# WORKSHEET G3 — Eigensolid Fixed Point
## Verifiable with any calculator. No English inside formulas.

---

## PART A: The Crossing Operator C (applied numerically)

**INPUT:** p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)
**Check:** 0.3+0.1+0.15+0.05+0.2+0.08+0.07+0.05 = 1.0 ✓

**FORMULA:**
```
C(p)₁ = C(p)₂ = (p₁ + p₂) / 2
C(p)₃ = C(p)₄ = (p₃ + p₄) / 2
C(p)₅ = C(p)₆ = (p₅ + p₆) / 2
C(p)₇ = C(p)₈ = (p₇ + p₈) / 2
```

**WORK:**
```
C(p)₁ = C(p)₂ = (0.3 + 0.1) / 2 = 0.4 / 2 = 0.2
C(p)₃ = C(p)₄ = (0.15 + 0.05) / 2 = 0.2 / 2 = 0.1
C(p)₅ = C(p)₆ = (0.2 + 0.08) / 2 = 0.28 / 2 = 0.14
C(p)₇ = C(p)₈ = (0.07 + 0.05) / 2 = 0.12 / 2 = 0.06
```

**OUTPUT:** C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

**Check:** 0.2+0.2+0.1+0.1+0.14+0.14+0.06+0.06 = 1.0 ✓

---

## PART B: Idempotence C∘C = C (verified numerically)

**INPUT:** C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

**WORK:**
```
C(C(p))₁ = C(C(p))₂ = (0.2 + 0.2) / 2 = 0.4 / 2 = 0.2
C(C(p))₃ = C(C(p))₄ = (0.1 + 0.1) / 2 = 0.2 / 2 = 0.1
C(C(p))₅ = C(C(p))₆ = (0.14 + 0.14) / 2 = 0.28 / 2 = 0.14
C(C(p))₇ = C(C(p))₈ = (0.06 + 0.06) / 2 = 0.12 / 2 = 0.06
```

**OUTPUT:** C(C(p)) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

**VERIFICATION:** C(C(p)) = C(p) ✓

**RESULT:** C∘C = C. One application reaches the fixed point.

---

## PART C: Image M ≅ Δ₃ (verified numerically)

**Image M:** All vectors with p₁=p₂, p₃=p₄, p₅=p₆, p₇=p₈.

**Map from Δ₃ to M:**
```
(q₁, q₂, q₃, q₄) ↦ (q₁/2, q₁/2, q₂/2, q₂/2, q₃/2, q₃/2, q₄/2, q₄/2)
```

**TEST:** (q₁, q₂, q₃, q₄) = (0.4, 0.2, 0.28, 0.12)
**Check:** 0.4 + 0.2 + 0.28 + 0.12 = 1.0 ✓

**WORK:**
```
φ(q) = (0.4/2, 0.4/2, 0.2/2, 0.2/2, 0.28/2, 0.28/2, 0.12/2, 0.12/2)
     = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)
```

**OUTPUT:** φ(0.4, 0.2, 0.28, 0.12) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

**VERIFICATION:** This equals C(p) from Part A. ✓

**Inverse map:**
```
ψ(p₁, p₂, ..., p₈) = (2p₁, 2p₃, 2p₅, 2p₇)
```

**TEST:** ψ(0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)
```
= (2×0.2, 2×0.1, 2×0.14, 2×0.06)
= (0.4, 0.2, 0.28, 0.12)
```

**VERIFICATION:** ψ(φ(q)) = q ✓ and φ(ψ(C(p))) = C(p) ✓

---

## PART D: Contraction (verified numerically)

**INPUT:** p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)
**INPUT:** q = (0.2, 0.2, 0.1, 0.1, 0.15, 0.1, 0.1, 0.05)

**From Part A:** C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)

**Compute C(q):**
```
C(q)₁ = C(q)₂ = (0.2 + 0.2) / 2 = 0.2
C(q)₃ = C(q)₄ = (0.1 + 0.1) / 2 = 0.1
C(q)₅ = C(q)₆ = (0.15 + 0.1) / 2 = 0.125
C(q)₇ = C(q)₈ = (0.1 + 0.05) / 2 = 0.075
```

**OUTPUT:** C(q) = (0.2, 0.2, 0.1, 0.1, 0.125, 0.125, 0.075, 0.075)
**Check:** 0.2+0.2+0.1+0.1+0.125+0.125+0.075+0.075 = 1.0 ✓

---

**STEP 1: Compute d_F(p, q)**

```
√(p₁q₁) = √(0.3×0.2) = √0.06   = 0.24495
√(p₂q₂) = √(0.1×0.2) = √0.02   = 0.14142
√(p₃q₃) = √(0.15×0.1) = √0.015 = 0.12247
√(p₄q₄) = √(0.05×0.1) = √0.005 = 0.07071
√(p₅q₅) = √(0.2×0.15) = √0.03  = 0.17321
√(p₆q₆) = √(0.08×0.1) = √0.008 = 0.08944
√(p₇q₇) = √(0.07×0.1) = √0.007 = 0.08367
√(p₈q₈) = √(0.05×0.05) = √0.0025= 0.05000
```

**Sum:**
```
S_pq = 0.24495 + 0.14142 + 0.12247 + 0.07071
     + 0.17321 + 0.08944 + 0.08367 + 0.05000
     = 0.97587
```

**d_F(p,q) = 2·arccos(0.97587)**

Type into calculator: `2 * arccos(0.97587)`

---

**STEP 2: Compute d_F(C(p), C(q))**

```
√(C(p)₁·C(q)₁) = √(0.2×0.2)  = 0.2
√(C(p)₂·C(q)₂) = √(0.2×0.2)  = 0.2
√(C(p)₃·C(q)₃) = √(0.1×0.1)  = 0.1
√(C(p)₄·C(q)₄) = √(0.1×0.1)  = 0.1
√(C(p)₅·C(q)₅) = √(0.14×0.125) = √0.0175 = 0.13229
√(C(p)₆·C(q)₆) = √(0.14×0.125) = 0.13229
√(C(p)₇·C(q)₇) = √(0.06×0.075) = √0.0045 = 0.06708
√(C(p)₈·C(q)₈) = √(0.06×0.075) = 0.06708
```

**Sum:**
```
S_CpCq = 0.2 + 0.2 + 0.1 + 0.1 + 0.13229 + 0.13229 + 0.06708 + 0.06708
       = 0.99874
```

**d_F(C(p), C(q)) = 2·arccos(0.99874)**

Type into calculator: `2 * arccos(0.99874)`

---

**STEP 3: Compare**

| Value | Calculator Input | Expected |
|-------|-----------------|----------|
| arccos(0.97587) | `arccos(0.97587)` | ~0.2197 |
| arccos(0.99874) | `arccos(0.99874)` | ~0.0502 |
| d_F(p,q) | `2*arccos(0.97587)` | ~0.4394 |
| d_F(C(p),C(q)) | `2*arccos(0.99874)` | ~0.1004 |

**VERIFICATION:** d_F(C(p), C(q)) ≈ 0.1004 < d_F(p,q) ≈ 0.4394

**CONTRACTION CONFIRMED:** C shrinks Fisher distance.

**Contraction ratio:** 0.1004 / 0.4394 ≈ 0.228

Type: `0.1004 / 0.4394` → ~0.23

---

## PART E: Strict Inequality (when p, q differ within a pair)

**INPUT:** p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)
**INPUT:** r = (0.1, 0.3, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)

p and r differ only within pair 1: (0.3, 0.1) vs (0.1, 0.3).

**C(p) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)**
**C(r) = (0.2, 0.2, 0.1, 0.1, 0.14, 0.14, 0.06, 0.06)**

C(p) = C(r) because C averages each pair.

**d_F(C(p), C(r)) = 0**

**d_F(p, r):**
```
√(0.3×0.1) = √0.03 = 0.17321
√(0.1×0.3) = √0.03 = 0.17321
√(0.15×0.15) = 0.15
√(0.05×0.05) = 0.05
√(0.2×0.2) = 0.2
√(0.08×0.08) = 0.08
√(0.07×0.07) = 0.07
√(0.05×0.05) = 0.05

Sum = 0.17321 + 0.17321 + 0.15 + 0.05 + 0.2 + 0.08 + 0.07 + 0.05
    = 0.94642
```

**d_F(p,r) = 2·arccos(0.94642)**

Type into calculator: `2 * arccos(0.94642)`

Expected: ~0.66 (significantly > 0)

**VERIFICATION:** d_F(C(p), C(r)) = 0 < 0.66 = d_F(p,r)

**STRICT INEQUALITY CONFIRMED:** When p, r differ within a pair, C collapses them completely.

---

## PART F: Information Loss (computed numerically)

**FORMULA:**
```
I_loss(p) = Σₖ₌₁⁴ sₖ · [ (p_{2k-1}/sₖ)·ln((p_{2k-1}/sₖ)/(1/2)) + (p_{2k}/sₖ)·ln((p_{2k}/sₖ)/(1/2)) ]
```

where sₖ = p_{2k-1} + p_{2k}.

**INPUT:** p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)

**Pair 1:** s₁ = 0.3 + 0.1 = 0.4
```
a = 0.3/0.4 = 0.75, b = 0.1/0.4 = 0.25
KL = 0.75·ln(0.75/0.5) + 0.25·ln(0.25/0.5)
   = 0.75·ln(1.5) + 0.25·ln(0.5)
   = 0.75·0.40547 + 0.25·(-0.69315)
   = 0.30410 - 0.17329
   = 0.13081
```
Term 1: s₁ × KL = 0.4 × 0.13081 = 0.05232

**Pair 2:** s₂ = 0.15 + 0.05 = 0.2
```
a = 0.15/0.2 = 0.75, b = 0.05/0.2 = 0.25
KL = 0.75·ln(1.5) + 0.25·ln(0.5)
   = 0.13081 (same as above)
```
Term 2: s₂ × KL = 0.2 × 0.13081 = 0.02616

**Pair 3:** s₃ = 0.2 + 0.08 = 0.28
```
a = 0.2/0.28 = 0.71429, b = 0.08/0.28 = 0.28571
KL = 0.71429·ln(0.71429/0.5) + 0.28571·ln(0.28571/0.5)
   = 0.71429·ln(1.42858) + 0.28571·ln(0.57142)
   = 0.71429·0.35668 + 0.28571·(-0.55962)
   = 0.25477 - 0.15989
   = 0.09488
```
Term 3: s₃ × KL = 0.28 × 0.09488 = 0.02657

**Pair 4:** s₄ = 0.07 + 0.05 = 0.12
```
a = 0.07/0.12 = 0.58333, b = 0.05/0.12 = 0.41667
KL = 0.58333·ln(0.58333/0.5) + 0.41667·ln(0.41667/0.5)
   = 0.58333·ln(1.16667) + 0.41667·ln(0.83333)
   = 0.58333·0.15415 + 0.41667·(-0.18232)
   = 0.08992 - 0.07597
   = 0.01395
```
Term 4: s₄ × KL = 0.12 × 0.01395 = 0.00167

---

**TOTAL:**
```
I_loss(p) = 0.05232 + 0.02616 + 0.02657 + 0.00167
          = 0.10672 nats
```

**In bits:** 0.10672 / ln(2) = 0.10672 / 0.69315 = 0.15396 bits

**VERIFY on calculator:**
```
0.05232 + 0.02616 + 0.02657 + 0.00167
```
Result: ~0.1067

---

## SUMMARY (all verified numerically)

| Claim | Verification Method | Result |
|-------|---------------------|--------|
| C(p) computed | 8 additions, 4 divisions | ✓ sums to 1.0 |
| C∘C = C | Apply C twice, compare | ✓ identical |
| Image M ≅ Δ₃ | Map (0.4,0.2,0.28,0.12) → C(p) | ✓ inverse works |
| Contraction d_F(C(p),C(q)) < d_F(p,q) | Calculator: arccos comparison | ✓ 0.1004 < 0.4394 |
| Strict inequality | p, r differ in pair 1 only | ✓ 0 < 0.66 |
| Information loss | 4 KL divergences, weighted | ✓ 0.1067 nats |