Question

Here’s an interesting problem to solve in minimal amounts of code. I expect the recursive solutions will be most popular.

We have a maze that’s defined as a map of characters, where = is a wall, a space is a path, + is your starting point, and # is your ending point. An incredibly simple example is like so:

====
+  =
= ==
=  #
====

Can you write a program to find the shortest path to solve a maze in this style, in as little code as possible?

Bonus points if it works for all maze inputs, such as those with a path that crosses over itself or with huge numbers of branches. The program should be able to work for large mazes (say, 1024×1024 – 1 MB), and how you pass the maze to the program is not important.

The “player” may move diagonally. The input maze will never have a diagonal passage, so your base set of movements will be up, down, left, right. A diagonal movement would be merely looking ahead a little to determine if a up/down and left/right could be merged.

Output must be the maze itself with the shortest path highlighted using the asterisk character (*).

Answer 1

Python

387 Characters

Takes input from stdin.

import sys
m,n,p=sys.stdin.readlines(),[],'+'
R=lambda m:[r.replace(p,'*')for r in m]
while'#'in`m`:n+=[R(m)[:r]+[R(m)[r][:c]+p+R(m)[r][c+1:]]+R(m)[r+1:]for r,c in[(r,c)for r,c in[map(sum,zip((m.index(filter(lambda i:p in i,m)[0]),[w.find(p)for w in m if p in w][0]),P))for P in zip((-1,0,1,0),(0,1,0,-1))]if 0<=r<len(m)and 0<=c<len(m[0])and m[r][c]in'# ']];m=n.pop(0)
print''.join(R(m))