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A* search algorithm: Difference between revisions
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Cost: 11
Route: (0, 0), (1, 1), (2, 2), (3, 1), (4, 1), (5, 1), (6, 2), (7, 3), (6, 4), (7, 5), (6, 6), (7, 7)
</pre>
=={{header|Python}}==
<lang Picat>% An A*-like algorithm is used in tabling
%
main =>
Maze = new_array(8,8),
Obs = [(2,4), (2,5), (2,6), (3,6), (4,6), (5,6), (5,5), (5,4), (5,3), (5,2), (4,2), (3,2)],
foreach ((R0,C0) in Obs)
Maze[R0+1,C0+1] = 100
end,
foreach (R in 1..8, C in 1..8)
(var(Maze[R,C]) -> Maze[R,C] = 1; true)
end,
search((1,1),(8,8),Maze,Cost,Path),
writeln(cost=Cost),
println([(R0,C0) : (R1,C1) in Path, R0 = R1-1, C0 = C1-1]).
table (+,+,+,min,-)
search(G,G,_Maze,Cost,Path) => Cost = 0, Path = [G].
search(S@(R,C),G,Maze,Cost,Path) =>
neibs(R,C,Neibs),
member(S1,Neibs),
S1 = (R1,C1),
search(S1,G,Maze,Cost1,Path1),
Cost = Cost1+Maze[R1,C1],
Path = [S|Path1].
neibs(R,C,Neibs) =>
Neibs = [(R1,C1) : Dr in [-1,0,1], Dc in [-1,0,1], R1 = R+Dr, C1 = C+Dc,
R1 >= 1, R1 <= 8, C1 >= 1, C1 <= 8, (R,C) != (R1,C1)].
</lang>
{{out}}
<pre>
cost = 11
[(0,0),(1,0),(2,0),(3,0),(4,0),(5,1),(6,2),(6,3),(6,4),(6,5),(6,6),(7,7)]
</pre>
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