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Listener 4061: 50-50 by Phi

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emcee | 21:34 Sat 21st Nov 2009 | Crosswords
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Good grief, where is everyone? This week's offering is by Phi, and, my word, I'm having a fair old struggle with it! Has anyone finished it yet?

Good luck everybody!

emcee
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You may well ask. I was busy, so didn't have the chance to make an early start. I, too, am struggling: cold solving has cracked about half the clues, and a possible arrangement of the top left-hand corner is starting to emerge. Still a long way to go, though, and I haven't started to think about A, B and C. Does anybody else have fond memories of Stephen Leacock's story about the A, B and C who used to feature in mathematics problems?
I can see a long hard struggle ahead. Phi is usually very approachable. but his 50th birthday seems to have toughened him up. There are so many imponderables - where does the answer grid start and finish, how many 50-50 clues are there, and then there are the A, B, C to contend with. I've solved a number of clues, but have only managed to enter the first down one and one or two nearby which are anyway probably in the wrong place.
wow that was a challenge and a half. Went off to bed last night with only 2 words entered (and a couple of other possibilities) was a record for me. Then words slotted into place and the symmetry and remaining entries were deduced. The finale was straightforward luckily with a very aesthetic grid.

In a sadistic way I really enjoyed this frustrating challenge, not one for the faint-hearted. Thanks and a happy 50th Phi (whenever it was).
Hi, just found this site by accident and registered. Equally frustrated but have made some progress in the basic layout (no 'reflective symmetry' as I can see as yet). Have worked out B is this helps.
welcome tilbee, always good to see a new pseudonym. If you have not come across this site before, then it is worth saying that we discuss the merits/downfalls of weekly listeners, but do not give the game away or give answers (gentle nudges only please).
We do however give much more explicit guidance (including answers, if requested) on other puzzles. So either way, be warned - and welcome! :)
well I'm just about exactly where AHearer is, half cold-solved, some pieces of grid falling into shape, and a hunch as to what the grid might look like. So far, fair as always for Phi, but much tighter and tougher than his usual standard. Should keep me going well into Sunday this time round - I hope not too far beyond!
I suspect my maths tuition came well before Stephen Leacock!
Very tough. Bit disappointed in the final grid, unless I'm missing something.
I'm still in the early stages, but am puzzled by the references to 'reflective symmetry' and 'the 'words of the length that appears only in the larger part of the grid' - aren't these two statments paradoxical? How can you have two equal haves that aren't equal?
I'm glad you experts think this is hard. I've solved 10 clues so far (which is about 9 more than I thought I could when I first looked at the puzzle!), and have been trying to see how they could fit into the grid following the symmetry rules. However, nothing quite works so I think I need to solve a few more.
faint-hearted reporting for duty - the cold solving itself wouldn't normally put me off, but the presence of definitions and wordplays all over the show did. Saw half a dozen or so of those in a couple of hours at which point I threw in the towel.

Belated thanks to sunny-dave for the link to Schadenfreude's CAM puzzle - this looks much more my cup of tea.

Regards to all - I'll follow the thread with interest.
A brutal struggle indeed. The toughest since I started a year or so ago. I needed to cold solve more than half the clues before it started to come together. That seems to be the tipping point. But undeniably fair, and a good tough challenge. Thanks and many happy returns, Phi.
That was a struggle and a half ! Probably the hardest of the year so far, and I'm still not sure if I like it or not.
I was right. It's a real struggle and I'm nowhere near getting to the bottom of it. But one point is worrying me - how can a grid have reflective symmetry if there are an odd number of both across and down clues? Surely reflective symmetry must mean both halves would look the same if a line was drawn down the middle, and that would mean a line through half a row or column, which is absurd. Or am I being impenetrably stupid? And can one assume that the reflectivity is both horizontal and vertical?
Tristram37, I think you need to reflect on your problem (sorry!) There is obviously a solution which satisfies the preamble.
Surely each across light could be reflected (diagonally) by an equal length down light (there are 23 of each in this puzzle)
Very tough indeed; definitely one of the hardest of the year. Diagramless puzzles usually require a lot of cold-solving before one can make headway in the grid and this was no exception.

I am sure I have the answer correct, but the wordplay in the clue that starts "Characters in Greece" troubles me...
-- answer removed --
Yes - why did I ever start these Listener things? Bobby Collins, I reported for duty right behind you with the faint-hearted but with lots and lots of help and encouragement, am labouring on and light has dawned - indeed, luminance. Like thebuzz999, I am puzzled by that one - but do have a fair idea of the solution that fits.
Well, it took a bit more of Sunday than I anticipated, but a satisfying conclusion at last. Generally very fair but wicked - definitely up among the year's three most tricky to crack. Many happy returns Phi (presumably belated)
Here's looking forward to a nice easy numerical one next week - it is one of those isn't it?

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