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Archive for March, 2010

Labour by Elint – struggling labour indeed!

Posted by shirleycurran on 26 March 2010

At first sight this one was none too auspicious for the ‘Horse with stripes (5)’ team. Those words ‘Labour’, ‘suffer’, ‘affected’, and ‘mutilations’ sounded ominous. ‘Eight

Labour, by Elint - Ashtanga

across entries suffer from 25′. Well, we soon identified the eight by simply counting lights in the grid.  Subtracting those from the stated word lengths produced the fact that the mutilations concerned five three-letter words, two four-letter words and one five.

We are learning to be more careful with our preliminary reading of the preamble. We could establish that eight extra words in down clues were going to define the shortened words. Elint is clearly a wise old setter masquerading under a new pseudonym, as he had meanly split his remaining fourteen down-clue hints into ‘the initial letters … taken alternately in clue order’. Ouff! No easy way into that!

This word play was difficult! Desultory solving filled Friday evening and half of Saturday and we had half the grid almost completed with no light appearing over the eastern hills. ICONIC, TROG, TARTY, GAUP, AGLOW, AGGRO, GNOMIC, EN FETE, ESTEEM, ELTCHI – no problem, but they took us nowhere. PROCRUSTEAN gave us our first breakthrough, as we then realized that the eight special clues did actually define the full-length word ‘Brutally ensuring compliance’. It was a short step from there to separating the word into PROTEAN and CRUS. ‘Variable’ stood out in 8d, as the definition of PROTEAN but we jumped to the conclusion that we were dealing in CRUS of our local type – first class wines. That didn’t sound like labour at all!

It was hours later that we spotted the CAMEMBERTS with some sort of wine-club MEMBER amongst the CATS but it took CLIMBABLE with its LIMB tangled in the CABLE to shift us off the French food thing and into appendages. We had to be looking for an arm and a leg, maybe a toe? It was a long, hard foot slog that finally led us to the PIN in SCOTCH PINES, the LEG in CYCLOPLEGIC, the FIN in CONFINES and the GAM in ALLOGAMY. As for TARMACKING – it seemed right, and intersected with OPAL, CAKEHOLE, DIVOT, ERNIE and SEG but I am still wondering what the Victorian was doing in the clue.

I frequently ask which comes first, the understanding of the misprint or extra word message or its implementation. As usual, we worked backwards. We found EARHART and BLOOMER by filling gaps in our words. AMELIA tied in neatly with that, and thus we had our remaining extra words: ‘mean’, for example, in 21d (Root mean square leaving thicko cross, scratching head). I have opted for IMPLANT as my solution to that but haven’t any idea why (Help, Denis?)

Amelia, of course, led to the appendix and ‘Struggling labour’ – back to where we started – rather legless again after the crus and camembert. We hunted all through Chambers, to complete the task of highlighting an eight-letter word (how neat in the context!) but, of course, no spiders or octopodi crept out. However, this crossword was so cleverly symmetrical that all that was needed was to link the two parts to produce a likely word, ASHTANGA – and it gave just the definition we needed [Sans, eight-limbed, referring to the eight stages of classical yoga]. No cherry trees to chop or wrens to batter to death, so I got my pencils out and did a bit of yoga.

Rewarding and very fair – thank you, Elint.

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Listener 4076: Labour by Elint

Posted by Gareth Rees on 26 March 2010

In some Listener puzzles it takes the penny a long time to drop, in others … well, the first clue I solved was 38a CAMEMBERTS. According to the rubric, this had to be “mutilated” to fit into a four-cell entry, and I said to myself, “I suppose I have to remove MEMBER to leave CATS: perhaps all eight mutilated entries are missing a limb?” And so it proved, though unravelling Elint’s intricate wordplay took a lot of time.

The excellent &lit clue for CAMEMBERTS should give a flavour of this intricacy: “Cheeses authentically made, originally, 250 miles west of Brest, roughly? No, east.” After considerable thought, and recourse to an atlas, I figured that it’s the initial letters of C[heeses] A[authentically] M[ade], “250” ⇒ E, “miles” ⇒ M, all placed to the left (“west”) of an anagram of BREST. And Camembert was originally made in the town of that name in Normandy, which is roughly 250 miles east of Brest in Brittany.

I was misled for a while by 9a, “In French, sept étoiles: this could translate to The Pleiades”, which looked very much like the hidden word SEPTET (especially since I already had the P from the crossing word OPAL). But this would be too easy for Elint: in fact it’s a composite anagram (ETOILES HEPTAD = TO THE PLEIADES).

I took a very long time to figure out 29d. “The Gunners like a pro ref” had to fit T_RTY, with “ref” the extra word. I could see that “like a pro” ⇒ TARTY, but it was only some time later that I saw “The” ⇒ T “Gunners” ⇒ artillery ⇒ ARTY.

The eight answers to be mutilated (and the extra words in the down clues that defined the resulting entries) were:

1a SCOTCH PINES − PIN = SCOTCHES (24d squashes)
6a CONFINE − FIN = CONE (13d figure)
12a TARMACKING − ARM = TACKING (4d fixing)
16a CLIMBABLE − LIMB = CABLE (27d wire)
31a ALLOGAMY − GAM = ALLOY (3d mixture)
35a PROCRUSTEAN − CRUS = PROTEAN (8d variable)
38a CAMEMBERTS − MEMBER = CATS (20d animals)
39a CYCLOPLEGIC − LEG = CYCLOPIC (35d giant)

The unclued keyword at 25a (from which the mutilated entries were suffering) proved to be AMELIA, “a rare birth defect marked by the absence of one or more limbs” (and not found in the OED!). The initial letters of the other twelve extra words in the down clues spelled BELAOROHMAERRT, which “taken alternately” yielded two pioneering women, [Amelia] BLOOMER and [Amelia] EARHART.

The final instruction was to “highlight in the completed grid a word (eight letters) which, etymologically speaking, might have described the puzzle before the mutilations caused by [amelia] occurred.” Since the unmutilated puzzle would have had eight limbs, I was expecting something like octopoid, but instead, the leading diagonal contained:

ashtanga, n. [< Sanskrit aṣṭāṅga eightfold (< aṣṭa- eight + āṅga limb, member)] Originally: a form of yoga based on eight principles and incorporating meditation, physical activity, and spiritual practices. In later use chiefly: spec. a system of physical exercise associated with this. [OED]

I’m not completely sure about the title. “Labour” has the sense “bodily exercise”, so it could serve as a loose definition of the highlighted word.

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Square-bashing by Arden

Posted by shirleycurran on 19 March 2010

There’s a scientific half to the ‘Stripey horse (5)’ team and, in desperation with Arden’s Square-bashing, I handed the pencil box over to him. Many stubs and heaps of paper later I joined him for the triumphal stages of squaring those numbers and finding that they locked neatly into each other, producing the LEFT/RIGHT that we had suspected from the start. Here’s his explanation in full, of how he went about this fearsome task. (Of course, you can skip the programmed bit!):

As usual I start by counting the occurences of the 20 letters in the clues. This gives T 32 times, I  23 times, S 19 times, R 16 times, E 13 times and so on. This makes it rather likely that T=1 and that I,S,R,E are ‘small’: 2,3,4,5….

The next step is to look at the 6  2-digit grid entries, 8a, 11a and 22d. Here there is a severe limit to the values of the entries (to be squared and have no leading zeros) to 4,5,6,7,8 or 9.Looks promising, but there are 7 letters involved (A,I,K,M,R,S and T). The most tightly fixed are I and T, appearing as TIT and T+I. The only possible values for T are immediately 1,2 and 3 (4 is too big when TIT is squared).

If T=3 then I=1 is the only option as TIT is at most 9. Also, if T=2, then I can only be 1 or 2 if TIT is at most 9: neither works for I+T (too small, and I cannot be 2 as well as T). If T=1 then the only possible values for I are 4,5,6,7 and 8. Can we say more?

Yes, looking at M/R.  As either T or I is 1, R cannot be 1 so is at least 2, so M is in the range 8-20. Also, R can be 2,3,4,5 only (R=6 cannot produced M/R at least 4).

What about MR/S? Not very helpful, but 20a is MR, 3 digits, 100-999 so 10 <= MR <= 31.6. As M is at least 8, R is now restricted to 2 or 3 only and thus MR is 16 at least, and this in turn restricts S (look at MR/S as 4-9) to range 4-20.

This is getting a bit complicated, as A and K are not yet considered. The former physicist resorts now to a programmed attack, as the possible combinations are not excessive. There are no longer 20x19x18x17x16x15x14 combinations, all entries must be integers and in range 4-9.

A 70 or so-line BASIC program running over the possible ranges of these 7 letters A-T checking these 6 entries to be the integers 4-9, each occurring just once, ran in a few minutes, checking 4.5 million combinations, producing just 2 options:

5 X%=0

6 Y%=0

10 FOR I%=4 TO 8

20 FOR A%=2  TO 20

30 IF A%=I% THEN GOTO 730

40 FOR K%=2  TO 20

50 IF K%=I% THEN GOTO 720

60 IF K%=A% THEN GOTO 720

70 FOR M%=2  TO 20

80 IF M%=I% THEN GOTO 710

90 IF M%=A% THEN GOTO 710

100 IF M%=K% THEN GOTO 710

110 FOR R%=2 TO 20

120 IF R%=I% THEN GOTO 700

130 IF R%=A% THEN GOTO 700

140 IF R%=K% THEN GOTO 700

150 IF R%=M% THEN GOTO 700

160 FOR S%=2 TO 20 STEP 2

170 IF S%=I% THEN GOTO 690

180 IF S%=A% THEN GOTO 690

190 IF S%=K% THEN GOTO 690

200 IF S%=M% THEN GOTO 690

210 IF S%=R% THEN GOTO 690

220 FOR T% = 1 TO 3 STEP 2

230 V1%=A%*S% \ K%

240 IF A%*S% MOD K% <> 0 THEN GOTO 680

250 IF ( V1% < 4 ) OR (V1% > 9 ) THEN GOTO 680

251 V1$=STR$(V1%)+” ”

255 VA$=V1$

260 V2%= M%*R% \ S%

270 IF M%*R% MOD S% <> 0 THEN GOTO 680

280 IF ( V2% < 4 ) OR (V2% > 9 ) THEN GOTO 680

282 V2$=STR$(V2%)+” ”

285 IF INSTR(VA$,V2$) > 0 THEN GOTO 680

286 VA$=VA$+V2$

290 V3%=M%/R%

300 IF M% MOD R% <> 0 THEN GOTO 680

310 IF ( V3% < 4 ) OR (V3% > 9 ) THEN GOTO 680

312 V3$=STR$(V3%)+” ”

313 IF INSTR(VA$,V3$) > 0 THEN GOTO 680

314 VA$=VA$+V3$

320 V4%=T%*T%*I%

330 IF ( V4% < 4 ) OR (V4% > 9 ) THEN GOTO 680

332 V4$=STR$(V4%)+” ”

333 IF INSTR(VA$,V4$) > 0 THEN GOTO 680

334 VA$=VA$+V4$

340 V5%=A%*R%*T%  \ S%

350 IF ( V5% < 4 ) OR (V5% > 9 ) THEN GOTO 680

352 V5$=STR$(V5%)+” ”

353 IF INSTR(VA$,V5$) > 0 THEN GOTO 680

354 VA$=VA$+V5$

360 IF A%*R%*T% MOD S% <> 0 THEN GOTO 680

370 V6%=I%+T%

380 PRINT STR$(I%)+STR$(A%)+STR$(K%)+STR$(M%)+STR$(R%)+STR$(S%)+STR$(T%)

390 IF ( V6% < 4 ) OR (V6% > 9 ) THEN GOTO 680

392 V6$=STR$(V6%)+” ”

393 IF INSTR(VA$,V6$) > 0 THEN GOTO 680

394 VA$=VA$+V6$

400 I$=STR$(I%)+” ”

410 ANS$=I$

420 A$=STR$(A%)+” ”

430 IF INSTR(ANS$,A$) > 0 THEN GOTO 680

440 ANS$=ANS$+A$

450 K$=STR$(K%)+” ”

460 IF INSTR(ANS$,K$) > 0 THEN GOTO 680

470 ANS$=ANS$+K$

480 M$=STR$(M%)+” ”

490 IF INSTR(ANS$,M$) > 0 THEN GOTO 680

500 ANS$=ANS$+M$

510 R$=STR$(R%)+” ”

520 IF INSTR(ANS$,R$) > 0 THEN GOTO 680

530 ANS$=ANS$+R$

540 S$=STR$(S%)+” ”

550 IF INSTR(ANS$,S$) > 0 THEN GOTO 680

560 ANS$=ANS$+S$

570 T$=STR$(T%)+” ”

580 IF INSTR(ANS$,T$) > 0 THEN GOTO 680

590 ANS$=ANS$+T$

600 PRINT “I A K M R S T possible answer “+ANS$

610 PRINT “AS/K “+STR$(V1%)

620 PRINT “MR/S “+STR$(V2%)

630 PRINT “M/R  “+STR$(V3%)

640 PRINT “TTI  “+STR$(V4%)

650 PRINT “ART/S”+STR$(V5%)

660 PRINT “I+T  “+STR$(V6%)

670 INPUT “Go on?”,KYB$

680 X%=X%+1

681 IF X% > 10000 THEN Y%=Y%+1

682 IF X% > 10000 THEN X%=0

685 NEXT T%

690 NEXT S%

700 NEXT R%

710 NEXT M%

720 NEXT K%

730 NEXT A%

740 NEXT I%

750 PRINT “Loop counts X Y “+STR$(X%)+” “+STR$(Y%)

A,I,K,M,R,S and T as 10,7,15,18,3,6,1 or  15,7,10,12,2,6,1

The first is disallowed by MR (20d) being 3 digits. So we have the first 7 letters! Cleverer chaps might do this with just brain and pencil, of course, but I can’t. Where next? The other 3-digit entry clues, 2d RET and 20d TUT, now limit E to 8,9,11,13,14 and U to 11,13,14,16,17,18,19,20.

It’s now a question of grinding on using the shortest entries (in digits) to restrict remaining values as much as possible, and then looking at possible intersections in the grids to fix values. It doesn’t take a lot of this, as for example a 4 digit grid entry 1000-9999 has only 68 clue values 32-99, and one ‘check intersection’ pins these quite tightly.

Simple, yes? Maybe it is for mathematicians and for Arden, but when I hear of four million possible combinations, I cringe and celebrate the fact that we now have three months before the next numerical one appears.

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Listener 4075 – Square-bashing by Arden

Posted by erwinch on 19 March 2010

Arden’s third numerical Listener since 2004.  The first, Fossil Beds, I found straightforward after a hesitant start while the second, The Latin Squire, involved finding a long Knight’s tour of the grid – this took me two hours and was a touch tedious.  The clues for Square-bashing form real words but there is no apparent theme to them while Mr and tryst each appear twice.
 
I shall only detail my route to the first crossing grid entries, which took longer than usual for these puzzles, and at first I kept on forgetting that we had to square the answers before entry:
 
11ac M / R (2) is in the range 4 to 9
20dn MR (3) is in the range 10 to 30
  
M / R = 8 / 2, 10 / 2, 12 / 2 or 14 / 2: R = 2
MR = 16, 20, 24 or 28
 
11ac TIT (2) and 22dn I + T (2) are both in the range 4 to 9: T = 1 or 3 but if T = 3 then I = 1
4dn SIR (4) is too small if I = 1: T = 1
8ac MR / S (2) is in the range 4 to 9: S = 3, 4, 5, 6 or 7
1dn S²T (4) is in the range 32 to 99: S = 6 or 7: {MR, MR / S} = {24, 4} or {28, 4}: M = 12 or 14: MR / S = 4
 
Available (green) clue answers for two-digit entries: 4 5 6 7 8 9
 
2dn TA + T (3) is in the range 10 to 31: A is in the range 9 to 20
22dn ART / S (2) = 2A / 6 (A = 15 or 18)  or 2A / 7 (A = 14)
But 2A / 7 (A = 14) = 4 so S = 6, S²T = 36, M = 12, M / R = 6 and MR = 24
8ac AS / K (2) = 90 / K (K = 10 or 18) or 108 / K (K = 6 or 12): A = 15, TA + T = 16, ART / S = 5, K = 10 and AS / K = 9 
 
4 5 6 7 8 9
 
11ac TIT (2) and 22dn I + T (2): I = 7, TIT = 7, I + T = 8 and SIR = 84
 
5dn E + LITIST (6) = E + 294L and is in the range 317 to 999: L = 3 and 15dn I + LK = 37
E + LITIST = E + 882  The third digit of the square must be 1 or 8 and the fourth digit cannot be 2, 3, 7 or 8 since perfect squares do not have these as a final digit (12ac).  There are only two fits: {E, (E + 882)²} = {6, 788544} or {13, 801025} but S = 6 so E = 13
 
This gave me my first crossing grid entries: 801025 (5dn) and 16 (8ac), which I chose to put in the right half of the grid.  After this, progress was steady and I had filled a relatively modest three sides of A4 with workings by the time just F and H remained to be assigned to 4 and 11 as determined by the 12-digit number at 1ac.  It is perhaps appropriate to remind solvers at this point that the calculator in MS Accessories can cope with 32-digit numbers and so had no trouble handling 12 digits:
 
 
1ac ADROIT + OUTPUTS – F³(OOL + S) (12)
= 32130 + 775200 – 873F³ = 751458 (F = 4) or -354633 (F = 11)
Squared = 564689125764 (F = 4) or 125764564689 (F = 11)
 
I thought this astounding.  As it happened I had the value for 1ac with F = 11 at the top of the grid and was pleased to find that this was correct after looking at the two values for 23ac: 
 
1?4041 = 184041 or 429²: 429 = 3 × 13 × 11 × 1 = LEFT
6?4656 = 614656 or 784²: 784 = 2 × 7 × 14 × 4 × 1 = RIGHT
 
This gave the final grid:
 
 
How on earth did Arden come up with this 12-digit perfect square?  Surely such things cannot have any practical use, or can they?  I used BBC Basic to see how common such numbers were up to 10 digits and then a friend showed me how to extend the range to 14 digits using MS Excel.  Any leading zeros such as 1600 / 0016 etc were ignored:
 
 
Well, not very common at all with just 0.059% of available 8-digit pairs and 0.003% at 12 digits.  I am assuming that one value for X will never pair with more than one value for Y.  Is there anyone able to extend the search to 16 digits plus?  Are examples only found where the split is an even number of digits so might we expect to see fifty pairs at 16 digits and then none again at 18?  I would doubt that there is an algorithm to generate these and suspect that they are merely one of many numerical oddities circulated among number puzzlers.  All digits in 1ac were checked  so Arden had a choice of two pairs without zeros but perhaps did not like those four sixes in a row in the alternative.  A Google search found 564689125764 on this Polish site but it appears to have no connection with Arden’s puzzle.  Of course, finding the numbers was only the beginning and you then had to write the clue using real words – I honestly would not know where to start.
 
To sum up, this was tremendous fun and dare I suggest benefiting from computer-aided setting?  Thank you Arden.
 

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Listener 4075: Square-bashing by Arden

Posted by Gareth Rees on 19 March 2010

This is my first numeric Listener (and Arden’s third), so I’m experiencing some trepidation. The trouble with numeric crosswords is that no clue stands on its own: instead there’s a delicate chain of reasoning. If you make an error—and it’s pretty easy to do so—then you might not discover that you’ve goofed until much later in the puzzle, and then there’s no way to find your mistake except by painstakingly checking your logic, from the beginning if necessary.

The puzzle is in left-and-right style, with two identical sides to the grid, connected only by 1 across. Each clue is in two parts, giving two answers, the squares of which are to be entered, one in each side of the puzzle. The letters from A to U (excepting J) represent the numbers from 1 to 20 in some order. There is no mention of whether the entries are in base 10. I guess I’ll proceed under the assumption that they are, and hope it becomes clear quickly if they are not.

The entries at 23a are unclued: I am required to deduce clues for them, and “write them in the correct order under the completed grid. These clues are both words with their letters combined entirely by multiplication into single products.” There’s the amusing admonition, “Alternatives to the obviously intended clues are not acceptable.”

How can I get started? Well, since every entry is different and none starts with a zero, the six 2-digit entries must be 16, 25, 36, 49, 64 and 81, in some order, squares of 4 to 9. The six clues are 8a AS / K : MR / S, 11a TIT : M / R, and 22d ART / S : I + T.

Consider TIT. T must be at most 3, and if T = 2 or T = 3, then I = 1. And T can’t be 2, or else I + T would be 3, which is too small. So either T = 1 and I ∈ {4,5,6,7,8} or else T = 3 and I = 1.

Consider M / R. 20d MR is a 3-digit entry, so 4 ≤ M / R ≤ 9 and 10 ≤ MR ≤ 31, hence 2 ≤ R² ≤ 7, hence R = 2 and M ∈ {8,10,12,14}.

Consider MR / S, and possible values for S. MR ∈ {16,20,24,28}, so S ∈ {3,4,5,6,7}. However, S can’t be 4, because otherwise MR / S = M / R and all entries (hence all answers) are supposed to be different. Also, 1d S²T is a 4-digit entry, so 32 ≤ S²T ≤ 99, and since T ∈ {1,3} that means that 11 ≤ S² ≤ 99. So S can’t be 3. The only remaining possibilities for S mean that MR / S = 4, which means that T can’t be 3 (because then I = 1 and T + I = 4 too). So T = 1 and since TIT can’t be 4 either, I ∈ {5,6,7,8}.

And now that I have T = 1, I can further apply 32 ≤ S²T ≤ 99 from 1d and deduce that S can’t be 5. So S ∈ {6,7}.

Consider ART / S. If S = 7, then the only possibility for A is 14, but that means ART / S = 4, but I already have MR / S = 4, so that’s no good. So S = 6 and M = 12 and A ∈ {12,15,18}. Now if A = 12, then ART / S = 4 = MR / S, and if A = 18, then ART / S = 6 = M / R. So A = 15.

Consider AS / K. Now AS = 90 so K ∈ {10,15,18} and AS / K ∈ {9,6,5}. But I already have ART / S = 5 and M / R = 6, so K = 10.

So of the 2-digit entries I have MR / S = 4, ART / S = 5, M / R = 6 and AS / K = 9. This leaves 7 and 8 for TIT and I + T, so I = 7.

5d E + LITIST is a 6-digit entry, so 317 ≤ E + 294L ≤ 999, so 297 ≤ 294L ≤ 996, so L = 3. This entry intersects 8a (which is 16 or 81) so its third digit is 1 or 8, and the only possibilities are E ∈ {13,17,19} and (E + LITIST)² ∈ {801025,808201,811801}. However, the fourth digit of 5d is the last digit of the square at 9a, so must be 0, 1, 4, 5, 6, or 9. So the only possibility is E = 13.

18a TITBIT is a 6-digit entry, so 317 ≤ 49B ≤ 999, so 7 ≤ B ≤ 20. And 21a BAB + E is another 6-digit entry, so 317 ≤ BAB + E ≤ 999, hence 21 ≤ B² ≤ 65, hence 5 ≤ B ≤ 8. Since 7 = S is taken, the only possibility is B = 8.

The two 21a entries are now (BAB + E)² = 946729 and (AM + BLE)² = 242064. The fourth digit of 21a is the second digit of 20d, which is (MR)² = 576 or (TUT)². So (MR)² intersects (BAB + E)² at the 7, leaving (TUT)² to intersect (AM + BLE)² at the 0. This means that U ∈ {10,20} and since 10 = K is taken, U = 20.

7a TA + INT is a 4-digit entry, so 32 ≤ TA + INT ≤ 99, hence 3 ≤ N ≤ 11, and the only remaining possibilities are N ∈ {4,5,9,11}, giving (TA + INT)²  ∈  {1849,2500,6084,8464}. The other entry for 7a is (B + URR)² = 7744. The second digit of 7a is the second digit of 2d, which is (RET)² = 676 or (TA + T)² = 256. So (RET)² crosses (B + URR)² at the 7, leaving (TA + T)² to cross (TA + INT)² at the 5. Hence N = 5.

Time to start writing in entries, even though at this stage I might be mistaken about which entry goes in which half. I have two areas of intersecting entries, so for the moment I’ll write both of them in, and if they should clash later on, I can swap them around. I’ll write the bottom-half entries in red so that I don’t accidentally make deductions involving both entries from both areas.

1d (SK + Y)² = _7_ _. This means that Y ∈ {9,16} and (SK + Y)² ∈ {4761,5776}, and 9a (K + AY)² ∈ {21025,62500}. But 21025 can’t fit on either side. So Y = 16.

9a (K + AY)² = 62500 goes on the left, so (C + AY)² = 6_ _6_, hence C = 18.

Since I know Y and C, I can write in 11d NEA + RCTIC, 16d TRYST, and 17d MI + CK. Then I can resolve on which side to put 8a AS / K : MR / S, 11a M / R : TIT, 18a ETER + NAL : TITBIT, and 5d E + LITIST. Let me see how things look now:

3d (DI + SCS)² = _0_5_ _ so D = 9.

5d (DON + OR)² = _ _84_ _ so O = 17.

12a (B + ENT)² = 5329 goes at the left, so (LE + G)² = 2_0_, hence G ∈ {8,9,10,11,12,13,14}. But all of these are taken except G ∈ {11,14}. And if G = 11, then LE + G = 50, but I already have TA + INT = 50 at 7a. So G = 14.

This lets me write in 19a BOTT, 11d KING + DOM, 6d BE + GINS : G + USTO. And that completes the long entry at 1a.

Let me see if that’s a square. Yes, 564689125764 = 751458². I suppose I better check the other possibility, just to be sure: 125764564689 = 354633². Oho! I spot a theme. So there’s no way to use the entry at 1a to resolve which side is which: it can only be the clues for 23a that resolve the grid.* Perhaps the clues are LEFT and RIGHT?

I’ll leave that possibility for later. For the moment, I have the entries 13d (GILT)² = 86436 : (TRYS + T)² = 37249 but I can’t write them in the grid, because I’ve entered the bottom half the wrong way round compared to the top half. Let me swap them, and put in 14a TIM + ING, 15d I + LK : M + AN and 17d KI + LO while I’m about it:

14a (PO + LITE)² = 3_ _216, so P = 19.

That gives me 10d IPS + O : P + URE too.

The last clue is 19a S + HIFT, and I know that H and F are 4 and 11 in some order, so (S + HIFT)² = 98596.

The two entries at 23a must be 614656 = 784² and 184041 = 429². Let me see if my guess above about the clues is correct. Suppose F = 11 and H = 4. Then LEFT = 429 and RIGHT = 784. So, let’s swap the sides—again!—and write in the clues.

The requirement that these clues should be “obviously intended” rules out the need to look for alternative possibilities, but for the record: if F = 11 and H = 4 they are ELF and FELT for 429 and BIG, GIB, GIRTH and GRITH for 784; if F = 4 and H = 11 they are HEL for 429 and BIG, GIB, FRIG, or GRIFT for 784.

* Now that I look back over my notes, I seem to have neglected the fact that there was a clue for 1 across:

ADROIT + OUTPUTS − F³(OOL + S)

Since F = 11, this yields −354633 as required. But if F had been 4, then this would have yielded 751458 instead. So the clue preserves the ambiguity about which side is which! Any time spent investigating it would have been wasted, at best, or could have been a trap for foolish solvers who assumed that all answers were positive numbers. (So it was a lucky mistake for me that I forgot all about it.)

The construction is very impressive. With normal crosswords you can generally complete the fill before writing the clues, but in numerical crosswords you can’t do that, because you need to make sure there’s a chain of reasoning that leads to the solution. So you have to go back and forth between the grid and the clues. But in this crossword it seems as if it would be easy to get backed into a corner where there was no way to write a clue that was both a real word, and that supported the next deduction. So I take my hat off to Arden: I’d love to see an explanation of how it was done.

Finding the ambiguous square for 1 across would have been straightforward, since there are only a few hundred thousand possibilities. Here’s a short Python program that checks them all:

import math
M = 10**6
for a in xrange(int(math.sqrt(10**11)), M):
    aa = a * a
    bb = aa % M * M + aa // M
    if aa < bb:
        b = math.sqrt(bb)
        if int(b) == b:
            print a, int(b), aa, bb

It only takes a few seconds to establish that there are eleven possibilities, and only two of them (underlined) are suitable for this puzzle: the remainder have zeroes in the squares which would result in leading zeroes in some of the down entries. I can see why Arden picked the first of these: the consecutive 6666 in the other is rather inelegant.

354633 751458 125764564689 564689125764
429069 454710 184100206761 206761184100
431372 895759 186081802384 802384186081
472188 712281 222961507344 507344222961
479301 669777 229729448601 448601229729
480135 786273 230529618225 618225230529
508650 906918 258724822500 822500258724
626419 873820 392400763561 763561392400
696438 942255 485025887844 887844485025
816471 945432 666624893841 893841666624
858138 909420 736400827044 827044736400

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