Listen With Others

Are you sitting comfortably? Then we'll begin

Letter Squares by Elap

Posted by shirleycurran on 18 December 2009

Nobody said it was going to be easy!

Do we do this for pleasure? When I say that the Junior Easy-clues 8X8 team started out not even knowing for sure what a perfect square was, the distance we had to travel will be evident. Friends reassured us that it was the square of an integer and we were underway – for almost four days of hard labour. ‘Solvers must deduce how many complete cells each answer occupies’, said the preamble, and we were warned about a hint, and ambiguities. Oh dear! Where to start?

If in doubt, count the letters! K and U were the most frequent, so was K = 4? (Later, it proved to be so.) Next, the clues with the shortest answers are the most restrictive so we looked at 18ac, 23ac, 18d, 24ac, 2ac and 14d. Almost at once the numerical restrictions showed that U = 36 and B = 49.

2ac revealed our first problem. 1312 could be entered in two, three or even four cells. However this cell-entry problem became an advantage when, at 24ac, the values of Y and D (restricted to 2 each) became unique, using the only permitted entry in the two cells imposed by three digits.

We clanked on until at 17ac, we found a very big range for W, anywhere from 25 to 7921, but it had to end in 24, to match the other entries. We produced four possible values but only one that could legally be entered. By the time that we had found that S = 25, the hint was emerging. We had sorted the values to keep track of which squares we had used and the hint KEYSUBMWRDG appeared. In that string, E and M were potentially interchangeable at that stage, but deciphering the hint fixed E and M and soon we had KEY SUBMIT X WORD.

Obviously, if we ever finished this hard labour, we were going to submit our X Word – we didn’t see much further than that at this point.

But then one of our typical red herrings loomed into view.  A splendid X seemed to be appearing in the cell entries with its top left at 13, and composed of the squares 25, 16, 9 and 4. We attempted to force this onto 25ac and wasted some time and tooth enamel looking for that X word.

The bottom three rows were the worst, with 25ac where F was between 3123 and 5658 and there were 20 possible squares, one of which had to produce 8???169. This was going to be very tedious with a calculator, so programming was resorted to and proved that F was 3249. 16d, involving C and N was the worst of all because of the enormous possible range of both C and N. Programming showed only two possibilities which were resolved by solving 27ac.

So there (after many hours) we had a complete grid of numbers and a hint: KEY SUBMIT X WORD and mystification. We had to ‘find what we must do’, ‘make changes to the grid’, and ‘resolve any ambiguities’. Was it possible that the KEY to converting those numbers to words was actually there in those fourteen letters? It strikes me as odd that the KEY should, in effect, be self-defining in that way, but it proved to be so.

When we gave 123456 etc. to KEYSUB etc. we had our first 14 letters of the 25 we imagined we needed – though, in fact, the word square that appeared used only six that were not in those 14 (C,N,P,A,V and H). Resolving ambiguities? Well, we could have had 24 in the penultimate square and 4 in the ultimate one – but that would have given us a rather strange RECEDVS as our final word. But perhaps it should have been left to our wits to solve that, rather than waving that smelly red herring at us.

Writing a word square is difficult enough, without the complex mathematical calculations leading to it. This was magic and led us to wonder how Elap managed to conceive of this miracle of compilation (and why he didn’t devote his time to Fermat’s Last theorem). I, for one, am glad we will be back to words next week (though I might live to regret those words).

Advertisements

One Response to “Letter Squares by Elap”

  1. erwinch said

    Tackling Fermat’s Last Theorem might well have been considered the easier option for Elap.  Did you see the footnote to the full solution on the Listener site?  Each letter’s assigned value could be decoded using the same key to give a Chambers entry.  So, H = 4|12|16|4 = sons and C = 12|25 = oh, etc.  This was the ultimate Nina!

    PS Excellent artwork once more.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

 
%d bloggers like this: