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@afflom
Created June 15, 2025 20:43
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Coherence Object Calculus Example in Python
# Computational demonstration of two foundational lemmas in Coherent Object Calculus (COC)
# 1. channel_decomposition_bijective for n up to 1,000,000
# 2. resonance_well_defined for all v in 0..255
# --- Helper definitions mirrored from the spec -------------------------------
def channel_decompose(n: int):
"""Return the base‑256 little‑endian byte list of a positive integer n (n>0)."""
if n <= 0:
raise ValueError("n must be positive")
lst = []
while n:
lst.append(n % 256)
n //= 256
return lst
def reconstruct(bytes_le):
"""Reconstruct an integer from its little‑endian base‑256 expansion."""
n = 0
for i, b in enumerate(bytes_le):
n += b * (256 ** i)
return n
# Constants α₁ … α₈
constants = [
1.0, # α₁ unity
1.839287, # α₂ Tribonacci
1.618034, # α₃ golden ratio
0.5, # α₄ adelic threshold
0.159155, # α₅ interference null
6.283185, # α₆ 2π scale transition
0.199612, # α₇ phase coupling
14.134725 # α₈ Riemann resonance
]
def bit_pattern(n: int, i: int) -> bool:
"""Return True if bit i of n is 1."""
return ((n >> i) & 1) == 1
def resonance(v: int) -> float:
"""Product ∏ constants[i] over active bits of v (0 ≤ v < 256)."""
prod = 1.0
for i in range(8):
if bit_pattern(v, i):
prod *= constants[i]
return prod
# --- Lemma 1: channel_decomposition_bijective (finite verification) ----------
MAX_N = 1_000_000
bijective_pass = all(reconstruct(channel_decompose(n)) == n for n in range(1, MAX_N + 1))
# --- Lemma 2: resonance_well_defined (all v in 0..255) -----------------------
resonance_positive_pass = all(resonance(v) > 0 for v in range(256))
# --- Output results ----------------------------------------------------------
print(f"Lemma channel_decomposition_bijective holds for all n in [1, {MAX_N}]: {bijective_pass}")
print(f"Lemma resonance_well_defined holds for all v in 0..255: {resonance_positive_pass}")
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