Compositions of dozen-scores.
Version of Sunday 27 January 2013.

§ 6d1. In algebra and analysis, the composition of unary operations is frequently studied. However, composition may easily be applied to binary operations such as the dozen-scores, yielding a structure analogous to the binary tree of computer science.

Needed are two variables a and b, along with a supply of operations V₀, V₁, and so forth; some of these V_n might be equal. For this purpose, infix notation is clearer than prefix, and the parentheses are significant because the dozen-scores are rarely associative. In the table of examples below, each insertion is shown in red for emphasis.

building trees
Here is an ordinary invocation of a binary operation: a V₀ b #1
Within #1, (a V₁ b) can be placed two ways: for the a: (a V₁ b) V₀ b #2a
for the b: a V₀ (a V₁ b) #2b
Within #2b, (a V₂ b) can be placed three ways: for the left a: (a V₂ b) V₀ (a V₁ b) #3a
for the right a: a V₀ ((a V₂ b) V₁ b) #3b
for the b: a V₀ (a V₁ (a V₂ b)) #3c
From #3c, examples of further successive substitutions are: a V₀ ((a V₃ b) V₁ (a V₂ b)) #4
((a V₄ b) V₀ ((a V₃ b) V₁ (a V₂ b)) #5
(a V₄ b) V₀ ((a V₃ b) V₁ ((a V₅ b) V₂ b)) #6
((a V₆ b) V₄ b) V₀ ((a V₃ b) V₁ ((a V₅ b) V₂ b)) #7

building trees
Here is an ordinary invocation of a binary operation:	a V₀ b	#1
Within #1, (a V₁ b) can be placed two ways:	for the a:	(a V₁ b) V₀ b	#2a
for the b:	a V₀ (a V₁ b)	#2b
Within #2b, (a V₂ b) can be placed three ways:	for the left a:	(a V₂ b) V₀ (a V₁ b)	#3a
for the right a:	a V₀ ((a V₂ b) V₁ b)	#3b
for the b:	a V₀ (a V₁ (a V₂ b))	#3c
From #3c, examples of further successive substitutions are:	a V₀ ((a V₃ b) V₁ (a V₂ b))	#4
((a V₄ b) V₀ ((a V₃ b) V₁ (a V₂ b))	#5
(a V₄ b) V₀ ((a V₃ b) V₁ ((a V₅ b) V₂ b))	#6
((a V₆ b) V₄ b) V₀ ((a V₃ b) V₁ ((a V₅ b) V₂ b))	#7

Note that in these constructions:

The left input of any V_n is either the output of an operation, or an explicit a.
The right input of any V_n is either the output of an operation, or an explicit b.

Thus excluded are the cases where an explicit a is a right input, or an explicit b is a left input. This is separate from the case where the output of one operation, being an input to the next, just happens to equal a or b; there is no restriction against this implicit case, and such a value may freely appear as either a left or right input.

In the parlance of binary trees, the a's and b's serve in the role of leaf nodes.

§ 6d2. In most cases the following compositions, which are the simplest nontrivial trees:

(a V₁ b) V₀ b
a V₀ (a V₁ b)

do not yield dozen-scores. For instance, below are two ordinary dozen-scores followed by four examples of compositions thereof. In the tables of those operations that are not dozen-scores, note that the colors are reversed and a delta serves as an upside-down nabla; these differences in presentation help prevent mistakes in interpreting the tables.

22∇ b
0 1 2 3 4
a 0 0 2 4 3 1
1 3 1 0 2 4
2 2 4 3 1 0
3 1 0 2 4 3
4 4 3 1 0 2

57∇ b
0 1 2 3 4
a 0 1 0 4 2 3
1 4 2 3 1 0
2 3 1 0 4 2
3 0 4 2 3 1
4 2 3 1 0 4

Δ b
0 1 2 3 4
a 0 3 2 1 1 3
1 4 4 2 2 1
2 1 1 4 0 0
3 0 3 3 4 4
4 2 0 0 3 2

Δ b
0 1 2 3 4
a 0 2 0 1 4 3
1 4 0 2 1 3
2 1 4 2 0 3
3 1 3 2 4 0
4 1 0 3 4 2

Δ b
0 1 2 3 4
a 0 1 1 1 3 0
1 0 2 4 4 4
2 3 3 2 1 3
3 4 0 0 0 1
4 2 4 3 2 2

Δ b
0 1 2 3 4
a 0 1 4 3 2 0
1 1 2 4 3 0
2 0 2 4 1 3
3 4 0 2 1 3
4 4 0 3 2 1

(a 57∇ b) 22∇ b a 22∇ (a 57∇ b) (a 22∇ b) 57∇ b a 57∇ (a 22∇ b)

Although the delta tables above may seem random, two things can be said:

When any two dozen-scores are composed in the pattern (a V₁ b) V₀ b, each column of the table will contain five different values.
When any two dozen-scores are composed in the pattern a V₀ (a V₁ b), each row of the table will contain five different values.

Some compositions exhibit partial or full cancellativity:

Δ b
0 1 2 3 4
a 0 2 0 4 1 3
1 3 2 0 4 1
2 1 3 2 0 4
3 4 1 3 2 0
4 0 4 1 3 2

Δ b
0 1 2 3 4
a 0 1 1 1 1 1
1 0 0 0 0 0
2 2 2 2 2 2
3 4 4 4 4 4
4 3 3 3 3 3

Δ b
0 1 2 3 4
a 0 2 0 4 3 1
1 2 0 4 3 1
2 2 0 4 3 1
3 2 0 4 3 1
4 2 0 4 3 1

104∇ b
0 1 2 3 4
a 0 2 0 4 1 3
1 1 3 2 0 4
2 0 4 1 3 2
3 3 2 0 4 1
4 4 1 3 2 0

a 85∇ (a 173∇ b)
FOC but not SOC (a 151∇ b) 117∇ b
SOC but not FOC a 63∇ (a 66∇ b)
SOC but not FOC a 145∇ (a 167∇ b)
FOC and SOC

If a and b vary independently throughout all values of S₅; and if V₀, V₁ and V_R are dozen-scores being sought:

The equation (a V₁ b) V₀ b = a V_R b has 2400 solutions. Operations V₀ and V_R will always come from the same quadraginta, and V₁ will come from row 1, 2, 5 or 6 of quadraginta A, displayed below with new coloring for convenience.
As suggested by symmetry, the equation a V₀ (a V₁ b) = a V_R b also has 2400 solutions. Again, operations V₀ and V_R will always come from the same quadraginta, but now V₁ will come from row 3, 4, 7 or 8 of quadraginta A.

quadraginta A
1 2 3 4 5
1 0∇ 192∇ 180∇ 129∇ 66∇
2 110∇ 59∇ 47∇ 239∇ 173∇
3 167∇ 101∇ 89∇ 26∇ 218∇
4 213∇ 150∇ 138∇ 72∇ 21∇
5 181∇ 128∇ 67∇ 1∇ 193∇
6 238∇ 172∇ 111∇ 58∇ 46∇
7 20∇ 212∇ 151∇ 139∇ 73∇
8 100∇ 88∇ 27∇ 219∇ 166∇

quadraginta A
	1	2	3	4	5
1	0∇	192∇	180∇	129∇	66∇
2	110∇	59∇	47∇	239∇	173∇
3	167∇	101∇	89∇	26∇	218∇
4	213∇	150∇	138∇	72∇	21∇
5	181∇	128∇	67∇	1∇	193∇
6	238∇	172∇	111∇	58∇	46∇
7	20∇	212∇	151∇	139∇	73∇
8	100∇	88∇	27∇	219∇	166∇

In the case of V₀ ≡ V₁, everything must come from quadraginta A. In each column of the table below are listed, as ordered pairs, all ten solutions for one of the two compositions. Each nabla is subscripted with the row and column numbers of where the operation is found in the quadraginta A table. This labeling makes the patterns more conspicuous.

(a V₀ b) V₀ b = a V_R b a V₀ (a V₀ b) = a V_R b
V₀, V_R V₀, V_R

0∇₁₁, 27∇₈₃ 20∇₇₁, 47∇₂₃
192∇₁₂, 100∇₈₁ 212∇₇₂, 110∇₂₁
180∇₁₃, 219∇₈₄ 151∇₇₃, 239∇₂₄
129∇₁₄, 88∇₈₂ 139∇₇₄, 59∇₂₂
66∇₁₅, 166∇₈₅ 73∇₇₅, 173∇₂₅
238∇₆₁, 138∇₄₃ 213∇₄₁, 111∇₆₃
172∇₆₂, 213∇₄₁ 150∇₄₂, 238∇₆₁
111∇₆₃, 72∇₄₄ 138∇₄₃, 58∇₆₄
58∇₆₄, 150∇₄₂ 72∇₄₄, 172∇₆₂
46∇₆₅, 21∇₄₅ 21∇₄₅, 46∇₆₅

Whenever V₀ and V_R come from the same column of the quadraginta table, it is column 5. Thus the distinguished quadraginta has a distinguished column. The idempotent dozen-scores 21∇ and 46∇ form the one pair that works both ways, as listed in the last row.

Given dozen-scores Z and Y:

if there exists a dozen-score X such that a X b = (a Z b) Y b, then there exists a dozen-score W such that a W b = (a Z^V b) Y^H b;
if there exists a dozen-score X such that a X b = (a Z b) Y b, then there exists a dozen-score W such that a W b = (a Z^V b) Y^V b;
if there exists a dozen-score X such that a X b = a Y (a Z b), then there exists a dozen-score W such that a W b = a Y^TH (a Z^TH b);
if there exists a dozen-score X such that a X b = a Y (a Z b), then there exists a dozen-score W such that a W b = a Y^TV (a Z^TH b);

and in each case, X and W come from the same quadraginta. Like many results in this report, this was revealed by a computer program that examined all possible cases.

If a Z (a Y b) = a X b then (a Y^T b) Z^T b = a X^T b, and this applies whether Z, Y and X are dozen-scores or other operations.

§ 6d3. When three dozen-scores are composed, the range of possibilities grow far beyond what two can offer. This section looks at the balanced tree configuration:

(a V₁ b) V₀ (a V₂ b)

In the following examples, some values occur more often than others within a table. Below each table is its formula, and its census:

Δ₀ b
0 1 2 3 4
a 0 2 2 2 2 2
1 2 2 2 2 2
2 2 2 2 2 2
3 2 2 2 2 2
4 2 2 2 2 2

Δ₁ b
0 1 2 3 4
a 0 1 1 3 1 4
1 3 1 4 1 1
2 4 1 1 3 1
3 1 3 1 4 1
4 1 4 1 1 3

Δ₂ b
0 1 2 3 4
a 0 0 0 2 4 2
1 4 2 0 0 2
2 0 2 4 2 0
3 2 0 0 2 4
4 2 4 2 0 0

(a 42∇ b) 24∇ (a 155∇ b)
25·2 (a 239∇ b) 39∇ (a 30∇ b)
15·1, 5·3, 5·4 (a 224∇ b) 10∇ (a 183∇ b)
10·0, 10·4, 5·2

A table with either of these censuses:

15·a, 5·b, 5·c (for instance Δ₁)
10·a, 10·b, 5·c (for instance Δ₂)

turns out to satisfy one of these constraints:

a Z b = (a + 1) Z (b + 2)
a Z b = (a + 2) Z (b + 1)

Elementarily, Δ₀ satisfies both.

There are three censuses that do not occur:

20·a, 5·b
15·a, 10·b
10·a, 5·b, 5·c, 5·d

However, operations that have five instances of each value are plentiful and come in a variety of patterns not easy to categorize:

Δ b
0 1 2 3 4
a 0 2 2 0 0 1
1 2 2 4 3 3
2 0 1 1 3 3
3 4 1 1 2 4
4 4 3 0 0 4

Δ b
0 1 2 3 4
a 0 4 0 4 1 4
1 3 2 0 2 2
2 3 3 4 0 3
3 3 2 4 1 0
4 0 2 1 1 1

Δ b
0 1 2 3 4
a 0 4 1 4 4 3
1 3 2 1 3 3
2 4 2 0 1 3
3 4 0 0 0 1
4 1 2 0 2 2

(a 58∇ b) 35∇ (a 78∇ b) (a 49∇ b) 38∇ (a 197∇ b) (a 40∇ b) 43∇ (a 60∇ b)

Δ b
0 1 2 3 4
a 0 4 2 3 3 3
1 2 2 2 3 1
2 1 2 0 1 1
3 4 0 0 0 1
4 4 4 0 3 4

Δ b
0 1 2 3 4
a 0 4 0 3 4 3
1 1 0 2 2 1
2 1 0 2 2 1
3 4 0 3 4 3
4 2 0 1 3 4

42∇ b
0 1 2 3 4
a 0 0 4 2 3 1
1 2 3 1 0 4
2 1 0 4 2 3
3 4 2 3 1 0
4 3 1 0 4 2

(a 27∇ b) 59∇ (a 125∇ b) (a 53∇ b) 68∇ (a 229∇ b) (a 66∇ b) 32∇ (a 110∇ b)

In the special case where a V₃ b = (a V₁ b) V₀ (a V₁ b), in other words where V₁ ≡ V₂, it turns out that:

If V₀ and V₁ are any given dozen-scores, then dozen-score V₃ will exist and be unique.
If V₀ and V₃ are any given dozen-scores, then dozen-score V₁ will exist and be unique.
If V₁ and V₃ are any given dozen-scores, the situation is more complicated:
- Half the time, there are two dozen-score solutions for V₀. For instance, if V₁ ≡ 5∇ and V₃ ≡ 12∇, then V₀ can equal 35∇ or 39∇.
- Otherwise there is no solution, not even an operation that is not dozen-score. An example is when V₁ ≡ 5∇ and V₃ ≡ 13∇.
Taken together, V₁ and V₃ define the principal diagonal of V₀, if they do not lead to a contradiction. If a valid principal diagonal (five different values) is obtained, two dozen-scores immediately follow.

This composition is well-enough behaved that it might deserve to be called the standard composition for dozen-scores and receive a special symbol:

a (V₀ ◊ V₁) b = (a V₁ b) V₀ (a V₁ b)

More generally, let T and U be tree expressions of the type in the "building trees" table above. Then what T ◊ U means is to substitute the expression "(a U b)" for each instance of a or b in T. Hence this unsurprising result:

a ((V₀ ◊ V₁) ◊ V₂) b =
a (V₀ ◊ V₁ ◊ V₂) b =
a (V₀ ◊ (V₁ ◊ V₂)) b =
((a V₂ b) V₁ (a V₂ b)) V₀ ((a V₂ b) V₁ (a V₂ b))

The relation among the first three members of the equation above can more succinctly be written:

(V₀ ◊ V₁) ◊ V₂ ≡ V₀ ◊ V₁ ◊ V₂ ≡ V₀ ◊ (V₁ ◊ V₂)

As in the case of unary operations, this binary composition is associative. It also works for asymmetric trees. For instance, if T and U are thus:

a T b = ((a V₂ b) V₁ b) V₀ (a V₁ b)
a U b = a V₃ (a V₄ (a V₅ b))

Then a (T ◊ U) b equals the following complicated result, with the U-substitutions in red:

(((a V₃ (a V₄ (a V₅ b))) V₂ (a V₃ (a V₄ (a V₅ b)))) V₁ (a V₃ (a V₄ (a V₅ b)))) V₀ ((a V₃ (a V₄ (a V₅ b))) V₁ (a V₃ (a V₄ (a V₅ b))))

It might be observed that supplying the same expression to the left and right leaf nodes of T turns it rather into a de facto unary function.

§ 6d4. Two more compositions involving three operations are in straight configurations:

((a V₂ b) V₁ b) V₀ b
a V₀ (a V₁ (a V₂ b))

which are equivalent if the transposes of the operations are taken, and if a and b are exchanged.

Each of the four columns in the table below is an example of building up, step by step, a composition in the straight configuration. Some of the intermediate and final results are dozen-scores, and others are not.

— first step —

216∇ b
0 1 2 3 4
a 0 4 2 0 1 3
1 0 1 3 4 2
2 3 4 2 0 1
3 2 0 1 3 4
4 1 3 4 2 0

88∇ b
0 1 2 3 4
a 0 1 4 2 0 3
1 0 3 1 4 2
2 4 2 0 3 1
3 3 1 4 2 0
4 2 0 3 1 4

173∇ b
0 1 2 3 4
a 0 3 2 1 0 4
1 1 0 4 3 2
2 4 3 2 1 0
3 2 1 0 4 3
4 0 4 3 2 1

59∇ b
0 1 2 3 4
a 0 1 0 4 3 2
1 4 3 2 1 0
2 2 1 0 4 3
3 0 4 3 2 1
4 3 2 1 0 4

a 216∇ b a 88∇ b a 173∇ b a 59∇ b
— second step —

Δ b
0 1 2 3 4
a 0 4 2 1 2 2
1 2 4 4 3 4
2 3 0 3 4 3
3 0 3 0 0 1
4 1 1 2 1 0

Δ b
0 1 2 3 4
a 0 4 4 4 4 4
1 3 3 3 3 3
2 2 2 2 2 2
3 1 1 1 1 1
4 0 0 0 0 0

80∇ b
0 1 2 3 4
a 0 1 3 4 0 2
1 0 2 1 3 4
2 3 4 0 2 1
3 2 1 3 4 0
4 4 0 2 1 3

172∇ b
0 1 2 3 4
a 0 3 2 1 0 4
1 0 4 3 2 1
2 2 1 0 4 3
3 4 3 2 1 0
4 1 0 4 3 2

(a 216∇ b) 126∇ b (a 88∇ b) 151∇ b (a 173∇ b) 224∇ b (a 59∇ b) 219∇ b
— third step —

Δ b
0 1 2 3 4
a 0 2 1 0 4 2
1 0 0 1 0 4
2 3 3 2 3 1
3 1 4 4 2 3
4 4 2 3 1 0

120∇ b
0 1 2 3 4
a 0 2 3 0 1 4
1 0 1 4 2 3
2 4 2 3 0 1
3 3 0 1 4 2
4 1 4 2 3 0

Δ b
0 1 2 3 4
a 0 2 1 3 2 1
1 1 0 1 0 4
2 3 2 0 3 0
3 4 3 4 1 3
4 0 4 2 4 2

96∇ b
0 1 2 3 4
a 0 2 0 1 3 4
1 1 3 4 2 0
2 4 2 0 1 3
3 0 1 3 4 2
4 3 4 2 0 1

((a 216∇ b) 126∇ b) 83∇ b ((a 88∇ b) 151∇ b) 91∇ b ((a 173∇ b) 224∇ b) 85∇ b ((a 59∇ b) 219∇ b) 87∇ b

When V₀, V₁ and V₂ are any dozen-scores, the possible censuses of the operations at the three steps forming the straight configuration are the same as in the balanced case above, namely:

25·a
15·a, 5·b, 5·c
10·a, 10·b, 5·c
5·a, 5·b, 5·c, 5·d, 5·e

More can be said about the following straight configuration:

a V_R b = ((a V₂ b) V₁ b) V₀ b

where V₀, V₁ and V₂ are dozen-scores, as usual. If V_R is a dozen-score, then V₁ must be from quadraginta A.

The following list of ten dozen-scores comes from rows 4 and 8 of quadraginta A:

213∇ 150∇ 138∇ 72∇ 21∇
100∇ 88∇ 27∇ 219∇ 166∇

When V₁ is in this list:

If V₀ ≡ V₁, then V_R is a dozen-score for all V₂, and V_R ≡ V₂.
If V₁ ≡ V₂, then V_R is a dozen-score for all V₀, and V_R ≡ V₀.
If V₂ ≡ V₀, little conclusion can be drawn.

If V₀ ≡ V₁ ≡ V₂, then there are ten solutions, all from the list of ten selections from quadraginta A above.

§ 6d5. Two further three-operation compositions are in angled configurations:

a V₀ ((a V₂ b) V₁ b)
(a V₁ (a V₂ b)) V₀ b

which themselves are equivalent if the transposes of the operations are taken, and if a and b are exchanged, as before.

However, no simple conversion between the straight and angled configurations is available. The complication is that:

Within the straight configurations, one of a and b appears thrice and the other only once.
Within the angled configurations, each of a and b appears twice.

The lack of conversion is underscored by the fact that the angled configuration yields several censuses that do not appear in the straight configuration; this fact also means that there will be no simple conversion betweeen the angled and balanced configurations. Here is a list of the nine censuses that occur in the angled configuration; note that within each census, all the populations are equal in modulo 5:

25·a
15·a, 5·b, 5·c
10·a, 10·b, 5·c
5·a, 5·b, 5·c, 5·d, 5·e
11·a, 6·b, 6·c, 1·d, 1·e
6·a, 6·b, 6·c, 6·d, 1·e
7·a, 7·b, 7·c, 2·d, 2·e
8·a, 8·b, 3·c, 3·d, 3·e
9·a, 4·b, 4·c, 4·d, 4·e

A few examples are:

Δ b
0 1 2 3 4
a 0 3 3 3 3 3
1 3 3 3 3 3
2 3 3 3 3 3
3 3 3 3 3 3
4 3 3 3 3 3

Δ b
0 1 2 3 4
a 0 1 1 1 1 1
1 1 1 1 1 1
2 0 0 0 0 0
3 3 3 3 3 3
4 1 1 1 1 1

Δ b
0 1 2 3 4
a 0 4 1 4 3 1
1 4 3 1 4 1
2 1 4 1 4 3
3 1 4 3 1 4
4 3 1 4 1 4

Δ b
0 1 2 3 4
a 0 4 3 3 1 1
1 4 0 0 1 4
2 2 0 1 1 2
3 0 0 3 3 2
4 4 4 3 2 2

59∇ b
0 1 2 3 4
a 0 1 0 4 3 2
1 4 3 2 1 0
2 2 1 0 4 3
3 0 4 3 2 1
4 3 2 1 0 4

a174∇ ((a 21∇ b) 0∇ b)
25·3 a153∇ ((a 212∇ b) 48∇ b)
5·0, 15·1, 5·3 a142∇ ((a 11∇ b) 59∇ b)
10·1, 5·3, 10·4 a185∇ ((a 234∇ b) 27∇ b)
5·0, 5·1, 5·2, 5·3, 5·4 a166∇ ((a 26∇ b) 59∇ b)
5·0, 5·1, 5·2, 5·3, 5·4

Δ b
0 1 2 3 4
a 0 0 0 1 3 0
1 0 3 3 3 4
2 3 1 2 3 3
3 1 1 1 3 0
4 1 0 3 3 3

Δ b
0 1 2 3 4
a 0 4 2 0 4 2
1 3 2 0 2 0
2 4 3 4 0 3
3 2 3 4 2 4
4 3 0 3 0 1

Δ b
0 1 2 3 4
a 0 3 4 0 4 4
1 3 2 2 4 2
2 3 1 3 0 3
3 4 2 4 4 3
4 2 2 1 2 3

Δ b
0 1 2 3 4
a 0 4 4 4 0 3
1 4 0 2 2 2
2 4 4 3 2 4
3 3 2 2 2 0
4 1 4 2 1 1

Δ b
0 1 2 3 4
a 0 0 2 1 2 2
1 3 2 2 4 2
2 1 0 1 3 1
3 2 0 2 2 3
4 4 4 0 4 3

a101∇ ((a 142∇ b) 107∇ b)
6·0, 6·1, 1·2, 11·3, 1·4 a103∇ ((a 158∇ b) 233∇ b)
6·0, 1·1, 6·2, 6·3, 6·4 a112∇ ((a 23∇ b) 104∇ b)
2·0, 2·1, 7·2, 7·3, 7·4 a124∇ ((a 214∇ b) 102∇ b)
3·0, 3·1, 8·2, 3·3, 8·4 a135∇ ((a 23∇ b) 109∇ b)
4·0, 4·1, 9·2, 4·3, 4·4

(a V₀ b) V₀ b = a V_R b	a V₀ (a V₀ b) = a V_R b
V₀, V_R	V₀, V_R
0∇₁₁, 27∇₈₃	20∇₇₁, 47∇₂₃
192∇₁₂, 100∇₈₁	212∇₇₂, 110∇₂₁
180∇₁₃, 219∇₈₄	151∇₇₃, 239∇₂₄
129∇₁₄, 88∇₈₂	139∇₇₄, 59∇₂₂
66∇₁₅, 166∇₈₅	73∇₇₅, 173∇₂₅
238∇₆₁, 138∇₄₃	213∇₄₁, 111∇₆₃
172∇₆₂, 213∇₄₁	150∇₄₂, 238∇₆₁
111∇₆₃, 72∇₄₄	138∇₄₃, 58∇₆₄
58∇₆₄, 150∇₄₂	72∇₄₄, 172∇₆₂
46∇₆₅, 21∇₄₅	21∇₄₅, 46∇₆₅