The Cornish Cliff Mystery

dudeney cliff puzzle

Police have chased two thieves to the Cornish coast. There they find that two sets of footprints depart the hard road and cross soft soil to the edge of a cliff, where they end. The cliff juts out 200 feet above sea-washed boulders. No one could survive a fall from this height, there is no way to descend the cliff, and there are no other footprints.

The police have proven that the footprints match those of the fleeing criminals. The small foot belongs to Marsh, who apparently takes relatively long strides, walking heavily on his heels. Lamson takes shorter strides, treading more on his toes and evidently following behind Marsh, as he sometimes treads over the smaller man’s footprints.

The two men did not walk to the cliff edge and then return to the road by walking backward in their own footprints — such precision over a course of 200 yards is impossible. Accordingly the inspector says he will report that the criminals, hopeless to escape justice, have hurled themselves to their death.

“Then you will make a fatal mistake,” says Henry Melville, a visiting member of the Puzzle Club. “The men are alive and in hiding in the district.” He proves to be right, but how did the men get away from the edge of the cliff?

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From the road, Marsh alone walked to the edge of the cliff, carrying Lamson’s boots in his hands. At the edge of the cliff he changed boots and walked backward to the road, carrying his own boots. “This little maneuver accounts for the smaller footprints showing a deeper impression at the heel, and the larger prints a deeper impression at the toe, for a man will walk more heavily on his heels when going forward, but will make a deeper impression with the toes in walking backwards. It will also account for the fact that the large footprints were sometimes impressed over the smaller ones, but never the reverse; also for the circumstance that the larger footprints showed a shorter stride, for a man will necessarily take a smaller stride when walking backwards.” From Henry Dudeney’s The Canterbury Puzzles, 1907.

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November 27, 2016December 4, 2016 | Puzzles

Mens et Manus

David Hagen offered this puzzle in MIT Technology Review in 2007. The MIT logo can be thought of as a slider puzzle. In the figure above, can you slide the tiles about so that the gray I can escape through the opening at top left?

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Review reader Brad Edelman offered this solution. First number the tiles 0 to 6, left to right, with pieces in the same column numbered in sequence from top to bottom. Now: 2 – down, down 4 – down 6 – right 3 – down, right, down, down 1 – right, right 2 – up, up, left 4 – left, down 3 – left 1 – down, right, down 2 – right, right 4 – up, up 3 – left 2 – down 5 – left 6 – up 1 – right 2 – right 3 – right 4 – right 0 – right, right, down, down 5 – left, left, left, down, down 4 – up, left, left, left, up The original puzzle was posed in the November/December 2007 issue; the solution was presented two issues later. At the time, readers made some interactive tools, in Word and Flash, to aid in the solving, but those don’t seem to be available any longer. If anyone knows of any, please let me know and I’ll add a link here. 12/06/2016 UPDATE: Reader Chuck Rubins has created a downloadable interactive version in PowerPoint. (Thanks, Chuck.)

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November 19, 2016December 18, 2018 | Puzzles

Point to Point

point to point 1

Here’s a triangle, ABC, and an arbitrary point, D, in its interior. How can we prove that AD + DB < AC + CB?

The fact seems obvious, but when the problem is presented on its own, outside of a textbook or some course of study, we have no hint as to what technique to use to prove it. Construct an equation? Apply the Pythagorean theorem?

“The issue is more serious than it first appears,” write Zbigniew Michalewicz and David B. Fogel in How to Solve It (2000). “We have given this very problem to many people, including undergraduate and graduate students, and even full professors in mathematics, engineering, or computer science. Fewer than five percent of them solved this problem within an hour, many of them required several hours, and we witnessed some failures as well.”

Here’s a dismaying hint: Michalewicz and Fogel found the problem in a math text for fifth graders in the United States. What’s the answer?

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Here’s one way to do it. Extend AD to CB and call the intersection E. The sum of two sides of any triangle is longer than the third side, so AD + DB < AD + (DE + EB) [because DB < DE + EB] AE + EB < (AC + CE) + EB [because AE < AC + CE] So AD + DB < AC + CB. (Another way is to extend BD to AC, rather than extending AD to CB, and then follow a similar argument.)

November 17, 2016November 27, 2016 | Puzzles · Science & Math

Dividing the Spoils

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?

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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.)

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November 10, 2016November 20, 2016 | Puzzles

The Shortest Road

shortest road - 1

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?

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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.

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November 8, 2016November 20, 2016 | Puzzles

Black and White

dawson chess puzzle

From T.R. Dawson, a logic problem posing as a chess puzzle. If pinned men do not check, how can White mate in two moves?

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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: 1. f3 Rd5 2. Qxe2# (same idea as above — 2. … Rb5 is ineffective) 1. f3 Bxa5 2. Kc8# 1. f3 Bxc7 2. Nxc7# 1. f3 Bxe8 2. Kxe8# 1. f3 Qxe7+ 2. Kxe7# 1. f3 Rd2 2. Bxd2# 1. f3 Rxd6+ 2. Qxd6# Dawson wrote, “The idea amounts to ‘discovering mate’ from c7.” He dedicated the puzzle to Eduard Birgfeld. From Caissa’s Wild Roses, 1935.

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November 5, 2016November 13, 2016 | Puzzles

Alternate Routes

How many pairs of prime numbers are there whose sum is 999?

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One. Any pair of numbers with an odd sum must have an even member, and there’s only one even prime. So there’s only one answer, 997 + 2 = 999.

November 3, 2016November 13, 2016 | Puzzles · Science & Math

The Donjon Keep Window

dudeney donjon keep window puzzle

In Henry Dudeney’s Canterbury Puzzles, Sir Hugh De Fortibus takes his chief builder to the walls of his donjon keep and points to a window there.

“Methinks,” he says, “yon window is square, and measures, on the inside, one foot every way, and is divided by the narrow bars into four lights, measuring half a foot on every side.”

“Of a truth that is so, Sir Hugh,” says the builder.

“Then I desire that another window be made higher up whose four sides shall also be each one foot, but it shall be divided by bars into eight lights, whose sides shall be all equal.”

Bewildered, the builder says, “Truly, Sir Hugh, I know not how it may be done.”

“By my halidame!” exclaims De Fortibus in pretended rage. “Let it be done forthwith. I trow thou art but a sorry craftsman, if thou canst not, forsooth, set such a window in a keep wall.”

How can it be done?

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“The illustration will show how this was to be done. It will be seen that if each side of the window measures one foot, then each of the eight triangular lights is six inches on every side.” “Of a truth, master builder,” said De Fortibus slyly to the architect, “I did not tell thee that the window must be square, as it is most certain it could never be.”

October 25, 2016November 6, 2016 | Puzzles

Black and White

billiards chess puzzle

Billiards chess is a variant of traditional chess in which the pieces carom off the sides of the board at right angles. In the diagram above, the white bishop at a2 controls the diagonal a2-g8 as in normal chess, but its attack also “bounces” from g8 to h7 and then back along the h7-b1 diagonal. Both bishops attack and move along these “bouncing” lines. How can White mate the black king in two moves?

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The king is already nearly corralled; White just has to arrange a final attack along the long dark diagonal. 1. Ka1 e2 2. Bb2 does the trick; if 1. … Ke5 then 2. Bg7#. By F.F.L. Alexander and E.P. White, from The Problemist, June 1932.

October 21, 2016October 30, 2016 | Puzzles

Speaking Volumes

A puzzle by A. Kozlov from the Soviet popular science magazine Kvant:

Watching a solar eclipse, a girl asks her father how much farther away is the sun than the moon.

He says, “As far as I remember, 387 times farther.”

She says, “Then I can figure out how much greater is the sun’s volume than the moon’s.”

He thinks about this and says, “I think maybe you can.” How can she do this?

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During a solar eclipse the moon covers the sun almost perfectly. This means that the ratio of the sun’s radius to the moon’s radius is the same as the ratio of these two bodies’ distance from Earth. So the sun’s volume is larger than the moon’s by a factor of about 387³.

October 20, 2016October 30, 2016 | Puzzles