 Deletion-contraction recurrence for the chromatic polynomial

Deleting an edge removes one colour constraint. Edge contraction describes exactly the newly admitted colourings that violate that constraint. If $e$ joins distinct vertices $u,v$ of a simple graph $G$, write $G-e$ for the graph with that edge deleted and $G/e$ for the graph with $u,v$ identified and $e$ removed. Parallel edges produced by the contraction may be reduced to one edge, since they impose the same condition on a proper vertex coloring. ThenTo see why, divide the proper vertex colorings of $G-e$ into two classes. Those with different colours at $u,v$ are exactly the colourings of $G$. Those with equal colours at $u,v$ correspond one-to-one to colourings of $G/e$: the merged vertex receives their common colour. Subtracting the second class leaves the first.

For the triangle, delete $AC$ to get a three-vertex path graph, or contract $AC$ to get two vertices joined by an edge after removing the redundant parallel edge. Consequently,At $t=3$, this says $6=12-6$. Of the twelve path colourings pictured above, six give $A,C$ the same colour; these are precisely the ones ruled out when we restore $AC$.

This recurrence is useful because it reduces a constrained counting problem to smaller graphs. A graph on $n$ vertices with no edges has $P_G(t)=t^n$, since every vertex chooses its colour independently. Repeated edge deletion and edge contraction reduce to this case and show that the counting function is a polynomial. The same recurrence gives its leading coefficients. The method is conceptually simple, although the number of recursive cases can still grow rapidly.