Shrinking Somas

Introduction
4 x 4 x 4 Puzzle Solution
Other Shrinking Soma Puzzles
5 x 5 x 5 Puzzle
6 x 6 x 6 Puzzle
K'nex Enclosing Boxes
Shrinking Soma Gallery
Return to SomaSite

Introduction

The "Shrinking Soma" (perhaps better called an expanding Soma) is a puzzle invented by Volker Latussek of the University of Würzburg in Germany that consists of a standard Soma set combined with a 5-sided enclosing box that provides a seemingly impossible representation of a larger cube. The canonical version of this illusion is a single Soma set arranged inside a 4 x 4 x 4 cubelet box such that all of the following are satisfied:

A flat surface of 16 cubelets is presented at the top of the box.
Soma pieces in the top level can be moved upwards for removal, but not in any other direction.
Soma pieces in lower levels cannot move in any of the six directions (up, down, left, right, front, back).

These criteria are referred to as the "anti-slide" requirement wherein covering the top of the box and shaking it vigorously in any direction will not dislodge a piece. If the 64-cubelet enclosing box is opaque, then the 27-cubelet Soma set has the appearance of having magically grown to fill it completely.

Inverted & Mirrored Puzzles

When solving Shrinking Soma puzzles, it's often convenient to work on it upside down, where the base of the puzzle consists of the cubelets that will end up in the top level of the enclosing box. Pieces are added so that they will be locked into position when the enclosing sides are present, as well as providing support for other pieces that are temporarily below them. When complete, the arrangement of Soma pieces can be put into its enclosing box all at once simply by inverting the box over the pieces and then turning it right side up while holding the pieces in place within the open side of the box.

In some cases, it can be useful to consider the mirror image of a Shrinking Soma solution, which is created by exchanging the 3 and 9 pieces and reversing the rows and columns in the solution.

Analysis of the 4 x 4 x 4 Puzzle Solution

Puzzle Guy has found a solution to this puzzle by working on it in its inverted form. An essential element of the solution is the combination of two pieces which forms a 2 x 2 support at Level 3 for portions of pieces in the top layer. The only combinations of two pieces that provide a supported 2 x 2 platform at Level 3 are T + V or Z + V, in each case with the V piece flat on Level 3. Each combination consumes 7 cubelets, which together with the 16 cubelets in the top level leaves 27 - 7 - 16 = 4 cubelets that protrude down from Level 4 and can lock the 2 x 2 support in place. The Y piece accounts for one of these protuberances, so there are three cubelets left to be apportioned amongst the 3, the 9 and two flat pieces. Therefore, the 3 and 9 pieces cannot both have a 2-cubelet protuberance (which supports another piece), but each has at least 1 cubelet protruding since it is non-flat. Now the top layer has 5 pieces and at most 4 of them can be supported by the 2 x 2 square, so one of them is supported by a 3 or 9 piece, which itself is supported by the square, and the square supports a total of 4 pieces.

The problem has been simplified into finding 4 cubelets in Level 3 that are adjacent to a 2 x 2 square and prevent it from moving laterally (together with the walls of the enclosing box). The two-piece support can be oriented so that a single cubelet on each side of the 2 x 2 square will suffice to block it. The only restriction is that a blocking cubelet or a wall must abut the T or Z piece on two sides of the square so that the piece cannot move outside of the square. The 2 x 2 square cannot exist in any corner of the 4 x 4 lateral array since that would require one of its four supported pieces in Layer 4 to be an isolated cubelet.

It was previously established that the cubelets protruding down into Layer 3 consist of a row of two cubelets as part of the 3 or 9 piece, plus two single cubelets, one from the Y piece and the other from the remaining 3 or 9 piece. It follows that only three sides of the square can be blocked. Therefore, and by symmetry, the only 2 x 2 square that needs to be considered is one centered along a wall. Identify cubelet positions in Layer 3 as follows:

By excluding the T + V or Z + V combination, arrangements of the remaining 20 cubelets in the top layer and the three blocks can each be tested for feasibility. Due to symmetry, only the following possibilities need to be examined:

Blocks Results

If T + V used for support If Z + V used for support

AB D F Impossible Impossible

AB D G 1 discarded solution Impossible

AB E F Impossible Impossible

AB E G Impossible 2 discarded solutions

BC D F Impossible 1 discarded solution

BC D G Impossible Impossible

BC E F 1 discarded solution Impossible

BC E G Impossible Impossible

CD A F * Impossible

CD A G Impossible 5 discarded solutions

CD B F Impossible 1 discarded solution

CD B G Impossible Impossible

DE A F 2 discarded solutions Impossible

DE A G Impossible Impossible

Blocks	Results
If T + V used for support	If Z + V used for support
AB D F	Impossible	Impossible
AB D G	1 discarded solution	Impossible
AB E F	Impossible	Impossible
AB E G	Impossible	2 discarded solutions

BC D F	Impossible	1 discarded solution
BC D G	Impossible	Impossible
BC E F	1 discarded solution	Impossible
BC E G	Impossible	Impossible

CD A F	*	Impossible
CD A G	Impossible	5 discarded solutions
CD B F	Impossible	1 discarded solution
CD B G	Impossible	Impossible

DE A F	2 discarded solutions	Impossible
DE A G	Impossible	Impossible

Discarded solutions have one or more unsupported pieces and so violate the anti-slide requirement. In the results indicated by *, there are four solutions, three of which have unsupported pieces. The one remaining possibility has the L piece supported by the 3 piece, which in turn is supported by the square. Only the T + V combination is used for the support. The vertical "trunk" of the T(ree) piece is against the wall and its "branch" in Layer 2 points in one of two directions. Therefore, under the assumption of a 2 x 2 square support in Layer 3, there are exactly two solutions to the 4 x 4 x 4 Shrinking Soma puzzle, both identical except for the orientation of the supporting T piece. These solutions are shown here and here.

Other Shrinking Soma Puzzles

Many other versions of this puzzle are possible if we allow rectangular boxes (cuboids) with different dimensions. Some of these puzzles are shown in the Shrinking Soma gallery. Generally speaking, the difficulty of the puzzle increases with the ratio of its box volume to that of the Soma set(s) used to solve it. In other words, the harder a puzzle, the lower its cubelet occupancy. For example, the original puzzle in the preceding section has an occupancy of 27/4³ = 42%. Obviously, any puzzle with an occupancy below 50% has more space than cubelets inside its enclosing box. Also, any 1-set solution can have its height doubled and a solution found with 2 Soma sets simply by inserting mirror images of the 1-set solution inside the larger box, such as in this example. However, a more efficient use of pieces can usually be found.

What Shrinking Soma cuboids can be created from the seven pieces in one Soma set? Obviously, the cuboid volume RCL ≥ 27 for any single-set puzzle. Since the Soma set contains non-flat pieces, the dimensions of the puzzle each are at least 2, and so the top-level area RC is a composite number. The Soma pieces are all non-convex polycubes, so each piece in the top level either has at least one of its cubelets in lower levels, or is wholly contained in the top level if it is a flat piece. In the latter case, the piece requires support in the next level from at least one cubelet in another piece. Therefore, the maximum area of the top level is the volume of the Soma set, less one cubelet for each piece: 27 - 7 = 20.

This requirement can be generalized to A cubelets in any level of a Shrinking Soma puzzle. The minimum number of additional cubelets f(A) that are needed for support in the level immediately below it is the minimum number of Soma pieces that comprise the A cubelets. The support cubelets can be tabulated as follows:

Cubelets A Minimum pieces f(A) Description

1-4 1 1 four-cubelet flat piece

5-8 2 2 four-cubelet flat pieces

9-12 3 3 four-cubelet flat pieces

13-15 4 All 4 flat pieces

16-18 5 All 4 flat pieces plus 1 non-flat piece

19-20 6 All 4 flat pieces plus 2 non-flat pieces

Each level in turn is supported by cubelets in the level below it, so the total minimum number of cubelets in the upper region of the puzzle is

A + f(A) + f(f(A)) + f(f(f(A))) + ...,

where the additions are repeated until the number of support cubelets in the last level is reduced to 1, or until all 27 cubelets in the Soma set are exhausted. This process is best illustrated by an example, say a puzzle with R = 2 rows and C = 8 columns. We have A = RC = 16 and

Level Cubelets Cubelets remaining Minimum support pieces

L (top) 16 27 - 16 = 11 f(16) = 5

L-1 5 11 - 5 = 6 f(5) = 2

L-2 2 6 - 2 = 4 f(2) = 1

L-3 1

Therefore, the upper portion of this puzzle consists of three levels comprising at least 23 cubelets. At most four cubelets are available for lower levels.

Turning our attention to the lower portion of the puzzle, what theoretical maximum height can be achieved with a given number of cubelets? The maximum height of a single Soma piece is 3 cubelets, achieved with the L, Z or T piece. The anti-slide criterion requires that each Soma piece be blocked laterally by at least one additional cubelet that is not part of the column of three cubelets. Therefore, each increase of 3 levels in a vacant area of the bounding box consumes at least 5 cubelets: 4 in the Soma piece plus a blocking cubelet. Any individual cubelets left over when all multiples of 5 are used could contribute to 1 or 2 more levels if they were part of an L piece, for example. These results are consolidated into the following table:

Cubelets c Levels L

1 1

2-4 2

5 3

6 4

7-9 5

10 6

... ...

and can be summarized with the mathematical expression

L = 3⌊c/5⌋ + min{(c mod 5),2},

where ⌊ ⌋ represents the floor function, and modulo arithmetic is used.

The results for the upper and lower regions of Shrinking Soma puzzles can be combined to define theoretical limits on the height (number of levels) in a puzzle with a given horizontal top-level area A:

R C Area A Height (L) range Utilization range

2 2 4 7 - 15 45% - 96%

2 3 6 5 - 13 35% - 90%

2 4 8 4 - 13 26% - 84%

3 3 9 3 - 11 27% - 100%

2 5 10 3 - 10 27% - 90%

2 6 12 3 - 10 23% - 75%

3 4

2 7 14 2 - 7 28% - 96%

3 5 15 2 - 7 26% - 90%

2 8 16 2 - 5 34% - 84%

4 4

2 9 18 2 - 5 30% - 75%

3 6

2 10 20 2 68%

4 5

R	C	Area A	Height (L) range	Utilization range
2	2	4	7 - 15	45% - 96%
2	3	6	5 - 13	35% - 90%
2	4	8	4 - 13	26% - 84%
3	3	9	3 - 11	27% - 100%
2	5	10	3 - 10	27% - 90%
2	6	12	3 - 10	23% - 75%
3	4
2	7	14	2 - 7	28% - 96%
3	5	15	2 - 7	26% - 90%
2	8	16	2 - 5	34% - 84%
4	4
2	9	18	2 - 5	30% - 75%
3	6
2	10	20	2	68%
4	5

5 x 5 x 5 Shrinking Soma Solution

A larger version of the original Shrinking Soma puzzle can be created using a 5 x 5 x 5 cubical box and 2 sets of Soma pieces, having an occupancy of 2(27)/5³ = 43%. The solution to this puzzle is derived from that of the 4 x 4 x 4 puzzle. The basic idea is to add a larger layer of cubelets from the second Soma set on top of the stable platform formed from the smaller puzzle. There are two challenges to overcome: 1) ensuring that new pieces are added around the periphery of the smaller puzzle to prevent it from moving around in the larger bounding box, and 2) coming up with enough cubelets to cover the 5 x 5 top level. In the following description, the puzzle is assumed to be in its inverted form for assembly, prior to inserting it into its bounding box and rotating.

To address the first challenge, it is easier to orient the 4 x 4 x 4 solution so that the wall adjacent to its T + V supporting structure is shared with the larger enclosing box. In that case, the number of blocking cubelets required in Level 4 is minimized when the fifth row and column are added opposite and to the right of the supporting structure. Specifically, there is one cubelet needed to block any of the three linear cubelets of the L piece, and one cubelet to block either of the two edge cubelets of the 3 piece. All other pieces of the 4 x 4 x 4 solution are blocked by these two pieces or by the boundary of the enclosing box:

The non-flat pieces in the second Soma set can be positioned to meet these blocking requirements, as long as each has at least one of its cubelets supported by the 4 x 4 platform in Level 4. The following allocation of pieces in the inverted top level produces the best results (the caret ^ denotes a raised cubelet in the inverted arrangement):

The Y piece is not required to block the 4 x 4 x 4 puzzle solution in its original form, but its raised cubelet will be needed as a block later.

Next, we'll look at filling out the 16 cubelets in the rest of the top level. The remaining four pieces of the second Soma set (L, Z, T and V) are all flat but together consist of only 15 cubelets, so we'll need to take one from the smaller puzzle solution to make up the shortage. This is challenging to remove without creating violations of the anti-slide requirement. After some trial and error, it was found that the L piece in the 4 x 4 x 4 solution could be replaced with the V piece from the second Soma set, creating a void next to the Z piece, shown with an asterisk in the diagram above. Since this location is not used for support by any of the periperal pieces in the preceding diagram, the void should not cause any problems with finding a solution. However, the void does allow the Z piece in Level 4 to move, which is why the Y piece in Level 5 is needed in the orientation shown above in order to block it. As a result, we have traded a V piece for an L piece and now have the required 16 cubelets needed for the rest of the top level. A solution to this irregular shape can be found using the two L pieces, the Z piece and the T piece. The solution for the inverted top level is shown here. It is then a simple matter to overlay it with the adjusted 4 x 4 x 4 solution to complete the inverted 5 x 5 x 5 puzzle and its normal orientation.

6 x 6 x 6 Shrinking Soma Solution

We can also find a solution to an even larger Shrinking Soma puzzle consisting of a cubical 6 x 6 x 6 box. The solution to this puzzle uses 4 sets of Soma pieces and is assembled by joining a solution of the 1-set 3 x 3 x 6 puzzle with its mirror images. Therefore, it has the same occupancy as the smaller puzzle: 4(27)/6³ = 50%. The mirror images ensure that the boundary pieces match up between individual solutions and the anti-slide requirement is maintained. The solution is shown here and has the general appearance of an abstract table with splayed legs.

K'nex Enclosing Boxes

If you have access to the K'nex building toy, it's convenient to create variously-sized enclosing boxes for Shrinking Soma puzzles. The basic K'nex parts consist of straight rods and angled connectors whose lengths are in a geometric progression with successive ratios of

; for example, the shortest (green) rod plus its connector ends is multiplied by

to get the next longer rod-connector combination (white) and so on. The various dimensions needed for the Shrinking Soma boxes can be fairly well approximated by multiples of these basic parts. To provide more accuracy, we have used half-white connections that are created from the smallest green rods oriented diagonally. For Soma pieces having 1-inch cubelets such as the ones here, the best-fitting dimensions are as follows:

Lateral cubelets Greens Whites Piece fit

2 1 0.5 Loose

3 0 1.5 Snug

4 0 2 Normal

5 0 2.5 Loose

6 0 3 Loose

7 2 2 Normal

Vertical cubelets Greens Whites

3 2 0

4 2 0.5

5 2 1

8 1 3

Here is a list of K'nex boxes with piece counts:

Puzzle K'nex connectors K'nex rods Other Total

R C L Volume Sets Piece fit Image Link Purple White Yellow Red Green White Blue

2 5 3 30 1 Loose Image Link 24 1 15 2 29 18 4 93

2 6 3 36 1 Loose Image Link 26 2 16 3 32 22 5 106

3 3 4 36 1 Snug Image Link 24 6 8 25 20 8 91

3 4 3 36 1 Normal Image Link 22 6 7 2 23 19 2 81

2 5 4 40 1 Loose Image Link 28 7 13 2 43 22 6 2 light gray connectors 123

2 7 3 42 1 Loose Image Link 26 4 16 2 34 22 8 112

3 3 5 45 1 Snug Image Link 28 6 8 21 26 12 101

3 5 3 45 1 Normal Image Link 24 4 12 18 26 8 92

3 4 4 48 1 Snug Image Link 26 8 10 29 25 9 107

4 4 4 64 1 Normal Image Link 24 9 16 32 36 8 125

4 4 5 80 2 Normal Image Link 32 9 12 16 48 16 133

5 5 4 100 2 Loose Image Link 32 16 16 49 40 12 165

5 5 5 125 2 Normal Image Link 36 16 12 37 48 22 4 dark gray connectors, 2 hinges 177

4 4 8 128 2 Normal Image Link 48 17 4 24 64 24 181

Cubelets A	Minimum pieces f(A)	Description
1-4	1	1 four-cubelet flat piece
5-8	2	2 four-cubelet flat pieces
9-12	3	3 four-cubelet flat pieces
13-15	4	All 4 flat pieces
16-18	5	All 4 flat pieces plus 1 non-flat piece
19-20	6	All 4 flat pieces plus 2 non-flat pieces

Level	Cubelets	Cubelets remaining	Minimum support pieces
L (top)	16	27 - 16 = 11	f(16) = 5
L-1	5	11 - 5 = 6	f(5) = 2
L-2	2	6 - 2 = 4	f(2) = 1
L-3	1

Puzzle								K'nex connectors				K'nex rods			Other	Total
R	C	L	Volume	Sets	Piece fit	Image	Link	Purple	White	Yellow	Red	Green	White	Blue	Other	Total
2	5	3	30	1	Loose	Image	Link	24	1	15	2	29	18	4		93
2	6	3	36	1	Loose	Image	Link	26	2	16	3	32	22	5		106
3	3	4	36	1	Snug	Image	Link	24	6	8		25	20	8		91
3	4	3	36	1	Normal	Image	Link	22	6	7	2	23	19	2		81
2	5	4	40	1	Loose	Image	Link	28	7	13	2	43	22	6	2 light gray connectors	123
2	7	3	42	1	Loose	Image	Link	26	4	16	2	34	22	8		112
3	3	5	45	1	Snug	Image	Link	28	6	8		21	26	12		101
3	5	3	45	1	Normal	Image	Link	24	4	12		18	26	8		92
3	4	4	48	1	Snug	Image	Link	26	8	10		29	25	9		107
4	4	4	64	1	Normal	Image	Link	24	9	16		32	36	8		125
4	4	5	80	2	Normal	Image	Link	32	9	12		16	48	16		133
5	5	4	100	2	Loose	Image	Link	32	16	16		49	40	12		165
5	5	5	125	2	Normal	Image	Link	36	16	12		37	48	22	4 dark gray connectors, 2 hinges	177
4	4	8	128	2	Normal	Image	Link	48	17	4		24	64	24		181