This post is now closed. Please use the ‘General Discussion’ page instead.
This entry was posted on Tuesday, February 17th, 2009 at 5:00 am and is filed under Archive.
You can follow any responses to this entry through the RSS 2.0 feed.
Both comments and pings are currently closed.
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.
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!
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.
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.
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.)
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.
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.
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!
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.
Re #115: Dear Barnaby, this is getter worse and worse now.
I feel the need to summarize.
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)
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.
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.
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 ….
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.
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.
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.
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.
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.
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!).
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.
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.
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.
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?
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.
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.
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.
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!
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.