From 49b77181b84c9e32001fb17e88c1c29362671b4c Mon Sep 17 00:00:00 2001
From: Allaun Silverfox <28494262+allaunthefox@users.noreply.github.com>
Date: Tue, 23 Jun 2026 05:29:40 -0500
Subject: [PATCH] =?UTF-8?q?math(worksheet):=20G2=5FWORKSHEET.md=20?=
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+# WORKSHEET G2 — Collision Breaking with Parse-Tree Features
+## Verifiable with any calculator and pencil. No English inside formulas.
+
+---
+
+## PART A: The Collision (count characters by hand)
+
+**Byte classes (8 buckets):**
+| Class | Bytes |
+|-------|-------|
+| 0 | control (0-31) |
+| 1 | punct-low (!-/ = 33-47) |
+| 2 | digits (0-9 = 48-57) |
+| 3 | punct-mid (:-@ = 58-64) |
+| 4 | upper (A-Z = 65-90) |
+| 5 | punct-high ([-` = 91-96) |
+| 6 | lower (a-z = 97-122) |
+| 7 | extended (123-255) |
+
+---
+
+**STRING 1:** "a+b=c" (5 characters)
+
+| Char | ASCII | Class |
+|------|-------|-------|
+| a | 97 | 6 |
+| + | 43 | 1 |
+| b | 98 | 6 |
+| = | 61 | 3 |
+| c | 99 | 6 |
+
+**Counts:** class 0=0, class 1=1, class 2=0, class 3=1, class 4=0, class 5=0, class 6=3, class 7=0
+
+**Probability vector F("a+b=c"):**
+```
+(0, 1/5, 0, 1/5, 0, 0, 3/5, 0) = (0, 0.2, 0, 0.2, 0, 0, 0.6, 0)
+```
+
+---
+
+**STRING 2:** "x+y=z" (5 characters)
+
+| Char | ASCII | Class |
+|------|-------|-------|
+| x | 120 | 6 |
+| + | 43 | 1 |
+| y | 121 | 6 |
+| = | 61 | 3 |
+| z | 122 | 6 |
+
+**Counts:** class 0=0, class 1=1, class 2=0, class 3=1, class 4=0, class 5=0, class 6=3, class 7=0
+
+**Probability vector F("x+y=z"):**
+```
+(0, 1/5, 0, 1/5, 0, 0, 3/5, 0) = (0, 0.2, 0, 0.2, 0, 0, 0.6, 0)
+```
+
+---
+
+**STRING 3:** "p/q=r" (5 characters)
+
+| Char | ASCII | Class |
+|------|-------|-------|
+| p | 112 | 6 |
+| / | 47 | 1 |
+| q | 113 | 6 |
+| = | 61 | 3 |
+| r | 114 | 6 |
+
+**Counts:** class 0=0, class 1=1, class 2=0, class 3=1, class 4=0, class 5=0, class 6=3, class 7=0
+
+**Probability vector F("p/q=r"):**
+```
+(0, 1/5, 0, 1/5, 0, 0, 3/5, 0) = (0, 0.2, 0, 0.2, 0, 0, 0.6, 0)
+```
+
+---
+
+**VERIFICATION:** All three vectors are identical.
+```
+F("a+b=c") = F("x+y=z") = F("p/q=r") = (0, 0.2, 0, 0.2, 0, 0, 0.6, 0)
+```
+
+**COLLISION CONFIRMED:** Three different equations → one probability vector.
+
+Type the counts into calculator. Verify they match.
+
+---
+
+## PART B: Parse-Tree Features (count by hand)
+
+**Node types for arithmetic expressions:**
+| Type | Code |
+|------|------|
+| variable | V |
+| binary_op_add | B+ |
+| binary_op_div | B/ |
+| binary_op_eq | B= |
+
+---
+
+**STRING 1:** "a+b=c"
+
+Parse tree: (a + b) = c
+Nodes: V(a), B+(+), V(b), B+(=) → wait, "=" is binary_op_eq
+
+Correct parse tree:
+```
+      B= 
+     /  \
+   B+    V(c)
+  /  \
+V(a) V(b)
+```
+
+**Node count:** V=3, B+=1, B==1, B/=0
+
+**Total nodes:** 5
+
+**Probability vector τ("a+b=c"):**
+```
+(V, B+, B=, B/) = (3/5, 1/5, 1/5, 0) = (0.6, 0.2, 0.2, 0)
+```
+
+---
+
+**STRING 2:** "x+y=z"
+
+Parse tree:
+```
+      B=
+     /  \
+   B+    V(z)
+  /  \
+V(x) V(y)
+```
+
+**Node count:** V=3, B+=1, B==1, B/=0
+
+**Probability vector τ("x+y=z"):**
+```
+(V, B+, B=, B/) = (3/5, 1/5, 1/5, 0) = (0.6, 0.2, 0.2, 0)
+```
+
+---
+
+**STRING 3:** "p/q=r"
+
+Parse tree:
+```
+      B=
+     /  \
+   B/    V(r)
+  /  \
+V(p) V(q)
+```
+
+**Node count:** V=3, B+=0, B==1, B/=1
+
+**Probability vector τ("p/q=r"):**
+```
+(V, B+, B=, B/) = (3/5, 0, 1/5, 1/5) = (0.6, 0, 0.2, 0.2)
+```
+
+---
+
+**VERIFICATION:**
+```
+τ("a+b=c") = (0.6, 0.2, 0.2, 0)
+τ("x+y=z") = (0.6, 0.2, 0.2, 0)
+τ("p/q=r") = (0.6, 0, 0.2, 0.2)
+```
+
+τ("a+b=c") = τ("x+y=z") ✗ (still collides!)
+τ("a+b=c") ≠ τ("p/q=r") ✓ (broken!)
+
+"a+b=c" and "x+y=z" have the same structure (addition, then equality).
+"p/q=r" has different structure (division, then equality).
+
+To break ALL collisions, we need a 3rd feature that distinguishes "a+b=c" from "x+y=z". These differ only in variable names, which is semantically irrelevant.
+
+**CONCLUSION:** τ breaks operator-type collisions (addition vs division) but NOT variable-name collisions (a+b=c vs x+y=z). This is CORRECT: the two equations are semantically equivalent (both are addition-then-equality).
+
+---
+
+## PART C: Product Fisher Distance (computed numerically)
+
+**FORMULA:** d²_F((p,r), (q,s)) = d²_F(p,q) + d²_F(r,s)
+
+**INPUT 1:** p = F("a+b=c") = (0, 0.2, 0, 0.2, 0, 0, 0.6, 0)
+**INPUT 2:** q = F("p/q=r") = (0, 0.2, 0, 0.2, 0, 0, 0.6, 0)
+**NOTE:** F is identical for these two strings.
+
+**INPUT 3:** r = τ("a+b=c") = (0.6, 0.2, 0.2, 0)
+**INPUT 4:** s = τ("p/q=r") = (0.6, 0, 0.2, 0.2)
+
+**STEP 1: Compute d_F(p,q)**
+```
+p = q = (0, 0.2, 0, 0.2, 0, 0, 0.6, 0)
+√(pᵢqᵢ) = pᵢ  (since p = q)
+Sum = 0 + 0.2 + 0 + 0.2 + 0 + 0 + 0.6 + 0 = 1.0
+d_F(p,q) = 2·arccos(1.0) = 2·0 = 0
+```
+
+**STEP 2: Compute d_F(r,s)**
+```
+r = (0.6, 0.2, 0.2, 0)
+s = (0.6, 0, 0.2, 0.2)
+
+√(r₁s₁) = √(0.6 × 0.6) = 0.6
+√(r₂s₂) = √(0.2 × 0) = 0
+√(r₃s₃) = √(0.2 × 0.2) = 0.2
+√(r₄s₄) = √(0 × 0.2) = 0
+
+Sum = 0.6 + 0 + 0.2 + 0 = 0.8
+```
+
+**WORK on calculator:**
+```
+2 * arccos(0.8)
+```
+
+**RESULT:** d_F(r,s) = 2 × 0.6435 = 1.2870
+
+**STEP 3: Product distance**
+```
+d²_F((p,r), (q,s)) = 0² + (1.2870)² = 1.6564
+d_F((p,r), (q,s)) = √1.6564 = 1.2870
+```
+
+**VERIFY on calculator:**
+```
+sqrt(0 + (2*arccos(0.8))^2)
+```
+
+**OUTPUT:** 1.2870
+
+**INTERPRETATION:** F alone gives distance 0 (collision). τ alone gives distance 1.2870 (distinguishes). Product gives 1.2870 (τ provides all the discrimination).
+
+---
+
+## PART D: Marginal Projection (computed numerically)
+
+**FORMULA:** π(p,r) = p
+
+**CLAIM:** d_F(π(x), π(y)) ≤ d_F(x,y)
+
+**TEST:** x = (p,r), y = (q,s) from Part C.
+
+**LEFT SIDE:** d_F(π(x), π(y)) = d_F(p,q) = 0
+**RIGHT SIDE:** d_F(x,y) = 1.2870
+
+**CHECK:** 0 ≤ 1.2870 ✓
+
+The projection does not increase distance.
+
+---
+
+## PART E: Another Example — "1+2" vs "3+4"
+
+**STRING 4:** "1+2" (3 characters)
+
+| Char | ASCII | Class |
+|------|-------|-------|
+| 1 | 49 | 2 |
+| + | 43 | 1 |
+| 2 | 50 | 2 |
+
+**F("1+2") = (0, 1/3, 2/3, 0, 0, 0, 0, 0) = (0, 0.333, 0.667, 0, 0, 0, 0, 0)**
+
+Parse tree:
+```
+  B+
+ /  \
+N(1) N(2)
+```
+Node types: N=2, B+=1, B==0, B/=0, V=0
+
+**τ("1+2") = (0, 1/3, 0, 0, 2/3) = (0, 0.333, 0, 0, 0.667)** 
+
+(using 5 types: V, B+, B=, B/, N)
+
+---
+
+**STRING 5:** "a+b" (3 characters)
+
+| Char | ASCII | Class |
+|------|-------|-------|
+| a | 97 | 6 |
+| + | 43 | 1 |
+| b | 98 | 6 |
+
+**F("a+b") = (0, 0.333, 0, 0, 0, 0, 0.667, 0)**
+
+Parse tree:
+```
+  B+
+ /  \
+V(a) V(b)
+```
+Node types: V=2, B+=1, N=0
+
+**τ("a+b") = (2/3, 1/3, 0, 0, 0) = (0.667, 0.333, 0, 0, 0)**
+
+---
+
+**VERIFICATION:**
+```
+F("1+2") = (0, 0.333, 0.667, 0, 0, 0, 0, 0)
+F("a+b") = (0, 0.333, 0, 0, 0, 0, 0.667, 0)
+
+τ("1+2") = (0, 0.333, 0, 0, 0.667)
+τ("a+b") = (0.667, 0.333, 0, 0, 0)
+```
+
+F differs (digits vs lowercase). τ differs (numbers vs variables). No collision.
+
+**Product distance:**
+
+STEP 1: d_F(F("1+2"), F("a+b"))
+```
+√(0×0) = 0
+√(0.333×0.333) = 0.333
+√(0.667×0) = 0
+√(0×0) = 0
+√(0×0) = 0
+√(0×0) = 0
+√(0×0.667) = 0
+√(0×0) = 0
+Sum = 0.333
+d_F = 2×arccos(0.333) = 2×1.231 = 2.462
+```
+
+STEP 2: d_F(τ("1+2"), τ("a+b"))
+```
+√(0×0.667) = 0
+√(0.333×0.333) = 0.333
+√(0×0) = 0
+√(0×0) = 0
+√(0.667×0) = 0
+Sum = 0.333
+d_F = 2×arccos(0.333) = 2.462
+```
+
+STEP 3: Product
+```
+d² = (2.462)² + (2.462)² = 6.061 + 6.061 = 12.122
+d = √12.122 = 3.482
+```
+
+**VERIFY on calculator:**
+```
+sqrt( (2*arccos(1/3))^2 + (2*arccos(1/3))^2 )
+```