How does a Breadth-First Search work when looking for Shortest Path?

Question

I\'ve done some research, and I seem to be missing one small part of this algorithm. I understand how a Breadth-First Search works, but I don\'t understand how exactly it wi

Answer 1

I have wasted 3 days
ultimately solved a graph question
used for
finding shortest distance
using BFS

Want to share the experience.

When the (undirected for me) graph has
fixed distance (1, 6, etc.) for edges

#1
We can use BFS to find shortest path simply by traversing it
then, if required, multiply with fixed distance (1, 6, etc.)

#2
As noted above
with BFS
the very 1st time an adjacent node is reached, it is shortest path

#3
It does not matter what queue you use
   deque/queue(c++) or
   your own queue implementation (in c language)
   A circular queue is unnecessary

#4
Number of elements required for queue is N+1 at most, which I used
(dint check if N works)
here, N is V, number of vertices.

#5
Wikipedia BFS will work, and is sufficient.
    https://en.wikipedia.org/wiki/Breadth-first_search#Pseudocode

I have lost 3 days trying all above alternatives, verifying & re-verifying again and again above
they are not the issue.
(Try to spend time looking for other issues, if you dint find any issues with above 5).

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More explanation from the comment below:

      A
     /  \
  B       C
 /\       /\
D  E     F  G

Assume above is your graph
graph goes downwards
For A, the adjacents are B & C
For B, the adjacents are D & E
For C, the adjacents are F & G

say, start node is A

1. when you reach A, to, B & C the shortest distance to B & C from A is 1

2. when you reach D or E, thru B, the shortest distance to A & D is 2 (A->B->D)

similarly, A->E is 2 (A->B->E)

also, A->F & A->G is 2

So, now instead of 1 distance between nodes, if it is 6, then just multiply the answer by 6
example,
if distance between each is 1, then A->E is 2 (A->B->E = 1+1)
if distance between each is 6, then A->E is 12 (A->B->E = 6+6)

yes, bfs may take any path
but we are calculating for all paths

if you have to go from A to Z, then we travel all paths from A to an intermediate I, and since there will be many paths we discard all but shortest path till I, then continue with shortest path ahead to next node J
again if there are multiple paths from I to J, we only take shortest one
example,
assume,
A -> I we have distance 5
(STEP) assume, I -> J we have multiple paths, of distances 7 & 8, since 7 is shortest
we take A -> J as 5 (A->I shortest) + 8 (shortest now) = 13
so A->J is now 13
we repeat now above (STEP) for J -> K and so on, till we get to Z

Read this part, 2 or 3 times, and draw on paper, you will surely get what i am saying, best of luck

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