Lecture 12 - 13 , ( Oct . 18 and 20 , 2011 ) : MST , Min - cost Bounded Degree ST

We know that any bfs is uniquely determined by n linearly independent tight constraints (where n is number of variables of the LP ). Since we have exponentially many constraints in this LP, a bfs may be satisfying many of them with equality (i.e. being tight). We want a “good” set of linearly independent tight constraints defining it. The notion of “good” here will be clear soon. First, observe that in any bfs, we can safely delete any edge e ∈ E with xe = 0 from the graph. So we can assume that every edge of G has xe > 0. Our goal is to show that there are at most |v| − 1 linearly independent tight constraints which implies that there are at most |v| − 1 non-zero variables. Since for any set S with size |S| = 2, the condition x(E(S)) ≤ |S| − 1 implies the value of that edge must be at most 1 and because x(E(v)) = |v| − 1 we get that all the |v| − 1 non-zero variables must have value exactly 1, i.e. the bfs is integral. Before we present the algorithm for this LP, we introduce some Lemmas first. Recall that: