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Archive for December, 2009

Resident by Hotspur, Geography lesson.

Posted by shirleycurran on 25 December 2009

Only five lines of preamble – just the job! Misprints (again!) a rhyming couplet and clashing letters. Hmmm. We set to with great enthusiasm and had soon reached our usual Friday solving total: ten clues solved. POINSETTIA (that obvious misprint spunge/spurge and the clear prompt that we needed to work out an anagram of saponite + it) appeared straight away, together with a few very promising long words RHOMBOIDAL and ASSONANTAL and ALMANDINES (after we had looked up ‘hoofer’ – Alma dines, taking in the N of new).

After last week’s long, hard labour with numbers, this was light relief and we were hooked (more of the hooker later!)  We kept going and the grid surprisingly filled up. The clashes began to appear, and the location of the clashing letters was soon evident as ‘OVER THE RIVER’ seemed to be in the left-right diagonal.

TRANS.., I muttered and Mr Clever beat me to it: ‘TRANSNISTRIA’, he proclaimed, ‘along the Dniester, between the Ukraine and Moldova. I had to get the Atlas out to complete my geography lesson (and have copied a bit of it into my grid – always enjoy the colouring in bit!) but, of course, solving was now rendered much easier. We had enough letters to insert UKRAINE and MOLDOVA and completion was in sight.

Oh how smug that all sounds! It is never so easy for the Easy clues 8X8 team. Lots of word-play held us up. Our rhyming couplet produced by the misprints was clearly something about trains but, apparently, all trains come to a halt in Transnistria – could it be SLOW LINE at the end of the couplet? Much later, it was Mr Clever who pointed out that TRAINS + TRAINS was a fine anagram of TRANSNISTRIA, and so trains had to ‘collide’ with other trains. Nice one! It was then that we understood that the first line must be, ‘Located on the farther side’. The best surprise of all was still to be understood – that title RESIDENT seemed rather irrelevant (there are not very many in Transnistria, are there?) until it was pointed out that it is an anagram of DNIESTER.

Our troubles were not over. Oh dear, the word-play! We had ICIER but, as usual, missed the French connections in the wordplay. We hear the language all day long so there must be some mental block about looking for it in Listener crosswords. Of course it was the grocer/épicier, losing the ‘ép’ from the middle ‘képi’ – his French hat (though our épicier is never seen in a képi!) Then there was DAE; it had to be half of ‘sundae’ but does that make ‘dish’ in Scotland? (We have some vague notion about ‘dish’ being to ‘cheat’ or ‘do’ you – perhaps Denis will explain!) We chose HOWLET for our minor strix, and that put a W for ‘wife’ into a ‘hole’ on T – but the word-play escapes us. Still, the grid filled up with the exception of one square. We had ?OER for the TART. Was she going to be a GOER, a DOER or just a simple Dutch HOER? And what had the gardener to do with it? That was our last struggle and after sleeping on her, the hooker appeared – not OK at all! We had our HOER.

We enjoyed this one thoroughly, Hotspur. It was just what we needed after the last few weeks. Thank you!

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A Year-End Message

Posted by Dave Hennings on 23 December 2009

I would like to take this opportunity to wish all contributors and visitors to Listen With Others a very merry Christmas and a happy, prosperous and all-correct 2010.

When I took over this site from Chris Lancaster in March, I had no idea whether it would run itself, or need a lot of work to maintain an active readership. I guess that it has been somewhere in between, with the many contributions from both solvers and setters relieving me of too much of a weekly commitment. I would especially like to thank Shirley Curran, whose weekly blogs have gone from strength to strength, as, hopefully, has her team’s Listener solving ability. I hope you have all found her posts entertaining, imaginative and, more recently, artistic. I would also like to mention Erwin, whose treatises on the quarterly numericals have been both informative and colourful.

For those of you who are fond of new year resolutions, may I suggest that “Doing an occasional LWO blog” would be a good one to add, even if it is at the bottom of the list.

With very best wishes,
Dave.

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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).

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4062: Elap’s Numbers and Letters (or 17 25 2 11)

Posted by Dave Hennings on 18 December 2009

I suppose I always approach numericals with a certain trepidation, although they have normally been relatively easy to solve. The exception for me was 4010, Euclid’s Algorithm by Aedites which consisted of a cicular grid and 145 clues! It coincided with a time when I was living in a hotel room for six weeks, and solving the Listener outside my normally cosy environment was difficult. I had already failed miserably very early in the year, so an all-correct was not on the cards. I hoped that Elap’s puzzle would not be too difficult … it would be unfair, I think, to be tripped up by a numerical, especially the last of the year (although one all-correct last year was caught out by Aedites).

It looked an interesting puzzle: no bars, and either one or two numbers in each square (1-25), so a bit of novelty there. Changes would be required after the grid was complete, with some sort of message appearing, and ambiguities to resolve. This last bit was the only thing that really worried me.

I found the way in fairly quickly: 18ac required K to be 4 or 9, and U 9, 16, 25 or 36, and 2ac being 4 digits required UU to be at least 920, ie 36. I am sure others will go into the full analysis of the solving method, so I will leave that to them. Suffice it to say that after I completed most of the grid I could sort the values of each letter and the first fourteen appeared as KEY SUBMIT XWORD, and replacing the number in each square resulted in a grid full of 7-letter across and down entries.

Identifying the likely words also meant that I didn’t have to solve the last few mathematical steps, which looked horribly like requiring a bit of quadratic equation solving.

I looked at the grid and felt pretty pleased with myself. There was just one problem … I hadn’t spotted any anomolies, and I suddenly wondered whether there were two solutions that were possible, which my corner-cutting had hidden from me! Perhaps one or two words had alternatives, differing by just one square. There was only one course of action that I could take … solve the whole bloody thing again, and this time right to the end. I don’t think I needed quadratics in my second solve, and managed to get C and N with just a bit of substitution between 8dn and 16dn. In the end, I only had one ambiguity in square 6 of the bottom row which could hold either 2 or 24. My original solution also proved to be correct.

Another excellent puzzle from Elap, although he will have to go a long way to give me more enjoyment than 3957, Eleven Across, 3879, Digital Diversion, or (especially) 3853, Sevens and Threes.

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Listener 4062: Letter Squares by Elap

Posted by erwinch on 18 December 2009

It is perhaps appropriate to start with an equation: an Elap Listener equals one of the highlights of the year.  To date this has always held true and it certainly did this time although getting the initial grid entries was straightforward with no concealed entrances to confound us.  We only had to look at 18ac:
 
K + KK + 2U (2)
 
Firstly, it was helpful to list all eligible perfect squares below 100:
 
4, 9, 16, 25, 36, 49, 64, 81
 
(On Radio 4’s Brain of Britain on 30th November there was the question: Which perfect square below 100 is equal to the sum of the previous two?  On the 7th December we had: What is the cube root of 216?  With just 10 seconds to answer my brain tends to seize up at this point.)
 
K could only equal 4 or 9 but if K = 9 then U = 4 and this doesn’t fit 2ac (KK + UU (4)) so K = 4, U = 36, 18ac = 9|2, 2ac = 1312 and 18ac could be entered into the grid.
 
18dn gave us our second entry:
 
2B (2) = 9? B = 49, 18dn = 9|8
 
From 14dn:
 
2P + Y + YYY (4) Y = 9 or 16
 
And 24ac:
 
D + K + 2Y (3) = 8?? = D + 22 or 36: D = 784, Y = 16, 24ac = 8|20, P ≤ 54²
 
9ac DY + UU + 2Y = 13|8|7|2
 
19dn G (4) = 220?: G = 2209, 19dn = 2|20|9
 
6dn B + S + 2SS (4) = ?2?2, ??2? or ?2?? (4, 9, 16, 25, 36, 49, 64, 81) S = 25, 6dn = 13|2|4
 
23ac E + KK + M (2) = E + 16 + M  so E + M ≤ 83 and E + M = 9 + 64 in some order: 23ac = 8|9
 
17ac BU + K + W (4) = 1768 + W = ??92: W = 324, 17ac = 20|9|2
 
10dn DKS + K + KR (5) = 78404 + 4R = 8??20 or 8?209 (81020 to 89209)
4R = 2616 to 10805
R = 654 to 2701
R = 26² to 51² but there was only one fit: R = 27² = 729, 10dn = 8|13|20
 
7ac BR + BRU + U (7) = 13|2|17|13 or 13|21|7|13
 
5ac G + 3I + II (5) = 2209 + 3I + II = ???13: {I, 5ac} = {121, 17213} or {196, 41213}
 
The first entry of 4ac (5) must therefore have two digits:
4ac 2K + KR + W + Z = 3248 + Z = ??172 or ??412
{Z, 4ac} = {13924, 17172} or {20164, 23412}, 2ac = 13|12
 
1dn D + EZ + UU + Z (6) = 2080 + (E + 1)Z = ??13?? or ?13??? with E = 9 or 64 and Z = 13924 or 20164
There was only one fit: E = 9, Z = 13924, M = 64, I = 121, 1dn = 14|13|20, 4ac = 17|17|2, 5ac = 17|2|13
 
At this point values had been assigned to thirteen letters and progress was steady until just A, C, L and N remained unknown and F = 3249 or 5329 with the grid looking like this:
 
 
None of the remaining clues had just one unknown so we had to resort to using simultaneous equations, which is unusual for these puzzles:
 
25ac = 8|14|2|16|9 (F = 3249) or 8|8|18|16|9 (F = 5329)
 
8dn C + II + 3N (5) = C + 14641 + 3N = 2??13 (21013 to 22513)
16dn C + CU + N + 2O (5) = 37C + N + 968 = 4814? (48141 to 48149) or 488?? (48810 to 48825)
16dn × 3 = 111C + 3N + 2904 = 144423 to 144447 or 146430 to 146475
 
Boiling all that down, 16dn × 3 – 8dn:
C = 1225, N = 1849, F = 3249, 25ac = 8|14|2|16|9, 8dn = 2|14|13, 16dn = 4|8|14|2, 27ac = 2|14|2|4 or 2|14|24
 
It was then straightforward to determine L from 15dn and A from 26ac and 21dn:
L = 71289, A = 5184, 15dn = 8|9|8|15, 26ac = 13|2|15, 21dn = 12|24|2
 
This gave us our final number grid with one ambiguity at 27ac:
 
 
We then had to sort the letters appropriately, according to their assigned values, and I imagine that we all went for the lowest first and saw the following hint appear:
 
KEY: SUBMIT XWORD (CNPGFAZJLVH)
 
My initial thought was the ubiquitous A = 1, B = 2, etc, which resulted in nonsense, but the second struck gold: K = 1, E = 2 … V = 24 and H = 25:
 
 
27ac should have been entered as 2|14|2|4.
 
A most satisfactory dénouement, as we have come to expect from Elap but are we expecting too much from our setters?  I wondered why Q was omitted from the puzzle but concluded that it would only have complicated matters.  Tacking it on to the end of the hint would have made the preamble difficult, with no 26 appearing in the number grid, and in any case having 25 letters was thematic.  I cannot begin to grasp the complexity of the construction yet the puzzle could be completed with just pen, paper and calculator – to my eyes at least, this was a breathtaking feat.
 
I consider 2009 to have been a vintage year for number puzzles.  Before this we had the dominoes and the baseball but my heart remains with the first of the year, Pentomino Factory.  I still think about it a lot and was even moved to add a footnote to the blog a couple of weeks ago.
 

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