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.”

(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.

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.

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.

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.

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.

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.)

Divide each figure into a grid of triangles. This shows that the square on the left occupies 2/4 of its triangle, while the square on the right occupies 4/9 of its triangle. So the square on the left is about 12.5 percent larger.

(Duane Detemple and Sonia Harold, “A Round-Up of Square Problems,” Mathematics Magazine 69:1 [February 1996], 15-27.)