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Jacob k ≥ epoedb nuto ntenme fvrie.Cery h sta the Clearly, vertices. of number the on induction by proceed We v − 2 eso hteeyconnected every that show We h rp bandfrom obtained graph the n xaconnected a fix and 3 pnigtes(yCye’ oml 3 ]o h ubro spanning of number the on 4] [3, formula Cayley’s (by trees spanning k k vr connected Every crmtcgah.Rcl htagahis graph a that Recall graphs. -chromatic or 1 = [email protected] l uhr eespotdb h UIAadi cec n Hu- and Science in Award MUNI the by supported were authors All . awgrmesCayley meets Hadwiger G k k crmtcgahcontains graph -chromatic yietfigvertices identifying by ,wihcvr h aeo h nuto.W assume We induction. the of base the covers which 2, = ∗ k k Th´eo Pierron k crmtcgahcnan tleast at contains graph -chromatic oosbtnot but colors crmtcgraph -chromatic G dmKabela Adam Abstract yrmvn vertex removing by k , crmtcgahcnan tleast at contains graph -chromatic 1 [email protected] k v ∗ − 1 v , . . . , k oos n14,Hdie [7] Hadwiger 1943, In colors. 1 K ∗ G crmtcgahcnan at contains graph -chromatic k efis hwthat show first We . samnr h conjecture The minor. a as ℓ v k n eoigalnewly all removing and n by and , , crmtci tadmits it if -chromatic ailKr´al’Daniel [email protected] e nvriyo War- of University ce, k 0Bn,CehRe- Czech Brno, 00 ≤ G/v sdrdt be to nsidered k 1] ftrue, If [11]. 6 ,6 –1 15] 8–11, 6, 5, , G k − edenote we , statement. 2 1 k k v . . . spanning − tr of cture ∗† tement 2 G ℓ can the and be assumed to be 2-connected, have minimum degree at least k − 1, and contain no Kk as a subgraph. The graph G is 2-connected. Indeed, suppose that G contains a vertex v for which G−v is disconnected. For every connected component of G−v, consider the subgraph of G induced by v and the vertices of the component. Note that at least one of the considered subgraphs has the same chromatic number as G, and hence this subgraph contains at least kk−2 spanning trees by the induction hypothesis. The statement follows by observing that the number of spanning trees of G is equal to the product of the numbers of spanning trees of the considered subgraphs. The minimum degree of G is at least k − 1. Suppose that G contains a vertex v of degree at most k −2. Note that the graph G−v is connected (since G is 2-connected) and its chromatic number is at least k (since G cannot be colored using k − 1 colors). By the induction hypothesis, G − v contains at least kk−2 spanning trees. Each of these trees can be extended to a spanning tree of G by adding an arbitrary edge incident with v, and the obtained spanning trees are all distinct. The graph G does not contain Kk. Suppose that G contains Kk as a k−2 subgraph (possibly G is Kk). By Cayley’s formula, this subgraph contains k spanning trees. The statement follows by noting that each of these trees can be trivially extended to a spanning tree of G and the resulting spanning trees are all distinct. We next choose an arbitrary vertex of G, say v. Let d be the degree of v and F be the set of all edges incident with v; in particular, |F | = d and d ≥ k − 1. By assumption, the graph G − v is connected, and its chromatic number is at least k − 1. Hence, G − v contains at least (k − 1)k−3 spanning trees by the induction hypothesis, and each of these trees can be extended to a spanning tree of G by adding an arbitrary edge of F . In this way, we obtain at least d(k −1)k−3 distinct spanning trees of G. We proceed by finding additional spanning trees of G (each containing at least two edges of F , and therefore distinct from those derived above). We distinguish two cases based on the value of d. We first consider the case d = k − 1. Since G does not contain Kk as a subgraph, the vertex v has two non-adjacent neighbors, say a and b. Note that the chromatic number of the graph (G − v)/ab is at least k; otherwise, G could be colored using k − 1 colors by taking a coloring of (G − v)/ab with k − 1 colors and modifying it as follows. Assign the vertices a and b the color of the vertex obtained by their identification, and assign v a color not appearing among its neighbors’ colors (there are at most k − 2 such colors). Moreover, (G − v)/ab is connected (as G − v is connected), and thus contains at least kk−2 spanning trees by the induction hypothesis. Every such tree of (G − v)/ab corresponds to a spanning forest of G − v with two components, one containing a and the other containing b. Each of these forests can each be extended to a spanning tree of G either by adding both edges av and bv, or by adding an arbitrary edge from the

2 set F \{av, bv} and adding either the edge av or the edge bv (the choice between av and bv is determined so that the resulting graph is a tree). This gives at least (d − 1)kk−2 distinct additional spanning trees. We conclude that G contains at least d(k − 1)k−3 +(d − 1)kk−2 spanning trees in total, which is at least kk−2 as d = k − 1. We next discuss the case d ≥ k. We show that the neighborhood of v can be assumed to contain two disjoint pairs of non-adjacent vertices. Suppose otherwise. Since G does not contain Kk as a subgraph, the neighborhood of v does not contain a complete graph on k − 1 vertices. Hence, d = k and there is a triple of pairwise non-adjacent neighbors of v, say a, b and c. Following the same line of arguments as presented in the previous case, we derive that the graph (G−v)/abc is connected and its chromatic number is at least k. By the induction hypothesis, (G − v)/abc has at least kk−2 spanning trees. Each of these trees can be extended to a spanning tree of G either by adding the three edges av, bv and cv, or by adding any edge from the set F \{av, bv, cv} and two edges among av, bv and cv so that the resulting graph is a tree. This yields at least (d − 2)kk−2 distinct additional spanning trees of G, and thus G contains at least k(k − 1)k−3 + (k − 2)kk−2 spanning trees in total, which is at least kk−2 as desired. Hence, we can assume that v has two disjoint pairs of non-adjacent neighbors, say a, b, and c,d. Since the chromatic number of G − v is at least k − 1, the chromatic number of each of the connected graphs (G − v)/ab and (G − v)/cd is also at least k − 1. By the induction hypothesis, each of these two graphs contains at least (k − 1)k−3 spanning trees. Every spanning tree of (G − v)/ab can be extended to a spanning tree of G either by adding both edges av and bv, or by adding any edge from the set F \{av, bv} and adding either av or bv. Similarly, each spanning tree of (G − v)/cd can be extended either by adding cv and dv, or by adding any edge from the set F \{av,bv,cv,dv} and adding either cv or dv. In this way, we obtain at least (d − 1)(k − 1)k−3 +(d − 3)(k − 1)k−3 distinct spanning trees of G (those obtained from the spanning trees of (G − v)/ab contain the edge av or bv but the latter do not). This together with the d(k − 1)k−3 initially constructed spanning trees implies that G contains at least (3d − 4)(k − 1)k−3 ≥ (3k − 4)(k − 1)k−3 ≥ kk−2 spanning trees, where the last inequality holds as k ≥ 3. We remark that a careful inspection of the proof yields that the only k- k−2 chromatic graphs with exactly k spanning trees are those obtained from Kk by iteratively adding pendant edges.

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