Ten pirates have 100 gold pieces and want to divide them according to the law of the sea, which says that the spoils go to the strongest. So they arrange themselves from weakest to strongest, P1, P2, …, P10. But these are democratic pirates, so they ask the strongest pirate to make a proposal as to how to divide the loot. All 10 pirates will then vote on it. If at least 50 percent of them support the proposal, then they’ll enact it and that’s that. Otherwise the proposer will be thrown to the sharks.
All pirates value their lives more than gold, all are rational, they cannot cut the gold pieces into smaller pieces, and no pirate will agree to a side bargain to share pieces. What proposal should the strongest pirate make in order to get the most gold?
This is easiest if we attack it from the end. When there are just two pirates left, P2 and P1, P2 should obviously propose that he keep all the gold for himself and give none to P1. The two of them will vote, and P2 will carry 50 percent of the vote and come away 100 gold pieces richer.
Now back up a step and consider the case with three pirates. P1 knows that if he doesn’t accept P3’s proposal then the situation will revert to the one we just considered, where he comes away empty-handed. So he should accept any proposal that gives him more than nothing. P3, who’s smart enough to see this, should propose giving 1 gold piece to P1 and nothing to P2.
Back up and consider four pirates. P4 will vote for his own proposal, of course, but he needs the vote of one other pirate to reach 50 percent, and he wants to spend as little gold as possible to get it. He proposes 99 gold pieces for himself, 0 for P3, 1 for P2, and 0 for P1. P2 is glad to vote for this, because if he doesn’t, P4 will go to the sharks and we’ll revert to the case of three pirates, where P2 gets nothing.
Back up one more step to five pirates. P5 needs two other pirates to vote for his proposal in order to have 50 percent of the votes. He proposes 98 gold pieces for himself, 0 for P4, 1 for P3, 0 for P2, and 1 for P1. P3 and P1 will vote for this, because otherwise they’ll find themselves in the four-pirate scenario, where they get nothing.
You can see where this is headed. In the original scenario, with ten pirates, P10 should propose 96 gold pieces for himself; 1 for each of the pirates P8, P6, P4, and P2; and none for the rest. By allocating just 4 gold pieces cannily, he can keep nearly all the treasure.
(Ian Stewart, “A Puzzle for Pirates,” Scientific American, May 1999, 98-99.)
Browsing the Post Office Guide in June 1891, Lewis Carroll discovered an ambiguity that produces “a very curious verbal puzzle” — he sent this pamphlet to friends and interested parties:
The Rule, for Commissions chargeable on overdue Postal Orders, is given in the ‘Post Office Guide’ in these words, (it is here divided, for convenience of reference, into 3 clauses)—
(a) After the expiration of 3 months from the last day of the month of issue, a Postal Order will be payable only on payment of a Commission, equal to the amount of the original poundage;
(b) with the addition (if more than 3 months have elapsed since the said expiration) of the amount of the original poundage for every further period of 3 months which has so elapsed;
(c) and for every portion of any such period of 3 months over and above every complete period.
You are requested to answer the following questions, in reference to a Postal Order for 10/- (on which the ‘original poundage’ would be 1d.) issued during the month of January, so that the 1st ‘period’ would consist of the months February, March, April; the 2nd would consist of the months May, June, July; and the 3rd would consist of the months August, September, October.
(1) Supposing the Rule to consist of clause (a) only, on what day would a ‘Commission’ begin to be chargeable?
(2) What would be its amount?
(3) Supposing the Rule to consist of clauses (a) and (b), on what day would the lowest ‘Commission’ begin to be chargeable?
(4) What would be its amount?
(5) On what day would a larger ‘Commission’ (being the sum of 2 ‘Commissions’) begin to be chargeable?
(6) What would be its amount?
(7) On what day would a yet larger ‘Commission’ begin to be chargeable?
(8) What would be its amount?
(9) Taking the Rule as consisting of all 3 clauses, in which of the above-named 3 ‘periods’ does clause (c) first begin to take effect?
(10) Which day, of any ‘period,’ is the earliest on which it can be said that a ‘portion’ of the ‘period’ has elapsed?
(11) On what day would the lowest ‘Commission’ begin to be chargeable?
(12) What would be its amount?
(13) On what day would a larger ‘Commission’ begin to be chargeable?
(14) What would be its amount?
(15) On what day would a yet larger ‘Commission’ begin to be chargeable?
(16) What would be its amount?
He followed up with this supplement later that month:
The trouble, as I read it, is that clause (c) is ambiguous. Presumably the postal authorities intended the general rule to be that a patron had three months to redeem a postal order, and beyond this would be charged a commission (here, 1 penny) for every three months that had elapsed since the deadline — including the last such period, which would not be prorated. Unfortunately, the phrase “every complete period” means exactly that — it refers to every completed period on the calendar. This sets the clock going twice as fast as intended. Our patron should owe 1d on May 1, 2d on August 1, and 3d on November 1. But with clause (c) worded as it is, she’ll owe 1d on May 1, 4d on August 1, and 6d on November 1. The final effect is that, beyond the first period, postal patrons who follow these rules will pay twice the intended commission.
I don’t know whether the post office ever learned about this. I imagine most patrons trusted them to do the math, and no one but Carroll recognized the ambiguity.
We want to build a road between two cities, A and B, that are separated by a river. We can build a bridge, but it must be perpendicular to the river’s banks, as shown. Where along the river’s length should we place the bridge if we want to minimize the total length of the road?
Reduce the river to a line of zero width along the bank nearest to A, and imagine moving City B up correspondingly to B’. Now a straight line drawn between A and B’ shows us where to put the bridge.
To prove this we can appeal to the triangle inequality, which says that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side. In the diagram below, the optimum path we’ve suggested goes A-P-Q-B. A bridge will have the same length no matter where we build it, so let’s omit that segment from our thinking. QB is parallel to PB’, so the length of our proposed road A-P-Q-B, minus the width of the river, is equivalent to AB’.
Now consider any alternative road, say A-P’-Q’-B. Again we can ignore the width of the river, and Q’B is parallel to P’B’, so the length of the alternative road A-P’-Q’-B, minus the width of the river, is equivalent to A-P’-B’. The triangle inequality says that AB’ must be shorter than AP’ + P’B’, so A-P-Q-B must be shorter than A-P’-Q’-B. We can apply a similar argument for any other proposed site for the bridge — bridge PQ must always produce a shorter road.
From Zbigniew Michalewicz and David B. Fogel, How to Solve It: Modern Heuristics, 2000.
The closing days of World War II witnessed a bizarre battle with some unlikely allies: American and German soldiers joined forces to rescue a group of French prisoners from a medieval castle in the Austrian Alps. In this week’s episode of the Futility Closet podcast we’ll follow the Battle for Castle Itter, the only time that Allies and Germans fought together in the war.
We’ll also dodge another raft of aerial bombs and puzzle over a bottled pear.
This week’s lateral thinking puzzles were adapted from the Soviet popular science magazine Kvant and the 2000 book Lateral Mindtrap Puzzles and contributed by listener Steve Scheuermann. We refer to this image in the second puzzle:
Please consider becoming a patron of Futility Closet — on our Patreon page you can pledge any amount per episode, and we’ve set up some rewards to help thank you for your support. You can also make a one-time donation on the Support Us page of the Futility Closet website.
Many thanks to Doug Ross for the music in this episode.
Hatebeak (above) is a death metal band fronted by a Congo African grey parrot named Waldo. To date they’ve released four albums, Beak of Putrefaction, Bird Seeds of Vengeance, The Thing That Should Not Beak, and Number of the Beak.
Bird Seeds of Vengeance was made with Caninus, a deathgrind band led by two pitbull terriers, Basil and Budgie. Unfortunately Basil died in 2011, and founding member Budgie died earlier this year, so the band has now retired.
“We are all lucky to have had her in our orbit for as long as we did,” the surviving members wrote after Budgie’s death. “She touched many lives, licked many faces, pushed many people out of beds, stole many slices of pizza, ate many soundguy burritos and, most importantly, inspired many to adopt from shelters instead of buying from pet shops or online breeders.”
The black rook on d4 pins the white rook on d6, so the black bishop on b6 is free to move, and this means that the white knight on c7 is pinned and not checking the black king. White plays 1. f3, threatening to take the e2 pawn with his queen. This would be mate — it’s no use blocking the queen’s check with the d4 rook, because this reverses the chain of pins — now the pinned d3 rook doesn’t check, so the d6 rook does, the b6 bishop doesn’t, and the c7 knight does, giving mate anyway! Likewise:
Here’s how the Union enciphered its messages during the Civil War. Abraham Lincoln sent this dispatch on June 1, 1863:
GUARD ADAM THEM THEY AT WAYLAND BROWN FOR KISSING VENUS CORRESPONDENTS AT NEPTUNE ARE OFF NELLY TURNING UP CAN GET WHY DETAINED TRIBUNE AND TIMES RICHARDSON THE ARE ASCERTAIN AND YOU FILLS BELLY THIS IF DETAINED PLEASE ODOR OF LUDLOW COMMISSIONER
The first word, GUARD, indicates the size of a containing rectangle and the paths on which the words must be laid out to decipher the message. In this case, they’ll go up the first column, down the second, up the fifth, down the fourth, and up the third. Also, just to confuse the Confederates, every eighth word after GUARD is a null and should be discarded. So we get:
FOR VENUS LUDLOW RICHARDSON AND
BROWN CORRESPONDENTS OF THE TRIBUNE
WAYLAND AT ODOR ARE DETAINED
AT NEPTUNE PLEASE ASCERTAIN WHY
THEY ARE DETAINED AND GET
THEM OFF IF YOU CAN
ADAM NELLY THIS FILLS UP
The last steps are to remove THIS FILLS UP, which is only there to fill out the block, and to replace a few code words:
VENUS = colonel
WAYLAND = captured
ODOR = Vicksburg
NEPTUNE = Richmond
ADAM = President of the United States
NELLY = 4:30 p.m.
That gives us the final message:
For Colonel Ludlow,
Richardson and Brown, correspondents of the Tribune, captured at Vicksburg, are detained at Richmond. Please ascertain why they are detained and get them off if you can.
The President, 4:30 p.m.
This system was such a valuable source of breaking news that Lincoln often visited the military telegraph office in the War Department, next to the White House, and would chat with the operators there. One of them, David Homer Bates, who was only 18 when the war started, remembered, “Outside the members of his cabinet and his private secretaries, none were brought into closer or more confidential relations with Lincoln than the cipher-operators, … for during the Civil War the President spent more of his waking hours in the War Department telegraph office than in any other place, except the White House. … His tall, homely form could be seen crossing the well-shaded lawn between the White House and the War Department day after day with unvaried regularity.”
“Those praised in a book take that praise, and more, as their due. What you meant as a gift is accepted as an obligation. In a second printing of one of his books, a writer listed the misprints in the first. Among them was the dedication.” — Baltasar Gracián