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The integers $1, 2, 3, \ldots, n^2$ are placed without duplication onto an $n \times n$ chessboard, with one integer per square. Show that there exist two (horizontally, vertically, or diagonally) adjacent squares whose values differ by at least $n + 1$.
(We assume $n > 1$.)

darij grinberg
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Gerard L.
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2 Answers2

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We assume $n > 1$. Let $C_a$ denote the integer placed in the square $a$ of the $n^2$-chessboard.

Assume $|C_x - C_y| \leq n$ for all squares $x$, $y$ that are horizontally, vertically, or diagonally adjacent. Pick the square $z$ where $C_z =1$.

A King can move to a square that is horizontally, vertically, or diagonally adjacent to his current square. Therefore a King needs at most $n-1$ moves to move from square $z$ to an arbitrary square $x.$ So $$|C_x - C_z| \leq n(n-1) = n^2 - n$$ obtain $$C_x \leq n^2 - n + 1 \tag {1}$$

Because the largest value placed on the chessboard is $n^2$, we can choose $C_x = n^2$, which contradicts $(1)$.

Hyperbolic Cake
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Mudream
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  • I don't understand: What is C? – miracle173 Jun 10 '17 at 02:56
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    The $C_{(a, b)}$ denote the cell in n × n chessboard. – Mudream Jun 10 '17 at 02:57
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    Is $C_{(a,b)}$ the integer placed on the square with coordinates (a,b)? – miracle173 Jun 10 '17 at 03:01
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    o is a rather bad variable name, it can be taken for 0. To move from Cx to Cy does not makes sense. x and y are the squares, so you can move from the square x to the square y, to move from square x to square y a queen needs at most 2 moves: one to get from x to the same row as y and the next one to get to y. The king is the chess piece that needs at most n-1 moves to get from square x to an arbitrary square y. – miracle173 Jun 10 '17 at 03:16
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A possible answer...........

This proof will be a kind of quasi-use of mathematical induction...

Let $x_{i,j}$ represent a point on the board (with i and j both in the range of 1 to n). Notice that the max difference possible between $x_{i,j}$ and n is given by the formula $n^{2}-x_{i,j}$, which can be factored into $$a)\; [(n+\sqrt(x_{i,j}))(n-\sqrt(x_{i,j}))]$$. Given that 1 will always be the smallest value, we can further change this into $$a_i)\;[(n+1)(n-1)]$$. Now, assume n=2 for a base case. Plugging this into the formula given, we have the claim is true. Let m be some arbitrary natural. Assume that the claim is true for m. Now consider $r=m+1$. After plugging this into formula $a_i$, we get that the distance between m+1 and 1 is $(m)(m+2)$. But notice that m is always greater than 1. So we have the proof of the claim.

Johnq
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  • What do you mean by "max difference possible between $x_{i,j}$ and n"? In your induction, what is the induction hypothesis and how do you apply it? This is completely unclear... – darij grinberg Dec 13 '23 at 19:50