You're trapped in a maze. There is a way out. Path junctions are all 3-way.
If you use the strategy of always taking the path going right, what will happen?
(Note: This problem is deliberarely vague.)
(In reply to
Depends. by TomM)
Sorry about the empty post.
What I was starting to say was that it depends on the maze. If you have a maze that has only one entrance and its walls are all connected (There is only one "piece" of the maze, then this strategy will eventually get you out.
If you have a maze with separate "In" and "Out" entrances, it will be in at least two pieces. But if it is only two pieces, the strategy will again eventually get you out, with the proviso that if at the start you were "inside" the "wrong" piece (or were between the two pieces, but chose the wrong direction for your first move [when your orientation was not yet determined by eliminating the path you arrived on], you will arrive at the "In" entrance first, and if you are not allowed to exit ther, you would therefore by necessity have to take the left fork. After this, always taking the right fork will lead you to the Exit.
If the maze has more "pieces" than entrances, some of the "pieces" are "inside pieces." In fact, the number of "outside pieces" equals the number of entrances [call this "E"], and the number of inside pieces [I] is equal to the totapl number of pieces [P] minus the number of outside pieces. (I = P - E).
If after making your first move, the wall on your right is part of an Outside piece, you will surely find the exit. If it is part of an Inside Piece, you will merely keep circling that piece and never escape.
So the probability of the strategy working is exactly the probabiliy of that piece being an outside piece (O/P) and the odds of getting out using that strategy are (O:I)
So, as I said, it depends on the maze.
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Posted by TomM
on 2002-06-25 20:48:05 |