More investigations

The set of the resulting numbers by only first step of the algorithm is always smaller than the set of all the N-figure numbers.

For example, in the 4-figure number, obviously the set of all the 4-figure numbers is from 0000 to 9999. There are 10000 elements in the set.

On the other hand, after one step of the above algorithm, we obtain only the following 44 numbers:

0000 *
5544 9810 9621 8532 7641 7641 *
8442 9954 5553 9981 8820 8532 
6444 9981 
8622 6543 8730 8532 
7632 6552 9963 6642 7641 
9531 8721 7443 9963 
7533 7641 
7551 9954 
9711 8532 
9441 9972 7731 6543 
8550 9972 
8640 8721 
9990 9981

We can expect the algorithm is a kind of contraction mapping which makes the number of elements of the set decrease.

The reason why is as follows.

In general case (p-adic, N-figures), L is given by

 

S is given by

 

With the expression (5) and (6), L-S is given by

 

The meaning of (7) can be understood by the following procedure:

(a)

Make a pyramidal arrangement of 1s as follows.

e.g. In the case of N=8,

        0001000
        0011100
        0111110
        1111111

In the case of N=4,

        010
        111

(b)

Calculate the difference of a pair of the numbers which are located in symmetrical places.

e.g. In the case of N=4, L=7641

The symmetrical pairs are (7,1) and (6,4) as follows:

                               +--+----->(6, 4)
                               |  |
                            7  6  4  1
                            |        |
                            +--------+-->(7, 1)

The differences are

(c)

Multiply (by difference numbers) the raws of 1-pyramid given in (a), then sum up the pyramid. By multiplying the number by p-1 value, you can get the result of the algorithm.

e.g. In the above case in (b),

      010 X (6-4) = 020       (6-4)=2  <=  6=(7-1)
+)    111 X (7-1) = 666
-----------------------
                    686  --------> 686*(10-1)=6174
                                        ~~=p

e.g. In the case of N=5, thus ,

      0110 X (b-d) = u     (b-d)<=(a-e)
+)    1111 X (a-e) = v
----------------------
                     A --------> A*(p-1)=L-S

(As this example, if N is odd number, the center value is not necessary for the calculation.)

The equation (7) or the above explained meaning of the equation looks very complicated compare to the original algorithm. But we can understand that the numbers which appear in the calculations after the second step always have the divisor p-1 ( therefore, in the case of decimal, the numbers are always divided by 9 (=p-1=10-1)). And the number of pairs is less than half of the former number of elements.

Based on (7), I calculate the several cases N=2,...,6 and p=2,...16. The result is drawn by ordered network in postscript. You can access the results by clicking the numbers which implies the number of attractors.

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base (p)	figures (N)
	2	3	4	5	6
binary (2)	1 2	1 2	1 2 3	1 2 3	1 2 3 4
triadic (3)	1	1 2	1 2	1 2	1 2
4-adic (4)	1 2	1 2	1 2 3	1 2	1 2 3 4
5-adic (5)	1 2	1 2	1 2	1 2	1 2
6-adic (6)	1 2	1 2	1 2	1 2 3	1 2 3 4 5
7-adic (7)	1	1 2	1 2	1 2	1 2
octal (8)	1 2 3	1 2	1 2 3	1 2 3	1 2 3 4
9-adic (9)	1 2	1 2	1 2 3	1 2 3	1 2
decimal (10)	1 2	1 2	1 2 <=7641	1 2 3 4	1 2 3 4
11-adic (11)	1 2	1 2	1 2 3	1 2	1 2
12-adic (12)	1 2	1 2	1 2 3	1 2 3	1 2 3
13-adic (13)	1 2	1 2	1 2 3 4	1 2	1 2
14-adic (14)	1 2 3 4	1 2	1 2	1 2 3	1 2 3 4 5
15-adic (15)	1	1 2	1 2 3 4 5 6 7 8	1 2	1 2 3
hexa (16)	1 2 3	1 2	1 2 3 4 5	1 2 3 4	1 2 3 4 5 6 7

For example, try to see the attractor number 2 of decimal N=4.