Yes, I did that and used a helper function to convert from {row,col} coordinates to integers: c2i = fn row, col -> row*71+col end. Thus my vertices are only identified by integers.
Now I get an answer of 269 steps but it is too high. I guess I will have to use my own stuff like you did.
I still think you’re missing the 0 and 70 rows and columns - it looks like you’re adding the four corners as extra cases to your graph but they’re entire rows and columns that need to be accounted for.
By the way, a simple BFS would suffice because all the weights are 1, thus Dijkstra effectively becomes BFS.
Thanks. Yes, that was the bug in the code for which no path at all could be found 
My bad… Now I get a wrong answer.
EDIT: After evaluating @Aetherus livebook, I found that it is 268 steps for part 1 instead of 269. Thus {0,0} is not considered a step. I am relieved that A* is working for directed graphs.
Yes, A* works for both directed and undirected graphs. The only restriction is that all the weights on all the edges must be positive.