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Calculating 'Diagonal Distance' in 3 dimensions for A* path-finding heuristic

By : Kyoshi
Date : November 24 2020, 03:41 PM
wish of those help It can be extended to 3D relatively easily. It does require finding the "middle" of 3 values, there is a trick for that given that we have the minimum and maximum.
code :
``````dx = absdiff(node.x, goal.x)
dy = absdiff(node.y, goal.y)
dz = absdiff(node.z, goal.z)
dmin = min(dx, dy, dz)
dmax = max(dx, dy, dz)
dmid = dx + dy + dz - dmin - dmax
``````
``````return (D3 - D2) * dmin + (D2 - D1) * dmid + D1 * dmax
``````

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Calculating distance in a 2D array, Non diagonal

By : CPBundrick
Date : March 29 2020, 07:55 AM
should help you out It's called Manhattan Distance and can easily be computed by:
code :
``````distance = abs(ydistance) + abs(xdistance)
``````

Finding the distance between two objects in a 2D array, not going Diagonal

By : Iqbal Hossain
Date : March 29 2020, 07:55 AM
This might help you You could take a look at the A* search algorithm which is widely used in games since it uses a heuristic to make performance improvements.
In your case, the heuristic could be the Manhattan Distance metric to calculate the distance between your grid elements. This distance metric takes into consideration the total of the differences in the X and Y coordinates (which unlike euclidean distance, does not allow for diagonal traversal).

By how much suboptimal can be path found with A* using heuristic which overestimates the remaining distance a little?

By : user2626684
Date : March 29 2020, 07:55 AM
like below fixes the issue A* returns when the partial answer it retrieves (which is the partial answer with smallest estimated total distance to reach the goal) has in fact reached the goal. Standard A* guarantees to find the correct answer because, by the definition of the heuristic, all the estimated total distances to reach the goal are lower bounds, so none of the other answers can do better.
Suppose the heuristic for an answer which in fact will end up with total distance T can be up to KT, where K > 1. If A* retrieves an answer with cost KT and thinks it has succeeded because the answer reaches the goal, then it might not be the best answer. We know that every partial answer still in the pool has cost at least KT, but because of the heuristic, an answer with heuristic KT might actually turn into an answer with total cost T < KT (but cannot turn into an answer with any cheaper cost). So with this sort of heuristic, you the answer returned by A* can be up to K times as expensive as the best answer, but no more.

A* search, grid, 8 directions, octile distance as heuristic, not finding the direct path

By : AleenaNeena
Date : March 29 2020, 07:55 AM
like below fixes the issue Your movement cost for diagonals is the same as for orthogonal steps.
A path going southeast, southeast, northeast, northeast is just as short as a path going east, east, east, east. Both have cost 4.
code :
``````int new_cost = cost_so_far[current] + 1;
``````

A* pathfinding - Euclidean distance heuristic behaving worse than Diagonal distance

By : Ahmad Faqih Al Fahmy
Date : March 29 2020, 07:55 AM
To fix the issue you can do The problem is that because the neighbours are all 8 adjacent grid points, and the cost between all of them is 1, euclidean distance is overestimating the cost between diagonal points.
Real distance between diagonal points: 1