00:01
We're asked to show that for given an undirected graph with n vertices, then no more than the floor of n over 2 to the k trees remain after the first step of solon's algorithm has been carried out, and the second step of the algorithm has been carried out k minus 1 times.
00:30
So, let g be a connected, weighted, undirected graph with n vertices, and let pk be the statement.
01:16
Solon's algorithm has no more than the floor of an over 2k trees remaining after the first step and the next k -minus -1 steps have been carried out.
02:24
And we'll prove this by induction.
02:27
So for the base case, we have that k is equal to 1.
02:40
As part of the algorithm, we'll construct a forest of n trees where each tree contains one vertex.
03:14
Then we know that every new tree that will be formed in the first step will contain at least two of the n trees.
03:51
This is because each of the n trees has to be connected to a new tree.
03:56
And so there are at most n over two new trees.
04:26
And since the number of trees is an integer, this means that the number of trees is an integer.
04:30
That there are no more than the floor of n over two trees.
05:00
And we see that in this case statement p1 is true.
05:08
For the inductive step, let p m be true for some m greater than or equal to 1...