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Suppose you are trying to prove result $X$ by induction and are getting nowhere fast. One nice trick is to try to prove a stronger result $X'$ (that you don't really care about) by induction. This h …

4-colour Theorem. Every planar graph is 4-colourable. This theorem of course has a well-known history. It was first proven by Appel and Haken in 1976, but their proof was met with skepticism because …

To answer the question it is important to disentangle the proof as follows. Theorem 1. Every minimum counterexample to the 4CT is an internally 6-connected triangulation. Theorem 2. If $T$ is a min …

Yes, such a subgraph always exist. Let $G$ be a planar triangulation. By the $4$-colour theorem, $G$ has a $4$-colouring. We let $H$ be the subgraph consisting of all edges with endpoints coloured …

The Berge-Fulkerson conjecture holds for planar graphs. Here is a proof. Let $G$ be a bridgeless cubic planar graph. The dual graph $G^*$ is a triangulation. By the Four Colour Theorem, $G^*$ has a …

Pach and Tóth proved that if $G$ is a $k$-planar graph (with $k \geq 1$), then $|E(G)| \leq 4.108 \sqrt{k} |V(G)|$. Thus, every $k$-planar graph has a vertex of degree at most $\lfloor8.216 \sqrt{k …

No. The set of all $\{0,1\}$-sequences is also a clique in $G$. Thus, $\chi(G) \geq 2^{\aleph_0}$. On the other hand, the set of all bounded real sequences has size $2^{\aleph_0}$, so $\chi(G)=2^{\ …

An old conjecture of Lovász and Plummer is that for every cubic graph $G$ with no cut-edge, the number of perfect matchings in $G$ is exponential in the number of vertices. Chudnovsky and Seymour pro …

Yes, every graph $G$ with maximum degree $\Delta$ can be partitioned into $k$ sets $X_1, \dots, X_k$ such that the maximum degree of $G[X_i]$ is at most $\lfloor \Delta / k \rfloor$ for all $i$. This …

According to this link (see point $15$), Matt DeVos and David Wood can prove that every digraph in which each vertex has an out-neighbour has a good $4$-colouring (I am using their terminology). Up …

The vertex-adding number can be arbitrarily large compared to the chromatic number. To see this consider a long odd cycle, $C_{2k+1}$. Then $\chi(C_{2k+1})=3$, but $a(C_{2k+1})=2k+1$. Note that $ …

Taking $F=K_{4n, 4n}$ does the trick. To see this, suppose we have coloured each edge of $K_{4n,4n}$ red or blue. Let $R$ and $B$ be the red and blue subgraphs of $K_{4n,4n}$. We may assume that $R …

As far as I know, your conjecture is an open problem, because Reed's conjecture is still open for $\omega=2$. For $\omega=2$, your conjecture is essentially equivalent to Reed's conjecture. That is, …

Well, it is possible that both (WH) and (H) are true, in which case (WH) implies (H). If on the other hand, you are asking if there is a short proof of (H) assuming (WH), then the answer is no. Fo …

Yes, for $X$ infinite the resulting graph also has chromatic number 2. To see this, just use the fact that a graph is bipartite if and only if it does not contain an odd cycle (this remains true for …