Listener 4686 Dice Nets By Arden

[HawkCrossword]

Either a) I haven't found the right thread, b) folks have found this one difficult, or c) nobody bothers reviewing numericals any more.

I had a brief conversation on Twitter with a numericals-hater yesterday, and it's clear that some people just don't enjoy them. I'm more of a fan than that, but I sometimes wish we could see some alternatives to fitting a jigsaw of squares, cubes, primes, etc into a grid - I probably enjoy the killer sudoku more than those.

But this one had a very interesting endgame, which made the donkey work worthwhile. The bars condition made the net-spotting a lot easier than it might have been, but it was fun, nonetheless. Many thanks, Arden!

[sunny-dave]

I've done the (hard) gridfill and identified a way into working out the nets - but can't actually summon up the necessary interest to finish it off.

I was actually introduced to The Listener by my liking for numericals and then progressed on to the word based puzzles - but that was long ago when numericals were an intellectual challenge requiring (at most) a pocket calculator and perhaps a list of primes.

The trouble is that easy access to coding resources and spreadsheets etc has resulted in an arms race - where puzzles are actually impossible unless the solver deploys all the big automated guns - and that's just not much fun.

Like many (most?) Listener solvers I now detest the numerical puzzles and only had a go at this when I'd finished most other weekend cerebral challenges and had a spare afternoon today.

Sorry if this sounds grumpy, Arden - the puzzle is a very decent challenge - just not one that I relished too much.

[Keepatit]

I quite enjoyed it. Trickier gridfill than your average numerical. But you raise an interesting point. 9d, a 7-digit triangular number. How are solvers expected to verify that with pen and paper? Needless to say, I used a spreadsheet. Could anyone who didn’t explain how?

[failedagain]

The nth triangular number is T = n(n + 1)/2 (for example, 10 objects form a triangle with rows of 1, 2, 3 and 4 objects; two such triangles, with one inverted, form a rectangle with 4 rows of 5 objects, or 5 rows of 4). For any suspected triangular number T, find the consecutive integers n and n + 1 that are either side of the square root of 2T. If plugging those numbers into n(n + 1)/2 doesn't give T, it isn't triangular.

[Keepatit]

Thank you kindly, Failedagain. That’s most instructive

[Scorpius]

I fully agree with sunny-dave. I quite enjoyed the Listener numericals when I started solving the Listener. My 'O' level maths and the Listener maths tables were all that I required. Now neither are sufficient in many cases, and there's only the occasional one that I enjoy.
This week's uses a grid that is at least three times the size of the average number puzzle, yet the clue set was as demanding as in a fairly hard puzzle, so it took ages to complete the grid. No wonder this thread didn't appear until after 5.00 pm on Sunday. I'd just filled the grid by then. The endgame had to wait until today. Even that was quite fiddly, and not my favourite pastime either, though the bars restriction did make it a bit easier, as HawkCCrossword says.

[Oyler]

I found the grid fill pretty straight forward. The 16ac/31ac relationship was particularly nice. The end game was spectacular as I didn't think that number of nets could be fitted in. But, of course, it was set by Arden, so they could! As regards the amount of computational assistance solvers use, that is up to them. If they want to write programs then so be it. However the Listener puzzles can all be solved without doing that. Having a comprehensive set of tables does help! Thanks Arden for the best numerical Listener of the year, in my opinion.

[Oyler]

I would be interested to know how solvers would have tackled L2309 Take Two or Three by Adam which appeared in Aug 1974. Solvers had to find triples which comprised only of two and three digit distinct integers such that the sum of any two as well as the sum of all three was a perfect square. Now the sum of two consecutive triangular numbers is a square, however if you go down that route you get nowhere. For example, 15, 21 and 28 look promising giving three squares but fails on 15+28=43. When the solution was published it came with a set of formulae that could be used to find the solutions. These formulae were obscure and required university level number theory. It seems that solvers had to be able to derive these formulae. Of course, nowadays a few lines of code gives the nine possible solutions in a few minutes.

[Dim16]

Hi HawkCrossword

I love most numericals, and am certainly loving this one because I am finding making progress satisfyingly difficult ie not at all easy, but also not requiring the use of a spreadsheet (yet). Have completed about half the grid (correctly I hope) and am looking forward to the challenge of the endgame because geometric jiggerypokery is not my thing.

[OzBilly]

I enjoyed it. A reasonably tough gridfill followed by an elegant endgame. I did use a spreadsheet to find the big numbers but I have no problem with that. The setters and solvers are generally aware of what tools are available but no specialised knowledge was required (beyond the definition of a triangular number which could be googled if necessary). One numerical puzzle each quarter suits me - I look forward to them at that frequency.

[Dim16]

If n is a triangular number then 8n + 1 has an integer square root. I found this very helpful.