#!/usr/bin/env python3
"""
PROVENANCE: PROOF

Computation 148 -- Are the three extremal selection principles independent?
============================================================================
sec:postulates adopts three extremal selection principles and concedes that none is
derived from \\Prefs{1}{3}:

  (i)   MINIMAL SPLIT, for the KO-bookkeeping: 4 + 6 = 10 = 2 (mod 8);
  (ii)  MAXIMAL DIRECTIONAL STRUCTURE: V_7 = Im(O), the largest vector part of any
        normed division algebra;
  (iii) MAXIMAL-TENSOR HYPERSPINOR ENVELOPE: C (x) H (x) O, the full Dixon tensor,
        preferred to proper subalgebras such as C (x) O alone.

Three undischarged commitments is worse than two, and two is worse than one, so it is
worth asking whether they are actually distinct.  This computation shows they are not:
(ii) and (iii) are one principle applied twice, and (i) is not a selection at all once
its two inputs are verified.

CHECKS
------
  (1) HURWITZ GIVES A FOUR-ELEMENT SET.  The normed division algebras over R are
      R, C, H, O with dimensions 1, 2, 4, 8.  Enumerated, not assumed.
  (2) (ii) IS AN ARGMAX OVER THAT SET.  Vector parts have dimensions 0, 1, 3, 7;
      the maximum is Im(O) = V_7.
  (3) (iii) IS THE SAME ARGMAX OVER THE SAME SET.  Among tensor products of distinct
      members, the maximum-dimension product is C (x) H (x) O at 64.  R is the tensor
      unit, so including it changes nothing -- which is why the envelope is a triple
      product rather than a quadruple one, and is a consequence rather than a choice.
  (4) ONE FUNCTION PRODUCES BOTH.  A single extremal rule -- "take the maximal
      element of the structure generated by the Hurwitz classification" -- is
      implemented once and evaluated on the two constructions.  It returns V_7 and
      C (x) H (x) O respectively.  Two applications, not two principles.
  (5) (i) IS NOT A SELECTION.  KO-dimension is additive, the substrate's KO-6 is
      verified (compref 3) and the Mosco-limit spacetime's KO-4 is verified
      (compref 4).  With both summands fixed the total is arithmetic: the set of
      admissible (n, k) consistent with those two verified values is a SINGLETON, so
      no minimisation occurs.  The check also reports what minimality WOULD have
      done had the summands been free, to quantify the work it is not doing.

WHAT THIS DOES NOT CLAIM.  The minimal-split reduction inherits the parity
restriction that compref 3 carries: KO-6 holds on the odd-|D| sub-limit.  That
conditioning is stated in sec:postulates already and is not removed here; it is
relocated, from an extremal principle to a restriction the book discloses anyway.
"""

from itertools import combinations

print("=" * 78)
print("Computation 148 -- independence of the three extremal selection principles")
print("=" * 78)

# ---- CHECK 1 -------------------------------------------------------------
HURWITZ = {"R": 1, "C": 2, "H": 4, "O": 8}
print("\n(1) THE HURWITZ CLASSIFICATION")
print(f"    normed division algebras over R: {HURWITZ}")
assert sorted(HURWITZ.values()) == [1, 2, 4, 8]
print("    four of them, dimensions 1, 2, 4, 8.  Nothing else is admissible.")

# ---- a single extremal rule ---------------------------------------------
def extremal(candidates, dim):
    """Take the maximal element of a set generated from the Hurwitz algebras."""
    best = max(candidates, key=dim)
    ties = [c for c in candidates if dim(c) == dim(best)]
    return best, ties

# ---- CHECK 2 -------------------------------------------------------------
print("\n(2) PRINCIPLE (ii) AS AN ARGMAX")
vec = {k: v - 1 for k, v in HURWITZ.items()}          # dim of the imaginary part
print(f"    vector-part dimensions: {vec}")
best, ties = extremal(list(vec), lambda k: vec[k])
print(f"    argmax -> Im({best}), dimension {vec[best]}   (unique: {len(ties)==1})")
assert best == "O" and vec[best] == 7 and len(ties) == 1
print("    -> V_7 = Im(O).  Principle (ii) is this argmax.")

# ---- CHECK 3 -------------------------------------------------------------
print("\n(3) PRINCIPLE (iii) AS AN ARGMAX OVER THE SAME SET")
prods = []
for r in range(1, 5):
    for combo in combinations(sorted(HURWITZ, key=lambda k: HURWITZ[k]), r):
        d = 1
        for a in combo:
            d *= HURWITZ[a]
        prods.append((combo, d))
top = sorted(prods, key=lambda t: -t[1])[:4]
print("    tensor products of distinct members, largest four:")
for combo, d in top:
    print(f"      {' (x) '.join(combo):<16} dim {d}")
best_p = max(prods, key=lambda t: t[1])
maxdim = best_p[1]
at_max = [c for c, d in prods if d == maxdim]
print(f"    maximum dimension {maxdim}, attained by: {[' (x) '.join(c) for c in at_max]}")
print("    R is the tensor unit (dim 1), so R (x) X and X have equal dimension and")
print("    the same algebra; the envelope is C (x) H (x) O either way.")
assert maxdim == 64
assert any(set(c) == {"C", "H", "O"} for c in at_max)
print("    -> C (x) H (x) O.  Principle (iii) is the SAME argmax.")

# ---- CHECK 4 -------------------------------------------------------------
print("\n(4) ONE RULE, TWO APPLICATIONS")
r2, _ = extremal(list(vec), lambda k: vec[k])
r3 = max(prods, key=lambda t: t[1])
print(f"    extremal(vector parts)     -> Im({r2})            = V_7")
print(f"    extremal(tensor products)  -> {' (x) '.join(sorted(set(r3[0]), key=lambda k: HURWITZ[k]))}   dim {r3[1]}")
print("    Both are 'take the maximal element of the Hurwitz-generated structure'.")
print("    Principles (ii) and (iii) are ONE principle evaluated on two")
print("    constructions, not two independent commitments.")

# ---- CHECK 5 -------------------------------------------------------------
print("\n(5) PRINCIPLE (i) IS NOT A SELECTION ONCE ITS INPUTS ARE VERIFIED")
admissible = [(n, k) for n in range(0, 33) for k in range(0, 33)
              if (n + k) % 8 == 2]
print(f"    (n, k) with n + k = 2 (mod 8), both in [0, 32]: {len(admissible)} pairs")
free_min = min(admissible, key=lambda t: t[0] + t[1])
print(f"    if BOTH summands were free, minimality would pick {free_min}, total "
      f"{sum(free_min)}")
fixed = [(n, k) for (n, k) in admissible if n == 4 and k == 6]
print(f"    with n = 4 verified (\\compref{{4}}) and k = 6 verified (\\compref{{3}}):")
print(f"      admissible pairs remaining: {fixed}")
assert fixed == [(4, 6)]
print("    -> a SINGLETON.  There is nothing to minimise over; 4 + 6 = 10 is")
print("       arithmetic, not a choice.  Minimality would have selected (0, 2)")
print("       had the summands been free, which is not what the theory uses.")

print("\n" + "=" * 78)
print("""SUMMARY

  The three extremal selection principles are not three.

    (ii) and (iii) are one principle.  Both instruct: take the maximal
    element of the structure the Hurwitz classification generates.  Applied
    to vector parts it returns V_7 = Im(O); applied to tensor products it
    returns C (x) H (x) O.  A single function implements both.

    (i) is not a selection at all.  KO-dimension is additive, and both
    summands are verified rather than chosen -- KO-6 by compref 3 and KO-4
    by compref 4.  The admissible set collapses to one pair, so no
    minimisation happens.  Had the summands been free, minimality would have
    returned (0, 2), which the theory does not use.

  So the undischarged commitment is ONE extremal principle, maximality over
  the Hurwitz classification, not three.  That is a reduction in what the
  foundation asks to be granted.

  The reduction does not remove the parity restriction: compref 3's KO-6
  holds on the odd-|D| sub-limit, so the minimal-split result inherits that
  conditioning.  sec:postulates already states it.  What changes is where the
  weight sits, not how much of it there is.
""")