Ramsey won't help: switching a black and white graph
Source: German TST 2004, exam VII, problem 3, by Eric Müller
June 1, 2004
linear algebramatrixvectorinductionalgorithmpigeonhole principlecombinatorics solved
Problem Statement
We consider graphs with vertices colored black or white. "Switching" a vertex means: coloring it black if it was formerly white, and coloring it white if it was formerly black.
Consider a finite graph with all vertices colored white. Now, we can do the following operation: Switch a vertex and simultaneously switch all of its neighbours (i. e. all vertices connected to this vertex by an edge). Can we, just by performing this operation several times, obtain a graph with all vertices colored black?
[It is assumed that our graph has no loops (a loop means an edge connecting one vertex with itself) and no multiple edges (a multiple edge means a pair of vertices connected by more than one edge).]