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Ponder This

March 2005

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Ponder This Challenge:

Puzzle for March 2005

This month's puzzle is attributed to Peter Manyakhin, and sent in by Max Alekseyev.

Consider a cube of dimension 6x6x6, such that each unit cell in it contains a bit 0 or 1.

From each row (column or stack) in the cube we can read six bits. These six bits up to direction of reading are called the "signature" of the row.
Examples of signatures are:
{000111, 111000} (i.e., 000111 and 111000 form the same signature);
{011101, 101110} (reading one direction it's 011101; from the other direction, 101110);
{101101} (this is a palindome, reading the same frontwards and backwards, so only needs one representative).
There are exactly 36 different signatures. 8 of them are palindromes (such as {101101}), and the other 28 are pairs of six-bit strings, one being the reversal of the other (such as {100111, 111001}).

Your problem: find such a cube such that all 36 rows (parallel to the x-axis) have different signatures; all 36 columns (parallel to the y-axis) have different signatures; and all 36 stacks (parallel to the z-axis) have different signatures. That is, each of the 36 possible signatures occurs exactly once in each of the three directions (x,y,z).

Please present the solution in plain text (no attachments), in the following format:

001001 101010 011110 110011 100001 000110
110001 100001 001001 010111 110011 011110
101010 011110 011111 111101 101010 110111
101001 111101 001111 011110 101110 011000
100110 011001 011001 110011 011110 100000
100011 011111 011111 110010 010100 011000

Each 6x6 block represents one layer of the cube (parallel to the x-y plane), and moving from block to block we move in the z direction. We see six lines, each having 36 digits, with a single space between each group of six digits. A standard format makes it easier for me to check, so no other format will be accepted. As usual, each solution should be your own work, though computer assistance is allowed. (After all, we sell computers!)

Extra credit for a "balanced" solution, where each octant has 13 or 14 1-bits.

The first 100 people who answer correctly will be listed. The answer will be posted a week after the 100th is received, or at the end of the month.

We will post the names of those who submit a correct, original solution! If you don't want your name posted then please include such a statement in your submission!

We invite visitors to our website to submit an elegant solution. Send your submission to the

If you have any problems you think we might enjoy, please send them in. All replies should be sent to:


Challenge: 03/01/2005 @ 08:30 AM EST
Solution: 04/01/2005 @ 08:30 AM EST
List Updated: 03/01/2005 @ 08:30 AM EST

People who answered correctly:

*Tapani Utriainen (03.06.2005 @08:02:16 PM EDT)
*Bertram Felgenhauer (03.07.2005 @06:33:45 PM EDT)
*Daniel Bitin (03.07.2005 @06:48:39 PM EDT)
John Tromp (03.19.2005 @12:54:15 PM EDT)
*Dharmadeep Muppalla (03.20.2005 @07:41:57 AM EDT)
Jack Rouse (03.22.2005 @08:45:05 PM EDT)
Raicho Tchalkov (03.26.2005 @05:36:14 PM EDT)
Luke Pebody (03.29.2005 @03:00:12 AM EDT)
Gary M Gerken (03.30.2005 @04:37:44 PM EDT)
Divyesh Dixit (03.31.2005 @06:31:35 PM EDT)

*Earned extra credit for balanced solution.

Attention: If your name is posted here and you wish it removed please send email to the