SOMA News 01 Dec 1998 E-Mail.

## SOMA Alternatives (Checkered, Couples, Balancing)

### [Go Top]     The Checkered SOMA

The idea for the Checkered Soma was published in the first issue of The Soma Addict as an alternative way of forming the cube and other figures. This idea was created by Piet Hein himself and put down on paper on June 20, 1970. Shown below is a diagram depicting the proper placement of the checkered cubes. Try making one and see if you can build the cube for an added challenge!

### [Go Top]     Couples

The following figures are known as "couples." A couple is made with just one Soma Cube, but some of the seven pieces are independent of one another as the figures stand separately. Try to make the following couples and see what other ones you can come up with.

Do note that the shape #P002 is doubling the figure size, and that the figures #P001 and #P003 can be combined to form the cube.

### [Go Top]     Piece Duplication

Above is a pair figure from the set: P0001-0025

Doubling each dimension of any piece will multiply the volume by 8. For any of the pieces comprised of 4 unit cubes, this increases the volume from 4 cubic units to 32 cubic units. For piece number 1, the volume increases from 3 cubic units to 24 cubic units. This equals the total number of unit cubes in the remaining 6 pieces. This poses a possible problem: Can you construct a figure similar to the number 1 piece but with all dimensions doubled, using the remaining 6 Soma pieces? The answer is yes, and the challenge is to construct this figure.

## [Go Top]     Balancing SOMA on a Pedestal

This interesting enigma was first published in the second issue of The Soma Addict. There is said to be a way to form the cube in which it can be stood atop a "pedestal" no wider than the bottom center cube. Can you find this solution?!?

Now, there are actually three ways of understanding this challenge.
 Make a 3x3x3 cube and balance it on a stick, touching only the center bottom cube. This chapter Make a 3x3x3 cube with a missing cube, having the missing cube protrude at the bottom as a balance pin. Balancing SOMA on its own piece Or, make another figure, having 1 (or 2) cube's protruding at the bottom, as balance pin's (No article about this - yet)

For historical reasons, the texts in this chapter do not destinguish between the two first methods.
And I have NO examples of the third method (Maybe you could find some ??)

### The Balancing Soma Cube

This interesting information about the Soma Cube was published in Martin Gardner's Mathematical Games column in July, 1969.

Martin Gardner notes in the article that there are 240 different ways the cube can be formed not counting rotations and reflections as being different. This figure was established in 1962 by John Conway and M.J.T. Guy, who were mathematicians at the University of Cambridge. In analyzing his work, Conway discovered the curious fact that only one of the 240 solutions allows the Soma cube to be balanced on a pedistal that touches only the central square of the cube's 3 by 3 unit base. This solution is diagrammed below. The central square of the bottom layer is the one that rests on the pedistal.

```TOP       MIDDLE    BOTTOM
677       557       524
667       364       524
311       314       322
```

The following additional solution to the balancing cube has been produced by Stuart Collins of Nottingham, UK in June, 1998 and refutes Conway's uniqueness assertion stated above!

Are there more solutions to this problem? Mr. Collins conjectures that any of the flat 4-cube pieces ( 2,3, and 4 ) might serve as the balance piece. It only remains to be shown that the 3 piece can serve this function.

```TOP       MIDDLE    BOTTOM
221       266       246
711       553       446
773       753       453
```

Now at 12.1.2000 My friend Courtney wrote to me:
"I borrowed a book from the Library about puzzles. It had a few pages on the Soma cube, and showed two new ways to construct the 3x3x3 cube so you can balance it on the bottom-center cube:
```551       466       226
751       451       426
773       733       423
```
and
```166       446       244
117       356       222
377       357       355
```
...I tried it both ways. It really works!
Now at 26.8.2002 Steven Mai wrote to me:
"Here are the solutions for the cube that can be supported on the centre of the base without the cube breaking up.
```/122/552/542
/116/376/544
/366/377/374

/322/552/542
/336/176/544
/366/177/174```

Now at 3.7.2006 Jim Waters wrote to me:
"I can now report more than DOUBLING the number of solutions found thus far:"
```Standardization of solutions by bottom piece:

Notice that when reflecting one solution onto another, pieces 5 & 6 must be swapped.

Solution #1:  #2:  #3:  #4:  #5:  #6:  #7:  #8:  #9:  #10: #11: #12: #13:

322  322  773  773  663  663  113  553  551  333  333  774  113
Top     311  337  711  733  633  611  517  753  751  735  711  744  766
341  377  221  223  223  221  577  773  773  775  771  143  776

552  552  763  761  771  773  433  466  466  466  436  766  413
Midt    376  566  663  661  671  673  466  453  451  455  466  553  463
446  117  255  255  255  255  557  711  733  711  755  113  755

572  452  463  461  471  473  223  226  226  226  226  226  223
Bottom  577  446  445  445  445  445  426  426  426  426  425  526  425
466  146  245  245  245  245  426  421  423  421  425  523  425

Bottom  .7.  4..  4..  4..  4..  4..  22.  22.  22.  22.  22.  22.  22.
piece   .77  44.  44.  44.  44.  44.  .2.  .2.  .2.  .2.  .2.  .2.  .2.
...  .4.  .4.  .4.  .4.  .4.  .2.  .2.  .2.  .2.  .2.  .2.  .2.

Reflections
#1:  #2:  #3:  #4:  #5:  #6:  #7:  #8:  #9:  #10: #11: #12: #13:

112  223  377  377  355  355  311  366  166  333  333  477  311
Top     412  733  117  337  335  115  716  367  167  637  117  447  557
333  773  122  322  322  122  776  377  377  677  177  341  577

552  266  357  157  177  377  334  554  554  554  534  557  314
Midt    476  556  355  155  175  375  554  364  164  664  554  366  354
436  711  662  662  662  662  766  117  337  117  667  311  667

572  264  354  154  174  374  322  522  522  522  522  522  322
Bottom  577  544  644  644  644  644  524  524  524  524  624  526  624
466  541  642  642  642  642  524  124  324  124  624  326  624

Bottom  .7.  ..4  ..4  ..4  ..4  ..4  .22  .22  .22  .22  .22  .22  .22
piece   .77  .44  .44  .44  .44  .44  .2.  .2.  .2.  .2.  .2.  .2.  .2.
...  .4.  .4.  .4.  .4.  .4.  .2.  .2.  .2.  .2.  .2.  .2.  .2.

```
Interestingly enough, John Conway's solution which he thought was unique was the ONLY solution which I had NOT already found!
It is #13 on the chart.
Note that Stuart Collins' cube is my #3;
The two ways found by Courtney are my #9 and #7 (see its reflection)
And the two found by Steven Mai are my #6 and #5 respectively.

When I first visited Science World BC in Vancouver BC Canada, they had a soma set and a pedestal
with the challenge and the claim that their exists only one solution, but without revealing what
it was; and I played with it and found my solution #1 (and notice piece 7 rests on the pedestal!!)
and I thought that was it.
An assembled a soma cube of my own, rested it on a pedestal at home, and challenged a friend of mine,
Dan Seafeldt, who found what I list as solution #2. Since then, I found the other solutions listed above.

Now at 17.11.2011 Michael Sundermann has sent me a mail reporting two more ways to make the balancing SOMA cube. (and its mirror image)
As he wrote: "I want to add two solutions where the 3-piece is at the bottom."
```Solution #1:  #2:

774  477
Top     712  117
112  122

744  447
Middle  655  655
662  662

345  345
Bottom  335  335
362  362

Reflections (pieces 5 and 6 must be swapped)
#1r  #2r

112  122
Top     712  117
774  477

552  552
Middle  566  566
744  447

352  352
Bottom  336  336
346  346
```
Solution #1 is (No.33 in N990201) and Solution #2r is (No.53 in N990201).

Interestingly, if you inspect the total number of solutions for the SOMA cube.
Found in N990201.HTM "Winning Ways for your mathematical plays"
You find that there are 23 ways to place the '3' piece such that the 'leg of the T' is at the center position of the bottom slice.
This of course raises the question. Are there more stable combinations.?
What if a solution with the '3' piece at the top is rotated 180 degrees.? etc.
In any case. Two more solutions is great.
Are there any more of them??

The way this list of 240 solutions was found is described at N030518.HTM "The complete SOMAP is found"

## A complete analysis of balanzing

At 15.11.2012 I received this interesting addition to the special shape of a balanzing SOMA.

Hartwig Beusch has added the following information telling us that 'all' pieces may be the supporting elements.
PS: Hartwig expanded the text a few days later, making it even more complete.
Hartwig wrote. (Translated from German)

In your web page [note: This page] you presented solutions for pieces 1, 3, 4 and 7.
Now using the mirrorimage of the piece 6, even the piece 5 can be the base of the balancing SOMA cube. This result did surprise me.
But following this, every piece can now be used as base to produce a balancing SOMA cube."
Here are the examples for:

Piece 1
```       2 2 3
5 3 3
5 5 3

2 7 7
5 7 4
6 6 4

2 7 4
6 1 4
6 1 1```
Piece 2
```       1 1 3   5 5 3   5 5 1   3 3 3   3 3 3   7 7 4   1 1 3   3 3 3   6 6 3   6 6 3   3 3 3   1 1 3   1 1 6
5 1 7   7 5 3   7 5 1   7 3 5   7 1 1   7 4 4   7 6 6   5 3 7   7 3 3   7 1 1   7 6 6   7 3 3   7 6 6
5 7 7   7 7 3   7 7 3   7 7 5   7 7 1   1 4 3   7 7 6   5 7 7   7 7 3   7 7 1   7 7 6   7 7 3   7 7 3

4 3 3   4 6 6   4 6 6   4 6 6   4 3 6   7 6 6   4 1 3   4 1 1   4 6 1   4 6 3   4 3 1   4 1 6   4 1 6
4 6 6   4 5 3   4 5 1   4 5 5   4 6 6   5 5 3   4 6 3   4 6 6   4 6 1   4 6 3   4 6 1   4 6 6   4 5 5
5 5 7   7 1 1   7 3 3   7 1 1   7 5 5   1 1 3   7 5 5   5 5 7   7 5 5   7 5 5   7 5 5   7 5 5   7 3 3

2 2 3   2 2 6   2 2 6   2 2 6   2 2 6   2 2 6   2 2 3   2 2 1   2 2 1   2 2 3   2 2 1   2 2 6   2 2 5
4 2 6   4 2 6   4 2 6   4 2 6   4 2 5   5 2 6   4 2 5   4 2 6   4 2 5   4 2 5   4 2 5   4 2 5   4 2 5
4 2 6   4 2 1   4 2 3   4 2 1   4 2 5   5 2 3   4 2 5   4 2 6   4 2 5   4 2 5   4 2 5   4 2 5   4 2 3```
Piece 3
```       7 7 4     4 7 7
7 1 2     1 1 7
1 1 2     1 2 2

7 4 4     4 4 7
6 5 5     6 5 5
6 6 2     6 6 2

3 4 5     3 4 5
3 3 5     3 3 5
3 6 2     3 6 2```
Piece 4
```       3 2 2     6 6 3     6 6 3     1 2 2     7 7 3     7 7 3     3 2 2     1 2 2
3 3 7     6 3 3     6 1 1     1 1 7     7 3 3     7 1 1     3 3 6     1 1 6
3 7 7     2 2 3     2 2 1     3 7 7     2 2 3     2 2 1     3 6 6     3 6 6

5 5 2     5 5 1     5 5 3     5 5 2     7 1 1     7 3 3     5 5 2     5 5 2
1 6 6     6 7 1     6 7 3     3 6 6     6 6 5     6 6 5     1 7 6     3 7 6
1 6 7     2 7 7     2 7 7     3 6 7     2 5 5     2 5 5     1 7 7     3 7 7

5 4 2     5 4 1     5 4 3     5 4 2     6 4 1     6 4 3     5 4 2     5 4 2
5 4 4     5 4 4     5 4 4     5 4 4     6 4 4     6 4 4     5 4 4     5 4 4
1 6 4     2 7 4     2 7 4     3 6 4     2 5 4     2 5 4     1 7 4     3 7 4```
Piece 5 (a mirror of piece 6)
```       1 7 7
1 1 7
2 2 3

4 4 7
6 6 3
2 5 3

6 4 4
6 5 5
2 5 3```
Piece 6 (a mirror of piece 5)
```       7 7 1
7 1 1
3 2 2

7 4 4
3 5 5
3 6 2

4 4 5
6 6 5
3 6 2```
and Piece 7
```       3 2 2
3 1 1
3 4 1

5 5 2
3 7 6
4 4 6

5 7 2
5 7 7
4 6 6```

The balanzing SOMA cube is possible for the pieces 1, -, 3, 4, 5, 6, 7 using the checkered constellation Satz-Ls. In these solutions the piece #2 have a black "hook" and is hanging.
Whereas the balanzing SOMA for piece #2 is possible, only using the checkered constellation Satz-Lw, where piece #2 have a white "hook" and carry the figure.

Using the Satz-Ls it is possible to build 240-21 = 219 SOMAcubes. The Satz-Lw allow 21 SOMAcubes.

The white "hook" of piece 2 is blocking the use of this piece as an edge of a SOMAcube.

So - in total, the pieces that carry our SOMAcube are capable of creating a total of:
(For pieces 1 through 7)    1 + 13 + 2 + 8 + 1 + 1 + 1 = 27 solutions (Not counting mirror solutions)
If we include mirrors, there are 52 solutions, as the pieces 5 and 6 exchange positions during the mirroring.

It is interesting that the piece#2 is either a carrier (in 13 cases) or is hanging (in 14 cases). When hanging, it will run along a vertical corner holding on at the top using its "hook"

Finally a few tips for the construction of the balanzing SOMA.

1. Place one of the pieces 1 through 7 on the center of the Base.
2. Place the following pieces so that they cannot fall, and settle in place.
3. Keep to the saying "Where shall it go, and how do we keep it there.

## [Go Top]     Balancing SOMA on its own piece

Back to the Balancing SOMA on a pedestal chapter.

Now one thing is to balance on an externally supplied pedestal, but what if the balancing object is one of the cubes forming one of the pieces.?

On 07.07.2015 - Matt Esser from USA Sendt me this interesting addition to the special shape of a balanzing SOMA.

Hi, I was recently inspired by puzzle 166 of your site. Your printed solution on your site does not balance however
```/SOMA166
;sscube
/553/533/113/...
/2.4/564/177/.7.
/266/264/274/...```
So, I challenged myself to find a way to balance the blocks with one center point down and one empty space in the otherwise 3x3x3 cube.
Attached are pictures that should allow you to see the solution.
 ```/SOMA166 ;sscube_balancing /.41/771/735/... /441/755/635/.3. /422/662/632/... ``` Balancing

I was at a house where they owned a SOMA cube. I do not own one, so I'm unable to further test.
I understand that this puts the empty square on the top row in a different spot than the middle
(so therefore, perhaps this is a different puzzle altogether), but it does balance with only a center square touching.
Let me know what you think!

Hi Matt: This is a very fine solution. The question of course is "What is a balancing SOMA".
Your idea of following up upon the #166, balancing on one center cube,
and - of course - leaving a hole somewhere... IS great.

Now - Can anyone find more of these figures ?

## More Balancing on the puzzle cubes

Now it is 28.08.2016 - A mail arrived from Hartwig Beusch <beuschdrbe@aol.com>
To my question "Can you find more of these figures?" the answer was simply "Yes, I can."

Ps. Note: These figures were originally drawn with the bottom upwards, it might be easier to build that way.
```         A1.     A2.     A3.     A4.     A5.

7 7 -   7 7 -   2 4 -   4 4 -   7 4 -
Top     2 7 4   4 7 2   2 4 4   2 4 4   7 7 2
2 1 1   1 1 2   1 1 4   2 1 1   1 1 2

6 7 4   4 7 5   2 6 6   6 7 7   4 4 5
Midt    6 5 4   4 6 5   5 5 7   6 5 7   7 6 5
2 5 1   1 6 2   1 7 7   2 5 1   1 6 2

6 6 4   4 5 5   2 3 6   6 6 7   4 5 5
Lower   3 3 3   3 3 3   5 3 6   3 3 3   3 3 3
2 5 5   6 6 2   5 3 7   2 5 5   6 6 2

- - -   - - -   - - -   - - -   - - -
Balance - 3 -   - 3 -   - 3 -   - 3 -   - 3 -
- - -   - - -   - - -   - - -   - - -```

How do I get such a balanced cube?
The carrier piece 3 "T" consists of the four individual cubes
```[ 3A | 3B | 3C ]
[ 3D ]```
The whole construction is resting on the pedestal made of cube 3D, and in the first Example 1:
3A carry piece 6
3B carry piece 5
3C carry piece 4

This is the principle!
If 3A, 3B or 3C, is not used as a carrier for three different pieces, there will be no balanced cube.

Mirroring is subject to other cubes. piece 5 is then changed into piece 6, and 6 to 5.
If the first 3 pieces are placed according to my principle, then the remaining can sometimes be placed in 2 various ways.

The resulting gap is always in a corner.
If the gap is placed in the middle of a side or in the center of the upper layer, then it is no longer possible to make a balanced cube.

In the next figures we can see:
B1 - B4 The gap may also occur in the lower layer on a corner.
B5 - B10 The gap may also occur in the uppermost layer on a corner.
B6 is identical to B5 , with the exception of the direction of pieces in the upper level.
B1 is the same as C2 except that the piece 7 is rotated.
```         B1.     B2.     B3.     B4.        B5.     B6.     B7.     B8.     B9.     B10.

4 2 2   1 4 7   4 2 2   1 7 7      2 2 4   2 4 -   1 4 7   2 2 7   1 1 4   1 7 7
Top     4 4 1   1 7 7   4 7 1   1 7 4      1 4 4   2 4 4   1 7 7   1 7 7   2 4 4   1 7 4
7 4 1   2 2 2   7 7 1   2 2 2      1 4 -   1 1 4   2 2 -   1 4 -   2 4 -   2 2 -

5 5 2   1 4 4   5 5 2   1 7 4      2 6 6   2 6 6   1 4 4   2 6 6   1 7 7   1 7 4
Midt    7 6 6   6 6 7   4 6 6   6 6 4      5 5 7   5 5 7   6 6 7   5 5 7   6 6 7   6 6 4
7 7 1   2 5 5   4 7 1   2 5 5      1 7 7   1 7 7   2 5 5   1 4 4   2 5 5   2 5 5

5 3 2   6 3 4   5 3 2   6 3 4      2 3 6   2 3 6   6 3 4   2 3 6   6 3 7   6 3 4
Lower   5 3 6   6 3 5   5 3 6   6 3 5      5 3 6   5 3 6   6 3 5   5 3 6   6 3 5   6 3 5
- 3 6   - 3 5   - 3 6   - 3 5      5 3 7   5 3 7   2 3 5   5 3 4   2 3 5   2 3 5

- - -   - - -   - - -   - - -      - - -   - - -   - - -   - - -   - - -   - - -
Balance - 3 -   - 3 -   - 3 -   - 3 -      - 3 -   - 3 -   - 3 -   - 3 -   - 3 -   - 3 -
- - -   - - -   - - -   - - -      - - -   - - -   - - -   - - -   - - -   - - -```
As mentioned, the crossbeam of piece 3 "T" is called 3A, 3B and 3C.
When we want to balance the cube, then we must select a different piece to place on the 3A, 3B, 3C cubes.
This selection must be from pieces 4, 5, 6 or 7 ("Z, A, B, P")
And pieces 4 and 7 may not occur at the same time. [Thorleifs note: Why not?]

This only allow us the combinations of 4-5-6 and 5-6-7.

A result is only obtained when the gap is selected as the cornerstone of the first or third level.
Gaps in the center of a level or the middle of a side can not be constructed.

We have 4 different positions for a selected gap in the lower level.
But there are 5 different solutions in the upper level.

Because the cube have 4 vertices (corners) in the lower and upper level,
then we find 4x4 + 4x5 = 36 different solutions.

I will now give the 5 solutions having the gap on the top left.
Note: C2 = B1 except for the direction of the 7 "P" piece.
```         C1.     C2.     C3.     C4.     C5.

4 1 1   4 2 2   7 4 1   7 2 2   7 7 1
Top     4 4 2   4 4 1   7 7 1   7 7 1   4 7 1
- 4 2   - 4 1   - 2 2   - 4 1   - 2 2

7 7 1   5 5 2   4 4 1   5 5 2   4 7 1
Midt    7 5 5   7 6 6   7 5 5   7 6 6   4 5 5
6 6 2   7 7 1   6 6 2   4 4 1   6 6 2

7 3 5   5 3 2   4 3 5   5 3 2   4 3 5
Lower   6 3 5   5 3 6   6 3 5   5 3 6   6 3 5
6 3 2   7 3 6   6 3 2   4 3 6   6 3 2

- - -   - - -   - - -   - - -   - - -
Balance - 3 -   - 3 -   - 3 -   - 3 -   - 3 -
- - -   - - -   - - -   - - -   - - -```
The pieces 1 and 2 ("v and L") are not suitable for the foundation, they will slide off the construction.
This can be realized by studying the patterns in the upper level.
. 1 1 or . 2 2 vertically and horizontally
This pattern occur together in the upper level, when a solution is found.

Now also the 4 solutions for a gap in the lower level:
```         D1.     D2.     D3.     D4.

7 4 1   2 2 2   2 2 2   7 7 1
Top     4 4 1   1 7 4   1 7 7   4 7 1
4 2 2   1 7 7   1 4 7   4 2 2

7 7 1   2 6 6   2 6 6   4 7 1
Midt    7 5 5   5 5 4   5 5 7   4 5 5
6 6 2   1 7 4   1 4 4   6 6 2

- 3 5   - 3 6   - 3 6   - 3 5
Lower   6 3 5   5 3 6   5 3 6   6 3 5
6 3 2   5 3 4   5 3 4   6 3 2

- - -   - - -   - - -   - - -
Balance - 3 -   - 3 -   - 3 -   - 3 -
- - -   - - -   - - -   - - -```

I find that constructing these cubes is particularly fun and quickly leads to success, if the specified rules are followed.

- submitted by many puzzlers (Mentioned in the text).
- User entries have been carefully edited by Thorleif Bundgård.

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