Hedge Funds puzzles, solved step by step

Light rope one at both ends and rope two at one end at the same moment; when rope one burns out, light rope two's other end, and it burns out at 45 minutes. Two flames always meet after burning 60 minutes of rope between them, so rope one takes 30 minutes however uneven it is. Rope two then has 30 minutes left, which two flames finish in 15.

Why does lighting both ends halve the time on an uneven rope?

Picture two people eating a long, uneven sandwich from opposite ends, each chewing through whatever is at their end. However the filling is spread, they meet once the whole sandwich has been eaten between them, and together they finish in half the time one would take. Two flames consume the rope's total burn time twice as fast, so a 60 minute rope lit at both ends is gone in 30 minutes, wherever the flames happen to meet. Length tells you nothing here; burn time is the only quantity you can trust.

Why is this really a question about information?

The rope hides where its time is stored, much as an order book hides how much size is waiting behind a price. The solution uses only what is known, the total burn time, and never what is not, how it is spread along the rope. Anyone who cuts a rope in half is assuming evenness that the first sentence ruled out. Say that out loud before you give the method: naming what you may not assume is half of a good answer.

Expect the follow-up. The same trick measures 15 minutes as an interval, the gap between rope one going out and rope two going out. Each rope lit from its second end at a known moment halves whatever burn time it has left, and chaining those halvings is how you reach times such as 52.5 minutes with a third rope. Walk through the chain in order, one lighting at a time.

The ten perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. Locker n is toggled once by each person whose number divides n, so its final state depends on how many divisors n has. Divisors come in pairs, d and n/d, which cancel out. Only a perfect square has an unpaired divisor, its square root, so only squares are toggled an odd number of times and end open.

What decides whether one locker ends open?

A light switch flipped an even number of times ends where it started; flipped an odd number of times, it ends the other way. Each locker is a switch, flipped once for every divisor of its number, so the question is which numbers from 1 to 100 have an odd number of divisors. Locker 12 is touched by persons 1, 2, 3, 4, 6 and 12, six times, and ends closed.

Why do only perfect squares have an odd number of divisors?

Pair every divisor d with n divided by d. For 12 the pairs are 1 and 12, 2 and 6, 3 and 4: six divisors, even. The pairing breaks only when a divisor is paired with itself, d = n/d, which happens exactly when n is a perfect square. For 36 the pairs are 1 and 36, 2 and 18, 3 and 12, 4 and 9, with 6 left over: nine divisors, odd, so locker 36 ends open. There are ten squares up to 100, so ten lockers.

Why does a fund ask a puzzle like this?

It tests whether you look for structure before you simulate. Walking through a hundred people toggling lockers is hopeless in an interview; turning it into a question about divisors takes one sentence and makes the answer obvious. The move, recasting a process as a property you can count, is the one that turns a messy trading rule into a quantity you can compute. Check it on a small case out loud: with 10 lockers, 1, 4 and 9 end open.

A proposes 98 for himself, 0 for B, 1 for C, 0 for D and 1 for E. Work backwards. With two pirates, D's own vote is half, so he keeps all 100. With three, C buys E with 1 coin. With four, B buys D with 1 coin. With five, A needs two votes beyond his own and buys the two pirates who get nothing in the four-pirate split, C and E, for one coin each.

Why start from the end?

Planning a train journey, you work back from when you must arrive, not forward from when you wake up. Each pirate votes by comparing the offer with what he would get if the proposal failed, so you can only price a vote once you know the next round's outcome, which means solving the smallest game first and working upwards. That method, backward induction, is the whole puzzle.

How does each round play out?

Two pirates, D and E: D proposes 100 for himself, and his own vote is half, so it passes. Three pirates: C needs one more vote and buys E, who gets nothing in the two-pirate game, for 1 coin: 99, 0, 1. Four pirates: B needs one more vote and buys D, who gets nothing in the three-pirate game: 99, 0, 1, 0. Five pirates: A needs two more votes and buys C and E, both empty-handed in the four-pirate game: 98, 0, 1, 0, 1.

What is the interviewer really testing?

Whether you reason about the alternative each party faces rather than about fairness. A vote costs exactly one coin more than what the voter gets if the deal fails, so the cheapest supporters are the ones with the worst fallback. The same logic runs through any negotiation: a creditor backs a restructuring plan when it beats their recovery in liquidation, and support is cheapest from those whose alternative is worst. State the assumptions: perfect rationality, and a pirate who is indifferent votes against, which is why one coin, not zero, is needed.

Both end up side by side in the middle, and customers are worse off. From the quarter points, a seller who steps inward keeps everyone behind them and wins beach from the rival, so both drift to the centre. There they still split customers half and half, but the average walk doubles from 125 m to 250 m. Competition moves the sellers to the spot that maximises share, not the one that serves customers best.

Why can neither seller stay at the quarter points?

Picture two petrol pumps on a highway. Each wants the drivers on its own side plus as many from the middle as it can reach first. A seller keeps every customer on the far side of them wherever they stand, so moving towards the rival only ever adds customers. From 250 m, A steps to 400 m against B at 750 m: the dividing line moves to the midpoint, 575 m, and A's share rises from 50% to 57.5%. B then responds the same way, and the dance ends only when both stand at 500 m.

What does the middle cost the customers?

With both sellers in the centre, a sunbather is on average a quarter of the beach away, 250 m. With sellers at 250 m and 750 m, nobody is more than 250 m away and the average is 125 m. The shares are identical in both setups; the only thing that changed is how far customers walk, so the stable outcome is strictly worse for them and no better for the sellers. This is the {term('Hotelling model', 'A model of competition on a line, set out by Harold Hotelling in 1929, in which rivals crowd towards the centre to win the middle ground.')}, and it is why rival shops cluster and why two parties often converge on the middle voter.

Say the equilibrium idea in one line: a pair of positions is stable when neither player can do better by moving alone, and the middle is the only such pair here. Then say where the model breaks: if customers stop buying when the walk is too long, or if prices can differ, the sellers have a reason to spread out again. Interviewers ask this to see whether you can reason about another player's best response, which is most of trading.

14 drops. With k drops available, the first egg should go from floor k: if it breaks, the second egg checks the k minus 1 floors below one at a time. If it survives, you have k minus 1 drops left, so the next gap is one smaller. k drops therefore cover k + (k minus 1) + ... + 1 = k(k + 1)/2 floors. 13 drops cover 91 floors, 14 cover 105, so 14 is the minimum: drop from 14, 27, 39, 50 and so on.

Why does binary search fail here?

Binary search assumes you can keep testing after a failure. With two eggs, the first break leaves you one egg, and one egg can only be used safely by walking up one floor at a time. Once the first egg breaks, every floor below it that has not been ruled out costs one drop of the second egg, so large jumps with the first egg are expensive. Dropping the first egg at floor 50 and seeing it break could cost 49 more drops. It is like searching for a leak with one spare pipe: once the first one bursts, you test the rest slowly.

How do you balance the worst cases?

Make every worst case take the same number of drops. If you allow k drops in total, the first drop should be from floor k, the next k minus 1 floors higher, the next k minus 2 higher, because each first-egg drop used leaves one fewer drop for the second egg's walk. With k = 14 the first egg goes from 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99 and 100. If it breaks at 27, the second egg tests 15 to 26: 2 plus 12 is 14 drops. The same count holds at every step.

Say what an interviewer wants beyond the number. The move is to fix the budget of drops and ask how many floors it can cover, rather than fixing the building and searching for a strategy. That reversal is what makes the problem easy, and it generalises: with three eggs and k drops, the floors covered are the two-egg coverage for each smaller budget, plus one per drop, added up, which is why three eggs need only 9 drops for 100 floors.

Weigh four against four first, then mix suspects with coins you already know are genuine so every later weighing splits the cases three ways. There are 24 possibilities, 12 coins each heavy or light, and three weighings have 27 outcomes. If 1 to 4 balances 5 to 8, weigh 9, 10, 11 against three good coins; if not, weigh 1, 2, 5 against 3, 6, 9. The third weighing settles what is left.

How do you know three weighings can be enough?

A game of twenty questions works because each yes or no halves what is left. A balance is better than a yes or no: it answers left heavy, right heavy or balanced. Three weighings give 3 x 3 x 3 = 27 outcomes, and there are 24 cases to tell apart, 12 coins each possibly heavy or light, so a procedure can exist only if every weighing splits the remaining cases into three near-equal groups. That counting sets the design: the first weighing must leave at most 9 cases on every branch.

What do you do after the first weighing tips?

Say the left pan was heavy: the odd coin is 1, 2, 3 or 4 and heavy, or 5, 6, 7 or 8 and light. Weigh 1, 2 and 5 against 3, 6 and 9, moving some suspects across and bringing in a known good coin, so each outcome points to a different small group. Left heavy again means 1 heavy, 2 heavy or 6 light: weigh 1 against 2, and a balance means 6. Right heavy means 3 heavy or 5 light: weigh 3 against a good coin. A balance means 4 heavy, 7 light or 8 light: weigh 7 against 8.

What if the first weighing balances?

Then coins 1 to 8 are genuine and the odd coin is among 9 to 12, still heavy or light. Weigh 9, 10 and 11 against three good coins: a tip tells you both that the odd coin is among the three and whether it is heavy or light, and a balance points to coin 12. After a tip, weigh 9 against 10: if the odd coin is heavy the heavier of the two is it, if light the lighter, and a balance means 11. After a balance, weigh 12 against a good coin to learn heavy or light.

17 minutes. 1 and 2 cross (2 minutes), 1 returns (1), 5 and 10 cross together (10), 2 returns (2), and 1 and 2 cross again (2). The total is 17. The obvious plan, with the fastest person escorting everyone, takes 19, because the 5 and the 10 each cost a separate crossing. Pairing the two slowest hides the 5 inside the 10.

Why is the obvious plan not the fastest?

The natural plan uses the quickest person as a shuttle: 1 walks each person over and comes back. That costs 2 + 1 + 5 + 1 + 10 = 19. Every slow person who crosses on a separate trip pays their own time in full, so the 5 and the 10 together cost 15 minutes of crossing. Think of sending two slow parcels in one courier van instead of two: the second rides along for free.

How does the 17-minute plan pay for pairing the slow two?

Pairing 5 and 10 costs 10 instead of 15, a saving of 5. But someone fast must already be waiting on the far side to bring the torch back afterwards. That setup costs one extra return by the 2-minute person and an extra crossing of the pair 1 and 2, so the net saving is 2 minutes: 19 down to 17. The general rule: pair the two slowest when twice the second-fastest time is less than the fastest plus the second-slowest, here 4 against 6.

How do you show 17 is the minimum?

Five crossings are unavoidable: three over and two back. If the 5 and the 10 cross separately, the forward trips cost at least 10 + 5 + 2 and the two returns at least 1 each, which is 19; if they cross together, the person who returns next must already be across, which forces the returns to be the 1 and the 2, and the best total is 17. Talking through that bound is what separates a remembered answer from a reasoned one, and it is what lets you handle the follow-up with different times.