Shift Arithmetic and Construction Complexity — Interactive Lesson

A teaching model for
a ⋆ b = a(b+1)

Starting amount a plus b additional groups of a. For nonnegative integer b literal repeated addition; real inputs extend formula algebraically. Custom operation, not ordinary multiplication.

Operation
a ⋆ b = a(b+1)
Algebra: real inputs. Trees: positive integers.
Companion
b⊕c = b+c+bc
Group on ℝ\{-1\}, identity 0. Log composition requires b,c>−1
Leaves
L(1)=1, S₁={1}
Recurrence uses proper divisors 1≤d<m only, no L(0). n leaves → [n, 2^{n-1}]

Checking construction data…

1. Intuition — Live Blocks

5
3
a⋆b = 20
For integer b ≥ 0, each block represents one group of a. At b = −1 the output is zero, so the starting amount cannot be recovered. Domain note: inverse b = c/a−1 requires a≠0; a = c/(b+1) requires b≠-1.

2. Elementary Algebra — Complete Statements

Domain for algebraic statements: real numbers unless stated otherwise. Construction model Sₙ, L(m): every leaf exactly 1, internal node ⋆, fully parenthesized binary trees, positive integer values, no shared subexpressions.

3. Companion Law ⊕ — Reversible Changes

b⊕c = b+c+bc = (1+b)(1+c)−1
Theorem: a⋆(b⊕c) = (a⋆b)⋆c = a(1+b)(1+c). Proof direct expansion.
Test composition

Group structure: ψ(b)=1+b, ψ(b⊕c)=ψ(b)ψ(c). Hence ⊕ on ℝ\{-1} is abelian group isomorphic to multiplication on ℝ\{0}. Identity 0, inverse of b is −b/(1+b) when b≠-1. At b=-1, no inverse — absorbing element (monoid).

Log: Restricting to b>-1 gives positive scale factors; log(1+b) converts composition to addition: log(1+(b⊕c))=log(1+b)+log(1+c) for b,c>-1.

Practice: 100 increased 20% then 30% → 156. Combined relative change 0.2⊕0.3=0.56. 25% increase undone by 20% decrease because 0.25⊕(-0.2)=0.

4. Expression Trees and Sₙ

S₁={1}, Sₙ = ⋃_{1≤k<n} { x(y+1) : x∈S_k, y∈S_{n-k} }. No associativity assumed — recurrence enumerates all bracketings.

Interactive builder — click a yellow leaf to grow: replace 1 with (1⋆1)

Click or focus a yellow leaf and press Enter or Space to split it. Scroll to explore large trees. Limit: 64 leaves; values remain exact.
First sets (verified)
S₄ has 5 binary bracketings → 4 distinct values because two bracketings yield 6. Missing 7 is reachability gap, not paradox. Try to make 7 with four leaves, then five.
Challenge hint and solution for 7

Four leaves produce only {4, 5, 6, 8}. With five, use 1 ⋆ ((1 ⋆ 1) ⋆ (1 ⋆ 1)) = 1 ⋆ 6 = 7. Since 7 is prime, its root must have left value 1 and right value 6.

5. Bounds and Extremal Trees

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For n-leaf trees: n ≤ value ≤ 2^{n-1}. For a root split k+ℓ=n, induction gives x(y+1) ≥ k(ℓ+1) ≥ n and x(y+1) ≤ 2^{k−1}(2^{ℓ−1}+1) ≤ 2^{n−1}. Right-grouped 1s attain n; left-grouped attain 2^{n-1}.

6. Prime Entrance

Theorem (positive integer tree values): If prime p = x(y+1) with x∈S_k, y∈S_{n-k}, positivity forces x=1 and y=p−1. Only single leaf evaluates to 1 by lower bound.

Consequence: L(p)=L(p−1)+1 for prime p. Attach left leaf 1 to optimal tree for p−1. Example: constructing 22 as 2⋆10 uses tree for 10, not 11. Hence L(23)=L(22)+1=8.

7. Minimum Construction L(m) — Complete Recurrence

L(m) = minimum unit leaves to build positive integer m.
Base: L(1)=1. Zero cannot be constructed from positive unit leaves.
For m≥2: L(m)= min_{1≤d<m, d|m} L(d) + L(m/d −1). Here d ranges over proper positive divisors of m; both child values are smaller positive integers.
Enter a whole number from 1 to 100. The expression uses the custom operation ⋆.

8. Computational Audit 1..100 — Verified Data

ratio undefined at m=1 (log₂1=0) — shown as —
mL(m)L/log₂mΔ = L-1-⌈log₂m⌉best d→eoptimal expr sketch
Verified max cost 11 attained only by 94 in 1..100. L(95)=10, L(47)=10, L(23)=8. These are exact values from divisor recurrence cross-checked with reachable-set enumeration S₁..S₁₁. Ranking favors small inputs; metric choice should be stated before interpretation. L(64)=7 ratio 1.166667.

9. Open Questions

• Which integers attain rounded lower bound 1+⌈log₂m⌉?
• How does excess Δ(m) grow? Not claimed from finite table.
• How many distinct values in Sₙ, which gaps persist?
• How many optimal trees per m?
Algebra can serve as teaching model for non-associativity, notation for relative changes, exact optimization puzzle. No computational advantage over established arithmetic demonstrated. Contribution is precisely scoped record of rule, consequences, reproducible observations.

10. Reproducible Reference Implementation

Python 3 • standard library only • ⋆ denotes custom operation