The Graphs of Stably Matchable Pairs

By a theorem of Dénes Kőnig, every regular bipartite graph can be decomposed into disjoint perfect matchings [21]. Arbitrarily order these matchings, and construct a partial preference system in which one side of the bipartition (the students) prioritizes each potential pairing of the graph according to this ordering, and the other side of the bipartition (the residencies) prioritizes the pairings according to the reverse ordering. With these priorities, every matching of the decomposition (as well as, potentially, some other matchings using the same edges) is stable. ∎

Let $G$ be an arbitrary bipartite graph. Add isolated vertices to one side of the bipartition of $G$, if necessary, to make the resulting graph become balanced. Let $s_{i}$ and $r_{j}$ be the vertices on the two sides of the bipartition of the resulting balanced bipartite graph. Construct a graph $G^{\prime}$ with twice as many vertices, by making a copy $s^{\prime}_{i}$ or $r^{\prime}_{j}$ of each vertex $s_{i}$ and $r_{j}$. For each edge $s_{i}r_{j}$ in $G$, make two edges in $G^{\prime}$, $s_{i}r_{j}$ and $s^{\prime}_{i}r^{\prime}_{j}$, and for each pair $(s_{i},r_{j})$ that is not an edge in $G$, make two edges in $G^{\prime}$, $s_{i}r^{\prime}_{j}$ and $s^{\prime}_{i}r_{j}$, as shown in Figure 3. The result is a regular balanced bipartite graph in which the vertices $s_{i}$ and $r_{j}$ induce a copy of $G$, and in which every vertex on one side of the bipartition is adjacent to exactly half of the vertices on the other side of the bipartition. ∎

It is an induced subgraph of a regular bipartite graph by Section 3.1, and this regular bipartite graph is a graph of stably matchable pairs by Section 3.1. ∎

If $G$ is to be a graph of stably matchable pairs, it is necessary that every non-isolated vertex of $G$ be matched in every stable matching. If $v$ is a degree-one vertex, its neighbor can have no other matches, so its neighbor must also be degree-one. Therefore, it is necessary for $G[1]$ to have a perfect matching.

If $G$ is a subcubic graph of stably matchable pairs of some stable-matching instance, and $v$ is a vertex of degree two or three in $G$, then one of the edges incident to $v$ must belong to the top matching $\top$ of the lattice of stable matchings, and a different one of the edges incident to $v$ must belong to the bottom matching $\bot$. Each degree-three vertex is incident to a unique edge that belongs neither to $\top$ nor $\bot$, so these edges form a perfect matching of $G[3]$.

If $G[1]$, $G[2]$, and $G[3]$ all have perfect matchings (Figure 4, far and middle left), then the subgraph formed by removing from $G$ the matched edges in $G[3]$ consists of isolated vertices and edges, and disjoint cycles of degree-two and degree-three vertices (Figure 4, middle right). Form two matchings $\top$ and $\bot$ consisting of the matched edges in $G[1]$ and alternating subsets of these disjoint cycles, choosing arbitrarily within each cycle how to alternate between edges of $\top$ (which we call top edges) and edges of $\bot$ (which we call bottom edges). Call the edges in the matching of $G[3]$ middle edges, as they are neither in $\top$ nor in $\bot$. Additionally, for a top edge or a bottom edge, we say that it is matched if it is part of the perfect matching of $G[2]$, and unmatched otherwise. Once this choice has been made, form a system of cycles in $G$ of two types (Figure 4, far right):

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  Upper cycles alternate between unmatched top edges and either middle edges (at degree-3 vertices of $G$) or matched bottom edges (at degree-2 vertices of $G$).

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  Lower cycles alternate between unmatched bottom edges and either middle edges or matched top edges.

These cycles are simply the connected components of a 2-regular graph formed by splitting each degree-three vertex and each middle edge into two copies, an upper copy and a lower copy, with the incident top edge going to the upper copy and the incident bottom edge going to the lower copy. In particular, they are simple cycles, with each top or bottom edge appearing in exactly one of these cycles and each middle edge appearing in one upper cycle and one lower cycle.

Partially order these cycles by making two cycles $C$ and $C^{\prime}$ comparable when they share one or more middle edges, with $C\leq C^{\prime}$ when $C$ is the lower cycle and $C^{\prime}$ is the upper cycle of the two, and by making all other pairs of cycles incomparable. Then this system of partially ordered cycles, together with the matchings $\top$ and $\bot$, form a rotation system for $G$, and the result that $G$ is a graph of stably matchable pairs follows from Section 2. ∎

A graph with an articulation vertex $v$ cannot be matching-covered, and so by Section 2 cannot be a graph of stably-matching pairs. For, if any component formed from the graph by the removal of $G$ has an even number of vertices, the edge or edges connecting that component to $v$ cannot be included in any perfect matching. And if all of these components have an odd number of vertices, only one of these components can have an edge to $v$ in a perfect matching, and the rest of these components cannot be perfectly matched. ∎

One direction is Section 3.3. In the other direction, let $G$ be a bipartite outerplanar graph, drawn in an outerplanar way (with all vertices belonging to the unbounded face). We prove by induction on the number of bounded faces that $G$ is a graph of stably matchable pairs for a rotation system in which the outer face of $G$ is the union of the top and bottom matchings, and (when $G$ has more than one bounded face) all bounded faces are rotations.

As a base case, if $G$ is a cycle graph, it has a rotation structure consisting of two disjoint perfect matchings (the top and bottom matchings of the rotation structure) and an empty set of rotations. As a second base case, if $G$ consists of two cycles sharing an edge, it has a rotation structure where these two cycles are both rotations, alternating between top edges and bottom edges except at their shared edge. One of these two rotations (the upper one in the partial order of the two rotations) has top edges adjacent to the shared edge, and the other one has bottom edges adjacent to the shared edge.

In the non-base case, the weak dual of $G$ (a graph with a vertex for each bounded face and an edge between bounded faces that share an edge in $G$) is a tree with at least one leaf $f$, a bounded face of $G$ that shares only one edge $e$ with the rest of the graph. By induction, the outerplanar graph $G^{\prime}$ formed from $G$ by removing $f$ (but keeping the shared edge $e$ in place) has a rotation structure in which the outer face of $G^{\prime}$ (including $e$) alternates between top and bottom edges, and each face is a rotation. We construct a rotation structure for $G$ by adding $f$ as a rotation to this structure, alternating between top and bottom edges. If $e$ was a top edge in $G^{\prime}$, we place $f$ above the neighboring rotation in the partial order, and choose the alternation of top and bottom edges of $f$ to have top edges adjacent to $e$. Otherwise, we place $f$ below the neighboring rotation, and choose the alternation of its edges with bottom edges adjacent to $e$. Both cases preserve all the defining properties of a rotation system. ∎

We consider the following cases:

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  If $ab$ is odd, the number of vertices is odd and no perfect matching can exist.

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  If $\min\{a,b\}=2$, the graph is outerplanar and realizability follows from Theorem 3.2.

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  If $\min\{a,b\}=3$, suppose for a contradiction that the grid has a rotation system. Consider the block $B$ of six squares at one end of the grid, separated by three edges from the rest of the grid. In any rotation system, the bottom matching includes an even number of these separating edges. A rotation that does not include these edges does not change the number of separating edges that are used. A rotation that uses the middle of these three edges either adds two to the number of separating edges that are used, or removes two of these edges from use, because (by the parity of the distance between the endpoints of the two separating edges that it uses) either both separating edges must be upper in the rotation or both must be lower. However, in any sequence of stable matchings progressing upward through the lattice of stable matchings from its bottom to its top, each of the three separating edges must be introduced exactly once, so there must be a rotation that introduces only one new edge. Such a rotation $R$ must necessarily use the outer two of the three edges separating $B$ from the rest of the graph; again, by the parity of the distance between the endpoints of these outer separating edges, one of them would be an upper edge of any such rotation and the other of them would be a lower edge. Within $B$, the path of $R$ uses an odd number of vertices, so in order for all vertices of $B$ to be matched, the two perfect matchings that are connected by $R$ must both include the middle separating edge of $B$ as one of their matched edge. The remaining five vertices of $B$ induce a path, which must be the path followed by $R$. Because the two corner vertices of this path have degree two, the four edges of this path must all belong to $\top$ or $\bot$. But this means that the rotation $R$ passes through the degree-three vertex at the center of the path using only edges of $\bot$ or $\top$, making it impossible for any other rotation to incorporate the third edge at that vertex into the rotation system. This contradicts our assumption that a rotation system exists, proving that no realization as a graph of stably matchable pairs is possible in this case.

  

  

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  If $a$ and $b$ are both even, we form a rotation system using a subset of the unit squares of the grid as rotations, as illustrated in Figure 5. If we associate a grid square with its lower left corner (the point of the square whose coordinates are smallest), then we choose the squares whose first coordinate is even and whose second coordinate as odd as the bottom elements of the partial order of rotations. We choose the squares with both coordinates even as intermediate elements in this partial order, and we choose the squares whose first coordinate is odd and whose second coordinate is even as the top elements of the partial order. We do not use squares with both coordinates odd as rotations. The top matching consists of horizontal edges of top squares, and vertical edges of intermediate squares that are not shared with top squares. The bottom matching consists of vertical edges of bottom squares, and horizontal edges of intermediate squares that are not shared with bottom squares.

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  In the remaining cases, the smallest odd dimension of the grid is at least five and the smallest even dimension is at least six. Figure 6 depicts a rotation system realizing the $4\times 5$ grid and Figure 7 shows how to extend this to all $(2i)\times 5$ grids for $i\geq 2$. Examination of each grid edge in the figures shows that they are all covered by two elements of the rotation system and top and bottom matching, one upper and one lower, and examination of each rotation shows that they all alternate between upper and lower edges, as required for a rotation system.

  Note that, in each of the resulting rotation systems, the top matching includes a perfect matching of the vertices on the bottom rows of the rotation system. This allows these rotation systems to be extended by any even number $2i$ of rows below these bottom rows, by using the square rotations from the bottom $2i+1$ squares of the pattern used in the previous case for even-by-even grids, maintaining the partial order among these rotations and placing them in the partial order above all of the rotations used for the top five rows of the grid.

In all cases with $ab$ even, the grid is realizable when $\min\{a,b\}\neq 3$ and unrealizable otherwise. ∎

We assume without loss of generality that neither $G$ nor $H$ has isolated vertices, as any such vertices lead in $G\mathop{\square}H$ to disconnected copies of $G$ or $H$ that can be handled separately.

A preference system realizing $G\mathop{\square}H$ from separate realizations of $G$ and $H$ can be derived by giving the students preferences that prioritize edges coming from $G$ over edges coming from $H$, and giving the residencies preferences that prioritize edges coming from $H$ over edges coming from $G$. However it is easier to verify that this does not lead to additional undesired matched edges if we describe more explicitly a rotation system realizing $G\mathop{\square}H$, derived from rotation systems for $G$ and for $H$.

Let $\top_{G}$, $\top_{H}$, $\bot_{G}$, and $\bot_{H}$ denote the top and bottom matchings of the rotation systems for $G$ and $H$. We form a rotation system whose top matching is $\top=V(G)\times\top_{H}$ (using the top matching in each copy of $H$, as preferred by the residencies) and whose bottom matching is $\bot=\bot_{G}\times V(H)$. For each rotation $R$ in $G$ we form $|V(H)|$ copies of $R$ in the rotation structure for $G\mathop{\square}H$, one for each vertex in $H$, and similarly for each rotation in $H$ we form $|V(G)|$ copies in the rotation structure for $G\mathop{\square}H$, one for each vertex in $G$. These copies are partially ordered in the same way they are in the rotation systems of $G$ or $H$, with two copies of rotations being incomparable when they correspond to different vertices in $G$ or $H$.

Finally, we add to these rotations the cycles of the 2-regular graph $(\top_{G}\times V(H))\cup(V(G)\times\bot_{H})$. Each such cycle is partially ordered above the copies of rotations from $G$ with which it shares an edge, and below the copies of rotations from $H$ with which it shares an edge, causing it to alternate between upper and lower edges (as required in a rotation system) and providing each of the edges of this 2-regular graph with a second rotation that it belongs to (as also required in a rotation system). ∎

From an arbitrary given finite distributive lattice, construct the partial order of its subset of join-irreducible elements, and let $H$ be the Hasse diagram of this partial order. That is, $H$ has a vertex for each join-irreducible element of the lattice, and an edge from the lower to the upper of each two comparable elements that have no other join-irreducible element between them.

For each vertex $v$ of $H$, construct a cycle graph $C_{v}$, of even length, with its edges labeled alternatingly as upper and lower and with its vertices labeled alternatingly as students and residencies. Each edge $u\rightarrow v$ in $H$ should be associated with one of the lower edges of $C_{v}$, and each edge $v\rightarrow w$ in $H$ should be associated with one of the upper edges of $C_{v}$, in such a way that the edges of $C_{v}$ that are associated with edges of $H$ in this way are not adjacent to each other in $C_{v}$. If $v$ has total degree $d_{v}$ in $H$, then $C_{v}$ can be constructed with length $4$ if $d_{v}=1$, or with length at most $2d_{v}+4$ if $d_{v}>1$.

Form a graph $G$ from the union of the cycles $C_{v}$ by, for each edge $u\rightarrow v$ in $H$, merging together the edges of $C_{u}$ and $C_{v}$ that are associated with $u\rightarrow v$ (at the same time merging the two student endpoints of these edges and the two residency endpoints of these edges). Then $G$ is subcubic, because each merged pair of endpoints produces a vertex of $G$ with degree three (one merged edge and its two neighbors in the two cycles from which it was merged) and each unmerged vertex of one of the cycles $C_{v}$ continues to have degree two. Form also a rotation system in $G$, in which the rotations are the cycles $C_{v}$, partially ordered by the partial order for which $H$ is the Hasse diagram, in which $\top$ is the matching formed by the unassociated upper edges of cycles $C_{v}$, and in which $\bot$ is the matching formed by the unassociated lower edges of cycles $C_{v}$. Both $\top$ and $\bot$ are perfect matchings, because each merged or unmerged vertex in $G$ has exactly one neighbor in $\top$ and one in $\bot$.

Because the subcubic graph $G$ has a rotation system, it follows from Section 2 that there is a system of preferences for which this rotation system describes all of the stable matchings. Because both the lattice of stable matchings for these preferences and the initially given finite distributive lattice can be described as the lower sets of isomorphic partial orders, they are isomorphic lattices. ∎

Let $G$ be a degree-matchable graph of stably matchable pairs. The degree-zero and degree-one vertices of $G$ are irrelevant for its lattices of stable matchings, so assume without loss of generality that there are none. Consider any rotation system for $G$; we wish to show that the lattice of stable matchings of this rotation system is isomorphic to the closures of a bipartite orientation, whose underlying undirected bipartite graph will be the comparability graph of the rotations. To see this, label the edges of $G$ as top, bottom, or middle, accordingly. Then each rotation must alternate between upper and middle edges, or between middle and lower edges, at the degree-3 vertices of $G$, and between upper and lower edges at the degree-2 vertices of $G$. Because $G$ is degree-matchable, every part of a rotation that passes through degree-2 vertices starts and ends on an edge with the same label, and does not allow the part of the rotation that passes through degree-3 vertices to change between upper–middle and middle–lower alternations. Therefore, the rotations can be partitioned into three classes:

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  the rotations with no middle edges, which form isolated vertices in the comparability graph of rotations,

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  the rotations in which every middle edge lies between two top edges, and

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  the rotations in which every middle edge lies between two bottom edges.

Each middle edge belongs to one rotation in the second class and one in the third. The rotations in the first and second classes are maximal in the partial order of rotations, and the rotations in the first and third classes are minimal. The lattice of stable matchings can be described as the lattice of lower sets of rotations, and these are just the closures in a directed graph on rotations having an edge from each rotation in the second class to each rotation sharing a middle edge with it in the third class. This graph is a bipartite orientation of the comparability graph of rotations.

In the other direction, suppose we are given a lattice of closures of a bipartite orientation of a graph $G$ and wish to find a corresponding subcubic degree-matchable graph $G^{\prime}$ of stably matchable pairs for an instance of stable matching with the same lattice. Because a bipartite orientation is always a transitively reduced directed acyclic graph, the join-irreducibles of its lattice of closures are in one-to-one correspondence with its vertices, and its underlying undirected bipartite graph is isomorphic under this correspondence to the comparability graph of its join-irreducibles. When we apply the construction of Theorem 4.1 to this lattice, we obtain a rotation for each vertex of $G$, with its edges labeled as top, middle, or bottom. The endpoints of middle edges of these rotations are glued to other rotations, producing degree-3 vertices in the resulting graph $G^{\prime}$ of stably matchable pairs, and these degree-3 vertices have the middle edges as a perfect matching. Because no vertex of $G$ has both incoming and outgoing neighbors in the bipartite orientation, each path of top or bottom edges in each rotation has an even number of degree-2 vertices, and these vertices can be perfectly matched using the edges of the rotation. Therefore, $G^{\prime}$ is degree-matchable. ∎

Let $C$ be a shortest cycle in the intersection graph of rotations and let $R$ be a rotation that is maximal among the rotations of $C$, according to the partial order of rotations. Let $S$ and $T$ be the two neighbors of $R$ in $C$. Then all of the edges shared by $R$ with $S$ and $T$ are lower edges in the alternation of upper and lower edges of $R$; in particular, no two of these shared edges can be adjacent. Because $C$ was taken to be shortest, by the observation above, the edges that $R$ shares with $S$ and $T$ are the only parts of $R$ shared with other rotations in $C$.

Let $H$ be the subgraph of $G$ induced by vertices in $C$, let $v$ be any vertex of an edge shared by two rotations of $C$, neither of which is $R$, and let $K$ be the connected component containing $v$ in $H\setminus C$. Then $v$ must have a path through $K$ to rotation $S$, and within $S$ to two endpoints of shared edges in $R$. Symmetrically, $v$ must have a path through $K$ to rotation $T$, and within $T$ to another two endpoints of shared edges in $R$. Therefore, $K$ is connected by $R$ via at least four edges to at least four distinct vertices of $R$. By combining a spanning tree within $K$ with these four connecting edges, we can find a tree in $G$ that is edge-disjoint from $R$ and has leaves at four vertices of $R$. Because $G$ is subcubic, this tree has no degree-four vertices, so we can contract it to a tree with exactly two degree-three internal nodes, still having four leaves at distinct vertices of $R$. Contracting the paths in $R$ between these four vertices produces a minor of $G$ consisting of a 4-cycle from $G$ and a tree with two internal nodes of degree three having these four vertices as its leaves. Depending on the order in which the tree leaves are attached to $R$, this minor can only be either $K_{3,3}$ or the graph of the prism. ∎

An outerplanar or series-parallel graph cannot have $K_{3,3}$ or the graph of a prism as a minor. Therefore, if a stable matching instance has a subcubic outerplanar or series-parallel graph as its graph of stably matchable pairs, by Section 4.3 the intersection graph of its rotations is a forest. Each edge of this forest corresponds to a pair of rotations that share an edge in the graph of stably matchable pairs, and can be oriented from the upper of the two rotations to the lower one. With this orientation, the lower sets of the partial order of rotations are the same as the closures of the orientation of this forest.

In the other direction, every oriented forest is a transitively-reduced acyclic graph, whose lattice of closures is the same as the lattice of lower sets of a partial order on its vertices in which $u\leq v$ when there is a directed path from $v$ to $u$. Applying the construction of Theorem 4.1 to this lattice produces a graph in which each vertex of the forest is represented as a rotation, and in which the shared edges between these rotations correspond to edges of the forest. Gluing cycles together on shared edges in the pattern of a forest always produces an outerplanar graph, with glued edges interior and unglued edges on the outer face of its outerplanar embedding, as can be proved by induction on the number of gluing steps. ∎

If a stable matching instance has a planar graph as its graph of stably matchable pairs, find a planar embedding of it, and represent each rotation of the instance as the curve through its vertices and edges in the embedding. Every two rotations whose curves intersect are comparable in the partial order of rotations; if they intersect at an edge this is immediate while if they intersect at a vertex they are part of the totally ordered subset of rotations that pass through that vertex. Therefore, we may orient the string graph of the rotations by directing each edge from rotation $R$ to rotation $S$ whenever $S\leq R$ in the partial order of rotations; with this orientation, the lower sets of the partial order of rotations coincide with the closures of the oriented string graph.

In the other direction, let $G$ be an oriented string graph. By doubling and perturbing each curve if necessary we may assume without loss of generality that each of the curves defining $G$ is closed (although possibly self-crossing), that each point of the plane belongs to at most two arcs of curves (either of two different curves or two arcs of the same curve), and that each point belonging to two arcs is a simple crossing point of these arcs. Whenever a curve crosses itself we may remove short lengths of the two crossing arcs, within a small disk surrounding the crossing and disjoint from the other curves, and replace the removed arcs by two non-crossing arcs while preserving the connectivity of the curve; one of the two ways of choosing how to connect these arcs will always produce a curve that remains connected. Repeating this transformation produces a collection of curves with the same intersection graph but with no self-crossings (Figure 10, left). Similarly, whenever two curves that belong to the same strongly connected component of $G$ cross, we may replace their crossing by two non-crossing arcs in a way that merges the two curves into a single curve; one of the two ways of choosing how to connect these arcs will merge them. Repeating this transformation produces a collection of curves with the same partial order on the strongly connected components of the oriented string graph but with at most one curve per component (Figure 10, right). After these transformations we may assume without loss of generality that $G$ is acyclic (although not necessarily transitively reduced) and that each vertex of $G$ is represented by a simple closed curve in the plane. Because $G$ is acyclic, the join-irreducibles of its lattice of closures (which should correspond to the rotations of a rotation system representing this lattice) also correspond one-for-one with the curves in this arrangement. Whenever two curves cross, they correspond to two distinct comparable rotations.

If any curve of this system of curves has no crossings, replace it by a cycle of four vertices, drawn with its edges along the curve. For the remaining curves, again remove a small disk surrounding each crossing point, partitioning each curve into a sequence of arcs connecting these disks. Along each arc of a curve $C$, draw a path of either three edges and four vertices (if the rotation corresponding to $C$ is above both of the rotations that it crosses at the ends of the arc, or below both of the rotations that it crosses) or two edges and three vertices (if the rotation is above one of the rotations that it crosses, and below the other one of these rotations, as illustrated in Figure 11 (center right). By this construction, if we were to add an edge in $C$ across each of the small disks, the result would be a cycle of even length; choose arbitrarily for each curve a bipartition of the vertices of this cycle into students and residencies.

At each of the small disks where a pair of curves cross, this construction creates two students and two residencies, in the cyclic order student–student–residency–residency around the disk. Merge the two students into a single vertex within the disk, merge the two residencies into a single vertex within the disk, and add an edge between them (Figure 11, far right). The result is a subcubic bipartite graph, embedded in the plane, and a rotation system for that graph whose rotations are the cycles following each curve of the string graph. The partial ordering on the rotations is the same as the ordering coming from the orientation of $G$. Therefore, it has a lattice of stable matchings that realizes the lattice of closures of the given oriented string graph. ∎

Consider the sequence of matchings produced by starting at $\bot$ and then applying the rotations of the rotation system, one by one, in an ordering given by an arbitrary linear extension of the partial order of rotations, until reaching $\top$. Then to cover all three edges at each of the degree-three vertices of $X$, the part of the matching within subgraph $X\cup P\cup P^{\prime}$ must change at least twice, passing through at least three distinct matchings, throughout this sequence. The first of these changes cannot pass through a degree-two vertex of $X$, because after a change through such a vertex its matched edge can never change again, and this would prevent the degree-three vertex at the other end of that matched edge from changing to its third match. The only other possibility is that this first change is generated by a rotation that passes through $P$, $P^{\prime}$, or both, causing this path to alternate between bottom and middle edges.

Symmetrically, the last change to the subgraph $X\cup P\cup P^{\prime}$, in this sequence of matchings, cannot pass through a degree-two vertex of $X$, because the edge that this degree-two vertex was matched to before the change could not have changed in any previous step, preventing its degree-three vertex at its other end from having changed enough times to cover its three incident edges. Therefore, this last change also passes through $P$ or $P^{\prime}$, alternating between top and middle edges. Since at least one of $P$ or $P^{\prime}$ must alternate between bottom and middle edges, and at least one of $P$ or $P^{\prime}$ must alternate between top and middle edges, exactly one of the two paths must alternate in each of these two ways. ∎

For $k$ Boolean variables $x_{1},\dots x_{k}$, there are $2^{k-1}$ complementary pairs of truth assignments, some of which cause $f$ to become true and some of which cause it to become false. We prove the result by induction on the number of pairs of truth assignments that cause $f$ to become false. As a base case, when there is only one forbidden truth assignment, we may use the not-all-equal gadget of Figure 16, with each of its terminal edges attached by a connector gadget to either an inverter gadget or an insulator gadget. The choice of whether to attach an inverter or insulator can be made by choosing an insulator for variables that have the same value as $x_{1}$ in the forbidden truth assignments, and an inverter for variables that have the opposite value to $x_{1}$. The not-all-equal gadget itself is unrealizable only for the all-equal truth assignment, and this choice of inverter gadgets instead causes the graph to become unrealizable only for the single forbidden truth assignment of $x$. The use of inverter and insulator gadgets at every terminal edge of the central not-all-equal gadget prevents rotation systems for this graph from having order relations between the rotations containing its terminal edges.

When $f$ has more than one forbidden truth assignment, choose one of them, let $f^{\prime}$ be the function forbidding only that truth assignment, construct a graph $G^{\prime}$ for it as above, and let $f^{\prime\prime}$ be the Boolean function formed from $f$ by changing its value to true on the forbidden assignment for $f$ and then complementing its variables in the same pattern that was used for the inverters in the construction of $G^{\prime}$. By induction, construct a graph $G^{\prime\prime}$ corresponding to the function $f^{\prime\prime}$, and embed $G^{\prime\prime}$ inside the central face of $G^{\prime}$, attached by connector gadgets to the inner terminal edges of the switch gadgets of $G^{\prime}$. Figure 17 shows this construction for the 2-in-4 function $f$, which forbids five of the eight complementary pairs of truth assignments to its four Boolean variables (the pair of all-equal assignments, and four pairs of assignments that set one of the four variables opposite to the other three). The figure omits the outermost layer of insulator gadgets, as they are unnecessary when this 2-in-4 gadget is incorporated into our overall reduction.

In the resulting graph $G$, each connection gadget, switch gadget, inverter gadget, and insulator gadget has two terminal edges, whose adjacent truth edges can be labeled in only one way consistent with the labeling at the outer terminal edges of $G$: at each connection gadget, switch gadget, or insulator gadget, the two terminal edges must have the same labeling at their incident truth edges, while at an inverter gadget the labels must be opposite. This labeling determines the structure of the rotations within all of these gadgets, leaving only the fuse gadgets undetermined. For a truth assignment that satisfies $f$, all of the cycles of switch and fuse gadgets in $G$ will have at least two consecutive switch gadgets with opposite states, breaking the potential cyclic ordering of rotations at the fuse gadget between these two switches and producing a valid rotation system for $G$ meeting all the conditions of the lemma. For a truth assignment that does not satisfy $f$, one of the cycles of switch and fuse gadgets, the one corresponding to the complementary pair including that assignment, will have all four of its switch gadgets in the same state. In this case, the only locally-valid edge labeling to the switch and fuse gadgets of this cycle produces a cyclically-ordered set of rotations, so there is no valid global rotation structure. ∎

To show that recognizing graphs of stably matchable pairs belongs to NP, we may either use a preference system for the stable matching instance as a witness, with a verification algorithm that constructs the rotation system from the preference system and checks that its graph equals $G$, or we may use a rotation system directly as a witness, with a verification algorithm that checks that it meets the definition of a rotation system and has $G$ as its graph.

To show that the problem is NP-hard, we find a polynomial-time many-one reduction from the known NP-hard monotone planar 2-in-4-SAT problem to the problem of the theorem. This reduction finds a planar embedding of the given monotone planar 2-in-4-SAT instance, replaces each variable of the instance by a term gadget, replaces each 2-in-4 clause of the instance by a 2-in-4 clause gadget according to Section 5.3 and Figure 17, and replaces each connection between a variable and a clause in the planar embedding of the instance by a connection gadget, attached to the terminal edges of the term and clause gadgets in the cyclic ordering given by the planar embedding.

When the 2-in-4-SAT instance has a satisfying truth assignment, this assignment may be reflected to a labeling of the truth edges of the term gadgets as being top or bottom edges, and this labeling propagated to the remaining gadgets as described in Section 5.3, producing a collection of rotations that are locally partially ordered and that have no nonlocal transitive relations in their ordering. The result is a valid rotation system for the reduced graph. Conversely, when the reduced graph has a rotation system, the truth edges of the term gadgets must be consistently labeled according to a valid truth assignment, and the unique consistent propagation of this labeling to the other gadgets, as described in Section 5.3, can only produce a value rotation system when this truth assignment satisfies all the clauses, so the starting 2-in-4-SAT instance is satisfiable.

We have described a polynomial-time many-one reduction from monotone planar 2-in-4-SAT to recognizing planar subcubic graphs of stably matchable pairs, and we have shown that recognizing arbitrary graphs of stably matchable pairs belongs to NP, so both the recognition problem for arbitrary graphs and its special case for subcubic planar graphs are NP-complete. ∎

By the binomial theorem, the number of structures consisting of a set of edges, together with a structure of type $T$ within that set, is

The bound $c_{M}\leq 1+c_{T}$ follows by considering the set of edges in such a structure to be the set complementary to a perfect matching.

To get the more precise bound of the lemma, we can list structures formed as a perfect matching plus a structure of type $T$ by the following algorithm: repeatedly find a vertex of minimum degree in the current graph, and loop over its incident edges. For each such edge $e$, recursively call the same algorithm in the subgraph formed by removing the two endpoints of $e$ and all their incident edges. If the recursion ever finds a vertex of degree zero, it returns without listing anything (as the choices made in this branch of the recursion do not lead to a perfect matching). Otherwise, the bottom-level calls to this algorithm (in which the remaining subgraph is empty) have each found a perfect matching, and list all structures of type $T$ (in an unspecified way) in the subgraph of removed but unmatched edges.

At a recursive call of this algorithm where the minimum degree is $i$, the number of unmatched removed edges that will be considered later in listing structures of type $T$ is at least $2i-2$, and the total number of removed edges is at least $2i-1$, explaining the two exponents in the formula of the lemma. (The cases where more than this number of unmatched edges are removed produce similar terms with greater exponents, but they can be omitted from the formula as they cannot produce the maximum value of the overall formula.) By choosing $c_{M}$ as upper-bounded by the formula of the lemma, it follows by induction on $m$ that the total number of structures found by this algorithm is as stated. ∎

This follows from a result of Alon and Friedman [1] according to which the number of perfect matchings in a graph with degree sequence $d_{i}$ is at most $\prod(d_{i})^{1/2d_{i}}$. In a graph with $m$ edges, there are $2m$ vertex-edge incidences, which can be grouped in various ways to vertices of varying degrees. According to the formula of Alon and Friedman, an incidence that participates in a vertex of degree $d$ contributes a factor of $(d!)^{1/2d^{2}}$ to the total bound on the number of matchings. This factor is maximized at degree three, for which it is $6^{1/18}$. If each vertex-edge incidence participates in a vertex of degree three, maximizing the contribution to the product from every incidence, then the total product is as stated. ∎

The bound on $c_{1}$ follows from Section 6.1, counting matchings plus a structure $T$ consisting of an arbitrary set of edges, with $c_{T}=2$. Calculation shows that the maximum value for the upper bound in the lemma is achieved at $i=5$, giving the value shown.

The bound on $c_{2}$ follows by using the same minimum-degree algorithm from the proof of Section 6.1 to choose a matching and a disjoint set of edges, and then applying Section 6.1 to bound the number of matchings in the set of edges that were not chosen to be included in the disjoint set of edges. Each matched edge or unmatched but unchosen edge contributes a factor of $c_{0}$ to the total number of structures found by this combination of bounds. When the minimum vertex has degree $i$ and all its neighbors, this produces a contribution per removed edge of $\bigl{(}ic_{0}(1+c_{0})^{2i-2}\bigr{)}^{1/(2i-1)}$, almost the same as if we were applying Section 6.1 with $c_{T}=1+c_{0}$. Here, the first factor of $c_{0}$ comes from the application of Section 6.1 to the edge that was chosen to be in the matching, and the factor of $(1+c_{0})^{2i-2}$ comes from applying the binomial theorem to the choice of edges to be part of the disjoint set and to the factor of $c_{0}$ that will be included when an edge is not chosen. When the minimum vertex has degree $i$ but some of its neighbors have higher degrees, the leading $ic_{0}$ factor in this formula is averaged over a greater number of edges each contributing a factor of $1+c_{0}$, leading to a more complicated calculation but a lower contribution per removed edge than for the case of equal degrees. Calculation shows that the maximum value for this upper bound is achieved for a vertex with neighbors of equal degree, at $i=4$, giving the value stated in the lemma. ∎

We will search for a sequence of perfect matchings, differing from one to the next in the sequence by a single rotation, that together cover all edges of $G$ and such that each edge of $G$ is rotated into a matching at most once and rotated out of a matching at most once. If such a sequence exists, we can produce a rotation structure that has the first matching in the sequence as its bottom matching, the last matching in the sequence as its top matching, and the rotations of the sequence as the rotations of the rotation structure, partially ordered consistently with the sequence order.

We search for this sequence as a path in a graph $\Gamma$ whose vertices are pairs $(M,S)$ where $M$ is a perfect matching and $S$ is any subset of the unmatched edges. (This is essentially a dynamic programming algorithm but we find it more convenient to describe it as a graph search.) In our search, we will use $M$ to represent the matching at the current endpoint of a partial sequence of matchings of $G$, and $S$ to represent the edges that have already been covered by matchings earlier in the partial sequence. We make an edge in $\Gamma$ from vertex $(M,S)$ to vertex $(M^{\prime},S^{\prime})$ when matchings $M$ and $M^{\prime}$ differ by a single rotation, when $M^{\prime}\cap S=\emptyset$, and when $S^{\prime}=S\cup(M\setminus M^{\prime})$. With this definition, paths in $\Gamma$ automatically correspond to sequences of matchings that never re-use any edges: an edge, once used, gets added to $S$ and its presence there and the requirement that $S$ be disjoint from $M^{\prime}$ prevent later matchings in the sequence from re-using it. Therefore, the sequences we seek are exactly the paths from any vertex of the form $(M,\emptyset)$ to any other vertex of the form $(M^{\prime},E(G)\setminus M^{\prime})$. We can find such a path by breadth first or depth first search of $\Gamma$, starting from the vertices $(M,\emptyset)$ and terminating whenever a vertex of the form $(M^{\prime},E(G)\setminus M^{\prime})$ is reached.

By Section 6.1, $\Gamma$ has $O(c_{1}^{m})$ vertices and $O(c_{2}^{m})$ edges. It remains to describe how to construct $\Gamma$ and search it in a space-efficient way, bearing in mind that we do not want to use the amount of memory it would require to store all the edges of $\Gamma$. We have enough memory that (by the assumption that machine words are large enough to store a single memory address) we can store a single matching $M$ or a single set of edges $S$ in a single word, and perform set operations like unions and intersections of these sets in constant time per operation. We will index the vertices of $\Gamma$ by consecutive integers in the range from $0$ to $|V(\Gamma)|-1$. Based on these indexes, we store the following information:

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  An array $A_{1}$, addressed by vertex index, of the pairs $(M_{i},S_{i})$ of structures associated with vertex $i$.

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  An array $A_{2}$, addressed by vertex index, of the predecessor of vertex $i$ in the first path to reach that vertex found by the search, or a special flag value to indicate that it has not yet been reached.

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  An array $A_{3}$ of indexes of vertices on the current frontier of the search, arranged either as a stack for depth first search or a queue for breadth first search.

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  An array $A_{4}$ of Boolean values indexed by sets of edges, true when that set of edges is a simple cycle and false otherwise.

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  An array $A_{5}$ of lists indexed by sets $S$ of edges, listing the matchings $M$ disjoint from $S$ in sorted order, together with the vertex index for $(M,S)$ for each matching.

Array $A_{4}$ takes space $2^{m}$ to store, and the remaining arrays (including the total length of the lists in $A_{5}$ take space $O(c_{1}^{m})$. Initializing $A_{1}$ can be accomplished in linear time using the structure enumeration algorithm of the proof of Section 6.1. Array $A_{2}$ is easily initialized to contain only flag values, and $A_{3}$ is initialized to list all vertices of the form $(M,\emptyset)$. Initializing $A_{4}$ can be done by a naive algorithm that tests each set for being a cycle, in time $O(m2^{m})$, smaller than our stated time and space bounds. $A_{5}$ can be initialized by using $A_{1}$ to count how many matchings are associated with each disjoint set in order to allocate storage for the lists within $A_{5}$, using a second pass over $A_{1}$ to place matchings into lists, and then sorting each list.

Each step of the search of $\Gamma$ is obtained by removing a vertex $(M,S)$ of $\Gamma$ from the search frontier $A_{3}$, terminating the algorithm if $M\cup S=E(G)$, otherwise looping through the neighbors of $(M,S)$ in $\Gamma$, and for each previously-unreached neighbor setting its predecessor in $A_{2}$ and adding it to the frontier $A_{3}$. It remains to describe how to find the outgoing neighbors of a vertex $(M,S)$ of $\Gamma$, since these are not explicitly represented by the data structures described above (and storing them explicitly would use too much space). We do this simply by using $A_{5}$ to find all of the matchings $M^{\prime}$ disjoint from $S$, performing set operations on $M$ and $M^{\prime}$ to find their symmetric difference, and using $A_{4}$ to check that this symmetric difference is a simple cycle. When it is, we have found a neighbor $(M^{\prime},S^{\prime})$ of $(M,S)$ in $\Gamma$ , with $S^{\prime}=S\cup(M\setminus M^{\prime})$. We can find the vertex index of this neighbor by a binary search in the list for $S^{\prime}$ of $A_{5}$. The bottleneck of the algorithm is performing these binary searches; each takes time $O(m)$ and the number of searches performed is $O(c_{2}^{m})$.

The alternative space and time bounds of $O(n!2^{m-n})$ and $O(n!^{2}2^{m-n})$, which are better for dense graphs, follow from the same algorithm (with a preprocessing stage that eliminates vertices of degree less than two) using a more naive analysis in which we count pairs of a matching and a disjoint set, or two matchings and a disjoint set, by bounding the number of matchings by $(n/2)!$ and the number of disjoint sets by two to the power of the number of unmatched edges. ∎

Rotation subsystems differ from rotation systems in allowing pieces of rotations that are paths rather than cycles, and in allowing the bag vertices to be unmatched in $\top$ and $\bot$. But because the root node of a good tree decomposition is assumed to be empty bag, these differences are not possible at that node. ∎

Because each piece of rotation represented within a rotation state uses at least one of the incident edges at one of the vertices of $B_{x}$, there can be at most $O(wd)$ pieces, and $O(wd)$ rotations. The partial order on rotations can be represented using $O((wd)^{2})$ bits of information, the top and bottom matchings in $G_{x}[B_{x}]$ require $O(w^{2})$ bits of information, and the information about which rotation pieces belong to which rotations, which bag vertices are endpoints of which pieces, and whether the incident edges are upper or lower, all take $O(wd\log(wd))$ bits of information. ∎

If $S$ is valid, it has a rotation subsystem for $x$, which cannot use vertex $v$ in any rotation or in the top or bottom matching because $v$ has no edges in $G_{x}$. Therefore, removing $x$ produces a rotation subsystem for $y$, showing that $S^{\prime}$ is valid. Conversely, if $S^{\prime}$ is valid, it has a rotation subsystem for $y$ Adding $v$ to this subsystem without changing its rotations or pieces produces a rotation subsystem for $x$, in which $v$ is unmatched, showing that $S$ is valid. ∎

If $S$ is valid, it has a rotation subsystem for $x$, which cannot have vertex $v$ as the endpoint of a piece of a rotation because $v$ does not belong to $B_{x}$. Because $x$ belongs to $F_{x}$, it must be matched in both $\top$ and $\bot$ in this rotation subsystem. The same rotation subsystem exists for $y$, and corresponds to a state $S^{\prime}$ that meets the conditions of the lemma. Conversely, if a state $S^{\prime}$ as described in the lemma exists, it has a rotation subsystem, which matches $v$ in both $\top$ and $\bot$. This matching is the only additional requirement for a rotation subsystem at $x$, so there is a rotation subsystem at $x$ showing that $S$ is valid. ∎

In each case $G_{x}=G_{x}[B_{x}]$, so rotation subsystems coincide with rotation states, and these are the only possible rotation subsystems for a single-edge graph. ∎

Validity of a rotation state at a node $x$ depends only on $G_{x}$ and $B_{x}$, and for the nodes described in the lemma we have $G_{x}=G_{x^{\prime}}$ and $B_{x}=B_{x^{\prime}}$. ∎

If $S$ is valid, it is associated with a rotation subsystem, and restricting that subsystem to $G_{y}$ and $G_{z}$ produces two subsystems with disjoint matchings $\bot$ and $\top$, each having a subset of the rotations with a restriction of the partial order on rotations, and each having pieces of rotations that are subsets of the pieces of rotations in the subsystem for $x$. It follows that the rotation states $S^{\prime}$ and $S^{\prime\prime}$ associated with those restricted rotation subsystems are valid and meet the conditions of the lemma. Conversely, if valid rotation states $S^{\prime}$ and $S^{\prime\prime}$ meeting the conditions of the lemma exist, then they each are associated with rotation subsystems for $y$ and $z$ that can be combined together in the obvious way to produce a rotation subsystem for $x$, showing that $S$ is valid. ∎

We can find a good tree-decomposition of width $O(w)$ in the stated time bound. A graph with width $O(w)$ has $O(wn)$ edges, so there are $O(wn)$ edge nodes in the tree-decomposition. We may eliminate any join nodes in which one of the two sides of the join has no edges, without change to the validity of the decomposition, leaving a decomposition that also has $O(wn)$ join nodes and (as for any nice tree decomposition) $n$ forget nodes. Removing the forget nodes and join nodes partitions the tree-decomposition into subtrees within which there can be at most $O(w)$ introduce nodes, so there can be at most $O(w^{2}n)$ introduce nodes total.

Because $G$ has carving width $w$, its treewidth and maximum degree are both $O(w)$, so by Section 6.2 the number of states at each node of this decomposition is exponential in $O(w^{4})$. By the lemmas above, we can compute the set of valid states at each node by an algorithm that examines each valid state at the child node (when there is one child) or each pair of valid states at the two children (when there are two children), in an amount of time per state or pair of states that is polynomial in $w$. Therefore the total time per node is bounded by this polynomial multiplied by the square of the number of states, which remains exponential in $O(w^{4})$. ∎