Two pentacube problems with S3 symmetry.           Torsten Sillke, 1999-01-22

They are based on the equation 145 = (11 choose 3) - (6 choose 3).
This gives a tetrahedron of length 9 minus a tetrahedron of length 4.


level 1     2          3         4        5       6      7     8    9 

* 
**          * 
***         **         * 
****        ***        **        * 
*****       * **       ***       **       * 
******      *  **      * **      ***      **      * 
*******     *   **     *  **     * **     ***     **     * 
********    *    **    *   **    *  **    * **    ***    **    * 
*********   ********   *******   ******   *****   ****   ***   **   * 


level 1     2          3         4        5       6      7     8    9 

* 
**          * 
***         **         * 
****        ***        **        * 
*****       ****       ***       **       * 
 *****      *****      ****      ***      **      * 
  *****      *****     *****     ****     ***     **     * 
   *****      *****     *****    *****    ****    ***    **    * 
    *****      *****     *****    *****   *****   ****   ***   **   * 

These problems where created by Michael Reid.


Solution:
We will ignore stability aspects in the solution.

In both figures we can trucate two tetrahedra of length 4.
Checkering the cubes shows that a 4:1 pentacube (50 or 51)
must be in each tetrahedron4. Therefore no third tetrahedron4
can be truncated and packed with pentacubes.

Packing the two tetrahedron4:
The following 28 subsets of pentacubes pack a tetrahedron4.

 11 33 37 51       12 31 33 51       12 33 51 82       31 34 51 60
 11 34 37 51       12 31 51 60       12 34 35 51       31 51 60 70
 12 30 31 50       12 32 34 51       12 34 40 51       32 33 51 60
 12 30 32 50       12 32 51 60       12 34 51 60       32 51 60 70
 12 30 33 51       12 33 36 51       12 34 51 82       33 34 51 61
 12 30 34 51       12 33 40 51       30 33 51 60       33 51 60 72
 12 30 51 60       12 33 51 60       30 34 51 60       34 51 60 71

There is only one subset (ignore reflection) that uses the 50: 12, 30, 31, 50
The following seven subsets can be combined with the first one

 11 33 37 51       33 51 60 72       32 33 51 60       33 34 51 61
 11 34 37 51       34 51 60 71       32 51 60 70

I select the first pair 12, 30, 31, 50 and 11 33 37 51.

       30 
       30 50         30 
       50 50 50      30 31      30 
       12 50 31 31   12 31 31   12 12   12 

       11
       11 37         37
       11 37 33      37 37      51
       11 11 33 33   51 33 33   51 51   51

A solution for the big component for both problems
  
  .  .  .  . 81 81 21 21 22  .        .  .  .  . 20 20 70 70 10
  .  .  . 90 90 81 20 20 22  .        .  .  . 61 61 20 90 90  .
  .  . 71 90 36 81 20 22 22  .        .  . 71 61 61 20 20  .  .
  . 34 71 90 90 20 20 42 42  .        . 34 71 32 21 40  .  .  .
 34 34 71 71 10 10 10 10 10  .       34 34 71 71 21  .  .  .  .

  .  .  .  .  . 81 21  . 22  .        .  .  .  .  . 70 70 42 10
  .  .  .  . 36 36  .  . 42  .        .  .  .  . 41 41 41 90 60
  .  .  . 41 36  .  .  . 42  .        .  .  . 61 41 35 35 13  .
  .  . 71 41  .  .  .  . 42  .        .  . 71 32 21 40 35  .  .
  . 34 34 41 72 80 80 80 40  .        . 34 34 32 81 40  .  .  .

  .  .  .  .  .  . 21 82 13  .        .  .  .  .  .  . 70 42 10
  .  .  .  .  . 36 21  . 13  .        .  .  .  .  . 35 90 90 60
  .  .  .  . 72 32  .  . 13  .        .  .  .  . 41 35 72 13 60
  .  .  . 72 72  .  .  . 60  .        .  .  . 21 21 82 82 82  .
  .  . 41 41 72 70 80 40 40  .        .  . 32 32 81 40 82  .  .

  .  .  .  .  .  .  . 82 13  .        .  .  .  .  .  .  . 42 10
  .  .  .  .  .  . 82 82 82  .        .  .  .  .  .  . 42 42 60
  .  .  .  .  . 32 32  . 61  .        .  .  .  .  . 72 72 13 60
  .  .  .  . 35 35  .  . 60  .        .  .  .  . 81 36 36 13 22
  .  .  . 70 70 70 80 40 60  .        .  .  . 81 81 40 82 13  .

  .  .  .  .  .  .  .  . 13  .        .  .  .  .  .  .  .  . 10
  .  .  .  .  .  .  . 61 61  .        .  .  .  .  .  .  . 22 22
  .  .  .  .  .  . 32 61 61  .        .  .  .  .  .  . 72 80 22
  .  .  .  .  . 35 32  . 60  .        .  .  .  .  . 36 72 80 22
  .  .  .  . 70 35 35 40 60  .        .  .  .  . 36 36 80 80 80

There are 1085 solutions for the big component of the first problem.

Annotation:
Checkerboard coloring is useful for the second problem.
The pentacubes have an excess of only one for this figure.
That means the figure has a black:white ratio of 88:57 
while the piece can deliver 89:56. Therefore all upto one
3:2 piece must fit the maximal possible number of black cubes.
This saves much computation time.