You have a cube made up with 6x6x6 cells, with each cell containing a bit (either 0 or 1). In each direction (row, column or stack), you can thus read a series of 6 bits. We define the couple of the series in both direction to be the signature of the row. For example, a row reading, from left to right, "000111" would have a signature "{000111,111000}".There are obviously 36 different signatures, out of which 8 are palindromes (such as "101101", whose signature is "{101101}" as it reads the same in both directions).The purpose is to fill the cube such that:- all 36 rows have different signatures;- all 36 columns have different signatures;- all 36 stacks have different signaturesAs a matter of practicality, I suggest presenting the results as :123456123456123456123456123456123456(first layer, closest on the z axis)123456123456123456123456123456123456(second layer)and so on...

I am not sure I understand the problem...Isn't it 64 signatures ? (2^6)...Quantro

quantro ... the problem is fine ... there will be only 36 possible signatures.(you can see the excel file attached to see all the 36 signatures .... see Cell Q28 onwards).I am tempted to use brute force ... but it's quite not do-able on a PC.There will be 36! x 2^28 possible scenarios to test. ~ 10^50 scenarios.(The excel also contains a scenario, where 4 rows and 4 columns do not have a unique signature (all stacks have unique signature)

Thanks bhutes... Now I get it.The problem is even harder than I thought, and like you said, brute force seems to be out of the way...Cube symmetry could help to reduce the number of cases, or find new constraints between digits, I'll try to give it a look.Quantro