a fixed signature Σ of [ G, Σ]. We will use the same notations ( P xy , t xy etc.)as in Lemma 2.2. Thus as proved in that lemma, parallel edges of K are subdivided samenumber of times. Furthermore, repeating the same argument modulo 2, we can conclude thatthe number of negative edges in P xy and the number of negative edges in P wz have same parityfor all { x, y, w, z } = { a, b, c, d } .Let u and v be two vertices from same part of G (thus any path connecting u and v haveeven length). We would like to prove that they are adjacent in [ G, Σ] (2 k − . If they belong to asame facial cycle, then the two paths connecting these two vertices in that (unbalanced) cyclesatisfy the conditions and we are done. Hence, assume without loss of generality that u ∈ P ab and v ∈ P cd .Removing the edges of the parallel paths P ad and P bc will result in a cycle of length 4 k − t ad containing u, v . This implies: ( t ua + t ac + t cv ) + ( t ub + t bd + t dv ) ≤ k − ⇒ min { ( t ua + t ac + t cv ) , ( t ub + t bd + t dv ) } ≤ k − . (4)Similarly by removing P ac and P bd we getmin { ( t ua + t ad + t dv ) , ( t ub + t bc + t cv ) } ≤ k − . (5)It remains to show that the two paths of equation (4) and (5) have different number ofnegative edges modulo 2. To see this note that union of any one of the two paths from (4) witha path from (5) covers a facial cycle exactly once and one a part of P ab or P cd twice. Since eachfacial cycle is unbalanced, our claim is proved.We are now ready to present our general construction. Theorem 3.3.
There exists a planar signed bipartite graph [ G, Σ] such that in [ G, Σ] (2 k − eachpart of G induces a clique of size at least k − .Proof. Consider a K on four vertices a, b, c and d . Let G be a subdivision of this K whereedges ab and cd each are subdivided into 2 k − G is a connected bipartitegraph and let V and V (cid:48) be its partite sets. Let Σ be the signature with the new edge incident to a (created by the subdivision of ab ) and the new edge incedent to c (created by the subdivisionof cd ) being negative. Thus the signed bipartite graph [ G , Σ ] is a subdivision of K in whichall the four faces are unbalanced cycles of length 2 k . Hence by Lemma 3.2 we know that eachof V and V (cid:48) induces a clique of order 2 k − G , Σ ] (2 k − .In the following we will build a sequence of signed graphs [ G i , Σ i ], for i ∈ { , , · · · , k − } ,such that each [ G i +1 , Σ i +1 ], i ≤ k −
2, contains [ G i , Σ i ] as a subgraph. Moreover, the signedgraph [ G i , Σ i ] is a bipartite planar graph with unbalanced-girth 2 k and partite sets V i , V (cid:48) i . Letus denote the clique number of the graph induced by V i (or V (cid:48) i ) from [ G i , Σ i ] (2 k − by f ( i ) (or f (cid:48) ( i )). Note that both the functions are strictly increasing and at the final step we will have f (2 k − , f (cid:48) (2 k − ≥ k − .
8e start with the following partial construction. Suppose [ G i , Σ i ] is built and let P = uv v · · · v r w be a maximal thread. Furthermore, assume that P is either part of a path oflength 2 k − a and b or part of a path of length 2 k − c and d .Since P is a thread, if we add a new edge uw in [ G i , Σ i ], the resulting graph will still be planar.So we add such an edge and subdivide it r times to obtain the new thread P (cid:48) = uv (cid:48) v (cid:48) · · · v (cid:48) r w .Also we assign signs of the new edges in such a way that the edges uv (cid:48) and v r w have a samesign, the edges v (cid:48) r w and uv have a same sign and the edges v (cid:48) i v (cid:48) i +1 and v r − i +1 v r − i have a samesign.Consider a planar drawing of the graph in which P and P (cid:48) form a facial cycle of length 2 r .In the face P P (cid:48) connect v and v (cid:48) r by a new edge. Subdivide this new edge 2 k − r − k − r − k . Choose signs of the edges of this new path in such a way thateach of the facial cycles containing the new thread is unbalanced.Let [ G (cid:48) i , Σ (cid:48) i ] be the resulting signed graph. We first note that [ G (cid:48) i , Σ (cid:48) i ] is also planar bipartiteof unbalanced-girth 2 k . Now suppose that the vertices of P , indexed by odd (or even) numbersare all part of a maximal clique in the graph induced by one partite set of G i in [ G i , Σ i ] (2 k − .Then the vertices of P ∪ P (cid:48) , indexed by odd (or even) numbers are all part of a maximal cliquein the graph induced by the corresponding partite set of G (cid:48) i in [ G (cid:48) i , Σ (cid:48) i ] (2 k − . This can be provedby similar logic used in the proof of Theorem 2.1. The only difference is that to prove the aboveclaim one needs to repeat the argument based on the parity of number of negative edges insteadof the parity of number of edges and use Lemma 3.2 instead of Lemma 2.2.Now we describe our general construction. At first we have [ G , Σ ] on 4 k − k − G , Σ ] (2 k − . We apply the previously mentioned construction on both the threads to build[ G , Σ ] which will have four maximal threads each of length 2 k − k − a, b or c, d , the greenthreads are not considered). There are two disjoint cliques, each of order (2 k −
1) + (2 k −
3) in[ G , Σ ] (2 k − , that is, f (2) = f (cid:48) (2) = (2 k −
1) + (2 k − k −
3, each a part of a path of length 2 k − a, b or connecting c, d .Continuing this construction, in general, f ( i ) = f (cid:48) ( i ) = (2 k −
1) + (cid:80) i − j =1 j − (2 k − j − i maximal threads of length 2 k − i − k − a, b or connecting c, d .Note that [ G i , Σ i ] at each step is a planar bipartite signed graph of unbalanced-girth 2 k .Therefore f (2 k −
1) = f (cid:48) (2 k −
1) = 2 k − k − (cid:88) j =1 j − (2 k − j −
2) = 2 k − k − k − (cid:88) j =1 j − k − (cid:88) j =1 j j − = 2 k − k − k − − − [(1 − k − ) − ( − k − k − ]= 2 k − k k − − k − k − + 2] − [1 − k − + k k − − k − ]= 2 k − . This completes the proof.
Corollary 3.4. If [ B, Π] is a minimal signed bipartite graph of unbalanced-girth k to which ev-ery planar signed bipartite graph of unbalanced-girth k admits a homomorphism, then | V ( B ) | ≥ k − and δ ( B ) ≥ k . roof. Let [ G, Σ] be the graph built in the previous theorem. Since [ G, Σ] is of unbalanced-girth2 k , by the assumption, it maps to [ B, Π]. Since [ B, Π] is also of unbalanced-girth 2 k , both[ B, Π] (2 k − and [ G, Σ] (2 k − are simple bipartite graphs and [ G, Σ] (2 k − → [ B, Π] (2 k − . Hencethere is a K k − in each part of [ B, Π] (2 k − which, in particular implies | V ( B ) | ≥ k − .To prove the lower bound on minimum degree, note that since [ B, Π] is minimal, there existsa signed bipartite planar graph [ G B , Σ B ] of unbalanced-girth 2 k whose mappings to [ B, Π]are always onto. Now consider the graph [ G, Σ] built in the previous theorem. Note that in[ G, Σ] (2 k − all the 2 k neighbours of a are adjacent to each other where a is one of the verticesof the K that we started our construction with. Therefore, in any mapping of [ G, Σ] to [ B, Π]image of a must be of degree at least 2 k .Now for each vertex x of [ G B , Σ B ] add a vertex disjoint copy [ G x , Σ] of [ G, Σ] and identifythe vertex a of [ G x , Σ] with x . Let [ G (cid:48) B , Λ B ] be the new graph. By the construction, [ G (cid:48) B , Λ B ]is also a signed bipartite planar graph of unbalanced-girth 2 k . Hence it maps to [ B, Π]. In anysuch mapping, by the choice of [ G B , Σ B ] and construction of [ G (cid:48) B , Λ B ], each vertex of [ B, Π] isan image of a in a mapping of [ G, Σ] to [ B, Π]. Thus each vertex of [ B, Π] has degree at least2 k . P. Seymoure has conjectured in [S75] that the edge-chromatic number of a planar multi-graphis equal to it fraction edge-chromatic number. It turns out that the restriction of this conjecturefor k -regular multigraph can be proved if and only if Conjecture 1.3 is proved for this valueof k = d −
1. This special case of Seymour conjecture is proved for k ≤ k = 4 , k = 6), [E11] ( k = 7)and [CES12] ( k = 8). Thus Conjecture 1.3 is verified for d ≤
7. Hence we have the followingcorollary.
Theorem 4.1.
For d ≤ the signed graph SP C d is the smallest consistent graph (both in termsof number of vertices and edges) of unbalanced-girth d + 1 which bounds all consistent planarsigned graphs of unbalanced-girth at least d + 1 . B. Guenin has proposed a strengthening of Conjecture 1.3 by replacing the condition ofplanarity with no ( K , E ( K ))-minor.For further generalization one can consider the following general question: Problem 4.2.
Given d and r , d ≥ r and d = r (mod 2) what is the optimal bound of unbalancedgirth r which bounds all consistent signed graph of unbalanced-girth d with no ( K n , E ( K n )) -minor? We do not yet know of existence of such a bound in general. For n = 3, consistent signedgraphs with no ( K n , E ( K n ))-minor are bipartite graphs with all edges positive, and, therefore,bounded by K . For n = 5 if the input and target graph are both of unbalanced-girth d + 1,then our work and Geunin’s extension of Conjecture 1.3 proposes projective cubes as the optimalsolutions. For d = r = 3, the answer would be K n − if Odd Hadwiger conjecture is true. Forthe case of n = 4 some partial answers are given by F. Foucaud and first author. For all othercases there is not even a conjecture yet. 10 eferences [CES12] M. Chudnovsky, K. Edwards and P. Seymour. Edge-colouring eight-regular planargraphs. Manuscript (2012), available at http://arxiv.org/abs/1209.1176v1.[DKK] Z. Dvoˇr´ak, K. Kawarabayashi and D. Kr´al’. Packing six T-joins in plane graphs.Manuscript (2014), available at http://arxiv.org/abs/1009.5912v3.[E11] K. Edwards. Optimization and Packings of T-joins and T-cuts. M.Sc. Thesis,McGill University (2011).[G05] B. Guenin. Packing odd circuit covers: A conjecture. Manuscript (2005).[G12] B. Guenin. Packing T-joins and edge-colouring in planar graphs. Mathematics ofOperations Research , to appear.[N07] R. Naserasr, Homomorphisms and edge-colorings of planar graphs.
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