Given a 5×5×5 cube of identical cubical cells (total 125 cells). In each cell there is a prisoner. There are doors from each cell to the adjacent cells (not diagonally). Their task is to move at same time from their cell to an adjacent cell so that no two prisoners end up in the same cell. They are not allowed to switch cells with an adjacent prisoner. Devise a plan for them to make the move.

  • $\begingroup$ so if I say: right, all move to the right cell? $\endgroup$
    – lois6b
    Commented Aug 10, 2016 at 6:04
  • $\begingroup$ To be honest, I failed to see how this kind of problems are suitable for Puzzling. This is a typical problem for a maths forum. $\endgroup$
    – rhsquared
    Commented Aug 10, 2016 at 6:05
  • $\begingroup$ Basically you could say it - but some can not move to the right. $\endgroup$
    – Moti
    Commented Aug 10, 2016 at 6:06
  • 4
    $\begingroup$ @Radoslav This may be said on many puzzles. Are you familiar with the famous 7 bridges puzzle? Are you familiar with cutting puzzles?... and I can go on and on. All encryption puzzles here are math puzzles... $\endgroup$
    – Moti
    Commented Aug 10, 2016 at 6:08
  • 6
    $\begingroup$ @Radoslav Reasons why this puzzle is better here than on math.SE: The real-life story. The reader must puzzle out what the real maths problem is, from the real-life story. An "aha!" idea is the key to the answer, but the reader must think about the puzzle in order to determine it and how to apply it to this particular puzzle. $\endgroup$
    – Rosie F
    Commented Aug 10, 2016 at 6:50

2 Answers 2


You can't.

Color them like a checkerboard - the top-left-front cell is black, and the ones adjacent to it are white, and the ones adjacent to those are black...

Each prisoner in a white cell must move to a black cell, and each prisoner in a black cell must move to a white cell. But there are more of one color than the other, since there are an odd number of cells. Therefore it's impossible.

  • 4
    $\begingroup$ I like how looking at it a certain way makes the answer clear. $\endgroup$ Commented Aug 10, 2016 at 12:20
  • $\begingroup$ This is a ridiculously elegant solution. I was thinking something about how you can't have an odd length cycle of shifting people in the described manner, so there was an odd man out. $\endgroup$ Commented Aug 10, 2016 at 20:41
  • $\begingroup$ But the puzzle never specified that everyone has to move... $\endgroup$
    – user24580
    Commented Aug 29, 2016 at 2:05

Confirmed incorrect... Ruddy "identical cubical cells"! Fair enough though...

Are all cells perfect cubes as well? I think I could solve this using a few other shapes (still forming a cube and still totaling 125) but I can't help thinking that it wouldn't be in-keeping with the spirit of the puzzle...

If each floor looked like this...

if each floor looked like this

This can probably still be

5x5x5 enter image description here

  • 1
    $\begingroup$ The first attempt does not result in a 5x5x5 structure. The second one somehow does. I like it! $\endgroup$
    – M.Herzkamp
    Commented Aug 10, 2016 at 12:31
  • $\begingroup$ Well... you may try to be smart but clearly the specifics lend to EQUAL cubes situated in a 3D structure each cell has access to single cells directly in perpendicular directions to the cell. Your first path is not relevant since it is a continuous path and not "discrete" - which forces the black/white or odd/even case. The second suggestion allows for access to two cells adjacent in same direction to a cell... $\endgroup$
    – Moti
    Commented Aug 10, 2016 at 15:52
  • 1
    $\begingroup$ @moti Without wanting to sound like I don't like this puzzle (I do - and I accept that is is mathematical and logical in nature) I'd argue that the original wording had some holes in it making it far from 'clear' that the "cells" (not originally specifically cubes) were all equal, and I'll quickly point out that a lot of the puzzles (the best ones in my opinion), here and elsewhere, hinge on unclear or unspecified information that incorages false assumption. $\endgroup$ Commented Aug 10, 2016 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.