Appendix C — The Reserve Rack {#appC}

The chapters used thirty-three of the canon's fifty puzzles; these seventeen complete the set. No chapter warms you up here, and no section title whispers the move in advance — which is exactly their value to the Chapter 16 regimen. Fresh meat: solve them cold.

★ 17 · The Riddle of the Sphinx

"What goes on four legs in the morning, two legs at noon, and three legs in the evening?"

The Sphinx is waiting.

Solution. Withheld — the one puzzle in this book whose answer you must earn by finishing it. It has stood at the gate since Chapter 1, and the colophon settles it on the book's final page. If the wait is unbearable, that is the point of the wait.

The Move — Change the timescale. When a literal reading yields a monster, re-scale time or re-bind a word — every noun in a riddle is a metaphor slot.

★★ 21 · The Rope Around the Earth

A rope hugs the Earth's equator (~40,000 km). You splice in one extra meter and float the rope uniformly above the surface. Can a cat slip under? And if instead you pull all the slack taut upward at one point, how high is the peak?

Commit to a guess — cat or no cat — before reading on.

Solution. The gap is the extra circumference divided by 2π: 1/(2π) ≈ 15.9 cm, all the way around. The cat strolls under. Better: the answer is independent of the planet's size — C = 2πr, so Δr = ΔC/2π, and the radius cancels; a marble gives the same 15.9 cm. The second question is just as rude (verified numerically): pulled taut at one point, the peak sits about 121 m above the surface (θ ≈ (3s/2R)^(1/3) with s = 1 m of slack; height ≈ Rθ²/2). Intuition fails in both directions.

The Move — Let the algebra veto the gut. Feelings scale with absolute size and the relation Δr = ΔC/2π doesn't; when a two-line derivation and intuition disagree, the derivation wins.

★★★ 23 · The Coin Rotation Paradox

Coin B rolls without slipping once around an identical, fixed coin A. How many times does B rotate about its own axis? Then: a circle of radius r rolls around a circle of radius 3r — how many rotations?

Get two coins and try it. Really.

Solution. Two, not one — roll a coin around another and watch: the head is already upright after half the journey. Rotation = rolling along the path + the orbit itself: contact-rolling contributes C_A/C_B = 1, and the orbit contributes 1 more. The general exterior formula is R/r + 1, so the radius-3 case gives 4, not 3. That exact question appeared on the May 1982 US SAT with "3" keyed as correct; three students proved the test wrong, ETS conceded, and every score was adjusted. The sky runs the same phenomenon: the sidereal versus the solar day — Earth rotates 366.25 times in a year of 365.25 days.

The Move — Separate spin from orbit. When motions compose, decompose them — and when a count is disputed, count a second, independent way and demand agreement.

★★★★ 24 · The Spider and the Fly

A room 30 ft long, 12 ft wide, 12 ft high. A spider sits on one 12×12 end wall, centered, 1 ft below the ceiling. A fly sits on the opposite end wall, centered, 1 ft above the floor. What is the spider's shortest walking path to the fly, along any surfaces?

Sketch the room before you trust yourself.

Solution. The obvious route — straight up, along the ceiling, straight down — is 1 + 30 + 11 = 42 ft. But distances on a box are straight lines on its unfoldings. Flatten the right five-face net — end wall, ceiling, side wall, floor, far end wall — and the path becomes the hypotenuse of a 32 × 24 right triangle: √(1024 + 576) = √1600 = 40 ft. (Intermediate unfoldings give 43.2 and 40.7; 40 is best, machine-checked.) A geodesic touching five of the six faces beats the route that looks direct. Dudeney, 1903 — Gardner called it one of the best puzzles ever devised.

The Move — Unfold the space. Change the representation until the answer becomes a straight line: reflect the billiard table, unroll the cylinder, take logs.

★★★★ 25 · The Haberdasher's Problem

Cut an equilateral triangle into as few pieces as possible that reassemble into a perfect square.

Scissors down. Think first.

Solution. Four pieces — and, astonishingly, they can be hinged in a chain that swings one way into the triangle and the other way into the square. Construction sketch: bisect two sides at D and E; place F on the third side so that EF equals the side of the target square (√area); perpendiculars complete the three cuts — the full compass-work is in Dudeney (1907). Existence was never in doubt: any polygon dissects into any equal-area polygon (Wallace–Bolyai–Gerwien theorem, 1807–1833); the game is minimality and elegance. Dudeney posed the problem on 6 April 1902, had the four-piece answer by 4 May, and showed a brass-hinged model to the Royal Society in 1905. In December 2024, Demaine, Kamata and Uehara proved that no three-piece dissection exists (arXiv 2412.03865) — the 122-year-old answer, finally certified optimal.

The Move — Re-partition the same area. Conservation guarantees a bridge between equal-area shapes and craft is finding the minimal one — proving you can't do better can be a century harder than doing it.

★ 27 · Gauss's Schoolroom Sum

Sum the integers from 1 to 100 — fast, in your head.

You have ten seconds, the story says.

Solution. Pair the ends: 1+100, 2+99, …, 50+51 — fifty pairs, each summing to 101: 5050. In general, n(n+1)/2 — the sum counted forward and backward at once is n copies of (n+1). About the legend: the schoolboy-Gauss anecdote (slate slapped down, "ligget se") is real in outline but embellished in detail — the earliest source (Sartorius, 1856) names no specific series, and the "1 to 100" specifics accreted later (Hayes, 2006). The method is ancient regardless.

The Move — Pair the extremes. Impose a symmetry the data doesn't advertise, and the computation collapses.

★★ 28 · The Fly Between the Trains

Two trains 20 miles apart approach each other, each at 10 mph. A fly leaves one train at 15 mph, flies to the other, turns instantly, flies back, and so on until the trains meet. How far does the fly fly?

Resist the urge to sum a series.

Solution. The trains close at 20 mph, so they meet in exactly 1 hour. The fly flies at 15 mph for that hour: 15 miles. No series needed — though the infinite zig-zag sum converges to the same answer, which is the joke in the (probably polished) von Neumann anecdote. Told the puzzle, he answered instantly. "Ah, you knew the trick." — "What trick? I summed the series."

The Move — Total the aggregate, skip the episodes. When a process is a mess of episodes but a global quantity — time, distance, energy — is clean, compute the global.

★★ 29 · Wine and Water

A glass of wine and a glass of water, same volume. Move one spoonful of wine into the water, stir (or don't — it doesn't matter), then move one spoonful of the mixture back. Is there more wine in the water, or more water in the wine?

Take a stance before the argument starts.

Solution. Exactly equal — regardless of stirring, spoon size, or how many round trips are made. Endstate argument: both glasses finish at their original volumes, so whatever volume of wine is missing from the wine glass must be occupying the water glass, and the space it left behind is filled by exactly that much water. Swap-symmetry settles the matter before any tracking of concentrations begins — the chessboard coloring's skeleton, in liquid form (a Rouse Ball and Gardner staple).

The Move — Conservation ends the argument. When the bookkeeping looks hairy, ask what the final totals must be and let conservation do all the arithmetic.

★★★ 30 · The Thousand Lockers

Lockers 1–1000, all closed. Student k (k = 1…1000) toggles every kth locker. After all 1000 students pass, which lockers stand open?

Try lockers 1 through 10 by hand first.

Solution (verified by simulation). Locker n is toggled once per divisor of n, so it ends open iff n has an odd number of divisors. Divisors pair up, d ↔ n/d — except when d = n/d, that is, when n is a perfect square. The open lockers are 1, 4, 9, …, 961: the 31 perfect squares ≤ 1000. The exception to the pairing is the answer.

The Move — Pair things up; interrogate the exceptions. For parity questions, find the involution and look hard at its fixed points.

★★ 31 · The Tower of Hanoi

Three pegs; n disks stacked by size on one peg. Move the whole tower to another peg, one disk at a time, never placing a larger disk on a smaller. Minimum moves? (The monks of the legend run n = 64.)

Three disks by hand teach more than an hour of staring.

Solution. T(n) = 2T(n−1) + 1: move n−1 disks aside, move the big disk, restack — so T(n) = 2ⁿ − 1, and the recursion is the algorithm. For the monks' 64 disks: 2⁶⁴ − 1 = 18,446,744,073,709,551,615 moves — at one per second, about 585 billion years (the wheat-on-the-chessboard number; doubling incarnate). The "ancient Brahmin legend" was invented wholesale by the puzzle's actual author, Édouard Lucas, who sold it in 1883 under the anagram-pseudonym "N. Claus de Siam" — marketing and mathematics in one box.

The Move — Recurse: trust the smaller self. Solve the problem assuming a solved smaller instance and let the structure carry you — the puzzle most worth feeling in the hands before ever writing a recursive function.

★★★★ 36 · The Two Envelopes

Two sealed envelopes, one holding twice the other's amount. You pick one — it holds X. The argument: "The other holds 2X or X/2 with equal chance; expected value (2X + X/2)/2 = 1.25X > X, so switch. But the same argument says switch back, forever." Where exactly does it break?

Find the precise sentence where the argument cheats.

Solution. The flaw is a variable shift. In the branch "other = 2X," the symbol X denotes the smaller amount; in "other = X/2," it denotes the larger. One letter, two meanings — the formula adds apples to oranges. Done properly: call the pair (Y, 2Y); whichever envelope is in hand, switching gains +Y or −Y with equal probability — expected gain 0. The seductive version silently assumes all amounts equally likely, a uniform prior on an infinite range, which is not a probability distribution; with any proper prior, the conditional expected gain integrates to zero. (For a specific observed amount, a proper prior can genuinely favor switching — that is Bayesian updating, not paradox.) Ancestry: Kraitchik's necktie paradox (1943); the modern treatment is Nalebuff (1989).

The Move — Pin the random variable. Write down what is random, what is observed, and what each symbol denotes in each branch — most probability "paradoxes" are notation crimes.

★★★★ 39 · Rubik's Cube and the Commutator

Not "solve the cube" — the design problem underneath it. Find move sequences that change only a little — cycle three corners, say — while leaving the rest of the cube's 4.3×10¹⁹-state chaos untouched. How does one even search for such a needle?

This one asks for a method, not an answer.

Solution. The commutator: for sequences A and B, perform A B A⁻¹ B⁻¹. If A and B touch mostly disjoint pieces, almost everything cancels, and the net effect is a small local change — a three-cycle of corners or edges, a pair of twists. Conjugation (C X C⁻¹) then aims the tool: carry the pieces you care about into position, apply the commutator, carry them back. Every human method is a library of conjugated commutators; that is why beginners' "algorithms" exist at all. The object's scale: 43,252,003,274,489,856,000 positions — yet God's number is 20 moves (Rokicki–Kociemba–Davidson–Dethridge, 2010, ~35 CPU-years), while human methods spend 50–60. The gap between what exists and what is findable, held in the palm.

The Move — Build commuting tools. Do, do, undo, undo: engineer operations whose interference is exactly the change you want, then conjugate to aim them — how one edits any system too complex to rebuild, codebases and organizations included.

★★ 40 · The Caesar Shift and the Frequency Attack

Decrypt: YMJ RFU NX YMJ YWJFXZWJ (a Caesar cipher — every letter shifted by the same amount).

Count letters before you count keys.

Solution. Only 25 keys exist, so brute force is instant — but the elegant attack is statistical. The ciphertext's most frequent letter is J (four times); English's most frequent is E; the hypothesis J = E means shift 5. Apply shift −5: THE MAP IS THE TREASURE. The same idea scales to general substitution ciphers, where 26! keys kill brute force: match letter-frequency fingerprints (E T A O I N …), then common digrams (TH, HE), and the cipher unzips. That is al-Kindi's 9th-century insight — statistics defeating secrecy a millennium before "statistics" was a word.

The Move — Attack the statistics. Encryption hides content, not structure; whatever must preserve function must preserve pattern, so find the pattern that couldn't be hidden.

★★★ 42 · Two Cryptic Clues

Solve as cryptic-crossword clues: (a) "Gegs (9,4)" (b) "HIJKLMNO (5)".

Say each clue aloud. Twice.

Solution. (a) SCRAMBLED EGGS — the clue demonstrates its answer: "gegs" is literally eggs, scrambled; the enumeration (9,4) confirms. (b) WATER — the letters run from H to O: "H-to-O" = H₂O. Both are definition-by-performance: the clue doesn't describe the answer, it enacts the instruction that produces it. Beneath them sits standard cryptic grammar: every fair clue contains a definition and a wordplay path — two independent proofs of one answer, the Ximenean contract (Ximenes, On the Art of the Crossword, 1966).

The Move — The surface hides an instruction. Read strange text twice — once as description, once as recipe — because sometimes the form, not the content, is the payload.

★★★ 43 · Nim

Heaps of counters — say 3, 5, 7. Players alternate removing any positive number of counters from one heap. Taking the last counter wins. Who wins, and how? (Warm-up: from 21 flags, players alternately remove 1–3; taking the last wins — who?)

Play the 21-flag warm-up against yourself first.

Solution (theorem machine-verified for all heaps ≤ 5,5,5). Chapter 9 derives this balance as a toolkit weapon; here is the pocket edition. Warm-up: multiples of 4 are losing positions — leave 20, 16, 12, 8, 4 and mirror. Full Nim: write the heap sizes in binary and XOR them — the nim-sum. Position (3,5,7): 011⊕101⊕111 = 001 ≠ 0, so the first player wins; the winning reply is 3 → 2, making 010⊕101⊕111 = 0. Strategy: always move to nim-sum 0. From 0, every move breaks the balance — the opponent must hand back a nonzero position — and from nonzero, a balancing move always exists. Bouton proved this completely in 1901, the first game ever solved; Sprague and Grundy (1935–39) then showed every impartial game is secretly Nim — one puzzle's invariant swallowed a genre.

The Move — Hold the balanced state. Identify the invariant class of dead positions and return the game to it every turn, until the opponent's freedom is irrelevant.

★★ 45 · Guess Two-Thirds of the Average

Everyone in a group privately picks a number in [0,100]. Closest to two-thirds of the group average wins. What do you pick?

Pick your number before you meet the theory.

Solution. Pure theory: any guess above 66.7 is dominated, since the average cannot exceed 100. If nobody plays dominated strategies, the average cannot exceed 66.7, so guesses above 44.4 die next; the elimination iterates all the way down to 0, the unique Nash equilibrium. Reality: in Nagel's 1995 experiments, and in newspaper contests with thousands of players, the winning number sits around 20–25 — real populations are a mixture of level-0 players (random, ~50), level-1 (guess 33), level-2 (guess 22), with few going deeper. The right play is not the equilibrium; it is one step deeper than the crowd, no more. Keynes' beauty-contest description of markets (1936) is this puzzle wearing a suit.

The Move — Iterate other minds (then stop). Model what others will do, including their model of you, but calibrate the depth of recursion to the actual population rather than an ideal one.

★ 47 · Fold Paper to the Moon

A sheet of paper is 0.1 mm thick. Fold it in half 42 times (grant the physical impossibility). How thick is the stack — a meter? A building? More?

A meter, a building, or more — commit.

Solution (computed). 2⁴² × 0.1 mm ≈ 4.4×10⁸ m — past the Moon (mean distance 3.84×10⁸ m). Twenty folds already exceed 100 m; thirty reach about 107 km. Physical reality check: the world record is 12 folds — Britney Gallivan, 2002, with a 1.2 km roll of tissue, after first deriving the loss function L = (πt/6)(2ⁿ+4)(2ⁿ−1), which shows why each fold costs quadratically more paper. Sibling legend: the wheat-doubling chessboard, 2⁶⁴ − 1 ≈ 1.8×10¹⁹ grains (told of Sissa by Ibn Khallikan, 1256).

The Move — Respect the exponent. Intuition extrapolates linearly and doubling punishes that within thirty steps: when a process multiplies — debt, virality, compute, compounding error — calculate, never eyeball.