Book Codes

Benedict Arnold encrypted his messages to the British Army using Blackstone’s Commentaries on the Laws of England. Arnold would replace each word in his message with a triplet of numbers representing the page number, line number, and word position where the word might be found in Blackstone. For example:

The 166.8.11 of the 191.9.16 are 129.19.21 266.9.14 of the .286.8.20, and 291.8.27 to be on 163.9.4 115.8.16 114.8.25ing — 263.9.14 are 207.8.17ed 125.8.15 103.8.60 from this 294.8.50 104.9.26 — If 84.8.9ed — 294.9.12 129.8.7 only to 193.8.3 and the 64.9.5 290.9.20 245.8.3 be at an 99.8.14.

British Army Major John André could then look up the words in his own copy of Blackstone to discover Arnold’s meaning:

The mass of the People are heartily tired of the War, and wish to be on their former footing — They are promised great events from this year’s exertion — If disappointed — you have only to persevere and the contest will soon be at an end.

The danger in using a book code is that the enemy can decode the messages if he can identify the book — and sometimes even if he can’t. In the comic strip Steve Roper, a reporter once excitedly telephoned the coded message 188-1-22 71-2-13 70-2-11 68-1-25 19-1-6 112-2-10 99-1-35. Reader Sean Reddick suspected that this message had been encoded using a dictionary, with each triplet of numbers denoting page, column, and word number. He never did discover the book that had been used, but by considering the ratios involved and consulting half a dozen dictionaries he managed to break the code anyway — he sent his solution to a nationally known columnist, who verified his feat when the comic strip bore out his solution. What was the message? (Hint: In the comic, the reporter mentions significantly that the plaintext message was given to him by “the delivery boy.”)

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WOMAN HERE HAS GIVEN BIRTH TO SEXTUPLETS. This appears in Dave Silverman’s article “Some Cryptographic Challenges” in the May 2008 issue of Word Ways; I’m not sure who the columnist was. Silverman’s own best guess was WISCONSIN MUSICAL MULTIMILLIONAIRE LEADS CONCERT UNDER PAVILION, which would have made for an entertaining comic.

March 2, 2015March 8, 2015 | Puzzles

One Two Three

Each point on a straight line is either red or blue. Show that it’s always possible to find three points of the same color in which one is the midpoint of the other two.

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Choose two points of the same color, say blue. Call these A and B, and let C be their midpoint. If C is blue then we’re done. If C is red then mark points D and E as shown such that AD = AB = BE. Now if D or E is blue then we’re done, because these produce the blue triples DAB and ABE. And if D and E are both red then we have DCE, a red triple. So there’s no way to assign colors to these five points without producing some monochromatic triple.

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March 1, 2015March 8, 2015 | Puzzles

There and Back Again

John and Mary drive from Westville to Eastville. John drives the first 40 miles, and Mary drives the rest of the way. That afternoon they return by the same route, with John driving the first leg and Mary driving the last 50 miles. Who drives the farthest, and by what distance?

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We don’t know the distance between Westville and Eastville, but suppose it’s only 50 miles. In that case John drives 40 miles altogether and Mary drives 60. If the distance is any greater than that, then the two drivers are dividing the extra miles equally between them, with Mary driving east and John west. So the total distance is immaterial — Mary drives 20 more miles than John.

February 27, 2015March 8, 2015 | Puzzles

The Sixth Cent

the sixth cent puzzle

You toss 6 fair coins, and I toss 5 fair coins. What is the probability that you get more heads than I do?

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Either you’ll toss more heads than I do or you’ll toss more tails than I do, but not both. (Right?) These outcomes are symmetric, so each has probability 1/2. UPDATE: A number of readers questioned the validity of this solution, but I think that’s due to a bad writeup on my part. I don’t know who devised the problem originally; I found it in the 1995 collection Five Hundred Mathematical Challenges, by Edward Barbeau, Murray Klamkin, and William Moser. Here’s their presentation: John tosses 6 fair coins, and Mary tosses 5 fair coins. What is the probability that John gets more “heads” than Mary? More generally, suppose John tosses n + 1 coins and Mary tosses n coins. Then, either John tosses more heads than Mary does or John tosses more tails than Mary does — but not both. (Check this out!) Since these two outcomes are symmetric, each occurs with probability 1/2. I think I confused the issue by failing to note that this reasoning addresses specifically the case of n coins vs. n + 1. I still haven’t had time to think about this carefully, but most of the people who’d objected wrote back later to change their minds. If you still think it’s incorrect, please do let me know, and I’ll post further updates here. Sorry for the confusion, and thanks to everyone who’s written in. UPDATE #2: Yes, the solution appears to be valid with this change. Thanks to everyone who contributed!

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February 23, 2015August 12, 2019 | Puzzles

Black and White

brownson chess problem

By O.A. Brownson Jr. White to mate in two moves.

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1. Qh8! and the queen will mate. From Chess Problems, 1876.

February 18, 2015March 1, 2015 | Puzzles

The Three Hats Game

Three players enter a room, and a maroon or orange hat is placed on each one’s head. The color of each hat is determined by a coin toss, and the outcome of one toss has no effect on the others. Each player can see the other players’ hats but not his own.

The players can discuss strategy before the game begins, but after this they may not communicate. Each player considers the colors of the other players’ hats, and then simultaneously each player must either guess the color of his own hat or pass.

The group shares a $3 million prize if at least one player guesses correctly and no player guesses incorrectly. What strategy will raise their chance of winning above 50 percent?

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A player who sees two hats of the same color (say, orange) guesses the opposite color (maroon). A player who sees hats of two different colors passes. There are 8 possible arrangements of hats: OOO OOM OMM MOO MMO MMM MOM OMO In six of these, two of the players see hats of different colors and so will pass, and the third player sees two hats of the same color and guesses the opposite color, winning. In the other two cases, all three players are wearing hats of the same color, so all guess the wrong color and lose. This strategy produces a win in six of the eight cases, so the players will win 3/4 of the time. From “Applications of Recursive Operators to Randomness and Complexity,” the 1998 UCSB doctoral thesis of computer scientist Todd Ebert.

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February 17, 2015February 22, 2015 | Puzzles

Red and Black

I propose a card game. I’ll shuffle an ordinary deck of cards and turn up the cards in pairs. If both cards in a given pair are black, I’ll give them to you. If both are red, I’ll take them. And if one is black and one red, then we’ll put them aside, belonging to no one.

You pay a dollar for the privilege of playing the game, and then we’ll go through the whole deck. When the game is over, if you have no more cards than I do, you pay nothing. But for every card that you have more than I, I’ll pay you 3 dollars. Should you play this game?

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No. Because the cards that are put aside are paired by color, the remaining cards must be divided equally between red and black. So at the end of the game you and I will always have the same number of cards. You’ll lose a dollar every time you play. From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.

February 14, 2015February 22, 2015 | Puzzles

Black and White

shinkman chess problem

By W.A. Shinkman. This is a self-mate in two moves: White makes a move, Black is allowed to make any legal reply, then White plays a second move that forces Black to checkmate him.

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1. Qd6! protects both white rooks and keeps Black’s bishop pinned. Black has only two options: 1. … Kc3 Rf3+ 2. Bxf3# 1. … Ke3 Rb3+ 2. Bxb3# From Benjamin Glover Laws, The Two-Move Chess Problem, 1890.

February 8, 2015February 15, 2015 | Puzzles

A Holiday Puzzle

New Year’s Day normally falls one week after Christmas: If Christmas falls on a Thursday, then New Year’s Day will fall on a Thursday as well. What is the most recent year in which Christmas and New Year’s Day fell on different days of the week?

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Trick question. Within any calendar year, Christmas and New Year’s Day always fall on different days:

February 6, 2015February 15, 2015 | Puzzles

Switching Visits

A prison warden greets 23 new prisoners with this challenge. They can meet now to plan a strategy, but then they’ll be placed in separate cells, with no means of communicating. Then the warden will take the prisoners one at a time to a room that contains two switches. Each switch has two positions, on and off, but they’re not connected to anything. The prisoners don’t know the initial positions of the switches. When a prisoner is led into the room, he must reverse the position of exactly one switch. Then he will be led back to his cell, and the switches will remain undisturbed until the warden brings the next prisoner. The warden chooses prisoners at his whim, and he may even choose one prisoner several times in a row, but at any time, each prisoner is guaranteed another visit to the switch room.

The warden continues doing this until a prisoner tells him, “We have all visited the switch room.” If this prisoner is right, then all the prisoner will be set free. But if he’s wrong then they’ll all remain prisoners for life.

What strategy can the prisoners use to ensure their freedom?

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If the prisoners knew that both switches were off at the start, then the solution would be fairly straightforward. One switch (say, the left one) will be used for counting, and the other is a meaningless dummy. One prisoner is chosen as the Counter. Each other prisoner is instructed to flip the left switch from off to on one time, at the first opportunity, but otherwise to flip the dummy switch on each visit. When the Counter enters the room and finds the left switch on, he turns it off; otherwise he too flips the dummy switch. When the Counter has found the left switch turned on 22 times, he knows that all prisoners have visited the room and can safely tell the warden so. Because the prisoners don’t know the initial states of the switches, this plan must be enhanced (otherwise, if the left switch starts in the on position, then the Counter will make his announcement prematurely). The solution is to ask each prisoner to flip the left switch from off to on twice in total, rather than once. The Counter will wait until he’s seen the left switch turned on 44 times altogether, and then tell the warden that all prisoners have visited the room. (At first I wondered why they couldn’t just stick with the first scenario but add 1 to the expected count to allow for the possibility that the left switch had started in the on position. But this won’t work: If the Counter is waiting until he’s found the left switch turned on 23 times, and if it started in the off position, then he’ll wait forever, because each of the 22 other prisoners will turn it on only once.)

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February 4, 2015February 15, 2015 | Puzzles