Posted by chatmeister on February 17th, 2009
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March 12th, 2009 at 1:19 pm
Nobody else apart from me and Rishi using the SB CC? Whatever. As well as grids for the SB CC I now also have those grids in Compiler Utility format. I’m working on repeating the exercise for Crossword Writer, but I’m not rushing as it’s a bit painful switching backwards and forwards between windows to verify the work as CW uses a large window that doesn’t resize and my screen is only laptop sized.
March 18th, 2009 at 12:14 am
I’ve just put a new crossword online if anyone’s interested: XWD008
March 18th, 2009 at 10:09 am
I find generally all this site helpful. I am pleased to see that even the best solvers debate and argue about the clues. I must have been living on another planet because many times I have solved crosswords and walked away mumbling that the setter should stop trying to be so convoluted. A good clue is almost always the simplest and yet taxes the brain to solve. Many times the setter seems to be child like in trying to be too smart by half and ending up with a totally twisted clue that only the setter can understand. Just my opinion!
March 19th, 2009 at 2:06 pm
Harry – “A good clue is almost always the simplest and yet taxes the brain to solve.” I don’t think you’ll find anyone here disagreeing with you on that. Convoluted clues are always less satisfying, unless they’re really, really clever.
Just had a look at Cryptica and seen there’s another good showing from the fifteensquared faithful in the clue-writing comp this week. Good efforts all round and particularly well done Geoff – a thoroughly deserving gold winner.
March 22nd, 2009 at 7:09 pm
And I fully agree.
Some anagrams are good, but this one was above average.
And, Smutchin, thank you for XWD008.
One more to enjoy!!
March 26th, 2009 at 7:41 pm
How many different grid patterns is it possible to get from a 15 by 15 square?
March 27th, 2009 at 11:39 am
We know that in crossword clues sometimes a component is given gratis.
Now is there a limit to the length of this component that is given for free?
In a clue in today’s G crossword we have to do no work to get the five-letter POINT.
Maybe if the surface reading of a clue is great, it does not matter even if a large part of the light is in the clue.
What is the general opinion?
March 27th, 2009 at 7:18 pm
Steven – if you will permit an asymmetrical grid I suppose the hypothetical limit would be 2 to the power of 225, which I reckon to be about 50,391,989,330,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, give or take a zillion.
Of course that would include huge numbers of practically unusable grids such as those only allowing one-letter words.
I can’t quite work out the math for calculating the number of symmetrical grids, but it would be a much smaller number because the usual kind of symmetrical grid is effectively one smaller grid repeated four times (plus the central column and row in a 15×15 configuration). It might be 2 to the power of 49 (for the repeated 7×7 grid), plus 2 to the power of 15 (for the central column), plus 2 to the power of 14 (for the central row), but I’m not 100% sure of that.
Still a very large number, in any case.
March 27th, 2009 at 8:17 pm
And then you could further reduce it by sticking to UK cryptic style. But yeah, no idea about the math either, keep getting lost in too many possibilities.
March 28th, 2009 at 12:52 pm
Is there a mathematician in the house? I’m starting to think that the number of symmetrical grids should be (2^49)*(2^15)*(2^14), i.e. 2^78 – is that right? (My error in #108 was adding rather than multiplying.)
March 28th, 2009 at 8:30 pm
Well here’s one, but I must admit, this is a tough question.
Dear Barnaby, you would indeed think that the number of possible grids would be 2^225, but there are some complications.
For instance, if the first row has only black squares, then the grid is not a 15/15 grid any more, but a 15/14.
The same can happen to the last row and the outer columns.
When you rule out the grids with the first row completely black, you’ll get 2^210 possibilities.
But then the second shouldn’t be completely black either ….
There are also 2^210 grids with the last row completely black, but unfortunately there is an overlap with the first situation.
And I haven’t even looked at the outer columns.
I find this an intriguing problem, I will surely spend some time on it.
If the grid must be symmetrical, you encounter the same problems.
Although I think the problem may be easier to solve.
A symmetrical grid is the result of a rotation about the square in the centre, either rotating the top left quarter or the left half of the grid.
So you have only to look at variations in a part of the grid.
That’s why I think it is probably easier)
But as I mentioned before, these outer rows make it complicated.
So, no solution yet, but I will give it a try!
March 28th, 2009 at 9:27 pm
Re #110, if you don’t bother about the outer rows, you will indeed have to multiply 2^49 by something, but not 2^15 and 2^14.
You are considering a rotation of the top left quarter, so the 8th column and the 8th row are not independent.
The first 7 squares of the 8th row (or column) will reappear.
So you have to multiply by 2^7, and finally by 2, because the one in the centre of the grid can either be black or white.
You will get 2^57.
Another way to look at it is to divide the grid in pieces of 8 by 7 + the one in the middle.
This will also give you 2^(7×8) x 2 (for the centre of the rotation) = 2^57.
But then, a symmetrical grid can also be obtained by rotating the complete left half about the centre.
Questions, questions ….
March 28th, 2009 at 10:12 pm
Thanks, Sil, in my fog of semi-understanding I had completely forgotten that there is rotational symmetry to the 8th column and row as well as to the four “mini-grids” in the corners.
I see your point regarding the first line, although I suppose in principle one might have (say) a 15×15 grid in which only three squares were blank, to form two intersecting two-letter words, and still term it a “crossword”. I’m not sure why one would want to, other than to stretch the boundaries.
Somewhere, I think, Ximenes laid down specifications for the number of unchecked letters and also perhaps for the relative number of black and white squares. If you could come up with the correct calculation taking these into account, that would be fascinating; I’m afraid it’s beyond me.
Incidentally, according to a quick Google, 2^225 is greater than the number of atoms in the world. So even using all available materials, it would only be possible to print a tiny fraction of those grids…which may be a good thing for the sanity of all of us!
March 28th, 2009 at 11:35 pm
My God, I am (again) involved in a matter of no importance …
When only three squares are blank (crossing), you will have a 3 by 3 crossword, and certainly not a 15 by 15.
And if you still want to call it a 15 by 15, despite all these black squares on the outside, there is just as much reason to call it a 25 by 25 with even more black squares.
Anyway, I never read anything about Ximenes or his rules.
Maybe I will do one day.
But having said that, I am a real Libertarian.
I come from a country where crosswords (called Cryptograms) are regularly free-formed and non-symmetrical.
For me the clues are the only important ingredients of a crossword, the grid is of no real importance.
(I know, in Britain this is like swearing in a church)
The crosswords I produced so far (for friends and relatives) were made with the free-form option of Crossword Compiler.
I have a lot of ideas for clues, and I put them in a database.
The free-form option gives me the opportunity to make optimal use of this database.
If I would use a pre-defined grid I can only use 10% of the clues in the database, so I have to find additional words & clues.
Some might say that this is the Great Thing of being a setter.
I do not agree.
First there is the clue, then there is the grid.
Quality above quantity.
There are times that I think I would cause a Revolution,if I would put my brainchilds on the Net (because of the grid).
But my God, it is only my hobby, so why should I stick to rules?
I just do it my way.
March 29th, 2009 at 7:55 am
Sil, we are in agreement on the relative importance of grid and clues.
But a three-blank printed 15×15 crossword would be a different beast from a three-blank printed 25×25 crossword; it would be trivial to gauge the “true dimensions”.
(That is, assuming all squares are the same size! Now there’s an idea for a twist on grid design that I don’t think I’ve ever seen.)
Or if you prefer, you could conceive of a six-blank crossword with two sets of three blanks in opposite corners, anchoring the 15×15 dimensions, as it were.
March 29th, 2009 at 12:08 pm
Re #115: Dear Barnaby, this is getter worse and worse now.
I feel the need to summarize.
1
If you allow ‘everything’ in a grid, including completely black rows or columns on the outside, one-letter words, isolated words, etc. , then the number of different 15×15 grids is not very hard to find, whether they have to be symmetrical or not (see # 108,111,112)
2
If you think (like I do) that completely black rows or columns on one of the outside borders of the grid, bring down the measures of the grid, then finding the number of different 15×15 grids is not so easy any more, even when you still allow ‘everything else’ like one-letter words and isolated words.
That is the first thing that I like to find out (just for fun).
Both in a symmetrical and a non-symmetrical situation.
3
In both 1 and 2 probably the vast majority of grids are of no use for a cryptic crossword.
It would be a real challenge to find out how many grids there are which are ‘useful’.
My starting point would then be situation 2 and symmetrical.
So, for instance, a grid with 6 blanks in opposite corners (Re # 115) would be perfectly acceptable as a 15×15 grid, although no one would ever use it.
A grid with 1 blank in two opposite corners would be a valid 15×15 grid when counting the number according to situation 2, but must be ruled out in this new situation 3, because we don’t want isolated one-letter words in the real world.
4
Next step could be: how many ‘useful’ grids are really useful?
Like: each word must be at least 3 or 4 letters, or: the number of words must be in between 25 and 30, or: we want a certain variety in word length.
But then we are at the point where probably the real Mathematician should stand up.
And I can assure you, that’s not me ….
March 29th, 2009 at 11:43 pm
Sil and Barnaby:
Fantastic stuff guys.
I appreciate the enthusiasm and must confess that my interest in posing the question was to see if it could be answered, not because I wanted to know how many combinations there are.
I’m no mathematician but even I could see it was a tricky one.
May I pick up on point 4 of comment 116 and suggest, the grid be symmetrical. Useful. Have a lowest letter count of three per word. No row or columns that are totally black. Contain no more than thirty words.
March 30th, 2009 at 10:01 pm
I think that the answer will be many thousands and so, for any useful realistic purposes, it is as good as infinite.
March 31st, 2009 at 7:37 am
Re 108-118:
Zzzzzzzzzzzzzzzzzzzzzzzz!
March 31st, 2009 at 9:08 pm
Re 119:
Well, Allan, I can perfectly understand that.
Although some people might say: Mmmmmmmmmmmm, or Aaaaaaaaaaah.
In #114 I said that I am (again) involved in a matter of no importance.
In a way it is, in another way it is not.
Because the problem in itself can be a challenging thing.
A puzzle.
To me, as a maths teacher who deals with these kind of problems (on a lower level) in everyday life, it is.
Besides that, I believe that it is mathematics that gives us the algorithms which lead to further understanding and development of e.g. finding useful grids.
We all take things for granted nowadays.
Mobile phones, I-pods, etc. etc.
But before computers can do their job (and one of their jobs could be finding grids), there is the human brain to put them on the right track.
Maybe (and probably) it has all been done before, but I find it challenging to think about questions like in #108.
If others do not, that’s fine by me.
And I can, as I said before, perfectly understand that.
April 1st, 2009 at 1:16 am
#120: A neat, well-reasoned-out, well-written response to a casually tossed off message (#119).
April 1st, 2009 at 1:32 pm
Here here!!!
Well said Rishi.
April 1st, 2009 at 3:51 pm
On a non-mathematical note, I wonder what’s happened to Paul (he of The Guardian and cryptica.co.uk)? He hasn’t updated the puzzle comp in ages…surely crossword setters aren’t allowed holidays?!
April 6th, 2009 at 6:52 pm
Way, way back to #71 and Pete’s mention of &Lit “flavours”…
For me it’s not quite spot on and the METHODIST CHURCH example is a clue which I would describe as demi-&Lit.
&Lit – What Tim Moorey calls an “all-in-one” clue, the wordplay alone offers a reasonable definition of the answer; no formal definition is given.
Semi-&Lit – The wordplay isn’t quite enough to work on its own (usually for grammatical reasons) so the setter adds sort of identifier outside the wordplay element, which could be as vague as “this”, “who”, “where” etc.
Demi-&Lit – The clue contains a formal, stand-alone definition, for which the wordplay happens to tie in very nicely.
April 6th, 2009 at 6:56 pm
Barnaby: re Cryptica
I don’t think there’s anything untoward. The website is updated by the agency which designed it and on occasion they may be a bit slow in responding to update instructions.
However, there is a major revamp in progress so, equally, it could be that no changes are being incorporated until the new version is launched.
April 8th, 2009 at 12:08 pm
Don’t know if the comments have goaded Paul into action or if it’s just coincidence, but there’s a new set of clue winners up on Cryptica. Well done, Sil!
April 8th, 2009 at 4:42 pm
Re #126: Thank you, Smutchin.
On the other hand, it looks like most regulars didn’t submit their clues this time (because of the problems with the website?). So, maybe, there was not much to choose from.
Having said that, of course, I really enjoyed it: 3 submitted, 3 mentioned. But especially, since 05/04/09 was my Birthday.
What a lovely present!!
Sorry, Chatmeister, this entry is completely off topic, I know, as are # 125 & 126, but for me the only way to reply to the previous one.
April 10th, 2009 at 11:12 pm
Just for a laugh, and to amuse (annoy?) those who are devoted to inter-clue references, I’ve bunged a puzzle on my site (click my name). It is not supposed to be any good as a crossword, just a bit of fun. It’s in Across Lite format (download sites all over the web!).
April 14th, 2009 at 7:31 am
In a thread elsewhere there has been a demand for a blog on the Everyman crossword.
While awaiting this website’s reaction to that request, I would like to mention that I have a blog on the same puzzle (http://dailydozen.blogspot.com/) except that it is not on the latest crossword.
That is because the crossword that appears in the magazine setion of the local paper is Everyman after a time-lag of some weeks (the puz on April 12 was 3256 whereas on the original Guardian site the puz on that day was 3263). The Guardian making its crossword free has opened the doors for countless solvers.
If the blog on Everyman is started here, well and good. (As some others too have volunteered, each week of the month one can handle it: I would be glad to blog one week). Otherwise, I can switch on to the latest puzzle in my own blog, but as it is a prize puzzle I would like to be told when it is safe to reveal the answers.
April 14th, 2009 at 5:29 pm
129/Rishi: The requirement for Everyman entries is that entries are “postmarked no later than Saturday”, so it’s safe to blog on the Sunday a week after publication. Everyman has been blogged here in the past, but the perception was that a tiny anount of comments meant that very few readers were interested. (See the Everyman category at top left)
My guess is that if you could find one or two people to share the work, those in charge here would let you revive the Everyman blog.
April 14th, 2009 at 5:51 pm
Thanks very much, Peter.
After I opened the PDF version and read the deadline for entries to the prize crossword, I decided that Sunday a week after publication would be fine.
If the blog is allowed to be started here with myself and one or two other volunteers, it’s fine. It would be nice because there is already a good readership. And readers outside the UK would appreciate explanations for British cultural references.
I do not agree with the perception that a tiny amount of comments meant that very few readers were interested: looked at it that way, the FT crossword has always had less than 10 comments.
April 15th, 2009 at 12:04 am
Rishi.
A blog for Everyman is a good thing.
Go for it!
April 16th, 2009 at 9:57 am
I hope someone will blog Everyman this weekend. There was one clue last week I got an answer for, but I don’t understand the clue
April 16th, 2009 at 11:37 am
Which clue was it Andrew? – obviously it would be wrong to post the answer – even tho’ you have it already – but it might be the same one that I didn’t get either. You could maybe just post the clue number?
April 16th, 2009 at 12:43 pm
I think it was 24 down Children on the radi0 etc. etc.
Pretty sure I got the right answer, but didn’t see the connection.
Hopefully the blog will elucidate all on Saturday!
April 16th, 2009 at 1:00 pm
Yep! Needed for beginners like me
April 16th, 2009 at 3:30 pm
I plan to start my own blog for the Everyman crossword on April 19.
I have written out the blog for No. 3263 (crossword for April 12)but it is on embargo now.
I shall put in a note here after I launch the blog on Sunday.
April 16th, 2009 at 5:29 pm
Andrew – same one as me.
I get the answer from what I assume is the definition.
But the rest of the wordplay – no bloomin’ idea apart from the first letter?
April 16th, 2009 at 6:36 pm
Yes. I agree. I’m looking forward to being enlightened!
April 17th, 2009 at 5:16 am
Andrew,
My blog will certainly explain that.
April 17th, 2009 at 5:22 am
My blog is separate from 15sqd and each Sunday it will be on the crossword for which the deadline for entries has expired.
I shall make my blog public on Apr 19th morning.
April 19th, 2009 at 5:23 am
The blog on the Everyman crossword is on this site itself. See under category Everyman.
April 21st, 2009 at 3:56 pm
In case anyone is interested, I put up an early attempt at setting a complete 15×15 puzzle here:
http://longair.net/blog/2009/04/21/cryptic-crossword-numpty-2/
… comments and suggestions would be very welcome – I’ll post my own criticisms with the answers later this week
April 28th, 2009 at 3:18 pm
Comment has been made elsewhere about two crosswords in today’s offerings having the same anagram fodder ‘Nero’s cousin’ for CONNOISSEUR.
I think if this had happened some decades ago before electronic aids were available, surprise might have been in order.
But now with anagram generators, it’s quite possible for setters to enter CONNOISSEUR and get several results: among them “cousin Nero’s” or “Nero’s cousin” are good candidates for nice clueing.
I think anagram clues, even for long phrases, have become quite sophisticated today compared to clues of this type in the Sixties or even Seventies. I always thought this was because of a wider choice of anagrams available to setters thanks to computer help.
One of the earliest books, The Nuttall Dictionary of Anagrams: A Pocket Aid for All Word Puzzles by A. R. Ball (date of publication not recorded anywhere) must have been prepared before the advent of PCs. This little book contains anagrams only for words probably of 10 letters and less. And ASTRONOMER is not among them. This might be because the book suggests only whole-word anagrams.
A later book, The Crossword Anagram Dictionary, compiled and devised by R. J. Edwards (Stanley Paul) and other similar books such as Anagram Finder (Oxford) are useful only for solvers, but with online apps even they might be just adorning the shelves.
April 28th, 2009 at 3:44 pm
Thanks, Rishi. Still, an impressive chain of coincidences had to happen for this to occur:
1. The setters chose the same word
2. The setters chose the same strategy for cluing the word (an anagram)
3. The setters choose the same anagram fodder (this, you are saying, is made more likely by the anagram software)
4. The puzzles, although in all likelihood prepared at different times, are published on the same day.
April 28th, 2009 at 3:56 pm
I don’t think points 2 and 3 are that coincidental. CONNOISSEUR is not a great word for breaking into a charade or container clue. There are some possibilities but looking for an anagram would seem to be a high probability. Given that an anagram is being sought, COUSIN is a very clear part of one and so NERO’S is pretty much the only use for the leftover letters.
April 28th, 2009 at 4:34 pm
At the risk of lowering the tone (yeah, I know, it had to be me!), one could try something along the lines of “Being knowledgable makes cousin snore”
April 28th, 2009 at 4:40 pm
Thanks, Derek. And I missed the Norse cousin! At the risk of slipping into a Monty Python sketch…and so NERO’S is pretty much one of a small handful of uses for the leftover letters. Now back to snoring!
April 28th, 2009 at 5:21 pm
Agentzero
You missed out “The setters used the same anagram indicator”.
I don’t think that #1 is a particular coincidence. There are many instances where the same word appears in different puzzles in the course of a few days or weeks. I put this down to the automatic grid-filling software that I am sure most setters use, for part of the grid at least.
If the same program is being used it will have the same vocabulary and I doubt that word selection is entirely random, though perhaps someone who has used this type of software could correct me if I’m wrong.
April 29th, 2009 at 2:54 pm
And the same anagram (Nero’s cousin) was in the Telegraph the day before