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hugo/content/posts/2014-12-19-matrix-puzzle.md

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+---
+lastmod: "2021-09-12T22:47:44.0000000+01:00"
+author: patrick
+categories:
+- mathematical_summary
+comments: true
+date: "2014-12-19T00:00:00Z"
+math: true
+aliases:
+- /mathematical_summary/matrix-puzzle/
+title: Matrix puzzle
+---
+
+I recently saw a problem from an Indian maths olympiad:
+
+> There is a square arrangement made out of n elements on each side (n^2 elements total). You can put assign a value of +1 or -1 to any element. A function f is defined as the sum of the products of the elements of each row, over all rows and g is defined as the sum of the product of elements of each column, over all columns. Prove that, for n being an odd number, f(x)+g(x) can never be 0.
+
+There is a very quick solution, similar in flavour to that [famous dominoes puzzle][Mutilated chessboard]. However, I didn’t come up with it immediately, and my investigation led down an interesting route.
+
+Preliminary observations
+===========
+
+It is easy to see that given a matrix of \\(1\\) and \\(-1\\), we have \\(f, g\\) unchanged on reordering rows and columns, and on taking the transpose. This leads to a very useful lemma: \\(f, g\\) are unchanged if we negate the corners of a rectangle in the matrix.
+
+The idea then occurs: perhaps there is a [normal form] of some kind?
+
+Specification of normal form
+========
+
+Given any four -1’s laid out at the corners of a rectangle, we may flip them all into 1’s without changing \\(f, g\\). Similarly, given any three -1’s on the corners of a rectangle, where the fourth corner is 1, we may flip to get a rectangle with one -1 and three 1’s.
+
+Repeat this procedure until there are no rectangles with three or more corners -1. (Note that we might get a different answer depending on the order we do this in!) A Mathematica procedure to do this (expressed in a very disgusting way) is as follows.
+{% raw %}
+    internalReduce[mat_] :=  Module[{m = mat},
+  Do[If[(i != k && j != l) &&
+     Count[Extract[m, {{i, j}, {k, j}, {i, l}, {k, l}}], -1] >
+      2, {m[[i, j]], m[[i, l]], m[[k, j]],
+       m[[k, l]]} = -{m[[i, j]], m[[i, l]], m[[k, j]], m[[k, l]]};
+    ], {i, 1, Length[mat]}, {j, 1, Length[mat]}, {k, 1,
+    Length[mat]}, {l, 1, Length[mat]}];
+    m]
+    reduce[mat_] := FixedPoint[internalReduce, mat]
+{% endraw %}
+
+Notice that columns which contain more than one -1 must not overlap, in the sense that no two columns with more than one -1 may have a -1 in the same row. Indeed, if they did, we’d have a submatrix somewhere of the form {{-1, -1}, {1, -1}}, which contradicts the “we’ve finished flipping” condition. Hence we may rearrange rows so that all -1’s appear together in contiguous columns.
+
+We may then rearrange columns so that reading from the left, we see successive columns with decreasingly many -1’s. Rearrange rows again so that they appear stacked on top of each other.
+
+![example of reduced matrix][reduced matrix]
+
+We’ve ended up with a normal form: columns of -1’s, diagonally adjoined to each other, followed by rows of -1’s. (The following Mathematica code relies on the fact that SortBy is a stable sort.)
+
+`normalform[mat_] := SortBy[Transpose@SortBy[Transpose@reduce[mat], -Count[#, -1] &], Count[#, -1] /. {0 -> Infinity} &]`
+
+We haven’t shown that it’s unique yet, and indeed it’s not. As a counterexample, {{-1,1,1,1,1}, {-1,1,1,1,1}, {1,-1,1,1,1}, {1,-1,1,1,1}, {1,1,1,1,1}} is transformed into {{-1,1,1,1,1}, {-1,1,1,1,1}, {-1,1,1,1,1}, {-1,1,1,1,1},{1,1,1,1,1}} by a rectangle-flip.
+
+This suggests a further improvement to the normal form: by flipping in this way, we may insist that any column of -1’s, other than the first, must contain only one -1. Indeed, if it contained two or more, we would flip two of them into the first column, rearrange so that all columns were contiguous -1’s again, and repeat.
+
+What does our matrix look like now? It’s a column of -1, followed by some diagonal -1’s, followed by a row of -1. We’ll call this the canonical form, although I’ve still not shown uniqueness.
+
+![example of matrix in canonical form][canonical matrix]
+
+Restatement of problem
+========
+
+The problem then becomes: given a matrix in canonical form, show that \\(f+g\\) cannot be 0.
+
+Notice that if the long column is \\(r\\) long, and there are \\(s\\) diagonal -1’s, and the long row is \\(t\\) long, and the matrix is \\(n \times n\\), then \\(f = -r-s+(-1)^t + (n-s-r-1)\\), \\(g = -t-s+(-1)^r + (n-s-t-1)\\).
+
+Hence \\(f+g = 2n - 2(r+2s+t+1) + (-1)^r + (-1)^t\\).
+
+Any choice of \\(r, s, t, n\\) with \\(r+s+1 \leq n; s+t+1 \leq n; r, t>1\\) yields a valid matrix. We therefore need to show that for all \\(r, s, t, n\\) we have \\(2(n-r-2s-t-1) + (-1)^r + (-1)^t \not = 0\\).
+
+Solution
+=======
+
+Reducing this mod 4, it is enough to show that \\(2(n-r-t-1) + (-1)^r + (-1)^t \not \equiv 0 \pmod{4}\\). But we can easily case-bash the four cases which arise depending on the odd-even parity of \\(r, t\\), to see that in all four cases, the congruence does indeed not hold.
+
+* \\(r, t\\) even: \\(2(n-1) + 2 = 2n\\), but since \\(n\\) is odd, this is not \\(0 \pmod{4}\\).
+* \\(r\\) even, \\(t\\) odd: \\(2n - 1\\), since \\(t-1\\) is even so \\(2(t-1)\\) is a multiple of 4. \\(2n-1\\) isn’t even even, let alone divisible by \\(4\\).
+* \\(r, t\\) odd: \\(2(n-1) + 2 = 2n\\) which is again not \\(0 \pmod{4}\\).
+
+Summary
+=======
+
+Once we had this canonical form, it was easy to find \\(f, g\\) and therefore analyse the behaviour of \\(f+g\\). Next steps: prove that canonical forms are unique (perhaps using the fact that \\(f, g\\) are invariant across forms, and showing a result along the lines that any two canonical forms with the same \\(f, g\\) must be equivalent). I won’t do that now.
+
+[Mutilated chessboard]: https://en.wikipedia.org/wiki/Mutilated_chessboard_problem
+[normal form]: https://en.wikipedia.org/wiki/Canonical_form
+[reduced matrix]: {{< baseurl >}}images/Matrices/matrix_reduced.jpg
+[canonical matrix]: {{< baseurl >}}images/Matrices/matrix_canonical.jpg