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 Vn 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.
|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:
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.
|(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:
Some compositions exhibit partial or full cancellativity:
|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 VR are dozen-scores being sought:
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 VR b||a V0 (a V0 b) = a VR b|
|V0, VR||V0, VR|
|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 VR 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 a Z (a Y b) = a X b then (a YT b) ZT b = a XT 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:
|(a 42∇ b) 24∇ (a 155∇ b)|
|(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:
There are three censuses that do not occur:
However, operations that have five instances of each value are plentiful and come in a variety of patterns not easy to categorize:
|(a 58∇ b) 35∇ (a 78∇ b)||(a 49∇ b) 38∇ (a 197∇ b)||(a 40∇ b) 43∇ (a 60∇ b)|
|(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:
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 —|
|a 216∇ b||a 88∇ b||a 173∇ b||a 59∇ b|
|— second step —|
|(a 216∇ b) 126∇ b||(a 88∇ b) 151∇ b||(a 173∇ b) 224∇ b||(a 59∇ b) 219∇ b|
|— third step —|
|((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:
More can be said about the following straight configuration:
a VR b = ((a V2 b) V1 b) V0 b
where V0, V1 and V2 are dozen-scores, as usual. If VR 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:
When V1 is in this list:
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:
A few examples are:
|a174∇ ((a 21∇ b) 0∇ b)|
|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
|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