00:04
We were asked to prove a theorem of oistine or, i think that's how he pronounced his name.
00:16
So we're given that g, which is the pair of ve, is a bipartite graph, and bipartition is v1, v2, and that a is going to be a subset of set v1.
00:50
We're asked to show that the maximum number of vertices of v1, that are the endpoints of a matching, equals the order of v1 minus the maximum for all subsets a of v of v of d of a, where defecency of a is the order of a minus the number of neighbors of the subset a.
01:32
So to prove this, let b be the maximum for all subsets a of v, this should in particular be, say, v1, of the deficiency of a, and let capital b be the subset of v1, such that b is equal to the deficiency of a, and let capital b be the subset of v1 such that b is equal to the deficiency of capital b.
02:57
And so it follows that b is equal to, by definition of deficiency, the magnitude of b minus the number of neighbors of the subset b.
03:19
And since nb is the number of neighbors of b, we have that a matching has at most nb edges with endpoints in b.
04:06
Because if we had more than nb edges with end points in b, well, that would be impossible, because then we would have more edges than we'd have neighbors.
04:32
And we have that any matching can have at most v1 minus b edges, with their end points in in the set v1 minus b, since v1 is the number of elements in v1 and b is the number of elements in b.
05:39
And so the total number of edges, this is going to be the number of neighbors of b plus and then v1 minus b.
06:16
And we have that doing some rearranging.
06:22
This is equal to v1 plus little b.
06:29
The deficiency of matrix b should be v1 minus b, my mistake.
06:49
And so it follows that a matching in g has at most v1.
06:54
Minus b edges.
07:17
And so the matching, the edges of a matching can touch at most v1 minus b vertices v1...