Every now and then Pickover tosses out an extremely obvious puzzle. Given the following sequence of numbers:

  1, 27, 125, 343, 729, 1331, 2197, 3375, x

what is x? Now, when I say “obvious” in this context I should clarify that the solution is relatively straightforward to the math-trained mind. Typically the solutions to such numerical puzzles are along the lines of “The latitude coordinates of the original U.S. colonies multiplied by the longitude coordinates of the same, then add 3.”

Anyway, that was today’s puzzle. The puzzle for April 4 was to find the numbers, A and B, that satisfy the following equation:

  AB + BA = A7

My first impulse was there should be some way to solve for A and B, maybe involving logarithms or something, until I noticed that there were 2 variables and only 1 equation which, per my math classes of long ago, proves difficult. So, a brute force program it was. Surprisingly, there is only 1 solution.

See [more] for the answers, I mean, if you really care.

I have a stack of puzzles to work through that should be brute-force-able using a combination of permutation logic and graph techniques. Eventually, I imagine I will be able to assemble a program toolkit to rip through an entire year of these puzzles automatically.

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I used to think that it would be too far-fetched to write a computer program to determine the solution(s) to one of Pickover’s more linguistic puzzles, such as the puzzle for 2006/03/22; find letters to fill in positions 1, 3, and 5 to form an actual word:

1 n 3 i 5 n

I decided to call in the help of a program I had only heard of — Ispell — that I previously had no use for because I’m the most awesome spehlurr I personally know. However, I researched it, installed it, and determined that I could use it to come up with a list of plausible puzzle solutions.

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Big Brain’s puzzle for the weekend of March 11-12 was essentially an anagram puzzle. Using the following 12 letters:

e c m a o n s a d e l k

form an 8-letter (English) word and a 4-letter (English) word. One word tastes good and the other carries stuff. Usually, I am at a loss to write a program that can re-sort letters and find what look to be valid words. However, I recently found an anagram generator. Go to the homepage of the GNU MP Bignum library, scroll way to the bottom and find “Demo 2 — Find Anagrams”. One day, I hope to understand how big numbers pertain to computing anagrams. Anyway, insert those 12 letters and you will find an impressive list of anagrams, including the correct answer for the puzzle! For added enjoyment, insert your name or those of your friends (the Guru’s name anagrams to “Hemimeridae larceny”).

Your reign of terror-logic will not last long, Pickover. We have technology on our side.

You thought those Pickover puzzles were inane? Well, you were right. But it does help to keep a programmer sharp on pure computer science skills. One reason I have been having such a good time reverse engineering the FAAD2 source is because it represents a bunch of little puzzles. Only they’re considerably more straightforward and grounded in reality than the Pickover puzzles. If these puzzles were on a Pickover calendar page, they would be worded along the lines of,

“There are 8,192 numbers stored in a table called iq_table. 0, 1, 2.5198420997897464, 4.3267487109222245, all the way up to 165113.4940829452. What is the common property that connects all of these numbers?”

The answer (thanks to Jindrich Makovicka for pointing this out on the ffmpeg-devel list) is that each number is its index (0..8191) raised to an exponent of 4/3: iq_table[index] = index^4/3. Here are some more puzzles:

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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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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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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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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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Cliff Pickover continues his torment of humanity with six puzzles each week (Saturday and Sunday are combined) with his 2006 day-by-day calendar:

As a good number of his puzzles can be solved by brute forcing through the entire solution space, modern computing technology can assist us greatly in mitigating Pickover’s reign of terror. Such is the case with the January 19 puzzle. The actual puzzle is explained a bit more “colorfully”. I’ll distill the mathematical principles. There is a grid of 8 squares laid out as a 4×2 matrix:

  1 2 3 4
  5 6 7 8

You have to place 8 letters in the 4 squares: (B G I O R T V Y) obeying the following rules: Neither R nor I can occupy square 1 or 5; neither V nor T can occupy squares 1..4. Y is left of R, V is left of T, R is above I, and O and B are on odd quares. I took the “p is left of q” type rules to mean that p to directly to the left of q.

Anyway, I deductively came up with a solution. But it seemed there might be more solutions. Sure enough, the official solution was different but challenged the reader to find out how many solutions there are. Fine! I will. Read on to find the solutions and the program to figure it out…

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