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26.3 Maximum bipartite matching


Run the Ford-Fulkerson algorithm on the flow network in Figure 26.8 (c) and show the residual network after each flow augmentation. Number the vertices in $L$ top to bottom from $1$ to $5$ and in $R$ top to bottom from $6$ to $9$. For each iteration, pick the augmenting path that is lexicographically smallest.

First, we pick an augmenting path that passes through vertices 1 and 6. Then, we pick the path going through 2 and 8. Then, we pick the path going through 3 and 7. Then, the resulting residual graph has no path from $s$ to $t$. So, we know that we are done, and that we are pairing up vertices $(1, 6)$, $(2, 8)$, and $(3, 7)$. This number of unit augmenting paths agrees with the value of the cut where you cut the edges $(s, 3)$, $(6, t)$, and $(7, t)$.


Prove Theorem 26.10.

We proceed by induction on the number of iterations of the while loop of Ford-Fulkerson. After the first iteration, since $c$ only takes on integer values and $(u, v).f$ is set to $0$, $c_f$ only takes on integer values. Thus, lines 7 and 8 of Ford-Fulkerson only assign integer values to $(u, v).f$. Assume that $(u, v).f \in \mathbb Z$ for all $(u, v)$ after the $n$th iteration. On the $(n + 1)$th iteration $c_f(p)$ is set to the minimum of $c_f(u, v)$ which is an integer by the induction hypothesis. Lines 7 and 8 compute $(u, v).f$ or $(v, u).f$. Either way, these the the sum or difference of integers by assumption, so after the $(n + 1)$th iteration we have that $(u, v).f$ is an integer for all $(u, v) \in E$. Since the value of the flow is a sum of flows of edges, we must have $|f| \in \mathbb Z$ as well.


Let $G = (V, E)$ be a bipartite graph with vertex partition $V = L \cup R$, and let $G'$ be its corresponding flow network. Give a good upper bound on the length of any augmenting path found in $G'$ during the execution of $\text{FORD-FULKERSON}$.

By definition, an augmenting path is a simple path $s \leadsto t$ in the residual network $G_f'$. Since $G$ has no edges between vertices in $L$ and no edges between vertices in $R$, neither does the flow network $G'$ and hence neither does $G_f'$. Also, the only edges involving $s$ or $t$ connect $s$ to $L$ and $R$ to $t$. Note that although edges in $G'$ can go only from $L$ to $R$, edges in $G_f'$ can also go from $R$ to $L$.

Thus any augmenting path must go

$$s \to L \to R \to \cdots \to L \to R \to t,$$

crossing back and forth between $L$ and $R$ at most as many times as it can do so without using a vertex twice. It contains $s$, $t$, and equal numbers of distinct vertices from $L$ and $R$—at most $2 + 2 \cdot \min(|L|, |R|)$ vertices in all. The length of an augmenting path (i.e., its number of edges) is thus bounded above by $2 \cdot \min(|L|, |R|) + 1$.

26.3-4 $\star$

A perfect matching is a matching in which every vertex is matched. Let $G = (V, E)$ be an undirected bipartite graph with vertex partition $V = L \cup R$, where $|L| = |R|$. For any $X \subseteq V$, define the neighborhood of $X$ as

$$N(X) = \{y \in V: (x, y) \in E \text{ for some } x \in X\},$$

that is, the set of vertices adjacent to some member of $X$. Prove Hall's theorem: there exists a perfect matching in $G$ if and only if $|A| \le |N(A)|$ for every subset $A \subseteq L$.

First suppose there exists a perfect matching in $G$. Then for any subset $A \subseteq L$, each vertex of $A$ is matched with a neighbor in $R$, and since it is a matching, no two such vertices are matched with the same vertex in $R$. Thus, there are at least $|A|$ vertices in the neighborhood of $A$.

Now suppose that $|A| \le |N(A)|$ for all $A \subseteq L$. Run Ford-Fulkerson on the corresponding flow network. The flow is increased by $1$ each time an augmenting path is found, so it will suffice to show that this happens $|L|$ times. Suppose the while loop has run fewer than $L$ times, but there is no augmenting path. Then fewer than $L$ edges from $L$ to $R$ have flow $1$.

Let $v_1 \in L$ be such that no edge from $v_1$ to a vertex in $R$ has nonzero flow. By assumption, $v_1$ has at least one neighbor $v_1' \in R$. If any of $v_1$'s neighbors are connected to $t$ in $G_f$ then there is a path, so assume this is not the case. Thus, there must be some edge $(v_2, v_1)$ with flow $1$. By assumption, $N(\{v_1, v_2\}) \ge 2$, so there must exist $v_2' \ne v_1'$ such that $v_2'\in N(\{v_1, v_2 \})$. If $(v_2', t)$ is an edge in the residual network we're done since $v_2'$ must be a neighbor of $v_2$, so $s$, $v_1$, $v_1'$, $v_2$, $v_2'$, and $t$ is a path in $G_f$. Otherwise $v_2'$ must have a neighbor $v_3 \in L$ such that $(v_3, v_2')$ is in $G_f$. Specifically, $v_3 \ne v_1$ since $(v_3, v_2')$ has flow $1$, and $v_3 \ne v_2$ since $(v_2, v_1')$ has flow $1$, so no more flow can leave $v_2$ without violating conservation of flow. Again by our hypothesis, $N(\{v_1, v_2, v_3\}) \ge 3$, so there is another neighbor $v_3' \in R$.

Continuing in this fashion, we keep building up the neighborhood $v_i'$, expanding each time we find that $(v_i', t)$ is not an edge in $G_f$. This cannot happen $L$ times, since we have run the Ford-Fulkerson while-loop fewer than $|L|$ times, so there still exist edges into $t$ in $G_f$. Thus, the process must stop at some vertex $v_k'$, and we obtain an augmenting path

$$s, v_1, v_1', v_2, v_2', v_3, \ldots, v_k, v_k', t,$$

contradicting our assumption that there was no such path. Therefore the while loop runs at least $|L|$ times. By Corollary 26.3 the flow strictly increases each time by $f_p$. By Theorem 26.10 $f_p$ is an integer. In particular, it is equal to $1$. This implies that $|f| \ge |L|$. It is clear that $|f| \le |L|$, so we must have $|f| = |L|$. By Corollary 26.11 this is the cardinality of a maximum matching. Since $|L| = |R|$, any maximum matching must be a perfect matching.

26.3-5 $\star$

We say that a bipartite graph $G = (V, E)$, where $V = L \cup R$, is $d$-regular if every vertex $v \in V$ has degree exactly $d$. Every $d$-regular bipartite graph has $|L| = |R|$. Prove that every $d$-regular bipartite graph has a matching of cardinality $|L|$ by arguing that a minimum cut of the corresponding flow network has capacity $|L|$.

We convert the bipartite graph into a flow problem by making a new vertex for the source which has an edge of unit capacity going to each of the vertices in $L$, and a new vertex for the sink that has an edge from each of the vertices in $R$, each with unit capacity. We want to show that the number of edge between the two parts of the cut is at least $L$, this would get us by the max-flow-min-cut theorem that there is a flow of value at least $|L|$. The, we can apply the integrality theorem that all of the flow values are integers, meaning that we are selecting $|L|$ disjoint edges between $L$ and $R$.

To see that every cut must have capacity at lest $|L|$, let $S_1$ be the side of the cut containing the source and let $S_2$ be the side of the cut containing the sink. Then, look at $L \cap S_1$. The source has an edge going to each of $L \cap (S_1)^c$, and there is an edge from $R \cap S_1$ to the sink that will be cut. This means that we need that there are at least $|L \cap S_1| - |R \cap S_1|$ many edges going from $L \cap S_1$ to $R \cap S_2$. If we look at the set of all neighbors of $L \cap S_1$, we get that there must be at least the same number of neighbors in $R$, because otherwise we could sum up the degrees going from $L \cap S_1$ to $R$ on both sides, and get that some of the vertices in $R$ would need to have a degree higher than $d$. This means that the number of neighbors of $L \cap S_1$ is at least $L \cap S_1$, but we have that they are in $S_1$, but there are only $|R \cap S_1|$ of those, so, we have that the size of the set of neighbors of $L \cap S_1$ that are in $S_2$ is at least $|L \cap S_1| - |R \cap S_1|$. Since each of these neighbors has an edge crossing the cut, we have that the total number of edges that the cut breaks is at least

$$(|L| - |L \cap S_1|) + (|L \cap S_1| - |R \cap S_1|) + |R \cap S_1| = |L|.$$

Since each of these edges is unit valued, the value of the cut is at least $|L|$.