Abstract
The Soma cube is a puzzle that fills a $3\times3\times3$ cube with one piece of three cubes and six pieces of four cubes. This paper counts all solutions, counts how each piece covers the two colours when the cube is coloured like a chessboard, and compares the combinations the colour count allows with those that actually have solutions. No new theorem or law is claimed.
First, there are 11520 solutions counting orientation and position, 240 up to rotation and reflection, and none is its own image: up to rotation alone there are $480$, and each of the $240$ classes up to rotation and reflection consists of $48$ solutions (Section 2).
Second, only the V, T and P pieces carry a colour difference: colouring $14$ even cells and $13$ odd cells, L, Z, A and B always cover $2$ of each colour, V has difference $1$, and T and P have difference $2$ (Section 3).
Third, the colour count allows two combinations: the total difference $+1$ can only be made with V at $+1$ and T and P splitting $+2$ and $-2$ in opposite directions, two ways in all (Section 3).
Fourth, all solutions lie in one combination and none in the other: all $11520$ solutions have T at $+2$ and P at $-2$; searching with T at $-2$ and P at $+2$ examined $140415$ nodes and found $0$ solutions. Restricted to the combination with solutions, the search used $\mathbf{0.395194}$ of the nodes of the unrestricted search (Section 3).
Fifth, the separator is whether one counts colour differences or actually searches by fitting the pieces: the colour count narrows the possible combinations to two but does not say which has solutions. That was settled by actually fitting the pieces in a search (Section 4).
Keywords: Soma cube, polycube, chessboard colouring, packing puzzle, symmetry, exhaustive search.
Scope (what is not claimed)
No new theorem or law is claimed: the seven Soma pieces, the fact that there are $240$ solutions ($11520$ counting orientation and position), and the idea of narrowing placements by chessboard colouring are all known. No reason is claimed: that the combination with T at $-2$ and P at $+2$ has no solution was only counted by search, and no explanation is given. Other shapes are not treated: building shapes other than $3\times3\times3$ and other sets of pieces are not treated. No method of solving is recommended: how people solve it and how hard it is are not discussed. Relation to earlier papers: Paper 925 showed that in peg solitaire an invariant sum narrows where the last peg can remain to five holes and that the peg can actually be left in all five. Here, of the two combinations the colour count allows, only one actually has solutions. Paper 847 showed that the count of shapes changes with whether reflections are identified. Here the number of solutions changes from $11520$ with everything distinguished to $480$ up to rotation and $240$ up to reflection as well. What is added: counting all solutions and finding the classes under symmetry and their sizes, counting the colour difference of each piece, listing the two combinations the colour count allows and separating, by search, the one with solutions from the one without, comparing the number of search nodes, and placing the separator on “counting colour differences, or actually fitting the pieces”.
1. Notation and the Question
Pieces: V is three cubes in an L shape, L, T and Z are four cubes in a plane, and A, B and P are V with one cube placed on top. A and B coincide in a mirror and not under rotation. Shapes of four cubes number $8$ when distinguished up to rotation and $7$ when reflections are identified; L, T, Z, A, B and P are the $8$ minus the straight line and the square (confirmed in the check).
Colour: colouring the $3\times3\times3$ cells by the parity of the coordinate sum gives $14$ even cells ($8$ corners and $6$ face centres) and $13$ odd cells ($12$ edge midpoints and the centre). The colour difference of a piece is “even cells covered minus odd cells covered”. Summed over all pieces it is $14-13=+1$.
Search: all placements of each piece are generated, and placements covering the lowest-numbered empty cell are tried in turn. All solutions are counted.
Question: how far does the colour count determine which combination the solutions lie in?
2. Number of Solutions and Symmetry
Computation 2.1: placements number V $144$, L $144$, T $72$, Z $72$, A $96$, B $96$ and P $64$. Counting orientation and position there were $11520$ solutions.
| identification | solutions |
|---|---|
| all distinguished | $11520$ |
| up to rotation ($24$) | $480$ |
| up to rotation and reflection ($48$) | $240$ |
Observation 2.2 no solution is its own image. Each of the $240$ classes up to rotation and reflection consists of $48$ solutions, and $11520\div48=240.000000$: a mirror swaps A and B, so a reflected solution is still a solution with the same seven pieces. Every solution is carried to $47$ others.
3. Colour Differences
Computation 3.1: how each piece covers the colours (even, odd).
| piece | coverage | colour difference |
|---|---|---|
| L, Z, A, B | only $2$, $2$ | $0$ |
| V | $2$, $1$ or $1$, $2$ | $+1$ or $-1$ |
| T | $3$, $1$ or $1$, $3$ | $+2$ or $-2$ |
| P | $3$, $1$ or $1$, $3$ | $+2$ or $-2$ |
Exactly half of the placements of V, T and P have a positive difference (V $72$ and $72$, T $36$ and $36$, P $32$ and $32$).
Observation 3.2 the colour count allows two combinations. To make the differences of V, T and P add to $+1$, V must be $+1$, and T and P split $+2$ and $-2$ in opposite directions: the allowed combinations are (V, T, P) $=$ ($+1$, $+2$, $-2$) and ($+1$, $-2$, $+2$).
Computation 3.3: combinations in the solutions, and searches restricted by combination.
| search | solutions | nodes examined | nodes relative to unrestricted |
|---|---|---|---|
| unrestricted | $11520$ | $861285$ | $1$ |
| V $+1$, T $+2$, P $-2$ | $11520$ | $340375$ | $\mathbf{0.395194}$ |
| V $+1$, T $-2$, P $+2$ | $0$ | $140415$ | $0.163030$ |
Observation 3.4 all solutions lie in one combination and none in the other. Looking at the $11520$ solutions one by one, every combination was ($+1$, $+2$, $-2$): the colour count alone also allows the other combination, but a search restricted to it examined $140415$ nodes and found $0$ solutions.
4. The Separator
Observation 4.1 the separator is whether one counts colour differences or actually searches by fitting the pieces. Counting colour differences fixes the direction of V and narrows T and P to two ways, which of them has solutions does not follow from the colour count; only fitting the pieces in a search showed $11520$ in one and $0$ in the other.
Observation 4.2 what is not said. The reason the combination with T at $-2$ has no solution, shapes other than $3\times3\times3$, and how people solve the puzzle are not treated. What is shown is that of the two combinations the colour count allows, only one actually has solutions.
5. Roll Call and Closing
- seven pieces and chessboard colour difference (Section 1): the total is $+1$.
- $11520$, $480$ and $240$ solutions (Computation 2.1): no solution is its own image.
- only V, T and P carry a difference (Computation 3.1): $\pm1$ and $\pm2$.
- two combinations allowed (Observation 3.2): T and P in opposite directions.
- $11520$ in one, $0$ in the other (Computation 3.3): node ratio $\mathbf{0.395194}$.
- colour difference, or actual fitting (Observation 4.1): the separator; Paper 925; Paper 847.
Closing: Soma cube solutions lie in only one of the two arrangements the colour count allows. There were $11520$ solutions counting orientation and position and $240$ up to rotation and reflection, and none was its own image: only V, T and P carry a chessboard colour difference, and two combinations make $+1$ with T and P in opposite directions, yet every solution had T at $+2$ and P at $-2$, and a search restricted to the other combination found $0$ solutions. The separator is whether one counts colour differences or actually searches by fitting the pieces. Placed among the earlier papers: in Paper 925 every place an invariant sum allowed was actually reached, whereas here half of the combinations the colour count allows had no solution. The “count changes with identification” of Paper 847 appeared as $11520$, $480$ and $240$. To be honest: the reason for the empty combination is not given. How far the colour count and the actual search agree, and where they part, and no further.
References
- Berlekamp, E. R., Conway, J. H., & Guy, R. K. (1982). Winning Ways for Your Mathematical Plays. Academic Press.
- Yamagishi, Y. (2026). “The Same Shape” Does Not Mean One Thing: The Count Changes With Whether Reflections Are Identified [Paper 847]. YOI: https://www.yamagishiyuuki.com/yoi/847
- Yamagishi, Y. (2026). In Peg Solitaire an Invariant Sum Narrows Where the Last Peg Can Remain to Five Holes, and It Can Actually Be Left in All Five [Paper 925]. YOI: https://www.yamagishiyuuki.com/yoi/925
Reference
Yamagishi, Y. (2026). Soma Cube Solutions Lie in Only One of the Two Arrangements the Colour Count Allows [Paper 944]. YOI: https://www.yamagishiyuuki.com/yoi/944
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 944. This paper is a preprint and has not been peer reviewed.