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  "body": "\n## Proof\n\nFor a [[graph|0JD5C82QMB]] $H$ on $V$ and $U \\subseteq V$, write $o(H - U)$ for the number of odd [[connected components|01VVP10MWJ]] of $H - U$. Say that $H$ satisfies *Tutte's condition* if $o(H - U) \\le \\abs{U}$ for all $U \\subseteq V$. Let $G$ be a finite graph with vertex set $V$.\n\n*The condition is necessary.*\n\n1. Let $M$ be a [[perfect matching|05ZPH8KF9K]] of $G$, and let $U \\subseteq V$. Let $K$ be an odd component of $G - U$.\n2. The edges of $M$ inside $K$ cover an even number of vertices of $K$. As $\\abs{K}$ is odd, some vertex $x_K \\in K$ is matched to a vertex $y_K \\notin K$.\n3. There are no edges between different components of $G - U$, so $y_K \\in U$.\n4. Distinct odd components give distinct vertices $x_K$. Since $M$ is a [[matching|0Z87BHAFTB]], they also give distinct $y_K$. So $K \\mapsto y_K$ is injective, and $o(G - U) \\le \\abs{U}$.\n\n*The condition is sufficient: setup.*\n\n5. Suppose that $G$ satisfies Tutte's condition but has no perfect matching. We derive a contradiction.\n6. For $U = \\varnothing$, the condition gives $o(G) = 0$. So every component of $G$ is even, and $\\abs{V}$ is even.\n7. Among all graphs on $V$ that contain every edge of $G$ and have no perfect matching, choose one, $G^*$, with the most edges. Then $G^* + uv$ has a perfect matching for any two nonadjacent vertices $u \\ne v$ of $G^*$.\n8. The graph $G^*$ satisfies Tutte's condition. Let $U \\subseteq V$. Each component of $G^* - U$ is a union of components of $G - U$, since $G$ is a [[subgraph|0P11X5HKAX]] of $G^*$.\n9. A union of even sets is even, so each odd component of $G^* - U$ contains an odd component of $G - U$. These are distinct for distinct components. So $o(G^* - U) \\le o(G - U) \\le \\abs{U}$.\n10. Let $S$ be the set of vertices of $G^*$ adjacent to all other vertices.\n\n*Case 1: every component of $G^* - S$ is a complete graph.*\n\n11. Let $K_1, \\dots, K_k$ be the odd components of $G^* - S$. By Step 8, $k \\le \\abs{S}$. Choose distinct $s_1, \\dots, s_k \\in S$ and vertices $x_i \\in K_i$.\n12. Match $x_i$ with $s_i$; this is an edge, since $s_i \\in S$. The set $K_i \\setminus \\set{x_i}$ is even and complete, so pair up its vertices by edges.\n13. Pair up the vertices of each even component in the same way.\n14. Let $S' = S \\setminus \\set{s_1, \\dots, s_k}$. Modulo $2$, $\\abs{V} = \\abs{S} + \\sum_{\\text{components}} \\abs{K} \\equiv \\abs{S} + k$. As $\\abs{V}$ is even, $\\abs{S'} = \\abs{S} - k$ is even.\n15. Any two vertices of $S$ are adjacent, so pair up the vertices of $S'$ by edges.\n16. Steps 12 to 15 give a perfect matching of $G^*$. This contradicts Step 7.\n\n*Case 2: some component $K$ of $G^* - S$ is not complete.*\n\n17. Choose nonadjacent vertices of $K$ and a shortest [[path|0JDE56HWQ8]] in $K$ between them. Let $a, b, c$ be its first three vertices. Then $ab$ and $bc$ are edges and $a, c$ are not adjacent, by minimality of the path.\n18. Since $b \\in K$, $b \\notin S$. So there is a vertex $d \\ne b$ not adjacent to $b$. Then $d \\ne a, c$.\n19. By Step 7, $G^* + ac$ has a perfect matching $M_1$, and $G^* + bd$ has a perfect matching $M_2$. Then $ac \\in M_1$ and $bd \\in M_2$, since $G^*$ has no perfect matching.\n20. Also $bd \\notin M_1$ and $ac \\notin M_2$, since these are not edges of $G^* + ac$ and $G^* + bd$ respectively.\n21. Let $D = M_1 \\mathbin{\\triangle} M_2$ be the [[symmetric difference|0Z49NTR1DZ]]. Each vertex lies on one edge of $M_1$ and one of $M_2$. So it lies on $0$ or $2$ edges of $D$, one from each matching.\n22. Hence $D$ is a disjoint union of [[cycles|07CT9RDY7Q]] whose edges alternate between $M_1$ and $M_2$. By Step 20, $bd \\in D$. Let $C$ be the cycle of $D$ through $bd$.\n23. At each vertex of $C$, both its $M_1$-edge and its $M_2$-edge lie on $C$. So edges of $M_1$ or $M_2$ not on $C$ have no end on $C$.\n24. *Case 2a: $ac \\notin C$.* Let $M = (M_2 \\setminus C) \\cup (M_1 \\cap C)$. The edges $M_1 \\cap C$ cover each vertex of $C$ once, and $M_2 \\setminus C$ covers each other vertex once, by Step 23.\n25. So $M$ is a perfect matching. It avoids $bd$, which is on $C$ and not in $M_1$, and it avoids $ac$, which is not on $C$ and not in $M_2$. So $M$ is a perfect matching of $G^*$, a contradiction.\n26. *Case 2b: $ac \\in C$.* Walk along $C$ from $b$ through $d$, and stop at the first vertex $u \\in \\set{a, c}$. This exists, since $a, c$ lie on $C$ and $b, d \\notin \\set{a, c}$. Let $P$ be the part of $C$ from $d$ to $u$.\n27. The path $P$ does not contain $ac$, since we stop on first reaching an end of $ac$. So $P$ reaches $u$ along the other $C$-edge at $u$, which is the $M_2$-edge of $u$.\n28. The first edge of $P$ is the $M_1$-edge of $d$, as the $M_2$-edge of $d$ is $bd$. So the edges of $P$ alternate, starting in $M_1$ and ending in $M_2$.\n29. The edge $ub$ is an edge of $G^*$ by Step 17. Let $C'$ be the cycle formed by $bd$, $P$ and $ub$. Its edges in order are $bd \\in M_2$, then $P$ alternating from $M_1$ to $M_2$, then $ub$.\n30. Let $N$ consist of $ub$ and the $M_1$-edges of $P$. These are every second edge of $C'$, so $N$ covers each vertex of $C'$ exactly once. No edge of $N$ is $ac$ or $bd$.\n31. Each vertex of $C'$ has its $M_2$-edge on $C'$: for $b$ this is $bd$, and for the other vertices it lies on $P$. So the edges of $M_2$ not on $C'$ have no end on $C'$.\n32. Let $M = (M_2 \\setminus C') \\cup N$. By Steps 30 and 31, $M$ is a perfect matching of $G^*$, since it avoids $bd$. This contradicts Step 7.\n33. All cases lead to a contradiction. So $G$ has a perfect matching whenever it satisfies Tutte's condition.\n\n## Idea\n\nA perfect matching must send one vertex of each odd piece of $G - U$ into $U$, which gives necessity. For sufficiency, add edges until any further edge would create a perfect matching. If the graph splits around its universal vertices into complete pieces, a matching is easy to build. Otherwise two matchings of slightly larger graphs combine along an alternating cycle into a matching of the graph itself.\n",
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