The Church Cipher

In 1775, an enciphered message was intercepted en route from colonial physician Benjamin Church to a British officer in Boston. Part of the message appears below. What does it say?

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It’s a simple substitution cipher: So this sentence reads, “Should Britain declare War against the Colonies, they are lost forever.” A court of inquiry found that Church had carried on a criminal correspondence with the enemy. He was jailed until 1778, when he sailed from Boston and was never heard from again. (From the National Park Service.)

November 5, 2020November 4, 2020 | History · Puzzles

Black and White

penrose chess problem

Physicist Roger Penrose devised this problem in 2017 to illustrate the difference between artificial and human intelligence. James Tagg, leader of the Penrose Institute, told the Telegraph, “A human looking at it for a short while will ‘see’ what white must, and more particularly, must not do, and use very little energy to decide this. But, for a computer, the puzzle requires an enormous number of calculations, far too many for even today’s supercomputers.” What is White’s insight?

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Penrose wrote, “If you put this puzzle into a chess computer it just assumes a black win because of the number of pieces and positions, but a human will look at this and know quickly that is not the case.” This is because most of Black’s pieces are entombed on the queenside; his bishops are free to move, but they can neither free the other black pieces nor checkmate the white king. Penrose explained: If both players understand the position and play perfectly there is no way either player can win – black can always keep its bishops together so the white king cannot pick them off, and the 50-move rule would result in a draw. Human players understand this and don’t need to calculate the position. However, we can see that Fritz (one of the better AI chess programs) looks ahead 20 moves and thinks black is winning. Fritz does not understand the nature of the position: It sees the material advantage and thinks it is ahead. The question is whether the human playing as white can use this lack of understanding to win. A small number of you ran the puzzle on chess programs and experimented with this. Although we maintain that chess programs (such as Fritz and Stockfish) do not ‘understand’ the position, they played correctly and achieved a draw despite believing they had a likely win. However, some programs set to a low level of skill were so convinced they should win, they refused the draw through repeated position. They accepted losing the bishops one after another until they blundered into a losing position, allowing the white pawn on the sixth rank to be advanced and turned into a queen. To make this actually happen requires a bit of trickery and a great deal of patience. “We know that there are things that the human mind achieves that even the most powerful supercomputer cannot but we don’t know why,” he added. “If we find out how humans differ from computers then it could have profound sociological implications. People get very depressed when they think of a future where robots or computers will take their jobs, but it might be that there are areas where computers will never be better than us, such as creativity.”

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Penrose wrote, “If you put this puzzle into a chess computer it just assumes a black win because of the number of pieces and positions, but a human will look at this and know quickly that is not the case.” This is because most of Black’s pieces are entombed on the queenside; his bishops are free to move, but they can neither free the other black pieces nor checkmate the white king. Penrose explained:

If both players understand the position and play perfectly there is no way either player can win – black can always keep its bishops together so the white king cannot pick them off, and the 50-move rule would result in a draw. Human players understand this and don’t need to calculate the position.

However, we can see that Fritz (one of the better AI chess programs) looks ahead 20 moves and thinks black is winning. Fritz does not understand the nature of the position: It sees the material advantage and thinks it is ahead. The question is whether the human playing as white can use this lack of understanding to win. A small number of you ran the puzzle on chess programs and experimented with this.

Although we maintain that chess programs (such as Fritz and Stockfish) do not ‘understand’ the position, they played correctly and achieved a draw despite believing they had a likely win. However, some programs set to a low level of skill were so convinced they should win, they refused the draw through repeated position. They accepted losing the bishops one after another until they blundered into a losing position, allowing the white pawn on the sixth rank to be advanced and turned into a queen. To make this actually happen requires a bit of trickery and a great deal of patience.

“We know that there are things that the human mind achieves that even the most powerful supercomputer cannot but we don’t know why,” he added. “If we find out how humans differ from computers then it could have profound sociological implications. People get very depressed when they think of a future where robots or computers will take their jobs, but it might be that there are areas where computers will never be better than us, such as creativity.”

November 2, 2020November 1, 2020 | Puzzles

Euphony

A charming puzzle from Crux Mathematicorum, December 2004:

If all plinks are plonks and some plunks are plinks, which of these statements must be true?

X: All plinks are plunks.
Y: Some plonks are plunks.
Z: Some plinks are not plunks.

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(Solution by Geneviève Lalonde, Massey, Ontario:) In this diagram the various regions are numbered. Since all plinks are plonks, there can be nothing in region 1 or 6. And since some plunks are plinks, the regions 6 and 7, considered together, cannot be empty. Combining these conclusions shows that region 7 can’t be empty. Now: Statement X says that regions 1 and 4 are both empty. We know nothing about region 4, so we can’t say whether this statement is true. Statement Y says that regions 5 and 7, considered together, can’t be empty. We know that region 7 is not empty, so this statement is true. Statement Z says that regions 1 and 4, considered together, can’t be empty. We’ve established that region 1 is empty, but again, we know nothing about region 4, so we don’t know whether this statement is true. Hence the only statement that must be true is Y.

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October 27, 2020October 26, 2020 | Puzzles

Podcast Episode 317: Lateral Thinking Puzzles

Here are six new lateral thinking puzzles — play along with us as we try to untangle some perplexing situations using yes-or-no questions.

See full show notes …

October 26, 2020November 2, 2020 | Podcast · Puzzles

Board Meeting

Prove that if seven darts land on a dartboard, there will always be two darts that are no farther apart than the distance of one radius.

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Image: Wikimedia Commons Divide the dartboard into six equal sectors. One of these must contain at least two of the darts, and these two cannot be farther apart than the length of a radius. Interestingly the same conclusion can be reached even if only six darts are thrown — see Tony Crilly and Stephen Suen, “An Improbable Game of Darts,” Mathematical Gazette 71:456 [June 1987], 97-100.

October 22, 2020October 21, 2020 | Puzzles

Black and White

immunity chess puzzle

Black is clearly lost. But there are two squares on which his king can never be checkmated, even if White is allowed to make consecutive moves and checks are ignored. What are they?

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b2 and g7. In order to trap the black king completely there, White would need at least one square beyond the edge of the board. (For example, if the black king is on b2, then White can mate by putting his king on b4, his queen on b3, and his bishop just “below” the b1 square.) From Stephen Addison, The Book of Extraordinary Chess Problems, 1989.

October 13, 2020October 11, 2020 | Puzzles

Pursuit

A problem from Sam Loyd’s Cyclopedia of Puzzles, 1914:

Here is the puzzle of Tom the Piper’s Son, who, as told by ‘Mother Goose,’ stole the pig and away he run. It is known that Tom entered the far gate shown at the top on the right hand. The pig was rooting at the base of the tree 250 yards distant, and Tom captured it by always running directly towards it, while the pig made a bee-line towards the lower corner as shown. Now, assuming that Tom ran one-third faster than the pig, how far did the pig run before he was caught?

Intriguingly, Loyd adds, “The puzzle is a remarkable one on account of its apparent simplicity and yet the ordinary manner of handling problems of this character is so complicated that solvers are asked merely to submit approximately correct answers, based upon judgment and common sense, just to see who can make the best guess. The simple rule for solving it, however, which will doubtless be quite new to our puzzlists, is based upon elementary arithmetic.” What’s the answer?

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Loyd’s answer: In the puzzle of Tom the Piper’s Son, who stole the pig, Mother Goose reports that Tom had to run 571 3/7 yards to catch the pig, while the porker ran three-quarters of that distance, which would be 428 4/7 yards. The simple rule for solving problems of this class is to halve the distance the man would have to travel to catch the pig in a straight line added to the distance he would go if they advanced toward each other. Tom is 250 yards from the pig, and as he goes one-third faster, in a straight line, he would catch him in 1,000 yards. If they run toward each other, Tom goes four-sevenths of the distance, viz.: 142 6/7 of the 250 yards, which, added to 1,000, equals 1,142 6/7 yards. Half of this equals 571 3/7 yards, which is the correct answer. Loyd doesn’t really explain his “simple rule,” but in the Pi Mu Epsilon Journal (Volume 6, Number 9, Fall 1978, page 545), Tom Apostol of Caltech shows that Loyd’s method is generally valid. The editors add, “The charm of this problem lies in the insidious way the stated result defies our intuitive notions. At first we suspect that the result, if true, might accidentally apply only to the special case involving the given distances and the relative speeds but then the denouement affirms the generality of Loyd’s method of solution.” In his edited collection of Loyd’s puzzles, Martin Gardner added, “The man’s path forms one of the simplest of ‘pursuit curves,’ the study of which forms an interesting branch of what one might call ‘recreational calculus.'”

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September 26, 2020September 25, 2020 | Puzzles · Science & Math

Black and White

Smyth-Helms 1915

In 1915 James Ferguson Smyth invited Hermann Helms to play a practice game at the Manhattan Chess Club. They had reached the position above when Helms, as Black, found a brilliant two-move mate. What is it?

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22. … Qg2+! 23 Kxg2 Rxg3# The whole game is here. Helms’s brilliancy was so inspiring that Anthony Santasiere produced an oil painting of the position, which for many years hung in the Marshall Chess Club.

September 22, 2020September 21, 2020 | Puzzles

Red Square

A problem from the 1999 Russian Mathematical Olympiad:

Each cell of a 50×50 square is colored in one of four colors. Show that there exists a square which has cells of the same color as it directly above, directly below, directly to the left, and directly to the right of it (though not necessarily adjacent to it).

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A 50×50 square contains 2500 cells. At least a quarter of these, 625 cells, must be of the same color, say red. Of these 625 red squares, at most 50 can be the uppermost red squares in their columns; at most 50 can be the lowermost red squares in their columns; at most 50 can be the leftmost red squares in their rows; and at most 50 can be the rightmost red squares in their rows. Together these make up at most 200 squares. The remaining 425 (or more) red squares have none of these distinctions, so each must have a red square directly above, directly below, directly to the left, and directly to the right of it. From Titu Andreescu and Zuming Feng, Mathematical Olympiads 1999-2000: Problems and Solutions from Around the World, 2002.

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September 20, 2020September 19, 2020 | Puzzles

A Perfect Square

An old problem from the Soviet Mathematical Olympiad:

Find the 4-digit number aabb that is a perfect square.

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The smallest 4-digit perfect square is 1024 (32²). The largest is 9801 (99²). And aabb = 1000a + 100a + 10b + b = 1100a + 11b = 11(100a + b). So 11 is one of the factors of aabb, and aabb is of the form (11k)². That leaves only six candidates to check (33², 44², 55², 66², 77², 88²), and it turns out the answer is 88² = 7744. (Thanks, Danesh.)

September 15, 2020September 14, 2020 | Puzzles