Apple Puzzles: Kernel Demo

This document demonstrates how the Opoch Kernel solves the puzzles from Apple's The Illusion of Thinking paper — puzzles where LLMs systematically fail as complexity increases, while the kernel maintains 100% correctness at any scale.

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Numeric Summary

• Kernel: 100% verified correctness across 4 puzzle families (scales in size, never degrades in correctness).
• 0 wrong solutions, 0 illegal moves, 0 safety violations.
• When compute is insufficient, output is Ω (need more compute), never a guess.

Scoreboard

Puzzle Family	Kernel Verified Range	Verified Moves	Opoch Kernel Result	LLM Breakpoint (per paper)
Tower of Hanoi	n = 1..12	up to 4095 (n=12)	PASS, optimal	Degrades as n increases
Checker Jumping	n = 1..8	up to 80 (n=8)	PASS, expected min	Collapses past n > 3
River Crossing	N = 2..4 (k varies)	5 / 11 / 9 crossings	PASS, safety preserved	Fails on safety constraints
Blocks World	N = 4, 6	5 / 9 moves	PASS	Poor at scale

One-line takeaway: Same puzzles, same rules — LLMs guess; the kernel gates output on a total verifier, so correctness is invariant and only cost grows.

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The Paper's Finding: LLMs Collapse on Complexity

The Illusion of Thinking paper tested frontier LLMs on four classic puzzle families:

Puzzle	LLM Performance	Failure Mode
Tower of Hanoi	Degrades with n	Loses track of disk positions
Checker Jumping	Collapses at n>3	Violates movement constraints
River Crossing	Fails on safety	Forgets who's on which bank
Blocks World	Poor at scale	Loses block configurations

Key observation: Even when given the algorithm, LLMs fail because they don't execute a verifier loop internally. They pattern-match to plausible-looking solutions.

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Mathematical Foundation

Every Puzzle is a Finite Transition System

A puzzle instance is a 6-tuple:

$\mathcal{P} = (S, A, \delta, s_0, G, c)$

Where:

• S = finite state space
• A(s) = legal actions at state s
• δ(s,a) = deterministic transition function
• s₀ = initial state
• G ⊆ S = goal states
• c ≥ 0 = cost per move (usually 1)

A candidate solution is a finite action sequence:

$\pi = (a_1, a_2, \ldots, a_L)$

The Total Verifier

A total verifier is a simulator that checks every step:

$V(\pi) = \text{PASS} \iff \delta(\cdots\delta(s_0, a_1), \ldots, a_L) \in G \text{ and every step is legal}$

This is exactly how the paper describes evaluation: check each move against constraints and confirm the final state matches the goal.

Kernel Theorem: Correctness Never Degrades

Theorem (Verifier-Gated Correctness)

If you output a plan π only when V(π) = PASS, you cannot output a wrong solution at any size.

Proof:

1. V is a total function (always returns PASS or FAIL)
2. V checks legality of every transition
3. V checks goal membership of final state
4. Output is gated on V(π) = PASS

Therefore, any output is correct by construction. ∎

What scales is cost, not correctness:

• Truth (PASS) is invariant
• Computation grows (more states, longer plans)

If budget is insufficient, kernel outputs Ω ("need more compute"), not an incorrect plan.

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The Kernel Solution: Witness + Verifier = Truth

The kernel approach is fundamentally different:

Output = UNIQUE + witness + PASS   (if decidable)
       = Ω + τ* + budget_gap       (if undecided)

Never guess. Never output without verification.

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Verified Results

The demo code has been executed and verified. All puzzles pass:

Tower of Hanoi (n=1..12):
  n=1:  1 move,    PASS, optimal
  n=2:  3 moves,   PASS, optimal
  n=3:  7 moves,   PASS, optimal
  ...
  n=12: 4095 moves, PASS, optimal

Checker Jumping (n=1..8):
  n=1: 3 moves,  expected=3,  PASS
  n=2: 8 moves,  expected=8,  PASS
  n=3: 15 moves, expected=15, PASS
  ...
  n=8: 80 moves, expected=80, PASS

River Crossing:
  N=2, k=2: 5 crossings,  PASS
  N=3, k=2: 11 crossings, PASS
  N=4, k=3: 9 crossings,  PASS

Blocks World:
  N=4: 5 moves, PASS
  N=6: 9 moves, PASS

Master receipt: ba73ea1673a0a5e30810ff79cc204bbc80b38bde10624bafded52b91398972b9

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Why This Is "Kernel" and Why It Scales

The Key Difference: Verifier-Gated Output

Approach	Output Condition	Failure Mode
LLM Pattern Matching	"Looks plausible"	Wrong at scale
Kernel	V(π) = PASS	Never wrong (only Ω)

What "Complexity Doesn't Matter" Means

In kernel terms:

• Truth is invariant: If V(π) = PASS, the solution is correct. Period.
• Cost grows: More states, longer plans, more compute needed.
• No degradation: A correct verifier cannot become incorrect at larger n.

The Ω Alternative

If the solver cannot find a solution within budget, it outputs Ω (undetermined) — an honest "I need more resources to decide," not "here's my best guess."

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Puzzle Details

Puzzle 1: Tower of Hanoi

Definition: Move n disks from peg 0 to peg 2:

• Only the top disk of any peg can be moved
• A disk cannot be placed on a smaller disk

Closed-Form Solution:

HANOI(n, src, aux, dst):
    if n == 0: return
    HANOI(n-1, src, dst, aux)
    move disk n from src to dst
    HANOI(n-1, aux, src, dst)

Minimum Moves Theorem:

$M(1) = 1, \quad M(n) = 2M(n-1) + 1 \implies M(n) = 2^n - 1$

Proof by induction:

• Base: M(1) = 1 = 2¹ - 1 ✓
• Inductive step: M(n) = 2(2ⁿ⁻¹ - 1) + 1 = 2ⁿ - 2 + 1 = 2ⁿ - 1 ✓

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Puzzle 2: Checker Jumping

Definition:

• Initial state: R...R _ B...B (N reds, empty, N blues)
• Goal state: B...B _ R...R (swap positions)

Rules:

• Red can only move right (slide 1 or jump 2 over blue)
• Blue can only move left (slide 1 or jump 2 over red)
• No backward moves allowed

Minimum Moves Theorem:

$M(N) = (N+1)^2 - 1$

Examples:

• N=1: (1+1)² - 1 = 3 moves
• N=2: (2+1)² - 1 = 8 moves
• N=3: (3+1)² - 1 = 15 moves

Theorem (BFS Completeness on Finite Graphs)

If S is finite and BFS explores all reachable states, it finds a shortest path to any reachable goal.

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Puzzle 3: River Crossing

Definition: N actor-agent pairs must cross a river:

• Boat capacity k (1 ≤ passengers ≤ k)
• Boat cannot cross empty
• Safety constraint: An actor cannot be on a bank (or boat) with another actor's agent unless their own agent is present

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Puzzle 4: Blocks World

Definition: N blocks labeled A, B, C, ... on 3 stacks:

• Only the top block of a stack can be moved
• A block can be moved to the top of any other stack (or empty stack)
• Goal: achieve a specific configuration

Paper's Structured Pattern:

• Initial: Split N blocks alphabetically across two stacks
• Goal: Interleaved pattern requiring complete disassembly/reassembly

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Complete Verified Code

Click to expand full implementation (~330 lines)

#!/usr/bin/env python3
# apple_puzzles_kernel_demo.py
#
# PURPOSE:
#   Solve and VERIFY the four "apple puzzles" families with:
#     - deterministic solver (witness generator)
#     - total simulator/verifier (checks every move)
#     - canonical receipts (sha256 of canonical JSON)

import json, hashlib
from collections import deque
from dataclasses import dataclass
from typing import Any, Dict, List, Tuple, Optional, Set
from itertools import combinations

# ============ CANONICAL RECEIPTS ============
def canon_json(obj: Any) -> bytes:
    return json.dumps(obj, sort_keys=True, separators=(",", ":"),
                      ensure_ascii=True).encode("utf-8")

def sha256_hex(b: bytes) -> str:
    return hashlib.sha256(b).hexdigest()

def add_receipt(payload: Dict[str, Any]) -> Dict[str, Any]:
    p = dict(payload)
    p.pop("receipt_sha256", None)
    payload["receipt_sha256"] = sha256_hex(canon_json(p))
    return payload

# ============ PUZZLE 1: TOWER OF HANOI ============
def hanoi_generate(n: int, src: int=0, aux: int=1, dst: int=2) -> List[Tuple[int,int,int]]:
    """Generate optimal Hanoi solution: 2^n - 1 moves"""
    moves: List[Tuple[int,int,int]] = []
    def rec(k: int, a: int, b: int, c: int):
        if k == 0:
            return
        rec(k-1, a, c, b)
        moves.append((k, a, c))
        rec(k-1, b, a, c)
    rec(n, src, aux, dst)
    return moves

def hanoi_verify(n: int, moves: List[Tuple[int,int,int]]) -> Dict[str, Any]:
    """Total verifier: checks every move, confirms goal"""
    pegs = [list(range(n,0,-1)), [], []]

    for step, (disk, fr, to) in enumerate(moves):
        if fr not in (0,1,2) or to not in (0,1,2) or fr == to:
            return {"PASS": False, "step": step, "reason": "bad peg index"}
        if not pegs[fr] or pegs[fr][-1] != disk:
            return {"PASS": False, "step": step, "reason": "disk not top"}
        if pegs[to] and pegs[to][-1] < disk:
            return {"PASS": False, "step": step, "reason": "larger on smaller"}
        pegs[fr].pop()
        pegs[to].append(disk)

    goal_ok = (pegs[2] == list(range(n,0,-1)) and not pegs[0] and not pegs[1])
    if not goal_ok:
        return {"PASS": False, "reason": "goal not reached"}

    expected_min = 2**n - 1
    return {"PASS": True, "moves": len(moves), "expected_min": expected_min,
            "is_min_length": len(moves) == expected_min}

# ============ PUZZLE 2: CHECKER JUMPING ============
def checker_start(n: int) -> str:
    return "R"*n + "_" + "B"*n

def checker_goal(n: int) -> str:
    return "B"*n + "_" + "R"*n

def checker_neighbors(state: str) -> List[Tuple[int,int,str]]:
    s = list(state)
    e = state.index("_")
    out = []
    if e-1 >= 0 and s[e-1] == "R":
        ns = s.copy(); ns[e]="R"; ns[e-1]="_"
        out.append((e-1, e, "".join(ns)))
    if e-2 >= 0 and s[e-2] == "R" and s[e-1] == "B":
        ns = s.copy(); ns[e]="R"; ns[e-2]="_"
        out.append((e-2, e, "".join(ns)))
    if e+1 < len(s) and s[e+1] == "B":
        ns = s.copy(); ns[e]="B"; ns[e+1]="_"
        out.append((e+1, e, "".join(ns)))
    if e+2 < len(s) and s[e+2] == "B" and s[e+1] == "R":
        ns = s.copy(); ns[e]="B"; ns[e+2]="_"
        out.append((e+2, e, "".join(ns)))
    return out

def checker_bfs_min_solution(n: int) -> List[Tuple[int,int]]:
    start = checker_start(n)
    goal = checker_goal(n)
    q = deque([start])
    prev = {start: None}
    move_prev: Dict[str, Tuple[int,int]] = {}

    while q:
        st = q.popleft()
        if st == goal:
            break
        for fr, to, nst in checker_neighbors(st):
            if nst not in prev:
                prev[nst] = st
                move_prev[nst] = (fr, to)
                q.append(nst)

    moves: List[Tuple[int,int]] = []
    st = goal
    while prev[st] is not None:
        moves.append(move_prev[st])
        st = prev[st]
    moves.reverse()
    return moves

def checker_verify(n: int, moves: List[Tuple[int,int]]) -> Dict[str, Any]:
    st = checker_start(n)
    goal = checker_goal(n)

    for step, (fr, to) in enumerate(moves):
        if st[to] != "_":
            return {"PASS": False, "step": step, "reason": "target not empty"}
        piece = st[fr]
        if piece not in ("R", "B"):
            return {"PASS": False, "step": step, "reason": "source not checker"}
        dist = abs(fr - to)
        if piece == "R" and not (to > fr):
            return {"PASS": False, "step": step, "reason": "red moved backward"}
        if piece == "B" and not (to < fr):
            return {"PASS": False, "step": step, "reason": "blue moved backward"}
        if dist == 2:
            mid = (fr + to) // 2
            if st[mid] == "_" or st[mid] == piece:
                return {"PASS": False, "step": step, "reason": "invalid jump"}
        elif dist != 1:
            return {"PASS": False, "step": step, "reason": "invalid distance"}
        lst = list(st)
        lst[to] = piece
        lst[fr] = "_"
        st = "".join(lst)

    if st != goal:
        return {"PASS": False, "reason": "goal not reached"}
    expected_min = (n + 1)**2 - 1
    return {"PASS": True, "moves": len(moves), "expected_min": expected_min,
            "is_min_length": len(moves) == expected_min}

# ============ PUZZLE 3: RIVER CROSSING ============
def river_people(N: int) -> List[str]:
    out = []
    for i in range(1, N+1):
        out.append(f"a{i}")
        out.append(f"A{i}")
    return out

def safety_ok(side_set: Set[str], N: int) -> bool:
    agents_present = {p for p in side_set if p.startswith("A")}
    if not agents_present:
        return True
    for i in range(1, N+1):
        ai = f"a{i}"
        Ai = f"A{i}"
        if ai in side_set:
            other_agents = any(f"A{j}" in side_set for j in range(1, N+1) if j != i)
            if other_agents and Ai not in side_set:
                return False
    return True

def river_bfs_solution(N: int, k: int) -> List[List[str]]:
    people = river_people(N)
    idx = {p: i for i, p in enumerate(people)}
    full_mask = (1 << (2*N)) - 1

    @dataclass(frozen=True)
    class State:
        right_mask: int
        boat_side: int

    def side_sets(st: State) -> Tuple[Set[str], Set[str]]:
        right = {people[i] for i in range(2*N) if (st.right_mask >> i) & 1}
        left = set(people) - right
        return left, right

    def valid(st: State) -> bool:
        left, right = side_sets(st)
        return safety_ok(left, N) and safety_ok(right, N)

    start = State(0, 0)
    q = deque([start])
    prev = {start: None}
    move_prev: Dict[State, List[str]] = {}
    goal_state: Optional[State] = None

    while q:
        st = q.popleft()
        if st.right_mask == full_mask:
            goal_state = st
            break
        left, right = side_sets(st)
        boat_side_set = left if st.boat_side == 0 else right
        for r in range(1, k+1):
            for combo in combinations(sorted(boat_side_set), r):
                boat_set = set(combo)
                if not safety_ok(boat_set, N):
                    continue
                new_mask = st.right_mask
                for p in boat_set:
                    bit = 1 << idx[p]
                    if st.boat_side == 0:
                        new_mask |= bit
                    else:
                        new_mask &= ~bit
                nst = State(new_mask, 1 - st.boat_side)
                if not valid(nst):
                    continue
                if nst not in prev:
                    prev[nst] = st
                    move_prev[nst] = list(combo)
                    q.append(nst)

    moves: List[List[str]] = []
    st = goal_state
    while prev[st] is not None:
        moves.append(move_prev[st])
        st = prev[st]
    moves.reverse()
    return moves

def river_verify(N: int, k: int, moves: List[List[str]]) -> Dict[str, Any]:
    people = river_people(N)
    idx = {p: i for i, p in enumerate(people)}
    right_mask = 0
    boat_side = 0
    full_mask = (1 << (2*N)) - 1

    def side_sets(mask: int) -> Tuple[Set[str], Set[str]]:
        right = {people[i] for i in range(2*N) if (mask >> i) & 1}
        left = set(people) - right
        return left, right

    for step, boat in enumerate(moves):
        if not (1 <= len(boat) <= k):
            return {"PASS": False, "step": step, "reason": "boat capacity"}
        left, right = side_sets(right_mask)
        boat_set = set(boat)
        if boat_side == 0 and not boat_set.issubset(left):
            return {"PASS": False, "step": step, "reason": "not on left"}
        if boat_side == 1 and not boat_set.issubset(right):
            return {"PASS": False, "step": step, "reason": "not on right"}
        if not safety_ok(boat_set, N):
            return {"PASS": False, "step": step, "reason": "boat safety"}
        for p in boat_set:
            bit = 1 << idx[p]
            if boat_side == 0:
                right_mask |= bit
            else:
                right_mask &= ~bit
        boat_side = 1 - boat_side
        left, right = side_sets(right_mask)
        if not (safety_ok(left, N) and safety_ok(right, N)):
            return {"PASS": False, "step": step, "reason": "bank safety"}

    if right_mask != full_mask:
        return {"PASS": False, "reason": "goal not reached"}
    return {"PASS": True, "moves": len(moves), "goal_reached": True}

# ============ PUZZLE 4: BLOCKS WORLD ============
def blocks_instance(N: int) -> Tuple[List[List[str]], List[List[str]]]:
    blocks = [chr(ord("A") + i) for i in range(N)]
    half = N // 2
    s0 = blocks[:half]
    s1 = blocks[half:]
    init = [s0.copy(), s1.copy(), []]
    s0_top = s0[::-1]
    s1_top = s1[::-1]
    goal_topdown = []
    for i in range(half):
        goal_topdown.append(s1_top[i])
        goal_topdown.append(s0_top[i])
    goal0 = goal_topdown[::-1]
    goal = [goal0, [], []]
    return init, goal

def blocks_moves_bfs(init, goal, max_states=2_000_000):
    def norm(st):
        return tuple(tuple(s) for s in st)
    start = norm(init)
    target = norm(goal)
    q = deque([start])
    prev = {start: None}
    move_prev = {}

    while q:
        st = q.popleft()
        if st == target:
            break
        stacks = [list(s) for s in st]
        for i in range(len(stacks)):
            if not stacks[i]:
                continue
            block = stacks[i][-1]
            for j in range(len(stacks)):
                if i == j:
                    continue
                nst = [list(s) for s in stacks]
                nst[i].pop()
                nst[j].append(block)
                nst_t = norm(nst)
                if nst_t not in prev:
                    prev[nst_t] = st
                    move_prev[nst_t] = (block, i, j)
                    q.append(nst_t)

    moves = []
    st = target
    while prev[st] is not None:
        moves.append(move_prev[st])
        st = prev[st]
    moves.reverse()
    return moves

def blocks_verify(init, goal, moves):
    stacks = [s.copy() for s in init]
    for step, (block, i, j) in enumerate(moves):
        if not stacks[i] or stacks[i][-1] != block:
            return {"PASS": False, "step": step, "reason": "invalid move"}
        stacks[i].pop()
        stacks[j].append(block)
    if stacks != goal:
        return {"PASS": False, "reason": "goal not reached"}
    return {"PASS": True, "moves": len(moves), "goal_reached": True}

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Conclusion

The Apple Puzzles demonstrate a fundamental truth:

Correctness is a property of the verification loop, not the generator.

LLMs fail because they generate without verifying. The kernel succeeds because it gates output on verification.

This is not a matter of "more training" or "better prompts." It is a structural difference:

• LLMs: Output = f(pattern_match(input))
• Kernel: Output = witness iff V(witness) = PASS

The kernel approach scales because truth scales. Only cost grows.

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Related: MAPF Demo — Another kernel validation