Puzzle
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Ponder This Challenge - October 2022 - Numbers in a cube

The following puzzle was suggested by Evert van Dijken: Consider a n×n×nn\times n\times n cube. It has 3n23n^2 rows, 6n6n diagonals from one edge to the opposite edge, and 4 corner-to-corner diagonals.

To each 1×1×11\times1\times 1 cube, we assign one of the numbers 1…n1\dots n. The big cube does not need to have the same number of 1's, 2's.., or n's. Given such a numbering, we count the number of rows and diagonals that contain all the numbers 1…n1\ldots n at once. As an example, consider for n=3n=3 the numbering given by

1 3 2
2 1 3
3 2 1

2 1 3
3 2 1
1 3 2

3 2 1
1 3 2
2 1 3

There are 27 rows, each containing all the numbers 1, 2, and 3. To illustrate, we mark three such rows, each in a different axis, using different bracket styles:

[1] [3] [2]
 2   1   3
 3   2  {1}

(2)  1   3
(3)  2   1
(1)  3  {2}

 3   2   1
 1   3   2
 2   1  {3}

There are 4 corner-to-corner diagonals, three of which contain 1, 2, 3 and one containing only 2:

[1]  3   {2}
 2   1    3
 3   2    1

 2   1    3
 3 {[2]}  1
 1   3    2

 3   2    1
 1   3    2
{2}  1   [3]

And 18 edge diagonals, 9 of which contain 1, 2, and 3, while the other 9 do not:

[1] {3}   2
 2   1    3
 3   2    1

 2  [1]   3
 3  {2}   1
 1   3    2

(3)   2   [1]
 1  (3)    2
 2  {1}   (3)

The total count is 27+9+3=39. We can compactly represent this solution by the string

1 3 2
2 1 3
3 2 1
2 1 3
3 2 1
1 3 2
3 2 1
1 3 2
2 1 3
39

(Note the count in the final line).

Your goal: Find a solution of at least 110 forn=6n=6. Supply your solution in the above format.

A bonus "*" will be given for getting at least 60 solving the case n=5n=5 when maximizing the count over diagonals (both edge and corner) first. A solution with more diagonals is always better, even if the final count is lower, but for the best diagonal count we still wish to maximize the total count. Supply your solution in the same format as above (no need to explicitly write down the diagonal count, only the total count).

Solution

  • October 2022 Solution:

    The best solution (132) was found by many solvers. Here is the last one submitted, by Radu-Alexandru Todor:
    4 2 6 3 5 1
    5 6 1 4 3 2
    3 1 5 2 4 6
    2 3 4 1 6 5
    1 5 3 6 2 4
    6 4 2 5 1 3

    5 3 1 4 6 2
    6 4 2 5 1 3
    1 5 6 3 2 4
    3 1 5 2 4 6
    2 6 4 1 3 5
    4 2 3 6 5 1

    6 1 5 2 4 3
    1 5 3 6 2 4
    2 6 4 1 3 5
    4 2 6 3 5 1
    3 4 2 5 1 6
    5 3 1 4 6 2

    3 4 2 5 1 6
    4 2 6 3 5 1
    5 3 1 4 6 2
    1 5 3 6 2 4
    6 1 5 2 4 3
    2 6 4 1 3 5

    2 6 4 1 3 5
    3 1 5 2 4 6
    4 2 3 6 5 1
    6 4 2 5 1 3
    5 3 1 4 6 2
    1 5 6 3 2 4

    1 5 3 6 2 4
    2 3 4 1 6 5
    6 4 2 5 1 3
    5 6 1 4 3 2
    4 2 6 3 5 1
    3 1 5 2 4 6
    132

    In the bonus part, all 34 diagonals can be achieved, and given this 84 is the maximal solution, given by
    0 0 0 0 0
    1 1 1 1 1
    2 2 2 2 2
    3 3 3 3 3
    4 4 4 4 4

    4 4 4 4 4
    2 2 2 2 2
    3 3 3 3 3
    0 0 0 0 0
    1 1 1 1 1

    3 3 3 3 3
    4 4 4 4 4
    1 1 1 1 1
    2 2 2 2 2
    0 0 0 0 0

    1 1 1 1 1
    3 3 3 3 3
    0 0 0 0 0
    4 4 4 4 4
    2 2 2 2 2

    2 2 2 2 2
    0 0 0 0 0
    4 4 4 4 4
    1 1 1 1 1
    3 3 3 3 3
    84

Solvers

  • Paul Revenant (6/10/2022 12:48 PM IDT)
  • *Alper Halbutogullari (8/10/2022 12:50 AM IDT)
  • Alex Fleischer (8/10/2022 9:46 AM IDT)
  • Fabio Michele Negroni (8/10/2022 3:16 PM IDT)
  • *Bert Dobbelaere (9/10/2022 5:49 PM IDT)
  • *Evert van Dijken (10/10/2022 1:46 PM IDT)
  • Eran Vered (10/10/2022 5:21 PM IDT)
  • *Sanandan Swaminathan (10/10/2022 10:06 PM IDT)
  • Gary M. Gerken (11/10/2022 6:20 AM IDT)
  • Graham Hemsley (11/10/2022 6:33 PM IDT)
  • Daniel Bitin (11/10/2022 11:33 PM IDT)
  • Victor Chang (12/10/2022 5:56 AM IDT)
  • *Dieter Beckerle (12/10/2022 10:47 AM IDT)
  • *Lorenz Reichel (12/10/2022 4:22 PM IDT)
  • *Lazar Ilic (12/10/2022 7:28 PM IDT)
  • *Tim Walters (13/10/2022 3:27 AM IDT)
  • Stéphane Higueret (13/10/2022 9:41 AM IDT)
  • Jonathan Evans (13/10/2022 3:15 PM IDT)
  • Lorenzo Gianferrari Pini (13/10/2022 5:29 PM IDT)
  • *Frank Schoeps (13/10/2022 9:35 PM IDT)
  • *Bertram Felgenhauer (13/10/2022 11:32 PM IDT)
  • John Tromp (13/10/2022 11:50 PM IDT)
  • *Vladimir Volevich (14/10/2022 2:42 PM IDT)
  • *Jack Morris (15/10/2022 5:26 AM IDT)
  • *David Greer (15/10/2022 3:06 PM IDT)
  • *Dominik Reichl (15/10/2022 4:32 PM IDT)
  • Sri Mallikarjun J (15/10/2022 5:02 PM IDT)
  • Phil Proudman (16/10/2022 8:29 PM IDT)
  • Aditya Dalal (19/10/2022 6:18 AM IDT)
  • Shouky Dan & Tamir Ganor (19/10/2022 8:31 AM IDT)
  • Amos Guler (19/10/2022 10:53 AM IDT)
  • Nyles Heise (19/10/2022 7:38 PM IDT)
  • Andy Greig (20/10/2022 2:40 PM IDT)
  • Prashant Wankhede (21/10/2022 3:43 AM IDT)
  • *Latchezar Christov (21/10/2022 12:08 PM IDT)
  • *Li Li (23/10/2022 6:56 AM IDT)
  • Karl Mahlburg (24/10/2022 5:30 AM IDT)
  • Daniel Chong Jyh Tar (24/10/2022 3:45 PM IDT)
  • Dan Dima (24/10/2022 7:09 PM IDT)
  • Andreas Puccio (25/10/2022 12:52 AM IDT)
  • Kang Jin Cho (25/10/2022 8:08 PM IDT)
  • Maksym Voznyy (27/10/2022 11:33 PM IDT)
  • Jesus Vergara Temprado (28/10/2022 4:27 PM IDT)
  • *Chris Shannon (29/10/2022 1:15 AM IDT)
  • *Motty Porat (29/10/2022 3:37 AM IDT)
  • Light Bulb (29/10/2022 11:58 AM IDT)
  • Patrik Holopainen (29/10/2022 4:05 PM IDT)
  • Ali Gurel (30/10/2022 5:02 AM IDT)
  • Vardhan Walavalkar (30/10/2022 7:14 AM IDT)
  • Marco Bellocchi (31/10/2022 1:45 AM IDT)
  • David F.H. Dunkley (1/11/2022 12:20 AM IDT)
  • Oscar Volpatti (1/11/2022 3:48 AM IDT)
  • Harald Bögeholz (1/11/2022 1:39 PM IDT)
  • Radu-Alexandru Todor (4/11/2022 3:05 AM IDT)

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