**Listener crossword 4075: Square-bashing by Arden (2010-02-27)**

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**^{1} 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 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

↩ A reader e-mailed me asking for a more detailed explanation of this deduction. The idea here is to divide the second inequality (10 ≤ MR ≤ 31) by the first (4 ≤ M / R ≤ 9), to eliminate M. It’s probably easiest to see how this works by inverting the first inequality and then multiplying. (Remembering that if A, B, C, and D are all positive, then A ≤ B and C ≤ D together imply AC ≤ BD.)

The tricky thing is that when you invert an inequality, the direction of the inequality changes: if A ≤ B, then

^{1}/_{A}≥^{1}/_{B}, and vice versa (assuming A and B are positive). So inverting 4 ≤ M / R ≤ 9, we get^{1}/_{9}≤ R / M ≤^{1}/_{4}. Multiplying by 10 ≤ MR ≤ 31, we get^{10}/_{9}≤ R² ≤^{31}/_{4}.Since R is an integer, R² must be an integer too, so it’s legitimate to round the

^{10}/_{9}up to 2 and the^{31}/_{4}down to 7.