Enzymind Labs

2007-08-30 Ted'd notes and correspondance on finding where and adjacent pair of symbols occurs:

You might check the math for that particular one, the sequence 11 might show up earlier, however i'm confident that for all cycles like that,for however many symbols, all pairs will occur.

I have some notes on this a while back when we were considering them.
An example for 5 symbols

starting with 'chunks that are n⁰, n¹, n2, etc we can write the following chart
a chunk is either of size 1, or 5 e.g. 12345, or 25 12345,23451,34512,45123,51234, etc...
i'm only going to write the beginning and ending of a sequence chunk, so 12345 becomes 15


0 5 25 125 625

1 15 14 13 12
2 21 25 24 23
3 32 31 35 34
4 43 42 41 45
5 54 53 52 51

This table can be used to construct the adjacencies
for order 0: 12 23, 34, 45, 51 (note that 51 doesn't occur in the first chunk of 5, but is part of the second, 3rd,4th,5th chunks of 5

for order 1 we get 52,13,24,35, and 41

we can make a new table of just adjacencies

12 52 42 32 22
23 13 53 43 33
34 24 14 54 44
45 35 25 15 55
51 41 31 21 11

All cycles of this form can generate tables like the above. There is an orderly progression of adjacencies and they 'synch' at some point (n^n)

the three symbol tables
chunks:
1 13 12
2 21 23
3 32 31
adjacencies

12 32 22
23 13 33
31 21 11

t


On Thu, 30 Aug 2007 13:37:12 -0700, MarkDow <dow@uoregon.edu> wrote:

Ah yes, I see the logic now. I'll have to check your math to convince
myself. I've wondered about the general question "how long does a
sequence have to be before all possible adjacent pairs occurs once.
Based on your calculation one needs a 512x512 tiling before including
the 1111 square at the center. The one I sent is 64x64 I think.

Theodore A. Bell wrote:

You may be right, but are you sure the 1111 square occurs? When? Nice

that you did this "computationally", in your head.

I'm not sure it occurs in your map,

but, if the sequence 11 occurs, so does the square as long as we have the same axes

in the base sequences, we have the following fundamental 'pairs'

12

23

34

41

but, since this is fractal, we also have pairs of high level sequences

1234 is followed by 2341 is followed by 3412 is followed by 4123 is followed by 1234

this creates the next level of adjacencies created by the last member of the first set and the first member of the next set

42

13

24

31

the next level is

1234234134124123

2341341241231234

3412412312342341

4123123423413412

creating the set of adjacencies

32

43

14

21

likewise the next level is

1234234134124123, 2341341241231234, 3412412312342341, 4123123423413412

...etc

giving:

22

33

44

11

so we get a 22 pair at the 64-65 boundary

and a 11 pair finally at 256-257 i think...

T

These solutions appears along the diagonal twice in this figure. The

smallest is :

aha!

1 4 1

4 2/3 4

1 4 1