Hall's marriage theorem

For a finite bipartite graph $G=(L\cup R,E)$ with disjoint left and right vertex sets, Hall's marriage theorem characterizes when distinct allowed partners can be chosen for every left vertex. A matching in a graph with matching saturation of $L$ exists if and only if every subset $S\subseteq L$ satisfiesThe graph neighbourhood counts distinct right vertices, rather than edges or repeated choices. This is the Hall condition. Its failure supplies a concrete Hall obstruction. Necessity follows because the chosen partners of $S$ are distinct members of $N_G(S)$; sufficiency is the surprising assertion that these collective counts rule out every obstacle.^[ref]

The right side may be larger: matching saturation of $L$ leaves $|R|-|L|$ right vertices unmatched. Only when $|L|=|R|$ does a matching in a graph saturating $L$ automatically become a perfect matching. For a specified subset $X\subseteq L$, apply the same criterion only to $S\subseteq X$, using the induced subgraph on $X\cup R$. The remaining left vertices need not be matched. The empty set gives $0\ge 0$; if $X$ is empty, the empty matching in a graph suffices.

Consider a scheduling model with three unit-duration jobs $a,b,c$ and four available slots $1,2,3,4$ on one machine. Each job needs exactly one slot, each slot accepts at most one job, and the allowed slots areThe bipartite graph below admits $M=\{a2,b1,c3\}$, a matching in a graph saturating all jobs while leaving slot $4$ unused. It is not a perfect matching of the seven-vertex graph. A greedy choice $a1$ initially blocks $b$, but the augmenting path in a matching $b,1,a,2$ replaces $a1$ by $b1,a2$; subsequent assignment of $c3$ succeeds. A bad partial choice is not an impossibility certificate.

There are eight subsets of $L=\{a,b,c\}$. Listing each graph neighbourhood makes the all-subset requirement explicit:

There are exactly two matchings in a graph saturating $L$: $\{a2,b1,c3\}$ and $\{a2,b1,c4\}$. Job $b$ forces slot $1$, then job $a$ forces slot $2$, leaving two choices for $c$.

The Hall condition requires enough distinct neighbours for every subset of the specified left vertex set in a finite bipartite graph. Checking the whole left set establishes only one inequality. Checking singleton sets establishes only that each left vertex has positive degree of a vertex. Overlapping lists can pass both tests and still fail to admit matching saturation of the left side.

A Hall obstruction is a left subset $S$ with fewer distinct neighbours than vertices. At most $|N(S)|$ members of $S$ can be matched, so at least $|S|-|N(S)|$ must remain unmatched. The pigeonhole principle proves failure without enumerating every proposed assignment.

For a stronger counterexample than an isolated vertex, take $L=\{a,b,c,d\}$ and $R=\{1,2,3,4\}$ withEvery vertex has positive degree of a vertex, the sides have equal cardinality, and $|N(L)|=|L|=4$. Every two-vertex left subset also passes the Hall condition. Nevertheless $S=\{a,b,c\}$ has only $N(S)=\{1,2\}$, so $2<3$ rules out matching saturation of $L$.

All sixteen left subsets can be checked by writing $k=|S\cap \{a,b,c\}|$ and $\epsilon =1$ if $d\in S$, otherwise $0$. ThenOnly $k=3,\epsilon =0$ violates the Hall condition. Every matching in a graph uses at most two edges incident to $\{a,b,c\}$ and at most one incident to $d$, so the matching number is three. There are exactly twelve size-three matchings in a graph: choose two of $a,b,c$ in three ways, assign them to $1,2$ in two ways, and choose $3$ or $4$ for $d$. Thus the obstruction identifies the bottleneck: adding availability only for $d$ cannot help; allowing $c$ to use slot $3$ permits $\{a1,b2,c3,d4\}$, a perfect matching.

In a unit-capacity assignment problem, left vertices represent demands and right vertices represent individually usable resources; an edge records compatibility. All constraints must be encoded by those allowed pairs and by using each resource at most once. Hall's marriage theorem decides feasibility. Costs belong to an additional mathematical optimization problem; precedence between jobs or requirements for consecutive slots need additional modelling.

The authored Notes on the design of a conference scheduler by David R. MacIver (Implementation, 7 August 2022) illustrates a Hall obstruction with four talks confined to three slots. With one talk per slot, that subset has size four and a graph neighbourhood of size three. An unused slot outside its graph neighbourhood does not remove the bottleneck. In the three-job example above, the same scheduling model instead has two complete assignments.

The allocation exercise in Graph Theory by Reinhard Diestel (Matching, Covering and Packing, printed p. 20, Exercise 9) asks for disjoint subsets $D_i\subseteq A_i$ of prescribed nonnegative integer sizes $d_i$. Replace demand $i$ by $d_i$ left vertices with allowed resource set $A_i$. Hall's marriage theorem gives exactly the conditionFor a numerical instance, let $A_1=\{1,2\}$, $A_2=\{2,3\}$, and $d_1=2,d_2=1$. The separate demands and their combined demand fit, and $D_1=\{1,2\},D_2=\{3\}$ works. Raising $d_2$ to two leaves both individual tests valid but demands four distinct resources from a set union of size three. This is the same collective obstruction with repeated demand vertices.

A system of distinct representatives for a finite indexed family of finite sets $(A_i{)}_{i\in I}$ chooses one element $x_i\in A_i$ per index, with all choices distinct. Form a bipartite graph with left vertex set $I$, right vertex set ${\bigcup }_{i\in I}{A}_i$, and edges $ix$ whenever $x\in A_i$. Matching saturation of $I$ is precisely a system of distinct representatives. Hence Hall's marriage theorem says that such choices exist exactly when $|{\bigcup }_{i\in J}{A}_i|\ge |J|$ for every subset $J\subseteq I$. Repeated equal sets still have separate indices and separate demands.

For sufficiency, take a matching in a graph of maximum size and suppose a left vertex $u$ is unmatched. Follow unused edges from left to right and matching edges back to the left. An unmatched reachable right vertex yields an augmenting path in a matching, contradicting maximal size. Otherwise let $S,T$ be the reachable left and right sets. Every member of $T$ is matched to a distinct member of $S\setminus \{u\}$, and every member of $S\setminus \{u\}$ is reached this way. Since $N(S)=T$, we obtain $|N(S)|=|S|-1$, contradicting the Hall condition. Finiteness ensures that a maximum-size matching in a graph exists.

The alternative Hall induction through a tight set uses mathematical induction: a proper nonempty subset with $|N(S)|=|S|$ splits off a balanced induced subgraph; without such a tight set, deleting one chosen edge's endpoints preserves the Hall condition on the smaller left side.

In the equal-part random bipartite graph with fixed edge probability $0<p<1$, rule out empty rectangles of side $\lceil \sqrt{n}\rceil$ and vertices with at least $n-\lceil \sqrt{n}\rceil$ nonneighbors. A union bound gives the displayed asymptotic estimate. Small vertex sets satisfy Hall's condition by the degree bound; intermediate sets do so because their omitted-neighbor rectangle would contain a forbidden square; large sets meet every vertex in the other part. Thus Hall's marriage theorem gives a perfect matching asymptotically almost surely. This does not mean probability one at any finite size.

Two maximum-cardinality $\epsilon$-separated subsets $A,B$ of a compact metric space have a bijection that moves each point by less than $\epsilon$. Join $a$ to $b$ when their distance is below $\epsilon$. If $S\subseteq A$ had fewer than $|S|$ neighbours, replacing these neighbours in $B$ by $S$ would create a larger separated set. The Hall marriage theorem therefore supplies a perfect matching. This allows empirical averages on a compact isometry group to be made approximately invariant without assuming a prior Haar measure construction.

In a finite bipartite graph, suppose every left vertex has positive degree and each edge joins vertices with $d(x)\ge d(y)$ from left to right. Then $|N(A)|\ge {\sum }_{x\in A}{\sum }_{y\sim x}1/d(y)\ge |A|$, so a matching covers the left side.

For finitely many Lebesgue measurable sets $A_i\subseteq [0,1]$ and nonnegative demands $b_i$, pairwise disjoint measurable subsets $B_i\subseteq A_i$ with $\lambda (B_i)=b_i$ exist exactly when $\lambda ({\bigcup }_{i\in I}{A}_i)\ge {\sum }_{i\in I}{b}_i$ for every index subset $I$. Necessity is additivity and monotonicity of Lebesgue measure. For sufficiency, partition the set union into membership cells $E_J$, send flow from a source through demand vertices $i$ to cells with $i\in J$, then to a sink. Source capacities are $b_i$, cell capacities are $\lambda (E_J)$, and intermediate capacities are $D={\sum }_i{b}_i$. Every cut of a flow network has capacity at least $D$ by the assumed inequalities. The max-flow min-cut theorem provides the allocations, and divisibility of Lebesgue measure turns each cell allocation into disjoint pieces. If $D=0$, all $B_i$ can simply be empty.

For Hall's theorem, either every proper nonempty set has one excess neighbour and one deletes an arbitrary edge's endpoints, or a tight set $S$ with $|N(S)|=|S|$ splits the problem into the induced graph on $S\cup N(S)$ and its complement.

In a $k$-regular bipartite graph, edge counting gives equal vertex-class sizes and $k|S|\le k|N(S)|$. Hall's theorem therefore supplies a perfect matching.