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Listener No 4501, Two Solutions: A Setter’s Blog by Quinapalus

Posted by Listen With Others on 27 May 2018

I first encountered the riddle used in Two Solutions in one of the books in Martin Gardner’s excellent Mathematical Diversions series. Its rhythm had stuck in my mind, and I had been thinking for a while that it might be possible to use it as the basis for a puzzle. Also, I’d been experimenting with ways to make it easier to construct grids where lights have two solutions in Qxw — with the current version it is possible but fiddly — and thought this might make a good test case.

I could see that the quadratic equation would have two distinct solutions, and on determining that they were complex, the idea of using the grid as an Argand diagram came naturally enough. Given that the coefficients were, I assume, chosen purely for metrical reasons it was good fortune that the roots can be plotted in a reasonable-sized grid with the side of a cell representing one unit. It was also fortunate that the real part of the roots is an odd multiple of one half so that they lie in the middle of a square horizontally, which makes the grid neater and the entries easier to check. In combination with the word ‘vertically’ in the instruction message it also gave some solvers some confidence that they were on the right track.

The complex roots of a polynomial with real coefficients always come in conjugate pairs: that is, pairs differing only in the sign of the imaginary part. This is because if you replace i with –i throughout a polynomial equation it will still be true; or as someone put it ‘how do you know that i is the positive square root of –1?’. It was therefore thematic for the grid to have up-down mirror symmetry. To avoid ambiguity it was important that there were no Xs in the grid other than those marking the roots; with a bit of work it also proved possible to make the grid pangrammatic.

Many solvers seem to like puzzles that have a little maths dust sprinkled on them, though I’m sure there are some—not too many I hope—for whom the endgame came as a slightly unpleasant surprise. To them I apologise (but only a bit). For the mathematical puzzles in the Listener series the editors use GCSE level as the reference for an acceptable level of difficulty. (Presumably if the same threshold were applied to the expected level of familiarity with, for example, English literature, setters would only be allowed to draw themes from Macbeth.) As far as I can establish this puzzle probably only just slips below the GCSE bar. Complex numbers are taught at that level, and so is the method of ‘completing the square’ for solving quadratics, which is convenient to use in this case as the coefficient of x² is 1. What is less clear is whether the combination of these two ideas falls within the syllabus.

It is unfortunate that ‘Argand diagram’ is not defined in Chambers but on the other hand there is an excellent explanation in Collins and of course in many online sources. All in all I was a bit apprehensive when I submitted the puzzle that the editors might reject it as demanding too much mathematics but in the end they let me get away with it. Nevertheless, to be on the safe side I arranged to be out of the country when the puzzle was published.

My thanks to the test solvers and the vetters for all the improvements they suggested, to the checker, and to the many solvers who gave feedback either online or along with their postal entries. Thankfully the response was mostly positive and there was no angry posse lying in wait for me at Heathrow. Your comments are all greatly appreciated.

Quinapalus.

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One Response to “Listener No 4501, Two Solutions: A Setter’s Blog by Quinapalus”

  1. Steve said

    As I’ve mentioned elsewhere I really enjoyed this one, so first: thanks for setting it! I particularly enjoyed the doubling-up of the names and titles, and the way in which the G/R used the double-letter feature in an entirely different way.

    I was wondering if at any point you considered a version of the puzzle where the gaps were both above and below the Xs, so that MOVE BOTH XS VERTICALLY required knowing the roots, and not just blindly shifting the Xs into the only free squares.

    Although that questions presupposes you intended the movement to be made without appreciation for the actual values, which is how I took it at the time, so a second question (inspired by my expectation of being marked wrong): I (mis)read the preamble as indicating that the instructed move, made in discrete units of whole squares, would result in a (reasonably) accurate plot of the solution. Reading other blog posts on this site I now think you intended that the solver aim for the proper points in the target squares to represent +/- 6.1, requiring knowledge of the roots after all. (Annoyingly I had the roots but was working in whole squares when I made the move.)

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