1. Choose Address N ≥ 1 or Coordinates (m, r)
Diagonal (V = 0)
Arithmetic Witness Presets:
| Input Address N | 25 |
|---|---|
| Chamber m | 7 |
| Previous Triangle Tm−1 | T_6 = 21 |
| Chamber Triangle Tm | T_7 = 28 |
| Row r = N − Tm−1 | 4 (1 ≤ r ≤ 7) |
| Complementary Column c = m + 1 − r | 4 |
| Centered Offset V = r − c = 2r − m − 1 | 0 (m + V = 7 odd) |
| Odd Legs (r̃, c̃) = (2r−1, 2c−1) | (7, 7), r̃ + c̃ = 14 = 2m |
| Centered Identity 2N = m² + V + 1 | 2(25) = 7² + 0 + 1 = 50 |
| Ceiling Inverse ⌈(√(8N+1)−1)/2⌉ | ⌈(√201 − 1)/2⌉ = ⌈6.5887⌉ = 7 |
| Nearest-Root Inverse ⌊√(2N)+1/2⌋ | ⌊√50 + 0.5⌋ = ⌊7.5711⌋ = 7 |
Why the Two Inverses Always Agree (Proposition 1.2):
Because Tm−1 < N ≤ Tm holds if and only if
Because Tm−1 < N ≤ Tm holds if and only if
(2m − 1)² < 8N < (2m + 1)² (strict inequalities since 8N ≡ 0 (mod 8) while odd squares are ≡ 1 (mod 8)), taking square roots yields
m − 1/2 < √(2N) < m + 1/2 with zero half-integer ties.