Breaking Eggs And Making Omelettes

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Beaten By A Puzzle

March 10th, 2006 by Multimedia Mike

It pains me to admit that I couldn’t even produce a computer program capable of solving a Pickover puzzle (update: But Aurel Jacobs could; read on). This is another graph traversal puzzle. The setup is that you have an 8×9 graph with 8 blocked nodes:

* * * * * * * *
* X * * * * * *
* * * X * X * *
* * * * * * * *
* * X * * * * *
* * * * * * * *
* X * * * * * *
* * * X * * * *
* * X X * * * *

Find a path that begins and ends on the same node that covers every unblocked square without crossing itself and moving only vertically or horizontally.

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Graph Traversal Puzzle

February 26th, 2006 by Multimedia Mike

Pickover came through with a slightly more involved brute force challenge. As usual I’ll spare you the flowery setup (though I did find it curious that it involved a character named Ben trying to reach one named Jennifer) and distill the essential details. Given this graph of values:

 1  2  3  4  5
 6  7  8  9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

find a path from any square in column 1 to any square in column 5, moving up, down, left, or right, such that the sum of all squares traversed tallies 90.

I liked writing this program because it is the most algorithmically interesting thing I have done in a long time, even though I originally solved the wrong problem. The way I originally read the problem, the task was to move from node 1 -> node 5, traversing up, down, left, or right, and find a path where the values sum to 90.

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Send More Geese

February 23rd, 2006 by Multimedia Mike

Pickover’s puzzle for the weekend of February 4-5 challenged the reader to find 8 numbers to stand in for the letters D, E, G, M, N, O, R, and S such that the equation SEND + MORE = GEESE would be true. Again, there are multiple solutions. I’m pleased to report that I found 3 orders of magnitude more solutions than the big brain himself.

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Another Pickover Puzzle

February 22nd, 2006 by Multimedia Mike

Cliff Pickover’s puzzle from 2006-02-22 queries the reader to come up with 3 pairs, or possibly more, of numbers, m and n, which satisfy the equation n! + 1 = m2. I didn’t think too hard about it. I just went straight to code which turned out to be a good thing because one of the solutions is quite large. Click more to see the reasoning I went through and the solutions output from my program.

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