Black and White

korolikov chess problem

A curiously ambiguous problem by V.A. Korolikov. Mate in 1. Either side can fulfill this easily — but which has the move?

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White’s knights began the game on squares of opposite color, but now they’re both on white squares. That means that, between them, they’ve made an odd number of moves in this game. By a similar argument White’s rooks and king have made an even number of moves in total, and we can see that White has made a single pawn move. White’s bishops haven’t moved, and his queen must have been captured on her home square. So White has made an even number of moves. The same considerations apply to Black’s position, but he’s made an extra pawn move, so altogether he’s made an odd number of moves. Since White moved first, this means it’s Black turn to move, and he can mate with 1. Nxc2#. (From Schach, 1957, via Noam D. Elkies and Richard P. Stanley, “The Mathematical Knight,” Mathematical Intelligencer 25:1 [December 2003], 22-34.)

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June 21, 2018June 21, 2018 | Puzzles

Close Quarters

Paulo Guarini di Forli offered this puzzle in 1512. On this tiny 3 × 3 board, which is the smallest number of moves in which the white knights can exchange places with the black ones?

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Henry Dudeney presented this solution in his 1917 Amusements in Mathematics; it’s now known as the “method of buttons and strings.” No knight will ever visit the center square, so we can ignore that; regard each of the remaining squares as a button and connect it by a string to each square that’s a knight’s move away. That produces diagram B. Now if we untangle the strings we get diagram C, a unique cycle that any knight must follow in making its way around this board. The four knights must follow this cycle, and no two knights may ever occupy the same square. The white knights start on squares 1 and 3, the black on 6 and 8. So to exchange the white and black knights, we must advance all four knights four steps along the circle, either clockwise or counterclockwise. Since the four knights make four moves each, the task is done in 16 moves. That’s the minimum. (Somewhat paradoxically, if three white knights oppose three black ones in the starting position, the minimum number of moves drops to 8.)

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June 18, 2018 | Puzzles

False or True?

A Russian coin-weighing puzzle:

You have 101 coins, and you know that 50 of them are counterfeit. Every true coin has the same weight, an unknown integer, and every false coin has the same weight, which differs from that of a true coin by 1 gram. You also have a two-pan pointer scale that will show you the difference in weight between the contents of each pan. You choose one coin. Can you tell in a single weighing whether it’s true or false?

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Yes. Put the coin in one pan, and put all 100 remaining coins in the other. Now suppose a true coin weighs t grams and a fake coin weighs f grams. If the coin is fake, then the difference between the weights will be 51t + 49f – f = 51t + 48f = 51t + 48(t ± 1) = 99t ± 48, which is always divisible by 3. If the coin is true, then the difference will be 50t + 50f – t = 49t + 50f = 49t + 50(t ± 1) = 99t ± 50, which is not divisible by 3. Another solution is to put aside the chosen coin and divide the remaining 100 coins into two piles. If the difference in weight between these piles is even, then the chosen coin is genuine; if it’s odd then it’s false. (From Sergei Aleksandrovich Genkin, Dmitrii Vladimirovich Fomin, and Ilia Itenberg, Mathematical Circles: (Russian Experience), 1996.)

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June 12, 2018June 11, 2018 | Puzzles

A Fractured Mystery

In 1934, Victor Gollancz published The Torquemada Puzzle Book, a miscellany of verbal puzzles by Edward Powys Mathers, who under the name Torquemada devised cryptic crossword puzzles for the Observer between 1926 and his death in 1939. At the back of the book was a short novel titled Cain’s Jawbone, which came with a unique twist:

Cain’s Jawbone, the bald narrative of a series of tragic happenings during a period of less than six months in a recent year, has met with an accident which seems to be unique in the history of the novelette. The pages have been printed in an entirely haphazard and incorrect order, a fact which reflects little credit on somebody. The author assures his readers, however, that while it is now too late for him to remedy the ordering of the pages, it is quite possible for them, should they care to take the trouble, to reorder them correctly for themselves. Before they attempt to do this, they may care to be assured that there is an inevitable order, the one in which the pages were written, and that, while the narrator’s mind may flit occasionally backwards and forwards in the modern manner, the narrative marches on, relentlessly and unequivocally, from the first page to the last.

The novel’s 100 pages had been printed and bound out of order. To solve the puzzle, the reader had to sort them into the correct order, read the story, solve the mysteries, and reveal the murderers. The task was so difficult that only two puzzlers solved it. Their names were printed in the Observer, but the solution to the problem was never revealed.

Last year the Laurence Sterne Trust got a copy of the book and has crowdfunded a new edition. For £30 you’ll get a box containing 100 jumbled pages, which you have to sort into a coherent mystery story, then identify six murderers and their six victims. The competition will run for 12 months from the date of publication, and the winner gets £1,000. As I write this 659 backers have signed up, putting the project well over its funding goal, but a publication date hasn’t yet been announced. You can find more info here and here.

Note: The announcement ends with a warning: “This is not a competition for the faint-hearted. The puzzle is phenomenally difficult.”

(Thanks, Sam.)

June 8, 2018 | Puzzles

Pole Position

A puzzle from James F. Fixx’s More Games for the Superintelligent, 1976:

A rope 150 feet long is strung between the tops of two flagpoles, each 100 feet high. At its lowest point the rope sags to within 25 feet of the ground. How far apart are the flagpoles?

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They’re touching. If the rope is 150 feet long then we have to double it to reach 75 feet.

June 1, 2018May 31, 2018 | Puzzles

Black and White

reshevsky chess problem

A sneaky puzzle by grandmaster Sammy Reshevsky. White to mate in two moves.

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1. Qa8! If the king goes to c4 then 2. Qe4 is mate. If he moves to the a-file then Bc5 traps him there. From the American Chess Bulletin, 1920.

May 31, 2018June 7, 2018 | Puzzles

Zigzag

From Lee Sallows:

zigzag puzzle

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(Thanks, Lee.)

May 30, 2018 | Puzzles

Unfinished

A puzzle by David L. Silverman:

On the back of an envelope you find an interrupted game of tic-tac-toe (noughts and crosses). You know that each player was an expert, which means that she never puts herself into a potentially losing position and that she always wins if her opponent gives her the opportunity. There are two Xs and two Os in the diagram, and it is impossible to tell whose move it is. Neglecting symmetry, what is the position?

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This solution is by Clayton W. Dodge of the University of Maine at Orono: Number the boxes 1-9 from top left to lower right, so that the first column is 1-4-7. Now: 1. The middle box must be empty. If it were marked, say with an X, then the board’s second X would have to be adjacent to it somewhere, and an O must complete that row, column, or diagonal. But now wherever we place the remaining O, the O player can immediately force a win by making a double threat if she has the move. X would be guilty of malpractice for allowing this disaster to come about, and we know she’s an expert player. So it must be X’s turn to move. But now we’ve deduced whose turn it is, and we’ve been told this is impossible. So this whole scenario cannot be the case — the center box must be empty. 2. No two similar marks can appear symmetrically around box 5, since then (if that player has the move) it leads to an immediate win or (if that player doesn’t have the move) we’re violating observation 1 above. 3. The same mark, say X, can’t appear in two adjacent corners, because then an O must appear between them, as well as in another box that will allow O to make two simultaneous threats if she has the move. 4. The same mark, say X, can’t appear in two middle-of-an-edge squares. If Xs appeared in, say, 2 and 8 we’d be violating observation 2. If they appeared in boxes 2 and 4, then one O must go in box 1 (to prevent the double threat) and another in box 6 or 8 to prevent X from winning with box 5. But then O can win with box 9. 5. The same mark, say X, can’t occupy two adjacent squares (say 1 and 2), since then an O must go in box 3 and another box that would allow O to make a double threat if she has the move. 6. Putting all this together, each player has marked one corner and one box opposite to it in the middle of an edge. There are four possibilities: If X has played to 1 and 6 and O to 2 and 9, then X can play to 4 and force a win. If X has played to 1 and 6 and O to 2 and 7, then X can play to 5 and force a win. If X has played to 1 and 6 and O to 4 and either 3 or 9, then expert play will always lead to a draw. So the envelope must bear one of these two diagrams (or its symmetrical counterpart): (From Pi Mu Epsilon Journal 5:10 [Spring 1974], 534.)

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This solution is by Clayton W. Dodge of the University of Maine at Orono:

Number the boxes 1-9 from top left to lower right, so that the first column is 1-4-7. Now:

1. The middle box must be empty. If it were marked, say with an X, then the board’s second X would have to be adjacent to it somewhere, and an O must complete that row, column, or diagonal. But now wherever we place the remaining O, the O player can immediately force a win by making a double threat if she has the move. X would be guilty of malpractice for allowing this disaster to come about, and we know she’s an expert player. So it must be X’s turn to move. But now we’ve deduced whose turn it is, and we’ve been told this is impossible. So this whole scenario cannot be the case — the center box must be empty.

2. No two similar marks can appear symmetrically around box 5, since then (if that player has the move) it leads to an immediate win or (if that player doesn’t have the move) we’re violating observation 1 above.

3. The same mark, say X, can’t appear in two adjacent corners, because then an O must appear between them, as well as in another box that will allow O to make two simultaneous threats if she has the move.

4. The same mark, say X, can’t appear in two middle-of-an-edge squares. If Xs appeared in, say, 2 and 8 we’d be violating observation 2. If they appeared in boxes 2 and 4, then one O must go in box 1 (to prevent the double threat) and another in box 6 or 8 to prevent X from winning with box 5. But then O can win with box 9.

5. The same mark, say X, can’t occupy two adjacent squares (say 1 and 2), since then an O must go in box 3 and another box that would allow O to make a double threat if she has the move.

6. Putting all this together, each player has marked one corner and one box opposite to it in the middle of an edge. There are four possibilities:

If X has played to 1 and 6 and O to 2 and 9, then X can play to 4 and force a win.
If X has played to 1 and 6 and O to 2 and 7, then X can play to 5 and force a win.
If X has played to 1 and 6 and O to 4 and either 3 or 9, then expert play will always lead to a draw.

So the envelope must bear one of these two diagrams (or its symmetrical counterpart):

(From Pi Mu Epsilon Journal 5:10 [Spring 1974], 534.)

May 23, 2018 | Puzzles

Podcast Episode 200: Lateral Thinking Puzzles

Here are five 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 …

May 14, 2018May 21, 2018 | Podcast · Puzzles

Two by Two

In poker, suppose you’re dealt a pair. Is the probability that your opponent also holds a pair higher, lower, or the same as it would be if you held nothing?

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It’s higher. If you’re holding a pair then you’ve left more pair combinations in the remaining deck than if you were holding a non-pair. A simple way to see this is to consider a deck of only 8 cards, four 2s and four 3s. If we remove a 2 and a 3, the remaining deck of 6 cards contains three pair combinations of 2s and three pair combinations of 3s, for a total of 6. But if we remove a pair of 2s, the remaining deck contains 7 pair combinations, one pair combination of 2s and six pair combinations of 3s. Here’s a more rigorous analysis of the effect in a real game — it’s small, but it’s there. (Thanks, John.)

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May 10, 2018May 9, 2018 | Puzzles · Science & Math