in ith iteration we are using at most i edges so in 1st iteration in the video example we can use only one edge so the verices which are not directly connected to souurce vertex there distance should be infinite but we are changing there distance in the first iteration itself.
Bellman ford algorithm
@deepakjumani09,
The statement you are using
n ith iteration we are using at most i edges so in 1st iteration in the video example we can use only one edge
is obviously incorrect, as proved by your contradiction itself.
Rather the correct statement is,
After the execution of i_{th} phase, the Bellman-Ford algorithm correctly finds all shortest paths whose number of edges does not exceed i.
So, as pointed out by your example, we may relax more than i edges in i_{th} phase, but only those will have shortest distance assigned, whose number of edges(in the shortest path) does not exceed I.
You can read more here (in the proof section).