'Understanding the AStar Algorithm
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If an adjacent square is already on the open list, check to see if this path to that square is a better one. In other words, check to see if the G score for that square is lower if we use the current square to get there. If not, don’t do anything.
Example:
parent (already traversed): O
branches: A, B, C
parent (working on): A
Branches: B, D
The open list contains, A, B, C and now D.
Now, the bold statements in the above quote are comparing which path with, path A to B? i.e. Comparison of A to B && O to B OR Comparison of A to B && A to D
Please clarify.
Solution 1:[1]
Well, if we are working on the node A, we are considering its neighbours. Say we are considering the B node now (which is in the openList).
The definition given tells us to compare the G value of B (previously computed when it was first added to open list when the working node was O) and the sum of the cost of reaching A from the begining (O) and the cost of reaching B from A.
Sources
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Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | tafa |