00:01
In this first problem, let us consider a directed graph that is g is equals to v comma e.
00:07
And consider the longest path that is v1, e1, v2, e2, so on up to vk, in, that is, lon, a bar, a bar, g.
00:22
Here, vi is belongs to capital v and ei belongs to capital e.
00:30
Now for all i and ei is an arc edge pointing vi minus 1 to vi.
00:45
Now since the out degree of vk is greater than 1, vk is adjacent to some vertex vk plus 1 in g.
00:53
Let ck be an arc point, that is ck be an arc point that pointing to vk to vk plus 1.
01:02
Now if vk plus 1 is not equals to v i, that is vk plus 1 is not equals to v i.
01:11
For any i in 1 2k, 1 2, 3, 4, so on after k, the path, this path would be 1 unit longer than longest path.
01:22
That is the same path in g, which is a contradiction...