2016-07-18 § Leave a comment
In the last post we saw that every 3×3 almost-magic square is a rearrangement of three three-term arithmetic progressions that have the same common difference. In other words, if we pick any three numbers x, y and z, and any common difference d, then the nine numbers
can be rearranged to make an almost-magic square
where all the rows and columns, and the leading diagonal, add up to x+y+z.
We’re interested in finding almost-magic squares where all the entries are square numbers, so that means we’re particularly interested in three-term arithmetic progressions of squares. Fortunately there is a well-developed theory of these things, which are closely related to Pythagorean triples.
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2016-07-13 § 1 Comment
A recent Numberphile video discussed an intriguing unsolved problem in number theory: is there a 3×3 magic square whose entries are all square numbers? (Matt Parker proposed a solution which doesn’t quite work: see the video for more. The “Parker square” even has its own Twitter account.)
It turns out that this question leads to some interesting mathematics, which I’m going to have a go at explaining. At the very least I hope to give some insight into why it’s a hard problem.
To reduce the risk that I’ll run out of time and energy before pressing Publish, I’m going to split this post into two or possibly three parts, which I’ll post separately. The first part, which you’re reading right now, gives some background on 3×3 magic squares. The second will explain how to find almost-magic squares of squares using a bit of number theory and computer algebra. By almost-magic I mean that the rows and the columns and one of the two diagonals have the same sum.
The third, if I write it, will be more speculative and look at some possible approaches to either finding the mythical square of squares or proving that it doesn’t exist.
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2015-04-26 § 14 Comments
2014-12-05 § 6 Comments
The UK Government Statistical Service recently released its good practice guidance for releasing statistics in spreadsheets. While this advice is clearly well-intentioned*, and parts of it are good, the overall effect is to encourage the release of data in formats that are difficult to process by computer. This is a disappointing retrograde step.
2014-08-22 § 5 Comments
What’s the shortest string that contains every possible permutation of ABCD somewhere inside it? As it happens, it’s 33 letters long: ABCDABCADBCABDCABACDBACBDACBADCBA. A string like this is called a minimal superpermutation.
So what’s the shortest string that contains every possible permutation of ABCDE? It was recently shown that 153 letters is the shortest possible, and that there are eight different superpermutations of this length.
Okay, what about ABCDEF? The answer is that nobody knows. Until this week the shortest known superpermutation of ABCDEF was 873 letters long:
ABCDEFABCDEAFBCDEABFCDEABCFDEABCDFEABCDAEFBCDAEBFCDAEBCFDAEBCDFAEB CDAFEBCDABEFCDABECFDABECDFABECDAFBECDABFECDABCEFDABCEDFABCEDAFBCED ABFCEDABCFEDABCADEFBCADEBFCADEBCFADEBCAFDEBCADFEBCADBEFCADBECFADBE CAFDBECADFBECADBFECADBCEFADBCEAFDBCEADFBCEADBFCEADBCFEADBCAEFDBCAE DFBCAEDBFCAEDBCFAEDBCAFEDBCABDEFCABDECFABDECAFBDECABFDECABDFECABDC EFABDCEAFBDCEABFDCEABDFCEABDCFEABDCAEFBDCAEBFDCAEBDFCAEBDCFAEBDCAF EBDCABEFDCABEDFCABEDCFABEDCAFBEDCABFEDCABACDEFBACDEBFACDEBAFCDEBAC FDEBACDFEBACDBEFACDBEAFCDBEACFDBEACDFBEACDBFEACDBAEFCDBAECFDBAECDF BAECDBFAECDBAFECDBACEFDBACEDFBACEDBFACEDBAFCEDBACFEDBACBDEFACBDEAF CBDEACFBDEACBFDEACBDFEACBDAEFCBDAECFBDAECBFDAECBDFAECBDAFECBDACEFB DACEBFDACEBDFACEBDAFCEBDACFEBDACBEFDACBEDFACBEDAFCBEDACFBEDACBFEDA CBADEFCBADECFBADECBFADECBAFDECBADFECBADCEFBADCEBFADCEBAFDCEBADFCEB ADCFEBADCBEFADCBEAFDCBEADFCBEADCFBEADCBFEADCBAEFDCBAEDFCBAEDCFBAED CBFAEDCBAFEDCBA
and it was thought that might be the shortest possible.
But we now know it isn’t, because I found a shorter one:
ABCDEFABCDEAFBCDEABFCDEABCFDEACBFDEACFBDEACFDBEACFDEBACFDEABCDFEAB CDAEFBCDAEBFCDAEBCFDAEBCDFAEBCDAFEBCDABEFCDABECFDABECDFABECDAFBECD ABFECDABCEFDABCEDFABCEDAFBCEDABFCEDABCFEDACBFEDCABFDECAFBDCEAFBDCA EFBDCAFEBDCAFBEDCAFBDECAFDBECADFBECADBFECADBEFCADBECFADBECAFDEBCAD FEBCADEFBCADEBFCADEBCFADEBCAFDECBAFDECABFDCEABFDCAEBFDCABEFDCBAEFD BCAEDFBCAEDBFCAEDBCFAEDBCAFEDBCAEFDBACEFDBAECFBDAECFBADECFBAEDCFBA ECDFBACEDFBACDEFBACDFEBACDFBEACDFBAECFDBAEFCDBAFECDBAFCEDBAFCDEBAF CDBEAFCDBAEFDCBEAFDCBEFADCBEFDACBEFDCABFEDCBAFEDCBFAECDBFACEDBFACD EBFACDBEFACDBFEACDBFAECBDFEACBDFECABDFCEABDFCAEBDFCABEDFCBAEDFCBEA DFCBEDAFCBEDFACBEDFCABDEFCBADEFCBDAEFCBDEAFCBDEFACBDEFCABDFECBADFE CBDAFECBDFAECBFDAECBFADECBFAEDCBFEADCFBEADCFEBADCEFBADCEBFADCEBAFD CEBADFCEBADCFEABDCFAEBDCFABEDCFABDECFABDCEFABDCFEADBCEFADBCEAFDBCE ADFBCEADBFCEADBCFEADCBFEDACFBEDACFEBDACEFBDACEBFDACEBDFACEBDAFCEBD ACFEDBACFEDABC
I’ve uploaded a short note about it to the arxiv.
2014-06-19 § 1 Comment
This document is an incomplete draft.
About two years ago I wrote about a category-theoretic treatment of collaborative text editing. That post is unique in the history of Bosker Blog in having been cited – twice so far that I know of – in the academic literature; so it’s a little embarrassing for me to have to explain that it is almost entirely wrong. The good news is that the core idea can be rescued, and the corrected story is quite interesting. Other writers on this subject seem to have made at least some of the same mistakes I did, so I hope this will be useful to at least a few other people too. « Read the rest of this entry »
2014-02-25 § Leave a comment
Suppose you have a lock of this sort that has n dials and k numbers on each dial. Let m(n, k) be the minimum number of turns that always suffice to open the lock from any starting position, where a turn consists of rotating any number of adjacent rings by one place.