Brain-Teaser 17: [Knight’s tour]
From The Sunday Times, 25th June 1961 [link]
In “Amusements in Mathematics” (Nelson, 1917), the late Henry Ernest Dudeney published a magic knight’s tour of the chessboard. That is to say, a knight placed on the square numbered 1 could, by ordinary knight’s moves, visit every square of the board in the ordered numbered, and the numbers themselves in each row and column added up to 260. Yet it was not a fully magic square, for the diagonals did not add to the same constant. After much trying Dudeney came to the conclusion that it is not possible to devise such a square complete with magic diagonals, but, as he said, a pious opinion is not a proof.
You are invited to try your skill in devising a magic knight’s tour of a square 7×7, with or without magic diagonals.
Dudeney’s Amusements in Mathematics is available on Project Gutenberg [link].
This puzzle was originally published with no title.
[teaser17]
Hugh Casement and 
S2T2 
Jim Randell 8:39 am on 3 January 2021 Permalink |
A 7×7 chessboard has 49 squares. So if we colour the corners black there will be 25 black squares and only 24 white squares. So the knight will have to start on a black square, and end on a black square. (So it is not possible for a tour to form a closed loop).
So, 1 is on a black square. The next square in the tour will be a white square (as a knight always moves to a square of the other colour). So, 2 will be on a white square, 3 on a black square, etc. When the tour is complete all the black squares will be odd numbers and all the white squares will be even numbers.
The rows/columns alternate between (4 black + 3 white) and (4 white + 3 black) squares, but the total of a row/column consisting of (4 black + 3 white) numbers will be (4 odd + 3 even) = even. And the total of a row/column consisting of (4 white + 3 black) will be (4 even + 3 odd) = odd. And we require the sum of each row and column to be the same (it should be T(49) / 7 = 175),
So it will not be possible for us to make a magic squares as the sums of the values in the rows and columns will necessarily alternate between even and odd, so they can’t all be the same.
Hence it is not possible to construct a magic knight’s tour on a 7×7 board.
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Hugh Casement 1:35 pm on 3 January 2021 Permalink |
Dudeney was by no means the first: see
https://www.mayhematics.com/t/1d.htm
https://mathworld.wolfram.com/MagicTour.html
It has been proved (by computer) to be impossible on an 8×8 board,
though there are many semi-magic tours.
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