Reputation: 32263
Given a grid of points I'm trying to find the path between two of them.
Like in this picture: I'd need to find the points for the yellow line:
What are the best methods / algorithms I can use?
Thanks
Upvotes: 2
Views: 1823
Reputation: 17213
You haven't exactly defined how you want to use diagonal lines so you will have to write the final function as you need it, i suppose taking the path with shortest length of those that use diagonals, noting that path from a>c is shorter than path a>b>c for a,b,c in path
grid = [[False]*16 for i in range(16)]
#mark grid of walls
def rect(p1,p2):
x1, y1 = p1
x2, y2 = p2
for x in range(x1, x2+1):
for y in range(y1, y2+1):
yield (x, y)
rects = [((1,2),(5,5)),
((5,5),(14,15)),
((11,5),(11,11)),
((5,11),(11,11)),
((4,7),(5,13)),
((5,13),(13,13))]
for p1,p2 in rects:
for point in rect(p1,p2):
x,y = point
grid[x][y] = True
start = (1,2)
end = (12,13)
assert(grid[start[0]][start[1]])
assert(grid[end[0]][end[1]])
def children(parent):
x,y = parent
surrounding_points = ((x1,y1) for x1 in range(x-1,x+2) for y1 in range(y-1,y+2) if x1>0 and y<15)
for x,y in surrounding_points:
if grid[x][y]:
#not a wall
grid[x][y] = False
#set as wall since we have been there already
yield x,y
path = {}
def bfs(fringe):
if end in fringe:
return
new_fringe = []
for parent in fringe:
for child in children(parent):
path[child] = parent
new_fringe.append(child)
del fringe
if new_fringe:
bfs(new_fringe)
bfs([start])
def unroll_path(node):
if node != start:
return unroll_path(path[node]) + [node]
else:
return [start]
path = unroll_path(end)
def get_final_path_length(path):
#return length of path if using straight lines
for i in range(len(path)):
for j in range(i+1,len(path)):
#if straight line between pathi and pathj
return get_final_path(path[j+1:]) + distance_between_i_and_j
Upvotes: 0
Reputation: 13213
Check out the A* algorithm
It's what's used in many video games for pathfinding problems, and can be built out to be very robust.
Upvotes: 3