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  "body": "\n## Statement\n\nLet $G$ be a finite [[graph|0JD5C82QMB]] that is cubic, that is, [[$3$-regular|0KA4S8R39M]], and that has no [[bridge|0D81KWH9ZM]].\nThen $G$ has a [[perfect matching|05ZPH8KF9K]].\n\n## Remarks\n\nThe proof uses [[Tutte's theorem|08J7R35B6Q]].\nIt is enough to show that, for each set $S$ of vertices, the graph $G - S$ has at most $\\abs{S}$ components with an odd number of vertices.\nIf $C$ is such an odd component, the degrees in $C$ add up to $3\\abs{C}$, which is odd, so an odd number of edges joins $C$ to $S$.\nThis number is not $1$, since $G$ has no bridge, so it is at least $3$.\nSince at most $3\\abs{S}$ edges end in $S$, there are at most $\\abs{S}$ odd components.\nFor example, the [[Petersen graph|0NR95CTFCZ]] is cubic and bridgeless, and its five spokes form a perfect matching.\nThe hypothesis on bridges is needed: there are cubic graphs with bridges that have no perfect matching.\nThe theorem also holds for cubic multigraphs without bridges, with the same proof.\n\n## Explanations\n\n### L1 human\n\nIn a cubic graph each vertex has exactly three edges.\nIf no single edge disconnects the graph, then the vertices can be paired off along edges, with every vertex in exactly one pair.\nThis is one of the oldest results of graph theory.\n\n### L2 human\n\nPetersen's theorem is a standard application of Tutte's theorem, by a counting argument on edges leaving odd components.\nRemoving a perfect matching from a bridgeless cubic graph leaves a $2$-factor, a disjoint union of cycles.\nThis is a starting point for edge colouring of cubic graphs.\nA cubic graph has chromatic index $3$ exactly when it has a perfect matching whose complement consists of even cycles.\n",
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