I wrote Elm before Small but Perfectly Formed (Listener 4176). Unfortunately (for me) Kea’s excellent Table-turning (Listener 4100), also involving a grid dissection, was published while I was in the middle of writing Elm, and I decided to postpone sending it in for vetting. Even so, it stayed in the pipeline for a while to ensure a seemly gap between the two puzzles.

The idea behind the puzzle had been the subject of many idle doodles. Sorted by longest side, the first three ‘primitive’ Pythagorean triples—ones that aren’t just multiples of other ones—are (3, 4, 5), (5, 12, 13) and (8, 15, 17). Now 5² is far too small for a Listener puzzle, and 17² is probably too big. But 13² is just in the Goldilocks zone.

Among the doodles I had a couple of dissections of the 12-by-12 square into three pieces that could be reassembled into the 13-by-13 square with the 5-by-5 in the corner. (Exercise for the reader: can it be done in two pieces?) Here they are, displayed, as seems traditional on this site, as animated GIFs.

First Dissection

Second Dissection

 
It’s nice to have solvers get the scissors out, but the problem is how to indicate the intricate pattern required concisely. I hit upon the vaguely apt idea of using deleted letters in entries to indicate where the scissors should go, taking advantage of the natural correspondence between the points where letters are omitted in lights and grid lines. Deleted letters give a medium for a message as well as making the clue answers longer, always a good thing.

Once the basic structure had been sorted out I got the computer to look for suitable bar patterns. Just as carpenters don’t like exposed end grain, I wanted to avoid bars on the edges of the grids or coinciding with cuts. A symmetric final grid would give solvers struggling to find the last few cuts an alternative angle of attack and anyway is more satisfying.

Trying some candidate grids with a slightly modified version of Qxw’s filler, I found that the first dissection was likely to be a much harder fill than the second, probably because there are more cut edges and hence a greater fraction of the answers have to be of the more constrained ‘skippy’ type. Nevertheless, even filling the second grid wasn’t easy: for example I couldn’t find a satisfactory fill with a message starting ‘cut along grid lines by deleted letters’ instead of ‘cut along grid lines by skipped letters’.

With the second dissection the top left of the grid is relatively unconstrained. The fill I had found placed several of the less friendly letters in the top and bottom right, and with a bit of rip-up-and-retry it was possible to mop up the rest in the top left and make the fill pangrammatic.

Permuting the skippy clues and interleaving them arbitrarily with the others to spell out the message gains some freedom in the grid fill. Not numbering the clues covers up the resulting untidiness; and writing the skippy clues so that they appear in alphabetical order compensates a bit for the extra difficulty at the cost of making life harder for the setter, which is as it should be.

It seems possible to write a clue for pretty much any word with pretty much any given letter as the correct form of a misprint (though it’s much harder if both misprint and correct form are specified). I could therefore present these in alphabetical order of answers, with the result that solvers who read the preamble carefully would be able to deduce the type of most of the clues from the outset.

The reaction of the vetters was fortunately largely positive, though the original preamble underwent drastic pruning and some clues were shortened to have a better chance of fitting on one line.

Early feedback on That Site and by e-mail indicated, as expected, that people were finding the puzzle a bit harder than average. But most seemed to think it was worth the effort, which is very gratifying. What I hadn’t expected was that many solvers completed Grid B first: since the grid wasn’t given I thought that almost all would leave it to the end.

Many thanks to the test solvers who persevered in the face of a draft that was far too difficult; to the vetters for all the improvements they made; and to all who commented on the puzzle after publication. I greatly value your feedback.

Quinapalus
 

I’m afraid this blog will be a little sketchy – it’s four or five years since I compiled the puzzle, and my memory of the process is not what it should be.

I can’t even remember how I hit on the idea of a pub cricket puzzle – I think I wanted to do something with pub names, and this seemed a good way. I played the game on car journeys as a child, and introduced it to my own children, though the growth of motorways and A-road bypasses has made it increasingly difficult. But I was aware that not everyone would be familiar with pub cricket, so the first thing I did was to google it.

The first entry that came up was Wikipedia, unhelpfully calling the game ‘car cricket’, which I’d never heard it called. But the game I knew came next, and – you’ll have to believe me here – at that time my search produced no reference to other versions involving drinking games or sexual harassment, which I’m afraid a number of solvers found.

The variety of pub names in Britain is such that I felt I needed some way of making them verifiable, so I checked in my Brewer’s (14^th edition, but I don’t think the chapter’s changed much), and sure enough that aptly-named tome had a reasonably long list, if far from comprehensive. Some solvers have pointed out that it gives Bear and Ragged Staff rather than just Bear and Staff, though I’ve certainly seen pubs called both, so apologies for that.

Next I needed a clue gimmick that could generate PUB CRICKET and a brief summary of the rules. A LEG IS A RUN NO LEGS IS OUT or some such. And I needed a final step. I decided to make it a game between across and down, and hit on the idea of a substitution to convey the x for y scoring format. I honestly can’t remember what order the next steps came in –I just know it was a very complicated process pairing the pubs to get scores that I could reflect in the grid, and getting the right number of clues that were not involved in producing pub names or scores to convey the message. Oenophile/xenophile was a very happy spot when trying to turn two into six.

I first thought of the title Innings, as a way of linking pubs and cricket, then moved on to Innings and Outings to reflect the fact that it was a driving game, thinking this would also allow me to construct an apt clue gimmick. But having constructed a grid around a 33-letter message, I decided to split the clues into three groups, in, out and “in and out” – less satisfactory in terms of the theme, but perhaps more challenging to solve.

This was the third advanced cryptic I set as Mohawk. One was rejected by the Listener and published in Crossword, another went in the Magpie, and this one sat in the Listener slush pile for the next few years, until I got a surprise email in December to say it was being considered for publication, and could I have a look at the following dodgy clues. It was quite hard to reimmerse myself in the mindset of all those extra/omitted letters and misprints after all this time, and to work through all the possible alternative solutions thrown up when you have three different types of clue. Roger and Shane did a lot to improve the clues and eliminate ambiguities.

I then had to re-solve the edited puzzle myself. Even though I knew what the endgame was going to be, it took me 45 minutes. I wondered, slightly in awe of my 2007 self, how I could have been so devious as to want to inflict this pain on my friends and fellow Listener solvers.

So it was lovely to get the puzzle into the Listener after a gap of a few years, and I’ve had a very gratifying response to it so far. It’s a lot of work putting together an advanced thematic puzzle, and as a dilettante debutante I have the utmost respect for all those setters who continually keep coming up with new ingenious ideas and putting them into practice.

This puzzle was developed from November 2008 to January 2009, and was submitted in January 2010.

I like my grids to be special in some way, and I was running out of ideas. Three years ago, I was about to squirt some water at a fat-ball-stealing squirrel when the idea struck; maybe at a subliminal level the SQU got translated into SQUares, who knows. The squirrel remained dry. Within two days all seven balls, including the plastic mesh, had been stealthily appropriated.

I wrote a program to produce filled grids with rotated eight-digit squares using my trusty 20-year-old 16-bit Pascal compiler. There were rather a lot of numbers to be fitted, and so I decided to restrict them to ones which had no more than two of any particular digit, but there were still more than 16000 of them. I had the refilled squirrel water near by in case my PC overheated.

The output was written to a huge text file, and I wrote a program which analysed it, and as part of this process I recorded the distribution of the digits in each grid.

I happened to notice that there were grids with no 7s in them and decided to make this a feature in some way – after all, seven is a nice number, isn’t it? During the preparation of the rotated squares, I noticed that there was a duplicated number in the file which was formed from rotations from two different squares, and I realised that if the bottom-left to top-right line itself formed a rotated square, then the first row could contain this special case so that two lines could form the missing 7. This could form the finale. What a stroke of luck! And there were seven fat-balls… perhaps, at a deeper level, that squirrel was trying to tell me something.

I therefore reran the program with it looking for a diagonal rotated square (running in either direction) being allowed for, and one of the grids had its only two zeroes at the bottom right. Although this was unlikely to cause any confusion, I decided to use this grid.

Since the grid contained rotated squares, I decided that the clues should also incorporate this theme.

But what form would the clues take? Since the theme was squares, I decided to base the clues on a set of geometric squares, and since rotation was part of the theme, the sides of the squares could be squares or rotated squares. Since the grid had no 7s in it, it made sense to avoid 7s in the sides, perimeters and areas of the squares too. I suppose I should have avoided 7s in the clues themselves, but I didn’t think of that at the time.

Since the theme was squares, it seemed right to make the number of squares a square too!

At some stage – I think it was when the next door neighbour’s cat caught a fat, waddling squirrel on our lawn – I realised that when SEVEN was appropriately rotated it could form EVENS, and decided that all references in the clues would be even.

It took a long time to produce the clues. I decided that no clue should refer to more than three references (otherwise they might look clumsy). It was important that there was to be a clear starting point.

I wrote a program which produced all the possible even values which appeared in the grid, the first row/column being deemed to follow the last row/column as part of the continuing theme (and because I needed as many possible clues as I could get because the evenness constraint restricted the number which would otherwise be available). I then wrote a program which used these to produce all the possible clues – hundreds of them! I decided that only addition and multiplication should be used because they place much tighter constraints on possible values during the solving process.

It was then a case of looking through the possible clues to find a clue which could be used as a starting point. The next step was to find another clue which used the information gleaned to lead to another deduction. And so on. And so on. And so on. There was a new generation of squirrels in the garden by the time I had finished, but at least I knew that there was a logical path available to solve the puzzle.

My wife has not put out any fat-balls for the last three years in a futile attempt to discourage the squirrerls, but now we’ve got another pest: a BADGER. Deep holes, ruined lawn… aaaagggghhhhh!

Elap
 

In these notes, □ denotes a digit and a full stop denotes multiplication (but in a few places, x has been used where that is clearer). ‘Side’ implies the length of the side.

The puzzle has been designed to be enjoyably solved by deduction, not by the defeatist, sledgehammer approach of a program or spreadsheet.

First of all, let’s highlight the cells which are known to contain an even digit – the preamble has stated or implied that all the clued numbers are even:

	Side	Perimeter	Area
Square 1:	C3 + G3 + V2	D2.H2 + J4	A6 + R6 + d4
Square 2:	D2 + h4	n4 + p4	S4.h4 + S5
Square 3:	L3	m3	Q4 + R4 + c4
Square 4:	N2.q3 + n3	b2.d4	d4.k3.r2
Square 5:	T2 + W4 + a2	i5 + j2	P8 + a8 + u7
Square 6:	T2 + t2	M2 + u3	F2.F2.j2
Square 7:	a2.a2	a2.j2	D4 + e2 + j5
Square 8:	e2 + g3	M2 + N4	E3 + K6 + f4
Square 9:	t2.t2	P3.t2	B7 + U5 + s4

 
Square 7: a2.a2, a2.j2, D4 + e2 + j5

Since the perimeter is four times the length of a side, j2 = 4.a2, and this means that a2 must be less than 25. Since, from Square 6’s area, j2 must be a square, a2 = 16 and j2 = 64. This leads to D4 + e2 + j5 = 65536, or D4 + e2 + 64□□□ = 65536. D4 therefore begins with 1:

Square 9: t2.t2, P3.t2, B7 + U5 + s4

P3 = 4.t2 = □□6, and so t2 = <□4 or □9. However, t2 must be even, and so t2 = □4. The area is t2⁴. The minimum value of B7 + U5 + s4 is 1061000 + 10000 + 1000 = 1072000, and the maximum is 9861898 + 99998 + 9888 = 9971784, ie t2⁴ = 1072000 to 9971784 (remember that the shaded squares must hold an even digit, meaning that the maximum possible value is 8). Thus t2 is in the range 34 to 56. Since we know that t2 ends in 4, the only possible values are 34, 44 and 54, corresponding to P3 = 136, 176 and 216. However, the first digit of P3 is even. Therefore, t2 = 54 and P3 = 216:

Square 5: T2 + W4 + a2, i5 + j2, P8 + a8 + u7

For the area we have 216□□□□□ + 16□□1□54 + 416□□1□. The minimum value of this is 21600000 + 16001054 + 4160010 = 41761064, and the maximum is 21698998 + 16991854 + 4169918 = 42860770, fixing the side in the range 6464 to 6546. This is T2 + W4 + a2, ie □□ + □□□4 + 16 = 6464 to 6546. Since the second digit of W4 is even, and t2 provides 4 as the last digit, we have W4 = 64□4.

Square 3: L3, m3, Q4 + R4 + c4

Since m3 is now known to end in 4, L3 must end in 6 (from m3 = 4.L3) – it can’t end in 1 because it must be even. From the size of the area (less than 30000), L3 must begin with 1.

Square 5: T2 + W4 + a2, i5 + j2, P8 + a8 + u7 again:

The minimum value of the side is 10 + 6404 + 16 = 6430 and the maximum is 98 + 6484 + 16 = 6598. The perimeter is therefore in the range 25720 to 26392. The second digit of i5 is therefore 5 or 6:

Square 2: D2 + h4, n4 + p4, S4.h4 + S5

The side is 1□ + □□□4 and the perimeter is 56□□□ + 4□□□ = 9000 to 11986 (n4 and p4 are both even, as is the third digit of n4). The side is therefore 2250 to 2998, and so h4 begins with 2.

Square 8: e2 + g3, M2 + N4, E3 + K6 + f4

The side is □□ + □□6 and the perimeter is □2 + 2216. The perimeter is therefore in the range 12 + 2216 to 92 + 2216, ie 2228 to 2308. This makes the side 557 to 577. The area is therefore 310249 to 332929. This means that K6 starts with 3. e2 + g3 is therefore □□ + □36 = 557 to 577. Therefore, g3 = 536.

Square 6: T2 + t2, M2 + u3, F2.F2.j2

The perimeter is □2 + 416, ie it’s in the range 428 to 508, making the side 108 to 126. The area is therefore 11664 to 15876 = F2².64, fixing F2² in the range 183 to 248. Since we know that F2 is even, we have F2 = 14. The side is therefore 112, giving T2 = 58, and the perimeter is 448, giving M2 = 32.

Square 5: T2 + W4 + a2, i5 + j2, P8 + a8 + u7 again:

The area is 216□□□32 + 16□□1854 + 416□□18, and this ends in 04. The side is 58 + 64□4 + 16, ie 64□4 + 74, which ends in 8, and the square of this value therefore ends in 04. The square root of a number ending in 04 must end in 02, 48, 52 or 98. W4 is therefore 6424 or 6474. However, we know that the third digit is even form n3, and so W4 = 6424. The perimeter is therefore 4 x (58 + 6424 + 16) = 25992, giving i5 = 25928:

Square 2: D2 + h4, n4 + p4, S4.h4 + S5 again:

The area is □□58 x 2□□4 + □□584 and so it ends in 6. D2 + h4 therefore ends in 4 or 6, meaning that D2 must end in 0 or 2. Since e2 cannot start with a zero, D2 must end in 2, and so we have D2 = 12.

Square 9: t2.t2, P3.t2, B7 + U5 + s4 again:

We have

B7	□□61214
+U5	□□□58
+s4	928□
	──────
	8503056

From this, B7 = 8461214, s4 = 9284, and we can deduce that U5 = 32558.

Square 2: D2 + h4, n4 + p4, S4.h4 + S5 again:

The side is 12 + 25□4, the perimeter is 5928 + 4□□□, and the area is 2558 x 25□4 + 25584. We therefore have (12 + 25□4)2 = 2558 x 25□4 + 25584. This gives h4 = 2544, producing a perimeter of 10224, and so p4 = 4296. This gives j5 = 64296, and from Square 7, e2 = 26.

Square 3: L3, m3, Q4 + R4 + c4 again:

h4 has supplied m3 = 544. This gives L3 = 136.

Square 8: e2 + g3, M2 + N4, E3 + K6 + f4

We have a side of 562, producing an area of 315844:

K6	3136□□
+f4	1□3□
+E3	214
	───────
	315844

Since f4 can’t end with 0 (otherwise Q4 would begin with 0), there must be a carry-over from the units column, and so the fifth digit of K6 must be 9. The second digit of f4 must then also be 9:

K6	31369□
+f4	193□
+E3	214
	───────
	315844

Square 3: L3, m3, Q4 + R4 + c4 again:

Since L3 is 136, the area is 18496. We therefore have □□32 + □322 + 4□□2 = 18496, and so c4 is 4□42.

Square 4: N2.q3 + n3, b2.d4, d4.k3.r2

Letting the first digit of b2 be d, we have 4(22.q3 + 592) = □4 x 1536, ie 88q3 + 2368 = 1536(10d + 4), or 11q3 = 1920d + 472. We know that d is even (since A6 is referenced), and so the only valid values are 2, 4, 6 or 8. d = 2 gives q3 = 392 (there is no point in trying the other values of d because adding small multiples of 3840 won’t create another number which is divisible by 11), resulting in a side of 9216. This gives b2 = 24. Do you see that it was not necessary to use a program or spreadsheet to get this result?

For the area, we have 1536.k3.r2 = 92162, and we know that the first digit of k3 is even, and that k3 and r2 are even, and that r2 is the same as the last two digits of k3. Let half of k3 be 100x + y, where x is 1, 2, 3 or 4 (representing half the first digit of k3) and y is in the range 1 to 49 (representing half of r2). Then we have

1536(100x + y)y = 92162 / 4 = 21233664,

ie (100x + y)y = 13824 where x = 1, 2, 3 or 4.

Is this the first time since our schooldays that we’ll need to solve some quadratic equations?

Let’s express 13824 (= 29. 33 ) as the product of two numbers which have a difference of a few hundred:

12 x	1152	Difference too large
16 x	864
18 x	768
24 x	576
27 x	512
32 x	432	This could be it – a difference of 400
36 x	384
48 x	288
54 x	256
64 x	216
72 x	192
96 x	144	Difference too small

Thus we have x = 4 and y = 32. Since half of k3 was 100x + y, we have k3 = 864 and r2 = 64 – and we didn’t need to solve any quadratic equations after all!

Square 8: e2 + g3, M2 + N4, E3 + K6 + f4 again:

f4 is now known to be 1938, leading to K6 = 313692.

Square 3: L3, m3, Q4 + R4 + c4 again:

We now have the first digit of Q4, and so the area (18496) = 8□32 + □322 + 4□42. Since the second digit of Q4 is the same as the first digit of R4, the only possibility is 8532 + 5322 + 4642, and so we have c4 = 4642, R4 = 5322 and Q4 = 8532.

Square 5: T2 + W4 + a2, i5 + j2, P8 + a8 + u7 again:

For the area we now have 216□8532 + 16□□1854 + 416□□18 = 64982 = 42224004:

P8	216□8532
+a8	16□□1854
+u7	416□□18
	──────
	42224004

It is easy from here to deduce that u7 = 4163618 and a8 = 16361854 (remember that u7 and a8 overlap). This gives P8 = 21698532.

Square 1: C3 + G3 + V2, D2.H2 + J4, A6 + R6 + d4

The side is 612 + □□6 + □4; the perimeter is 12.□6 + □□□6, and the area is 816□□2 + 532216 + 1536 – this must be a perfect square. We therefore have 1349754 + 10k is a square, where k is the two missing digits (ie in the range 0 to 99). The square root of 1349754 must be only a little less than the side, and is 1161.789, and so the side must be 1162. This gives an area of 1350244, making k = 49. This gives A6 = 816492.

Considering the side, we have 612 + □□6 + □4 = 1162, ie □□6 + □4 = 550, and for the perimeter we have 12.□6 + □□□6 = 4648. Let J4 = 1000a + 100b + 10c + 6, where a, b and c are 1 to 9 (none of them can be zero because each digit starts a clued number). Then we have

1) For the side: (100b + 10c + 6) + □4 = 550,

ie (100b + 10c) + □4 = 544,

ie (10b + c) + □ = 54,

2) For the perimeter: 12.(10c + 6) + (1000a + 100b + 10c + 6) = 4648,

ie 100a + 10b + 13c = 457

From 2) c must be 9 for the LHS to end in 7. This leads to 10a + b = 34, and so a = 3 and b = 4, in turn leading to J4 = 3496 and V2 = 54.

The clues have now been solved, but six cells are still empty:

We now have to look for the common property. Since we have been told that the sides of the squares share this property, let’s list them:

Square 1:	1162
Square 2:	2556
Square 3:	136
Square 4:	9216	(Perfect square)
Square 5:	6498
Square 6:	112
Square 7:	256	(Perfect square)
Square 8:	562
Square 9:	2916	(Perfect square)

Some of these numbers are square – but only three of them.

What is going on?

• The theme of the puzzle is squares
• A feature is that the grid entries sometimes wrap round to the beginning of the row or column.
• There are no zeroes or sevens in the grid
• The preamble: “…The final grid should show what would otherwise be missing from it”

Ah ha! The side of each square is either a perfect square or a rotated version of one – and there aren’t any zeroes or sevens there, either!

112	121 rotated
136	361 rotated
256	Already square, and 625 rotated
562	256 or 625 rotated
1162	2116 rotated
2556	5625 rotated
2916	Already square
6498	8649 rotated 3 digits to the right
9216	Already square

Each row and column of the grid is also a square or the rotation of a square.

Let’s start with the second row. The square could end with 84, 61, 21, or 4□, and so the possibilities are 8461214□, 61214□84, 214□8461 or 4□846121. Only the last one works: 48846121 = 69892. Now the first two columns can be tackled. By approaching row 7 next, and then the last two columns, we arrive at this (the bars indicate where the perfect squares begin, but were not required to be entered):

So there are some zeroes in the grid after all!

The preamble said: “After solving the clues, the grid must be completed so that every row and column shares a thematic property, given the right starting point. One of them is doubly thematic; solvers are to draw a straight line through this number’s digits.”

We’ve found the common property, but one of them is “doubly so”.

Investigation reveals that the first row is the rotation of two different squares – 81649296 = 90362 and 92968164 = 96422, and so we can draw a line through the digits:

The preamble also said

“A seventeenth number in the grid which has the same property as the rows and columns must also have a line drawn through its digits, so that the final grid then shows what would otherwise be missing from the puzzle.”

We therefore need to find another rotated square in the grid. But what is missing from the puzzle? Seven. There are no 7s in the grid (there weren’t any zeroes either, but a couple have appeared now).

For the final grid to show a 7, the top right / bottom left diagonal should have a line through it, and further investigation reveals that 12645136 (which reads in the direction bottom left / top right) is 35562.

We can therefore draw a line through these digits to complete the puzzle:

The double-rotation feature of the first row was echoed in the sides for Squares 7 and 8.

The puzzle’s title was a rotation of “seven”.

Elap.
 

Some OD comments …

12th February 2011, 9pm … so, I’ve received this e-mail from RO. “What do you think the attached grid might suggest …” it says.

I was instantly transported back to those mystery object rounds from ‘Ask the Family’ in which a close-up slice of something familiar is immediately recognised by the Potters of Slough as a rotary clothes line. On this occasion, having eliminated bottle opener and cheese grater, I get no further and am forced to concede.

A swift follow-up encloses the same diagram, now embellished with the letter pairs CC-UU-BB-EE and the word TESSERACT and tenders an invitation to assist in a joint effort. Would I like to join Gridmeister RO in producing a puzzle? – Oh, come on, is the Earth round? – Done deal!

13th February 2011, 2.33pm … yes, check out the elapsed time — a full grid lands in my in-tray. Unbelievable. Multi-symmetrical, good word lengths, no other duplicated adjacent letters on the main diagonal. Go RO

16th February 2011 and the email ping-pong has been hectic as we try to decide on the messages and how to generate them, as we’ll need a good few letters. RO has tweaked the grid and found a sequence from which we can generate the second instruction in wordplay missing a letter but a neat route to the long instruction eludes us. At last I try the old dog-walking trick and come back with the idea of a message pointing to ‘letters x & y in these clues’, something we think is worth a try.

1st March 2011. Two weeks of some tricky clue writing / mutual checking / cleaning and polishing, and a final draft is ready for two brave souls to test (thank you both). The first to respond ventures to suggest that it’s a ‘brilliant conception, very well executed’ and ‘a classic Listener contender’, so our target is endorsed.

… time passes …

27th August 2011. We hear from the first vetter – all seems well, a few clues to massage (mainly to eliminate link words) but we seem to have acceptance. A year or so later we have a publication date. The rest is now history.

But what about the real history. Where on earth did the idea for that original ‘attached grid’ come from in the first place – hope it wasn’t the product of someone not paying attention in class.

For that, over to RO …

Many school days were spent doodling odd shapes, but one that firmly sticks in the mind is the simple way of drawing a 3D “transparent” cube. Two overlapping squares, then join the four corners of each. This might appear simple, but could I transfer that idea into a Listener crossword?

Working out the grid outline was straightforward enough and I thought that the points between the letters C-U-B-E could be used to get the central four lines (thereby creating the main two squares), but creating a concise and logical instruction that will lead the solver to draw the remainder of the cube appeared not to be; I was going to need some help. Perhaps if the puzzle stopped at drawing the inner square and the four diagonal lines it might have led to an easier task, but spotting the word tesseract in Chambers, it felt as if the word was meant to be for this puzzle. The instruction therefore just became even harder to fathom out as it had to be done twice.

OD was going to be the perfect choice to help me work this conundrum out. Not only a great clue writer, but even more so, someone who knows the limit of acceptability to the solver. It was obvious there had to be two letters from each clue/answer, which limits the amount of gimmicks available. First and last letters of extra words was not challenging enough, and original with misprinted letters in definitions was too much of a challenge. So OD suggested we use the clues themselves and came up with a succinct instruction.

A criticism may be that the grid was carte blanche, but it was thought that with all those lines it would look too messy adding bars and so it was decided to remove them. However, in order to make solving easier, but naturally the setting harder, the creation of double mirror symmetry, still giving the standard 180 degree rotation was chosen. The grid fill didn’t take too long even with the restriction that no two letters could be adjacent on the main diagonal apart from C-U-B-E, but this wasn’t punishment enough.

I wondered if the second gimmick for the final instruction to copy the shape centrally could appear in the grid in the correct order. Luckily with a small amount of juggling I managed to alter the grid and we could then start on the clues.

Fixing the clues to the second instruction meant that the misprinted clues were also fixed, which forces one into frantically searching around for that often elusive definition. I think we managed to avoid the forced misprint on the whole, one of my bugbears, and the crossword was finally complete.

OD has already alluded to our test solvers and The Listener editors, but I thought I would add a comment regarding the feedback we have now received.

JEG is one incredible individual with the effort he puts into his analysis. He recreated all the different grid patterns that were submitted as well as the usual comments from each solver. This time he had to do it twice, which must have taken an age. One fact that we notice from this analysis is the slight difference between the definition of a tesseract in Chambers (cube within a cube) and the precise representation of what would seem a 4D structure. It was hard enough getting my head round the Klein bottle. We therefore apologise to the solvers who were confused whether they had to submit a complete tesseract or not. The dictionary’s definition, as the prime reference, was what we used and the instruction we felt was precise enough to lead to two cubes, one inside the other. From the feedback solvers have given us considerable praise, so we are glad most enjoyed.

Rood