Jushbox is a puzzle that consists of a box containing nine stacks of pieces, arranged in
a 3×3 grid. Each stack is supported underneath by a spring, very similar to coin
dispensers. Each stack can hold up six pieces. The top piece of any non-empty stack can
slide to the top of any adjacent stack, provided that adjacent stack does not already hold
6 pieces.
At the start, every stack has three pieces. They come in 3 colours (yellow, red, and
blue), and the three columns of stacks each have a different colour. You can set yourself
various objectives, such as changing the order of the coloured columns, or rearranging the
pieces so that the rows are different colours instead of the columns, or making the layers
different colours. This is generally not very difficult, so you may try to minimise the
number of moves you need for these tasks. For each stack, only the colour of the top piece
is visible, so the puzzle apparently trains your memory.
There is no simple formula for calculating the number of positions of this puzzle generally. Suppose the 27 pieces were all of the same colour. We need to find out how many ways there are to distribute 27 pieces into 9 stacks which each contain any number between 0 and 6 pieces inclusive. This can be calculated recursively. Let N(p,s) be the number of ways we can distribute p pieces into s stacks with at most 6 in any stack. The following equation then holds:
N(p,s+1) = N(p,s)+N(p-1,s)+N(p-2,s)+N(p-3,s)+N(p-4,s)+N(p-5,s)+N(p-6,s)
This counts the number of ways we to make s+1 stacks by counting all the positions that are possible before we added the 0 to 6 pieces that form the last stack. Using the trivial boundary conditions that N(0,0)=1 and N(p,0)=0 for non-zero values of p we can then find the value of N(27,9) by building the following table:
Stacks | ||||||||||
---|---|---|---|---|---|---|---|---|---|---|
pcs | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
2 | 0 | 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 | 45 |
3 | 0 | 1 | 4 | 10 | 20 | 35 | 56 | 84 | 120 | 165 |
4 | 0 | 1 | 5 | 15 | 35 | 70 | 126 | 210 | 330 | 495 |
5 | 0 | 1 | 6 | 21 | 56 | 126 | 252 | 462 | 792 | 1,287 |
6 | 0 | 1 | 7 | 28 | 84 | 210 | 462 | 924 | 1,716 | 3,003 |
7 | 0 | 0 | 6 | 33 | 116 | 325 | 786 | 1,709 | 3,424 | 6,426 |
8 | 0 | 0 | 5 | 36 | 149 | 470 | 1,251 | 2,954 | 6,371 | 12,789 |
9 | 0 | 0 | 4 | 37 | 180 | 640 | 1,876 | 4,809 | 11,152 | 23,905 |
10 | 0 | 0 | 3 | 36 | 206 | 826 | 2,667 | 7,420 | 18,488 | 42,273 |
11 | 0 | 0 | 2 | 33 | 224 | 1,015 | 3,612 | 10,906 | 29,184 | 71,127 |
12 | 0 | 0 | 1 | 28 | 231 | 1,190 | 4,676 | 15,330 | 44,052 | 114,387 |
13 | 0 | 0 | 0 | 21 | 224 | 1,330 | 5,796 | 20,664 | 63,792 | 176,463 |
14 | 0 | 0 | 0 | 15 | 206 | 1,420 | 6,891 | 26,769 | 88,852 | 261,891 |
15 | 0 | 0 | 0 | 10 | 180 | 1,451 | 7,872 | 33,390 | 119,288 | 374,808 |
16 | 0 | 0 | 0 | 6 | 149 | 1,420 | 8,652 | 40,166 | 154,645 | 518,301 |
17 | 0 | 0 | 0 | 3 | 116 | 1,330 | 9,156 | 46,655 | 193,880 | 693,693 |
18 | 0 | 0 | 0 | 1 | 84 | 1,190 | 9,331 | 52,374 | 235,348 | 899,857 |
19 | 0 | 0 | 0 | 0 | 56 | 1,015 | 9,156 | 56,854 | 276,872 | 1,132,677 |
20 | 0 | 0 | 0 | 0 | 35 | 826 | 8,652 | 59,710 | 315,918 | 1,384,803 |
21 | 0 | 0 | 0 | 0 | 20 | 640 | 7,872 | 60,691 | 349,840 | 1,645,791 |
22 | 0 | 0 | 0 | 0 | 10 | 470 | 6,891 | 59,710 | 376,160 | 1,902,663 |
23 | 0 | 0 | 0 | 0 | 4 | 325 | 5,796 | 56,854 | 392,848 | 2,140,866 |
24 | 0 | 0 | 0 | 0 | 1 | 210 | 4,676 | 52,374 | 398,567 | 2,345,553 |
25 | 0 | 0 | 0 | 0 | 0 | 126 | 3,612 | 46,655 | 392,848 | 2,503,053 |
26 | 0 | 0 | 0 | 0 | 0 | 70 | 2,667 | 40,166 | 376,160 | 2,602,341 |
27 | 0 | 0 | 0 | 0 | 0 | 35 | 1,876 | 33,390 | 349,840 | 2,636,263 |
Each number in the table is the sum of the number directly to its left and the six numbers above that. So there turn out to be 2,636,263 configurations for the 27 pieces in the box, ignoring colours.
There are 27 pieces, 9 of each colour, so for each of the stack configurations there are 27!/9!3 possible arrangements of the colours. This gives a total of 2,636,263 · 27!/9!3 = 600,734,296,146,484,500 positions.
This puzzle is easy to solve. You merely have to solve the bottom layer of a stack first, and only then the middle layer and finally top layer. This is done by emptying the stack, and then putting the pieces in to the empty stack in the correct order. Any solved stacks can still be used as repositories for spare pieces.
The number of positions of the puzzle as a whole is too large to calculate optimal solutions for, but it is easy to do so for three stacks that form a row. This gives some fairly short solutions for some of the transformations between nice positions. Some of these are shown below.
Swapping two adjacent columns
You can swap two adjacent stacks in 12 moves. This is very straightforward: Move all three
middle pieces of a row to the right, move all three left pieces to the middle, and then the first three
pieces from the right all the way to the left. This swaps the left two stacks of a row.
You can do this once in each row to swap left two adjacent columns in 36 moves.
Swapping two adjacent stacks cannot be done in fewer than 12 moves if you only do moves within
one row. It may be that using moves between rows can reduce the number of
moves to fewer than 36, but I don't believe that to be the case.
Cyclically permuting all three columns
Simply use the method above to swap the left two columns, and then repeat it in
mirror image to swap the right two columns.
Again, if you only allow moves within each row, then the 72 moves cannot be reduced. This
time it is definitely possible to shorten the move sequence by using moves between rows, as
follows:
This uses a total of 18 moves instead of 24, giving a total of 54 moves to permute the three columns. I do not know if this can be improved further.
Swapping the two outer (non-adjacent) columns
If you only use moves within one row, then it takes 22 moves to swap the two outer
stacks, so 66 moves to swap the two outer columns. Here are the 22 moves.
Again however, you can do this more easily and in fewer moves by temporarily putting one piece in another row.
This uses a total of 14 moves instead of 22, giving a total of 42 moves to swap the outer columns. I do not know if this can be improved further.
Changing three columns into three layers
If you only use moves within one row, then it takes 20 moves to rearrange the 9
pieces into three layers, giving 60 moves to make layers in the whole box. Here are the 20 moves:
Again however, you can do this more easily and in fewer moves by temporarily putting one piece in another row.
This uses a total of 18 moves instead of 20, giving a total of 54 moves to change columns to layers. I do not know if this can be improved further.