1.7 Matchings
1.7.4 Perfect Matchings
It’s a perfect ending.
— Sophie, in Anastasia We end this section on matchings by discussing perfect matchings. Recall that a perfect matching is a matching that saturates the entire vertex set of a graph. What kinds of graphs have perfect matchings? One thing that is clear is that a graph must be of even order in order to have a chance at having a perfect matching. But being of even order is certainly not enough to guarantee a perfect matching (look back at Figure 1.105).
We do know thatK2n,C2n, andP2n have perfect matchings. The following result regarding perfect matchings in bipartite graphs is a corollary to Hall’s The- orem. The proof is left as an exercise (Exercise 5).
Corollary 1.57. IfG is a bipartite graph that is regular of degreek, then G contains a perfect matching.
It seems very natural to think that the more edges a graph has, the more likely it is that the graph will have a perfect matching. The following theorem verifies this thought, to a degree.
Theorem 1.58. IfGis a graph of order2nsuch thatδ(G) ≥n, thenGhas a perfect matching.
Proof. LetG be a graph of order2nwithδ(G) ≥ n. Dirac’s theorem (Theo- rem 1.22) guarantees the existence of a Hamiltonian cycle,C. A perfect matching ofGis formed by using alternate edges ofC.
In 1947 Tutte [269] provided perhaps the best known characterization of graphs with perfect matchings. A number of proofs of Tutte’s Theorem have been pub- lished since then. The proof that we present is due to Anderson [7].
A definition first: Given a graphG, defineΩ(G)to be the number of connected components ofGwith odd order. Also, defineΣ(G)to be the number of con- nected components ofGwith even order.
Theorem 1.59 (Tutte’s Theorem). LetGbe a graph of ordern ≥ 2.Ghas a perfect matching if and only ifΩ(G−S)≤ |S|for all subsets ofSofV(G).
Proof. We begin with the forward direction. LetGbe a graph that has a perfect matching. SupposeS is a set of vertices and thatO1,O2, . . .,Ok are the odd components ofG−S. For eachi, the vertices inOican be adjacent only to other vertices inOi and to vertices inS. SinceGhas a perfect matching, at least one vertex out of each of theOi’s has to be matched with a different vertex inS. If k >|S|, then someOiwould be left out (Figure 1.117). Thus,k≤ |S|.
O
1S
? O
2O
kFIGURE 1.117.
For the reverse direction of the theorem, suppose that|S| ≥Ω(G−S)for all S. In particular, ifS=∅, thenΩ(G− ∅)≤0. This implies that there are no odd components ofG—every component ofGis even. More generally, we make the following claim.
Claim A. For any proper subsetS,|S|andΩ(G−S)are either both even or both odd.
LetCbe some component ofG. We know from earlier thatChas even order. If an even number of vertices is removed fromC, then the number of odd components remaining must also be even. If an odd number of vertices is removed fromC, then the number of odd components remaining must be odd. Since this is true for every component ofG, it is true for all ofG. Hence Claim A is proved.
We now proceed by induction onn, the order of the graph. Ifn = 2, thenG isK2, which certainly has a perfect matching. Suppose now that the result is true for all graphs of even order up ton, and letGbe a graph of even ordern. We now have two cases.
Case 1. Suppose that for every proper subsetS,Ω(G−S)<|S|. (That is, the strict inequality holds.) Claim A implies that|S|andΩ(G−S)have the same parity, so we can say in this case that for all subsetsS,Ω(G−S)≤ |S| −2. Let uv∈E(G), and consider the graphG−u−v(a graph with two fewer vertices thanG). We would like to apply the induction hypothesis toG−u−v, so we need the following claim.
Claim B. For all subsetsSofV(G−u−v),Ω(G−u−v−S)≤ |S|.
If Claim B were not true, thenΩ(G−u−v−S1)>|S1|for some subsetS1. But since|S1|=|S1∪ {u, v}| −2, we getΩ(G−u−v−S1)>|S1∪ {u, v}|, and this contradicts the assumption in this case. Claim B is proved.
Since Claim B is true, we can apply the induction hypothesis to G−u−v.
That is, we can conclude thatG−u−vhas a perfect matching. This matching, together with the edgeuv, forms a perfect matching ofG. Case 1 is complete.
Case 2. Suppose there exists a subset S such that Ω(G−S) = |S|. There may be a number of subsetsSthat satisfy this condition—suppose without loss of generality thatS is a largest such set. LetO1,O2,. . .,Okbe the components ofG−Sof odd order.
Claim C.Σ(G−S) = 0. That is, there are no even-ordered components of G−S.
Let Ebe an even ordered component of G−S, and letxbe a vertex ofE.
The graphG−S−xhas exactly one more odd component thanG−S. Thus,
|S∪ {x}| = Ω(G−S−x). But this means thatS∪ {x}is a set larger thanS that satisfies the assumption of this case. Since we choseS to be the largest, we have a contradiction. Therefore there are no even-ordered components ofG−S.
Claim C is proved.
Claim D. There exist verticess1,s2,. . .,sk∈Sand verticesv1,v2,. . .,vk, where for eachi vi∈Oi, such that{v1s1, v2s2, . . . , vksk}is a matching.
For eachi ∈ {1, . . . , k}, define the setSi to be the set of vertices in S that are adjacent to some vertex inOi. Note that if Si = ∅ for some i, thenOi is
completely disconnected from anything else inG, implying thatGitself has an odd component. Since this contradicts our assumption in this case, we can assume that eachSi is nonempty. Furthermore, our initial assumptions imply that the union of anyrof theSi’s has size at leastr. Thus, the condition in Theorem 1.52 is satisfied, implying that there exists a system of distinct representatives for the family of setsS1,S2,. . .,Sk. If we let these representatives bes1,s2,. . .,sk, and their adjacencies in theOi’s bev1,v2,. . .,vk, then Claim D is proved.
The situation inGis depicted in Figure 1.118, wherek=|S|.
O1
O2 S
Ok
FIGURE 1.118.
At this point, each vertex inShas been matched to a vertex in anOi. The goal at this point is to show that eachOi−vihas a perfect matching.
LetW be some subset of vertices of (the even-ordered)Oi−vi. Claim E.Ω(Oi−vi−W)≤ |W|.
IfΩ(Oi−vi−W)>|W|, then, by Claim A,Ω(Oi−vi−W)≥ |W|+ 2. But then,
Ω(G−S−vi−W)≥ |S| −1 +|W|+ 2
=|S|+|W|+ 1
=|S∪W ∪ {vi}|. This contradicts our assumption, and thus Claim E is proved.
Since Claim E is true, eachOi −vi satisfies the induction hypothesis, and thus has a perfect matching. These perfect matchings together with the perfect matching shown in Figure 1.118 form a perfect matching ofG, and so Case 2 is complete.
We conclude this section by considering perfect matchings in regular graphs. If a graphGis1-regular, thenGitself is a perfect matching. IfGis2-regular, then Gis a collection of disjoint cycles; as long as each cycle is even,Gwill have a perfect matching.
What about3-regular graphs? A graph that is3-regular must be of even order, so is it possible that every3-regular graph contains a perfect matching? In a word, no. The graph in Figure 1.119 is a connected3-regular graph that does not have a perfect matching. Thanks to Petersen [221], though, we do know of a special class of3-regular graphs that do have perfect matchings. Recall that a bridge in a graph is an edge whose removal would disconnect the graph. The graph in Figure 1.119 has three bridges.
FIGURE 1.119.
Theorem 1.60 (Petersen’s Theorem). Every bridgeless,3-regular graph contains a perfect matching.
Proof. LetGbe a bridgeless,3-regular graph, and suppose that it does not contain a perfect matching. By Tutte’s Theorem, there must exist a subsetSof vertices where the number of odd components ofG−Sis greater than|S|. Denote the odd-ordered components ofG−SbyO1,O2,. . .,Ok.
First, eachOimust have at least one edge intoS. Otherwise, there would exist an odd-ordered,3-regular subgraph ofG, and this is not possible, by Theorem 1.1.
Second, sinceGis bridgeless, there must be at least two edges joining eachOito S. Moreover, if there were only two edges joining someOitoS, thenOi would contain an odd number of vertices with odd degree, and this cannot happen.
We can therefore conclude that there are at least three edges joining eachOito S. This implies that there are at least3kedges coming intoSfrom theOi’s. But since every vertex ofShas degree 3, the greatest number of edges incident with vertices inSis3|S|, and since3k >3|S|, we have a contradiction. Therefore,G must have a perfect matching.
It is probably not surprising that the Petersen of Theorem 1.60 is the same person for whom the Petersen graph (Figure 1.63) is named.
Petersen used this special graph as an example of a3-regular, bridgeless graph whose edges cannot be partitioned into three separate, disjoint matchings.
Exercises
1. Find a maximum matching of the graph shown in Figure 1.119.
2. Use Tutte’s Theorem to prove that the graph in Figure 1.119 does not have a perfect matching.
3. Draw a connected,3-regular graph that has both a cut vertex and a perfect matching.
4. Determine how many different perfect matchings there are inKn,n. 5. Prove Corollary 1.57.
6. Characterize whenKr1,r2,...,rkhas a perfect matching.
7. Prove that every tree has at most one perfect matching.
8. LetGbe a subgraph ofK20,20. IfGhas a perfect matching, prove thatG has at most 190 edges that belong to no perfect matching.
9. Use Tutte’s Theorem to prove Hall’s Theorem.