Geek asks and answers

The complexity of the trajectory search algorithm

Problem:

Change the A * algorithm we optimized for fewer turns. Now the algorithm will find the path from (a, b) to ANY SURFACE TO (x, y), where (x + 1, y) or (x-1, y) is preferable when possible.

My decision:

  • Run the original A * algorithm from (a, b) to (x, y).
  • Find the last coordinate in the path passing through (x-1,) or (x + 1,), if any.
  • If there is a straight vertical line available from * to y in this coordinate plane, change the path following this line.

Visual demo: (path from S to E, where X is not available)

......S           .....S  
.     X           .    X
.          =>     .     
.                 .
E                E.

However, I am not sure that my solution will work in some cases, for example:

......S            .....S  
.     X            .    X
.X        ???     X.     
.                  .
E                E..

Can anyone think of a solution to this problem?

. A * , , .

+3
2

A * Dijkstra algorithm, [ [ ].

, (x,y), [A * ], , node. , h'(tile) = max { h(tile) - 1 , 0}, .

A *:

  • A *, [ , ]. d.
  • A* f(tile) , d. , d, - , d. [ ].
  • "" . , .

, !

+2

, (x+1,y) (x-1,y). , 2 , 2 .

0

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