Arithmetic representation of symbols according to certain rules

[jbriggs444]

View attachment 303541

This is amazing thank you very much.
These rules are the same local rules of Penrose tiling. No wondering the number 5 pops up.

You are correct that the cycle size of five seems to be consistent. I had not expected that.

C:\Users\John\Documents>cipher.pl
String length is 20
Generation limit is 17.388384878619 << Tuned the estimate further downward.
Generation Cipher Content
0 A
1 B
2 C
3 DE
4 FCHI
5 GADEJD
6 CBFCHIDFC
7 DECGADEJDFCGADE
8 FCHIDECBFCHIDFCGADEC
9 GADEJDFCHIDECGADEJDF
10 CBFCHIDFCGADEJDFCHID
11 DECGADEJDFCGADECBFCH
12 FCHIDECBFCHIDFCGADEC
13 GADEJDFCHIDECGADEJDF
14 CBFCHIDFCGADEJDFCHID
15 DECGADEJDFCGADECBFCH
16 FCHIDECBFCHIDFCGADEC
17 GADEJDFCHIDECGADEJDF
18 CBFCHIDFCGADEJDFCHID

 

[jbriggs444]

*BLUSH*. This is actually a cycle period of 4.

Danged fencepost errors.

 

[Adel Makram]

Now you found the period of the cycle is 4. Is it possible to run your program to find the minimum length of symbols in the same generation that repeats itself? I know that Penrose tiling is not periodic so there may be no period. But I just want to study the statistical transition from one symbol to another in the same generation. So, having that “wavelength” will allow me to draw exact figures of the Markov matrix.

 

[jbriggs444]

Now you found the period of the cycle is 4. Is it possible to run your program to find the minimum length of symbols in the same generation that repeats itself? I know that Penrose tiling is not periodic so there may be no period. But I just want to study the statistical transition from one symbol to another in the same generation. So, having that “wavelength” will allow me to draw exact figures of the Markov matrix.

Let me try to understand the request.

Instead of going vertically down the rows (generation after generation) looking for repeated rows, you want to sweep horizontally across a large string size looking for periodicity in the rows?

So basically, I need to iterate forward enough generations to reach the stable cycle, pick a row and then look for a cycle length in that row.

As is my usual habit, I will proceed a step at a time, working incrementally toward the goal. We can go for 1,000,000 length strings and iterate enough times to get to the eventual cycle. That much is a pretty easy tweak to the code.

If I can trust the resulting code, we get a one million character string at generation 31. I looked for periodicity in generation 32 and found none. That is, no leading sustring of length up to 500,000 was repeated after a period of between 1 and 500000.

The million character string in generation 32 was strongly patterned, but not periodic.

 

[Adel Makram]

Instead of going vertically down the rows (generation after generation) looking for repeated rows, you want to sweep horizontally across a large string size looking for periodicity in the rows?

Yes, this is correct.

 

[pbuk]

Is it possible to run your program to find the minimum length of symbols in the same generation that repeats itself?

I think if you are going to make any progress in this subject you should learn to do these things for yourself. If you don't like the look of @jbriggs444's Perl code, try it in Python: you can run it for yourself without installing anything at https://replit.com/@pbuk/StingyInfiniteNlp#main.py, and if you sign up for a repl.it account you can even try some edits.

# See https://www.physicsforums.com/threads/arithmetic-representation-of-symbols-according-to-certain-rules.1016467/

replacements = {
    'A': 'B',
    'B': 'C',
    'C': 'DE',
    'D': 'FC',
    'F': 'GA',
    'G': 'C',
    'E': 'HI',
    'H': 'J',
    'I': 'D',
    'J': 'D',
}

def transform(input):
    ''' Transform the input according to the rules '''
    output = ''
    for char in input:
        output += replacements[char]
    return output

truncateAt = 10

# Set the initial conditions.
current = 'A'
history = []

# This concise loop does all the work!
while not current in reversed(history):
    history.append(current)
    current = transform(current)[:truncateAt]

# Report the results.
cycle = len(history) - history.index(current)
count = len(history)
history.append(current)
print('Cycle length', cycle, 'after', count, 'iterations')

if truncateAt <= 100:
    for index, item in enumerate(history):
        if index == count - cycle or index == count:
            print(index, item, '<<<')
        else:
            print(index, item)