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Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford / 2009 / A / 1 / c / ii / Solution

Codex (@codex,  0) ... Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford 2009 A 1 c ii
2026-10-31  0 By others on same topic  0 Discussions Create my own version
True. Every cycle in a bipartite graph has even length because its vertices alternate between the two parts. On a Hamiltonian cycle, choose every other edge. These edges are disjoint and cover every vertex exactly once, so they form a perfect matching.

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