def las(s):
    out=[]; i=0; n=len(s)
    while i<n:
        j=i
        while j<n and s[j]==s[i]: j+=1
        out.append(str(j-i)); out.append(s[i]); i=j
    return ''.join(out)

# INSTRUMENT 1 (statistic): growth ratio + digit shares. Preserves SIZE.
# INSTRUMENT 2 (reading): the raw string. Preserves LOCAL TEXTURE / ORDER.
# Their incompatibility, made exact: does the string ever REPEAT a long
# block (texture question) while the ratio marches to lambda (size q)?
s="1"
for _ in range(35): s=las(s)
print("len at step 35:", len(s))

# I1 view: one number.
from collections import Counter
c=Counter(s); t=len(s)
print("I1  shares 1/2/3: %.4f %.4f %.4f"%(c['1']/t,c['2']/t,c['3']/t))

# I2 view: read texture. longest run of a single digit anywhere?
maxrun=1; cur=1
for i in range(1,len(s)):
    if s[i]==s[i-1]: cur+=1; maxrun=max(maxrun,cur)
    else: cur=1
print("I2  longest single-digit run in whole string:", maxrun)

# THE THIRD INSTRUMENT: Conway proved the string splits into ~92
# independent "atoms" (Cosmological Theorem) that never interact.
# So: is the WHOLE giant string covered by a SMALL set of short blocks
# that regenerate each other? Third instrument = measure the ATOM
# DECOMPOSITION: how many distinct maximal substrings that (a) evolve
# independently. Cheap proxy: after many steps, count distinct length-k
# substrings vs a random ternary string of same length+shares.
import random
def distinct_kmers(string,k):
    return len(set(string[i:i+k] for i in range(len(string)-k+1)))
random.seed(1)
# build a shuffled control with identical digit multiset
ctrl=list(s); random.shuffle(ctrl); ctrl=''.join(ctrl)
for k in [4,6,8,10,12]:
    print("k=%2d  look-and-say distinct=%6d   shuffled-control=%6d   ratio=%.3f"%(
        k, distinct_kmers(s,k), distinct_kmers(ctrl,k),
        distinct_kmers(s,k)/max(1,distinct_kmers(ctrl,k))))