  SOMA News 2 April 2012 E-Mail.

# Parity and Symmetry

Recently Doug Caine contacted me about an interesting topic. The question of parity and symmetry
It all started with my publication of the problem:

# Dear SOMA puzzler. We have a problem.

From one of your puzzle friends (Irina from Germany) I have heard of a problem.
The problem is to build the figure shown here, using pieces from the SOMA+Plus set.
The SOMA+Plus is sometimes called the "Herzberger Quader"
The figure shown has 35 cubes.
As the sum of all SOMA+plus pieces are 40, and this figure uses 35, it follows that 5 cubes are left.
Now 5 can only be reached by the 2' piece and one of the 3' pieces.
Leaving us to build the figure using all 8 pieces of 4' plus 1 piece of 3 (But which??)
SO: The question is:
Can anyone find the solution. Or, alternatively prove it to be impossible.?? The problem The pieces ``` 5 4 3 2 1 4 3 2 1 3 2 1 2 1 1```

We start by counting the "BLACK WHITE" cubes in each layer, from top-to-bottom:
1 black
2 black /1 white
4 black /2 white
6 black /4 white
9 black /6 white

For a total "BLACK/WHITE"-ratio of 22/13, which is a total parity of 22-13 = 9.
Now, the 8 tetra-cubes can provide, at most, 18_BLACK/14_WHITE cubes, while either tri-cube will provide 2_BLACK/1_WHITE, for a total of 20_BLACK/15_WHITE.
Therefore, the figure is impossible, because the set of available shapes will fail to provide enough black cubes.

Now - This was a very fine proof that the figure is not possible.

More about Parity can be found here:
Newsletter: 2000.12.03 The parity Proof. (Jotun's proof)
Newsletter: 2001.01.10 Parity and Centerness Applied to the SOMA Cube

More about Symmetry can be found here:
Newsletter: 1999.12.01 The 48 Symmetries of SOMA

# More about Parity and Symmetry

In my correspondence with Doug Caine he mentioned the interesting aspects of symmetry,
so here are Doug's thoughts that:

“Almost any shape that has an axis of mirror-symmetry will have a zero-parity, (especially if that axis "slices" BETWEEN the cube-layers, rather than THROUGH them)”

I will combine this with my solution to the challenge (at the top of this page).

"#" = "BLACK"
"O" = "WHITE"

Consider the following shape, rendered as a series of numbers, where each number is the amount of cubes stacked in that location:

``` 5 4 3 2 1
4 3 2 1
3 2 1
2 1
1```

...the sum of these 15 numbers (35) is the total volume of this shape. Next, we want to apply a "3-D Checkerboard"-coloring to this shape, so that every pair of cubes that share an entire square face are always of opposite colors. Each of the 5 layers might then look like this: ``` --- # - 5th_Layer (top) --- O # - 4th_Layer # --- # O # - 3rd_Layer O # # --- O # O # - 2nd_Layer # O # O # # --- # O # O # - 1st_Layer (bottom) O # O # # O # O # # ```

The resulting totals:
```        # /  O
-------------
5th =  1 /  0
4th =  2 /  1
3rd =  4 /  2
2nd =  6 /  4
1st =  9 /  6
-------------
22 / 13```

If we subtract the TOTAL_WHITE from the TOTAL_BLACK, the result (22-13=9) is the PARITY of the model.
If we had started with the colors reversed, the parity would then be (13-22=-9),
So I'll just refer to the "parity magnitude" instead, which is ALWAYS a positive number.

Now, 6 of the 8 tetra-cubes will always produce 2_#'s when colored in this manner, (6*2=12) & the remaining 2 tetra-cubes can produce (at most) 3_#'s each. (2*3=6)
The remaining volume requires a single tri-cube, which provides (at most) 2_#'s, no matter which tri-cube shape we choose.
Therefore, our 9 shapes are able to provide (at most): 12+6+2 = 20
..but our shape contains 22_#'s, not 20 - which means that it cannot be solved with the puzzle-shapes we've been given!
Since we were able to prove that 20_#'s is the maximum, that leaves 15_O's. (35-20=15)
This means that our maximum possible parity magnitude for these puzzle-pieces is (20-15=5), & any other shape which we try to make with a parity magnitude of 6-or-more will be impossible to build with these puzzle-pieces.

Most shapes that we try to solve will have an axis of symmetry (or 2, or more).
Such an axis might be either ROTATIONAL or MIRROR symmetry.
Parity can almost always be determined quickly for shapes that have mirror symmetry on an axis that is parallel to the cube-layers.

A simple example: ``` # O # O O # O #_________This axis-of-symmetry splits the figure BETWEEN 2 rows of cubes. # O # O O # O #```

This rectangle (only 1 layer thick) has an equal amount of each color, so the parity=0.
We can see that the design on one side of the axis is mirror-reversed on the other side, which has the effect of cancelling-out whatever parity might accumulate on just that one side. ``` # O # O O # O # # O # O---------This axis-of-symmetry splits the figure THROUGH a row of cubes. O # O # # O # O```

This rectangle also has parity=0, but only because the CENTER-row itself has zero-parity.
(in an example this simplistic, a more useful method would be to use a VERTICAL axis of mirror-symmetry)

Now let's look at a more 3-D example: ``` 1 2 1 3 2 1 4 3 2 1 5 4 3 2 1_________ 5 4 3 2 1 4 3 2 1 3 2 1 2 1 1```

Everything ABOVE this shape's axis is duplicated as a mirror-image BELOW, so this shape has zero-parity. ``` 1 2 1 3 2 1 4 3 2 1 5 4 3 2 1 6 5 4 3 2 1--------- 5 4 3 2 1 4 3 2 1 3 2 1 2 1 1```

Now in this case, the 5 rows ABOVE are matched-&-cancelled by the 5 rows BELOW, but the 6 stacks in the CENTER-row need further examination: ``` 6 5 4 3 2 1 O # O O # O # O # O O # O # O # O # O # O - Center_Row Parity=(9_# - 12_O = -3)```

So without even calculating the VOLUME, we can see that the parity magnitude is 3.

Submitted by Doug Caine <cainedm@hotmail.com>
Edited by Thorleif Bundgaard <thorleif@fam-bundgaard.dk>

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