Comments
-15

It is decidedly so.

Signs point to yes.

Better not tell you now.

Outlook not so good.

On nitor → Presenting Tourist Facts, 5 years ago
+3

video proof
+8

it's at least as cheap to go i->j+1->j+2 as i->j+2 because you get to subtract $$$c_{j+1}$$$. There's always a path from j+1 to j+2 because of either edge 3 or a combination of the edges from 1 and 2.