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Archive for June, 2011

Listener 4139: Nutmeg’s Unsettled Spell (or What Site is This?)

Posted by Dave Hennings on 17 June 2011

This is only the second Listener from Nutmeg, the first being 3971 Franglais in March 2008. I have become a bit remiss in solving Inquisitors and EVs recently, and a number of Nutmeg’s in those series have passed me by. However, I did solve EV 942 Block Capitals by Nutmeg (mainly because I blogged it for fifteensquared), and a thoroughly enjoyable and entertaining puzzle it was.

And so onto this week’s offering which involved letters being swapped between entries and clues, 50% going one way, and 50% the other. I can tell you now that this device made my head swim, but that’s par for the course these days.

Starting with the across clues, I guessed that 1ac had an extra letter – waiver, and 5ac had one missing trifling. However, neither answer was forthcoming. That privilege went to 23 Top monk shuns holy books in scripture lessons — this teacher wouldn’t, an excellent clue. 37 ARDENT (heart), and that was it for the acrosses, two out of 26! OK, my concentration wasn’t tip-top — did I mention it was 5:30am and I’d been awake since 4?! The down answers were just as reluctant to show themselves, only 3 ENTOMBED (harmed) and 26 (Wilde).

After another hour or so, I’d got only about a dozen more clues, but at least 1ac CHEERIO, 1dn CLEANSER and BEGINNING gave much to go on in the top left. And looking at the swapped letters, I saw that they started ITW.B.LE.., and it didn’t take long to get (My spelling is Wobbly. It’s good spelling but) it Wobbles, and the letters get in the wrong places. Milne. Thus WINNIE-THE-POOH went in at 5dn. Getting that so early in the puzzle helped a lot, and the top left and bottom right corners were finished fairly quickly. The other two quadrants took a bit more teasing, and resulted in the whole puzzle taking longer than average to solve.

Of course, longer than average for me is probably longer than longer than average for you … or not! Much as I try to log significant chunks of time spent solving the Listener, I invariably forget to make a note of some shorter periods at my desk. In order to improve this, I have now downloaded an app for my iPhone that will help me log solving times for LWO and for the EV puzzles that I blog. Now all I’ve got to do is remember to start the timer when I sit down for a session, and, just as importantly, stop it when I finish. Why I think that will work better than adding to the notes that I make anyway beats me, but I’ll give it a go.

Anyway, “longer than average” here means about 4 hours.

A fine puzzle from Nutmeg, and a couple of final points to make about it. Firstly, I know some Listener solvers (maybe even a lot of them) are tired of extra letters, extra words, misprints, etc. However, I think that here the device is appropriately thematic, and even the most stubborn missing/extra letter hater will appreciate its use. Secondly, I know how easy it can be to sit back after finishing a puzzle, satisfied with a job well done, oblivious to how hard the grid construction must have been. Unless I’m mistaken, this must have been fiendish difficult, with every entry, in order, requiring a specific letter to be inserted or omitted, 50% of each.

And if anyone says that they sussed the theme as soon as they saw the title … 😯

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Listener 4138: The X-factor by IOA

Posted by erwinch on 10 June 2011

A third Listener from IOA whose first eleven years ago was not numerical.  It was entitled IEEOO and only occurrences of the five vowels in answers were entered into the grid – hence the real title: Listener Crossword.  Also, I thought at the time that this would probably account for the pseudonym IOA but although you can form any number of names from three vowels I did not know of any candidates that fitted.  It is a bit different now though and we can probably rule out Vicky Pollard but I wonder, might IOA be Simon Shaw?
But back to the present and since they had double clues we knew immediately that A, R and V would be 1, 5 and 13 in some order.
From K. = H we know that K must be a two-digit entry (4, 12, 17 or 23) except that K = 23 is too big for R. = OKKK.  K = 17 will also be too big if O is greater than 2.
From W. = (WW – XXXXX)O we see that XXXXX < 676 (26 × 26) so X = 2 or 3
From U. = UUUU – UUU + UU + K only two values of U. are found that fit at the corresponding grid entry: U = 7 or 9
If U = 7 then U. = 2107 + K = 2111, 2119 or 2124
If U = 9 then U. = 5913 + K = 5917, 5925 or 5930
L. = UV so V cannot equal 1: V = 5 or 13
From V. = OOOOU – O – U
If O = 2 then V. = 103 or 133
If O = 3 then V. = 557 or 717
If O = 4 then V. = 1781 or 2291
If O = 6 then V. = 9057 after which V. is too big (greater than four digits) for further values of O
But O. = QU so O. cannot be 2dn (four digits): O = 3, 4 or 6
From V. = (VXX + A)VXXX with V = 5 or 13; X = 2 or 3; A = 1, 5 or 13
If V = 5 then 5ac or 5dn will end with zero and 15ac or 7dn will begin with zero (prohibited) so V = 13
If X = 2 then V. = 5512 or 5928
If X = 3 then V. is too big: X = 2, 13ac = 557, O = 3, U = 7, L. = 91, L = 4, 12, 17 or 23; A = 1 or 5; 13dn = 5512 or 5928 and K = 4 or 12
Now looking at R. = OKKK: {R., K} = {192, 4} or {5184, 12}
From R. = HHH + XXXXX = HHH + 32 knowing that H > 9 this must have four digits so the R. above (1dn or 5ac) = 192; H = 11 or 12 and K = 4, 7dn = 2111
R. = 1ac (1??1) or 5dn (1???) = HHH + 32 = 1363 (H = 11) or 1760 (H = 12) but cannot end with zero so 5dn = 1363, 5ac = 192, R = 5, H = 11,  A = 1 and 13dn (V.) = 5512
So, the assignments to date are as follows:
A. = (D + D + V)DD = 900 (D = 6) then 1856 (D = 8) so this must be the four digit value 1ac (???1) and D < 16
Only D = 9 fits so 1ac = 2511
1dn (A.) = (B + U)(B + U) + BU = 2??
{1dn, B} = {211, 6} or {281, 8}
2dn (X.) = ((B + U)(B + U) + BU)S = 5??5
If B = 6 then 2dn = 211S (S = 25)
If B = 8 then 2dn = 281S (no fit for S)
B = 6, 2dn = 5275 and S = 25
9ac (D.) = DDKK + S = 1321
Y. = BBBKKR – BK = 17256 so Y = 10 or 19
11dn (H.) = WW – HH + HH = WW – 113 = 5?? (Y = 10) or ??7 (Y = 19)
{11dn, W, Y} = {563, 26, 10} or {287, 20, 19}
W. = (WW – XXXXX)O = 1932 (W = 26) or 1104 (W = 20) but 1104 does not fit at 20dn so W = 26, 26ac = 1932, Y = 10 and 11dn (H.) = 563 
N. = ON = 3N so N must be a two-digit entry and only N = 17 fits so 17ac (N.) = 51
We now have values assigned to fourteen letters and the grid looks like this:
I did not attempt to deduce the instruction at this point but with hindsight I see that I could have found the probable value of E (= 21).
We can see from 3dn (O.) = QU = 7Q = 1?2 that Q = 16 is the only fit giving 3dn = 112
Z. = HH = 121 and since all answers are different Z = 8
M. = BBBB + RRRR – AAAA – OOOO = 1839, which can only fit in the grid at 22ac and also places 91 (L.): M = 22, L = 23
And so it went until just one letter and one number remained unassigned: J = 12
This gave the completed grid:
I made a meal of interpreting the bottom row but the instruction was soon found:
Shade evens but rub out all of odds – ergo X
Giving the solution:
So, a straightforward puzzle, which required just two sides of A4 for working using a pencil and calculator only.  My enthusiasm for the Listener in general has waned somewhat since it peaked during the nineties – I blame the Internet – but it remains equally as strong when it comes to the numerical puzzles.  IOA certainly had the harder role as setter but I enjoyed every minute of solving this – thank you.

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The X Factor by IOA

Posted by shirleycurran on 10 June 2011

Ah. The numerical one, again. Not a favourite, but in principle it is based on logic, has a solution, and only one solution. The title suggests a common factor, ‘X’, perhaps, but with 2 digit entries at 4d, 12a, 17a and 23d, maybe unlikely. An X somewhere?

The way in seems clear, with A,R and V being 1, 5 and 13 in some order, and V not too likely to be 1 seeing clue L. Clue W limits X to either 2 or 3 as X^5 <= 676, most likely 2 as we divide by it in clue J. Clue K limits H to 10-26, and, in turn, this means that OKKK is the 3-digit answer to clue R. As Clue K is one of 4, 12,17 or 23, we now have K=4. Clue U which depended only on U was next and led (after much calculation) to U=7 or 9 only being acceptable in the grid…


It was possible to get about 10 letters before a real point of indecision was reached: what was S? A wrong calculation first prompted S=15 and B=8 and then after correction, a wrong choice led to a final contradiction evaluating clue G.


Here the fact that we were on a painting trip to Ibiza and lacked scrap paper caused a slight incident as I was writing all over our check-ins for the return flight- fortunately on the reverse sides.


(The other non-mathematical numpty here – I was in a greater panic over the missing return tickets than over the moans and groans being generated by the blind numerical avenues – it took a sift through a heap of scribbled calculations to find them – that’s what it’s like being married to a physicist!)


Our host then showed us how to use the Open Office spread-sheet  (Excel-like, but free) to evaluate all the clues, based on what was known of A-Z at any time.  This avoided much error-prone calculator and scribble working – just as well, as the digit 7 on our calculator was now responding very erratically or not at all – the island’s very hot climate was taking effect.


Progress speeded up dramatically and the grid was soon filled satisfactorily. Now, what was the coded message? The only reasonable partition of the top line and the first digit of line 2 being SHADE was very encouraging. although the final instruction (SHADE EVENS BUT RUB OUT ALL OF ODDS – ERGO X)  removed most of the hard-won digits! At least this one will be easy for Mr Green to correct.


We posted our grid as early as possible, assuming a slowish post. Just as well, as a day later we were blocked from returning to our host’s home for 3 days as forest fires burnt over the north part of the island.
Numericals will be associated for some time now with the roar of the Canadair fire-fighting seaplanes flying just over the hotel where we were kindly received during the worst of the fire.


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Listener 4138: The X-factor by IOA (or Please, not Simon Cowell!)

Posted by Dave Hennings on 10 June 2011

It’s not often that The Listener catches me by surprise (!), but I really wasn’t expecting a numerical puzzle this week. Until this year, it was always on the last Saturday of February, May, August and November, but now it’s the week before that. Anyway, for whatever reason, I’d forgotten it was coming and was sort of disappointed, although that isn’t to say that I don’t normally enjoy these mathematical sojourns.

IOA, according to my records, has set two previous Listeners. The first, in 2000, was a ‘normal’ crossword, and the second, in 2004, was a numerical. Here we had a straightforward 7×7 grid with the clue numbers, as well as the clues themselves, represented by letters. The endgame would involve decoding the grid and following the instruction revealed. I assumed that the code would be based on the values of each letter used in the solution, or, less likely, using A=1, B=2, etc.

The first point of entry for me was given by the clue numbers , since 1, 5 and 13 were both across and down in the grid, so equated, in some order, to A, R and V which each had two clues. Clue U UUUU – UUU + UU + K was the first proper clue to attempt to solve. Trying all values of U from 2 to 26, only U = 7 or 9 gave a value that would fit those clues’ entry lengths. Thus U = 7 gave an entry of 2107 + K and would fit 7dn (4); U = 9 gave 5913 + K and would fit 9ac (4).

Using clues A, V and X, I soon had A = 1, X = 2 and V = 13, leaving R = 5. From clue R, K = 4, and I really was on my way.

I think this was the first numerical for some time where I didn’t end up having to solve the puzzle for a second or third time due to some problem, perceived or real. I remember the last numerical in February where the preamble stated “ambiguities in four cells must be resolved” and I had found only two (in fact there was only one!). That led to a resolve, but here there was no such problem, and I soon decoded the instruction: Shade evens but rub out all of odds. Ergo X. Not for the first time in a Listener, a lot of hard work disappeared as I erased most of my solution!

So, I think it fair to say, a relatively easy numerical from IOA, so thanks for that, especially given what was to come two weeks later!

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Ifor’s Frightened Catherine, No 4137

Posted by shirleycurran on 3 June 2011

A couple of years ago the numpties downloaded one of these blank grids and said, “Well, how amusing, they’ve forgotten to print the bars!” (That was our first dose of them; the maze where we had to find our way in then find the thread, string, rope etc. that would have led us out – but we didn’t and are still fumbling around in the dark in that maze.) We are not so naive these days – just a mite weary of cartes blanches.

It is a couple of weeks since we spotted ‘Catherine’ in the Tall’n’s devilish concoction and thought we were in for a celebration of the royal wedding. Perhaps this was it, just a couple of weeks late, though why would Catherine be frightened? (Or perhaps I should say “Why not?”)

Next we noticed that there were no clue lengths. Ah well, we could divide the grid into two horizontally, since there was 180° symmetry, so every horizontal clue we solved would give us the length of another. But no, Ifor had thought of that and there was an extra clue hidden in there somewhere that wasn’t to be entered.

Just to complete our joy, twelve clues consisted of definitions leading to two words differing by one letter. Well, there was, seriously, a bit of joy there since we only had to look for 12 shorter clues and we were on our way. One numpty red herring, ‘Make fast approach’ led to TIE and NIE (when, of course, we realized later that it was RUN-UP) but our system worked well and we soon (with Mrs
Bradford’s help – that’s where Dave is going to use the rest of his red pencil stub one of these days when we have the cover of her book to colour in as the final move!) had most of those pairs:

Demon from the Dark Continent AFRIT/AFRIC, Object to small creature PROTEST/PROTIST, Set of three minimum, LEASH/LEAST, Scotch whisky, for example, might be near Glaswegian, SCRUNT/STRUNT (and, with his RICE BEER and ICE BEER at 26dn, yet another Listener compiler joins the compiler oenophiles club!) and so on.

We have been seeing a lot of  Ifor lately, in the Magpie, EV, IQ and with that fearsome one currently on the Crossword Centre message board. His clues always strike us as fine and fair and this time was no exception.  We soon had enough clues to have a stab at filling a grid, and when SLIDING-SEAT and PORTERHOUSE were in place, we were away.

A pattern emerged with all those twin definition clues having their ambiguous letters in a couple of columns. That, of course, rendered the remainder of the grid fill easier and we had a complete grid with only a couple of doubts about word play by mid evening.

The scientific numpty immediately spotted CENTIGRADE and FAHRENHEIT in what looked like two parallel thermometer tubes and the title clearly anaground (nice word!) to those words with just a missing A – we wondered whether Ifor had slipped up. (Silly, of course not – the Editors would have picked that up – think again!)

Of course, we had slotted in TANGRAM for ‘Puzzle that’s opening with anagram (one answer going missing)’ Fine wordplay: T = the opening of ‘that’, anagram with one A missing makes ANGRAM and a tangram is a puzzle, so that left us clue 39 as the extra one. Odd! ‘Fit it to cavity’?

I started to try to make sense of that ‘cryptic’ clue and had a sort of muddled F(ahrenhe)IT I'(n) T(o) C(entigrade) A'(t) VI + TY – forty – when light dawned.

Clearly, since MINUS FORTY is where those two scales ‘converge’ and those words were going to ‘resolve the ambiguity of which clue to ignore’, we had to remove clue 40. I was up the vae with no spoon with my tangram.

Ifor’s brilliance was suddenly apparent. (Shackleton, Kea, Phi, Samuel and all those stars up there in the Listener clouds had better watch it!) That same clue, ‘Puzzle that’s opening with anagram, (one answer going missing)’  now described exactly what we had in front of us.  A puzzle opening with an anagram of CENTIGRADE FAHRENHEIT (the title) with just the A (Answer – all 46 words of it) missing. Dazzling genius!

We still had to decide which word should go down and which should go up, and discover which two clue numbers had to be left in (yes, left in, since we had been inserting bars and numbers as we went along, how else does one solve such a white cat?) What was this ‘somewhat outdated corresponding couple’? It had to be a point at which the two scales corresponded. If we left Fahrenheit going upwards from 41 and Centigrade going downwards from 5, we had that correspondence. Google tells me (yes, I am supposed to know) that 41 Fahrenheit = 5 Celsius, and that even solves the question of why ‘somewhat outdated’ – Centigrade to Celsius.

Ifor, you’re a star!

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