Black and White

friedländer chess problem

From Louis Friedländer’s 1919 collection Chess Poetry. White to mate in two moves.

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After 1. Nd2! one of the black knights must relax its guard of b3 or c4, and White can mate.

September 11, 2019 | Puzzles

Black and White

van dehn chess puzzle

A remarkable thematic chess puzzle by Bodo Van Dehn, 1951. White to move and win.

The solution is 10 moves long, but all Black’s moves are forced. (That’s a very valuable hint.)

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1. a8=N d3 2. Nb6 cxb6 3. c7 b5 4. c8=N b4 5. Nd6 exd6 6. e7 d5 7. e8=N d4 8. Nf6 gxf6 9. g7 f5 10. g8=N# Four knight underpromotions, and three knight sacrifices! You can see this played out on the excellent ChessNetwork YouTube channel here.

August 21, 2019August 20, 2019 | Puzzles

A Flea’s Journey

https://commons.wikimedia.org/wiki/File:Euclid_Tetrahedron_4.svg — Image: Wikimedia Commons

A flea sits on one vertex of a regular tetrahedron. He hops continually from one vertex to another, resting for a minute between hops and choosing vertices without bias. Prove that, counting the first hop, we’d expect him to return to his starting point after four hops.

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This appeared originally in the Journal of Recreational Mathematics in July 1969, posed by David L. Silverman. Benjamin L. Schwartz of McLean, Va., sent in three different arguments, the simplest of which is purely heuristic. Imagine that the flea is on a dodecahedron (20 vertices), and that he hops around it for a long period of time, say 10,000 hops. He’ll occupy each vertex for approximately the same time, in this case 500 visits. “Then obviously, his average time between calls at that vertex is 20, the number of vertices.” More formal proofs are possible; see R. Robinson Rowe, “Random Hops on Polyhedral Edges and Other Networks,” Journal of Recreational Mathematics 4:2 [April 1971], 124-130, as well as “Solutions to Problems” in that issue.

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August 17, 2019 | Puzzles

The Magic Total

magic total

Each of the 36 numbers in this table is the sum of the numbers at the head of its column and at the left of its row. For example, 3 = 2 + 1 and 13 = 5 + 8. The six bold numbers have been chosen so that each of them falls in a different row and a different column. The underlined numbers were chosen in the same way. But each of these two sextets produces the same total: 16 + 6 + 5 + 14 + 8 + 8 = 8 + 10 + 7 + 8 + 10 + 14 = 57. In fact, any six numbers chosen in this way will produce the total 57. Why is this?

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Each number in the table is just the sum of its row and column headers, so choosing six numbers in this way is equivalent to just adding up all the headers: (7 + 2 + 3 + 6 + 7 + 5) + (2 + 9 + 8 + 3 + 4 + 1) = 57. From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.

August 7, 2019August 6, 2019 | Puzzles

Home Again

https://commons.wikimedia.org/wiki/File:Knights_tour_solution_Warnsdorff.png — Image: Wikimedia Commons

On a regular 8 × 8 chessboard, a wandering knight can visit each square once and then return to his starting square. Show that he can’t do this on an m × n board if m and n are both odd.

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Any pair of squares that are a knight’s move apart are of different colors — one is white, one black. But any “knight’s tour” on our m × n board will visit mn squares, and mn is odd (since it’s the product of two odd numbers). That means that the knight’s starting and ending squares will be of the same color, and hence can’t be a knight’s move apart.

August 2, 2019 | Puzzles

Black and White

healey chess problem

By Francis Healey. White to mate in two moves.

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1. Qa7! and the queen or rook will mate. From A Collection of Two Hundred Chess Problems, 1866.

August 1, 2019August 1, 2019 | Puzzles

Black and White

pelle chess problem

This problem, by Charles Pelle, won third prize among the Union des Problémistes de France in 1946. How can White mate in two moves?

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1. Nb3! Black must capture a knight, and the queen will mate.

July 12, 2019July 12, 2019 | Puzzles

Simple Enough

I think this appeared originally in Ed Southall’s Twitter feed. What fraction of the square is shaded?

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The answer is 1/3, but there’s a beautiful variety of solutions in the discussion.

July 5, 2019July 4, 2019 | Puzzles

“A Chess Packing Problem”

In 2006 Martin Gardner asked: Can you arrange the 16 non-pawn pieces in a standard chess set on a 5 × 5 board so that no piece attacks a piece of the opposite color? As in a conventional game, the two bishops of each color must stand on squares of opposite colors.

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Cameron Byer, a second grader at Calvary Christian Academy in Harrisonburg, Va., sent this solution: The problem was originally proposed and solved by Mamikon Mnatsakanian in Soviet Science & Life in 1976, Issue 2. (Andy Liu, “Problem Section,” Math Horizons 14:2 [November 2006], 40-42.)

June 25, 2019 | Puzzles

The Dark Half

A puzzle from the 1997 Ukrainian Mathematical Olympiad:

Cells of some rectangular board are coloured as chessboard cells. In each cell an integer is written. It is known that the sum of the numbers in each row is even and the sum of numbers in each column is even. Prove that the sum of all numbers in the black cells is even.

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To begin with, notice that the sum of the numbers in the white cells in any given row or column must have the same parity (odd or even) as the sum of the numbers in the black cells. That’s necessary because we know that the sum of the whole row (or column) is even: If the white total is even, then the black total must be even, and if the white total is odd then the black total must be odd. Since that’s so, and since we’re interested only in parity, then in evaluating the black squares over the board as a whole we can make that substitution — in a given row or column, we can substitute the parity of the white sum for the parity of the black sum. Then let’s calculate the sum of the board’s black cells by considering each column in turn, and we’ll choose to sum the black cells in the odd-numbered columns and the white cells in the even-numbered columns. But that’s just the same as summing every second row on the board, and we already know that each of those totals is even. So the final sum must be even. (Via R.E. Woodrow, “Olympiad Corner,” Crux Mathematicorum 29:2 [March 2003], 87-103.)

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June 25, 2019June 24, 2019 | Puzzles