Lecture #7 discussion

1. Perfect Matchings

The proof of integrality of the non-bipartite perfect matching polytope was a bit fast. Let me give some more details.

Recall that the LP is defined by the variables {x_e} for {e \in E}, and the constraints:

\displaystyle  \begin{array}{rl}    x(\partial v) = 1 \qquad \qquad & \forall v \in V \\   x(\partial S) \geq 1 \qquad \qquad & \forall S \subset V, |S| \text{   odd} \\   x \geq 0 \qquad \qquad &   \end{array}

Note that in the second set of constraints, we can just consider {S} of size at least {3} and at most {|V|-3}: if {|S| = 1} or {|V|-1}, then the first set of constraints already implies {x(\partial S) = 1}. So just focus on

\displaystyle  \begin{array}{rl}    x(\partial v) = 1 \qquad \qquad & \forall v \in V \\   x(\partial S) \geq 1 \qquad \qquad & \forall S \subset V, |S| \text{   odd }, 3 \leq |S| \leq |V|-3\\   x \geq 0 \qquad \qquad &   \end{array}

Let us call this perfect matching polytope {PM}. We’ll call the first set of equalities the vertex constraints, and the second set the odd-set inequalities.

Clearly, every perfect matching is a feasible solution to this LP. What the next theorem says is perfect matchings are precisely the vertices of this LP. (We already saw this for bipartite graphs, in Lecture~3.)

Theorem 1 (Perfect Matching Theorem)

Every vertex of this LP is integral.

Suppose not, and suppose there exists graphs for which there is a fractional vertex. Consider a minimal counterexample {G = (V,E)} (minimizing the sum of {|V| + |E|}, say), and some vertex solution {x} that is not integral. Clearly, {|V|} must be even, else it will not satisfy the odd set constraint for {S = V}. First, the claim is that {G} cannot have a vertex of degree~{1}, or be disconnected (else we’ll get a smaller counterexample) or be just an even cycle (where we know this LP is indeed integral). Being connected implies that {|E| \geq |V|-1}, and neither being a cycle nor having a degree-{1} vertex implies that {|E| \neq |V|}. So {|E| > |V|}.

Recall there are {|E|} variables. So any vertex/BFS is defined by {|E|} tight constraints. If any of these tight constraints are the non-negativity constraints {x_e \geq 0}, then we could drop that edge {e} and get a smaller counterexample. And since at most {|V|} tight constraints come from the vertex constraints. So at least one odd-set constraint should be tight. Say this tight odd-set constraint is for the odd set {W \subseteq V} with {|W| \geq 3}: i.e.,

\displaystyle  x(\partial W) = 1

Now consider the two graphs {G/W} and {G/\overline{W}} obtained by contracting {W} and {\overline{W}} to a single new vertex respectively, and removing the edges lying within the contracted set. Since both {W} and {\overline{W}} have at least {3} vertices, both are smaller graphs.

Now {x} naturally extends to feasible solutions {y} and {z} for these new graphs. E.g., to get {y}, set {y_e = x_e} for all edges {e \in E \setminus \binom{W}{2}}. Note that if the set {W} got contracted to new vertex {\widehat{w}} in {G/W}, then the fact that {x(\partial W) = 1} implies that {y(\partial \widehat{w}) = 1}, and hence {y} is a feasible solution to the perfect matching polytope for graph {G/W}. Similarly, {z} is a feasible solution to the perfect matching polytope for graph {G/\overline{W}}.

By minimality of {G}, it follows that the perfect matching LP is integral for both {G/W} and {G/\overline{W}}: i.e., the vertices of the perfect matching polytope for these smaller graphs all correspond to perfect matchings. And that means that

\displaystyle  y = \sum_i \lambda_i \cdot \chi_{M_i},

where {\chi_{M_i}} is the natural vector representation of the perfect matching {M_i} in {G/W}, for values {\lambda_i \geq 0, \sum_i \lambda_i = 1}. Also, {\lambda_i}‘s can be taken to be rational, since {y} is rational, as are {\chi_{M_i}}. Similarly, we have a rational convex combination

\displaystyle  z = \sum_i \mu_i \cdot \chi_{N_i},

where {N_i} are perfect matchings in {G/\overline{W}}. Since {\lambda_i, \mu_i} are rationals, we could have repeated the matchings and instead written

\displaystyle  \begin{array}{rl}    y &= \frac1k \sum_i \chi_{M_i} \\   z &= \frac1k \sum_i \chi_{N_i}   \end{array}

Finally, we claim that we can combine these to get

\displaystyle  x = \frac1k \sum_i \chi_{O_i}

where {O_i}‘s are perfect matchings in {G}. How? Well, focus on edge {e = \{l,r\} \in \partial W}, with {l \in W}. Note that {y_e = z_e = x_e}. If we look at {k} matchings {M_i} in the sum for {y}: exactly {x_e} fraction of these matchings {M_i} — that is, {kx_e} matchings — contain {e}. Similarly, exactly {kx_e} of the matchings {N_i} in the sum for {x} contain {e}. Now we can pair such matchings (which share a common edge in {\partial W}) up in the obvious way: apart from the edge {e}, such an {M_i} contains edges only within {\overline{W}} and matches up all the vertices in {\overline{W}} except vertex {r}, and {N_i} contains edges only within {W} and matches up all the vertices in {W \setminus \{l\}}. And {e} matches up {\{l,r\}}. Hence putting together these perfect matchings {M_i} and {N_i} in {G/W} and {G/\overline{W}} gets us a perfect matching {O_i} for {G}.

So {x} can be written as a convex combination of perfect matchings of {G}. Hence, for {x} to be an extreme point (vertex) itself, it must be itself a perfect matching, and integral. This gives us the contradiction.

1.1. Max-Weight Matchings

We didn’t get to this, but suppose you want to write an LP whose vertices are precisely (integral) matchings in {G}, not just the perfect matchings. Here is the polytope Edmonds defined.

\displaystyle  \begin{array}{rl}    x(\partial v) \leq 1 \qquad \qquad & \forall v \in V \\   \textstyle \sum_{e \in \binom{S}{2}} x_e \leq \frac{|S| - 1}{2} \qquad   \qquad & \forall S \subset V, |S| \text{   odd } \\   x \geq 0 \qquad \qquad &   \end{array}

Clearly, all matchings in {G} are feasible for this LP. Moreover, one can use the Perfect Matching Theorem above to show that every vertex of this polytope is also integral.


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