# Archive for March, 2015

## Magic Cuboid by Brimstone

Posted by shirleycurran on 13 March 2015

Fig. 1 The Numpty version

Yes, one of the Numpties, even with a reasonable pass in O level maths, takes a week off when there is a numerical, and the other Numpty puts in a few grumbling hours with calculator, pencil, paper and a glass of the good stuff (well, Brimstone obviously didn’t qualify for the Listener Setters’ Tipsy Crew with his completely alcohol-free clues so someone has to!) – see fig. 1, sheaves of scribbled notes finally leading to the desired solution.

Fortunately there is an able friend who made less of a pig’s ear of it and kindly sent me his two and a half pages of neatly rationalised solution (see fig. 2).

Fig. 2 An expert’s solving path

Many thanks to Brimstone. The consensus is that this was nowhere near so tough as the preamble initially suggested it would be and that the rotating of the square slices was very satisfying.

Here’s the polished solution path:

Magic Cuboid by Brimstone
Entries in the original grid will be shown with spaces to indicate the cell boundaries. For example if a 3-cell entry were 12345, I would show it as one of 1 23 45, 12 3 45 or 12 34 5.
(a,b) denotes the numbers between a and b. E.g. (1,9) mean 1 thru 9.
[r,c] means the entry in row r, column c.
16a > 2*(F^7)^2 = 2*F^14. So F^14 < 50000 => F<3. F=2
6d = 16 + (V-2)^2 + 4V^6 < 10000 => V^6<50000 => V<7. So V = 3 or 5. If V=5, 18a=48V^2=1200 but 00 is not a valid cell entry. V=3. 18a=4 32 or 43 2. 16d = 3 42 or 34 2. These two entries must match so 18a= 43 2 and 16d=34 2. 6d=29 33.
16a = 324U^2 + 32768 and starts with 34 so is either in (34000,34999) or (340000,349999). If 16a <= 34999, then 324U<=(34999-32768)=2231 => U^2<6 => U=2 which is F. So 16a is in (340000,349999) =>
324U^2 is in (307232, 317231) => U^2 is in (948,979) => U is in (30,32) => U=31, 16a=34 41 32.
5a =4Q^2 + 9Q^2 = 13Q^2. 3d = 36 + Q^4. Since 11^4>10000, Q<11 => Q=5 or 7. If Q=5, 5a=325 and 3d=661 which do not match. So Q=7, 3d=24 37, 5a=6 37.
1d=1 48 or 14 8. So 1a=5J^2 is either in (100,199) or (1400,1499) => J^2 is in (20,39) or (2800,2999) =>
J is in (4,6) or (52,54). J=5 or 53. J=53 would make 8a much too big. J=5, 1a=1 25, 1d=1 48, 4a=28 12, 8a=14 10, 10a= 8 30 (not 83 0 because 0 is not a valid cell value), 12d=2 27 or 22 7, 15d=4 36 or 43 6.

Note that we now have the first three cells in column 1: 1, 48 and 29. So the magic square total in each case is 78. So 15d=4 36 (not 43 6).
10d=4K^2 + 162 is in (800,899) => 4K^2 is in (638,737) =>K^2 is in (160,184) => K=13, 7a=13 19,
10d=8 38, 17a=18 40 36.
6a = 9 + 25E^2 + 250000 is in (290000,299999) =>25E^2 is (39991,49990) => E^2 is in (1600,1999) =>
E is in (40,44). E=41 or 43. If E=43, 2d=22 91, but no cell entry is more than 51. E=41, 6a=29 20 34,
2d=21 23.
In column 1 we have [4,1]=33 and [5,1]=14. So the [6,1] = 31 (to give the magic total). Hence 9a starts with 31 => 9a=4 + 9R^2 + 2025 < 3200 => 9R^2 < 1171 => R^2 < 131 => R<12 => R=11. 9a=31 18, 11a=27 11 22, 12d=22 7 (not 2 27).
[8:5]=34 and [9:5]=2, so cell [7:5]=78-34-2=42. So 14a = 169 + 3969 + (2M+7)^2 + 8100 is in (42007,42997) (the last cell is 7 from 12d)=> (2M+7)^2 is in (29762, 30759) => 2M+7 is in (172,175) => 2M is in (165,168) => M=83, 13a=4 43 38 or 44 3 38, 14a=42 16 7.
If 13a=4 43 38, row 8 column 1 would have to be 56 which is too big. 13a=44 3 38.
The rest of the cells can be easily deduced from the magic square constraint.
1 25 21 45 17 26 24
48 33 23 28 12 6 37
29 20 34 5 49 46 17
33 50 19 13 19 36 47
14 10 51 35 32 31 9
31 18 8 30 27 11 22
44 3 38 20 42 16 7
16 35 4 15 34 41 32
18 40 36 43 2 21 39

If a column is abcdefghi, the three possible rotations of the magic square are gdahebifc, ihgfedcba and cfibehadg. Of course the middle cell (e) is constant – row 5 is fixed. Since 1 must remain in the top left cell, column 1 is not changed. The 9 in column 7 is unchanged. So all of the other single digit cells must change. These seven cells add to 190 (200 – 1 – 9); there should be two primes; two squares and three non-square multiples of 4. We get these possibilities – the ones marked in red are not valid because they are not prime, square, multiples of 4 or they are squares that are multiple of 4.
2 -> 49,17,42
3 -> 40,50,25
4 -> 8,23,19
5 -> 45,20,43
6-> 36,41,11
7 -> 39,17,24
8 -> 23,19,4
The two squares are 49 and 24. One of the primes goes where the 6 is. So 4->23, 8->19 will not work. Thus 4->8 and 8->23. 5 and 7 get multiples of 4: 5->20; 7->24
So there are two possibilities for the values in the original cells with single digit
1 49 25 8 20 41 24 23 9-> 200 Winner!
1 49 25 8 20 11 24 23 9-> 170

1 20 38 43 42 21 17
48 18 19 15 19 41 22
29 40 21 20 17 16 39
33 33 4 30 34 11 37
14 10 51 35 32 31 9
31 35 23 13 12 36 32
44 25 36 5 2 46 24
16 50 8 28 27 6 47
18 3 34 45 49 26 7

Magic Cuboid by Brimstone
Entries in the original grid will be shown with spaces to indicate the cell boundaries. For example if a 3-cell entry were 12345, I would show it as one of 1 23 45, 12 3 45 or 12 34 5.
(a,b) denotes the numbers between a and b. E.g. (1,9) mean 1 thru 9.
[r,c] means the entry in row r, column c.
16a > 2*(F^7)^2 = 2*F^14. So F^14 < 50000 => F<3. F=2
6d = 16 + (V-2)^2 + 4V^6 < 10000 => V^6<50000 => V<7. So V = 3 or 5. If V=5, 18a=48V^2=1200 but 00 is not a valid cell entry. V=3. 18a=4 32 or 43 2. 16d = 3 42 or 34 2. These two entries must match so 18a= 43 2 and 16d=34 2. 6d=29 33.
16a = 324U^2 + 32768 and starts with 34 so is either in (34000,34999) or (340000,349999). If 16a <= 34999, then 324U<=(34999-32768)=2231 => U^2<6 => U=2 which is F. So 16a is in (340000,349999) =>
324U^2 is in (307232, 317231) => U^2 is in (948,979) => U is in (30,32) => U=31, 16a=34 41 32.
5a =4Q^2 + 9Q^2 = 13Q^2. 3d = 36 + Q^4. Since 11^4>10000, Q<11 => Q=5 or 7. If Q=5, 5a=325 and 3d=661 which do not match. So Q=7, 3d=24 37, 5a=6 37.
1d=1 48 or 14 8. So 1a=5J^2 is either in (100,199) or (1400,1499) => J^2 is in (20,39) or (2800,2999) =>
J is in (4,6) or (52,54). J=5 or 53. J=53 would make 8a much too big. J=5, 1a=1 25, 1d=1 48, 4a=28 12, 8a=14 10, 10a= 8 30 (not 83 0 because 0 is not a valid cell value), 12d=2 27 or 22 7, 15d=4 36 or 43 6.

Note that we now have the first three cells in column 1: 1, 48 and 29. So the magic square total in each case is 78. So 15d=4 36 (not 43 6).
10d=4K^2 + 162 is in (800,899) => 4K^2 is in (638,737) =>K^2 is in (160,184) => K=13, 7a=13 19,
10d=8 38, 17a=18 40 36.
6a = 9 + 25E^2 + 250000 is in (290000,299999) =>25E^2 is (39991,49990) => E^2 is in (1600,1999) =>
E is in (40,44). E=41 or 43. If E=43, 2d=22 91, but no cell entry is more than 51. E=41, 6a=29 20 34,
2d=21 23.
In column 1 we have [4,1]=33 and [5,1]=14. So the [6,1] = 31 (to give the magic total). Hence 9a starts with 31 => 9a=4 + 9R^2 + 2025 < 3200 => 9R^2 < 1171 => R^2 < 131 => R<12 => R=11. 9a=31 18, 11a=27 11 22, 12d=22 7 (not 2 27).
[8:5]=34 and [9:5]=2, so cell [7:5]=78-34-2=42. So 14a = 169 + 3969 + (2M+7)^2 + 8100 is in (42007,42997) (the last cell is 7 from 12d)=> (2M+7)^2 is in (29762, 30759) => 2M+7 is in (172,175) => 2M is in (165,168) => M=83, 13a=4 43 38 or 44 3 38, 14a=42 16 7.
If 13a=4 43 38, row 8 column 1 would have to be 56 which is too big. 13a=44 3 38.
The rest of the cells can be easily deduced from the magic square constraint.
1 25 21 45 17 26 24
48 33 23 28 12 6 37
29 20 34 5 49 46 17
33 50 19 13 19 36 47
14 10 51 35 32 31 9
31 18 8 30 27 11 22
44 3 38 20 42 16 7
16 35 4 15 34 41 32
18 40 36 43 2 21 39

If a column is abcdefghi, the three possible rotations of the magic square are gdahebifc, ihgfedcba and cfibehadg. Of course the middle cell (e) is constant – row 5 is fixed. Since 1 must remain in the top left cell, column 1 is not changed. The 9 in column 7 is unchanged. So all of the other single digit cells must change. These seven cells add to 190 (200 – 1 – 9); there should be two primes; two squares and three non-square multiples of 4. We get these possibilities – the ones marked in red are not valid because they are not prime, square, multiples of 4 or they are squares that are multiple of 4.
2 -> 49,17,42
3 -> 40,50,25
4 -> 8,23,19
5 -> 45,20,43
6-> 36,41,11
7 -> 39,17,24
8 -> 23,19,4
The two squares are 49 and 24. One of the primes goes where the 6 is. So 4->23, 8->19 will not work. Thus 4->8 and 8->23. 5 and 7 get multiples of 4: 5->20; 7->24
So there are two possibilities for the values in the original cells with single digit
1 49 25 8 20 41 24 23 9-> 200 Winner!
1 49 25 8 20 11 24 23 9-> 170

1 20 38 43 42 21 17
48 18 19 15 19 41 22
29 40 21 20 17 16 39
33 33 4 30 34 11 37
14 10 51 35 32 31 9
31 35 23 13 12 36 32
44 25 36 5 2 46 24
16 50 8 28 27 6 47
18 3 34 45 49 26 7

## In the Event of Fire by Flying Tortoise

Posted by shirleycurran on 6 March 2015

We downloaded this totally unsymmetrical and very small grid with its very few clues and brief preamble and muttered anxious comments. “This is probably going to be tough!” We discussed what one does in the event of fire – raise the alarm, get out fast, leave behind … etc. (of course, a flying tortoise would have an advantage over a conventional one!) but came nowhere near what was finally revealed when we had filled the grid.

I did a speedy check to confirm that Flying Tortoise was still a full member of the compilers’ tippling club and found only a rather sad bit of proof ‘A gargle less good when liquid is sour beer’ (A GARGLE less G* = ALEGAR – fine clue even if the  surface reading is somewhat unpalatable!)

More surprising than the mini, rather unusual grid were the clue lengths. We soon saw that we somehow had to fit a ’15, two-word clue into a light that had room for five letters. ‘Bang up defector in nick, caught for swindling’ was a generous compound anagram that gave us CONFIDENCE TRICK and cross-checking letters soon suggested that the CONFIDENCE was not going to be entered into the grid. Even worse, we had a seven-letter word to fit into a single cell and the same with a five-letter word

This pattern continued as generous clues, coupled with the words they crossed, suggested which words or elements of words were superfluous, like ‘Early pea hybrid in period of adjustment’ giving LEAP YEAR of which only the LEAP was confirmed by intersecting letters.

We solved happily, laughing at a lovely clue, ‘Vow to imprison regressive Communist in new devious cover for one’s lies’ (I DO + RED< in NEW*) It was the fine definition that caused the smile – EIDERDOWN!

Then our grid was full and we gazed with perplexity at the words that seemed to be extra. We had (leap) YEAR followed by N(owhere) = YEARN,  (a)LONG, (omert)A followed by IM(agine) = AIM,  CONFIDENCE (trick), (sor)EX + PECT(in) = EXPECT, (gliom)AS + SUME(rian) = ASSUME and that very elusive potato DESIRE(e) (poor SIRE, drowned in the DEE!), then to confound matters, we had a double clue – two 10 acrosses leading to BELIE and F(ealty) = BELIEF.

CONFIDENCE, LONG, YEARN, IMAGINE, ASSUME, BELIEF, AIM and EXPECT. Head scratch, head scratch. What do they have in common? – something, obviously, and they have all been lost or removed.

There was a far longer grid-stare than usual and we tried anagramming all the letters that circled the grid. Then I gave myself a healthy kick. I studied Dante years ago and perhaps the best known line from the Divine Comedy is ‘ogni speranza lasciate, voi ch’entrate!‘ Obviously we were leaving behind or abandoning synonyms of HOPE. ‘ALL HOPE ABANDON YE WHO ENTER HERE’

Dante, guided by Virgil, told us that that injunction was around the portal of the Inferno, so it was now clear where our letters had to go and which ones we were to rearrange. A quick check proved that that was indeed what we were being instructed to do – reorganise three sides of our grid.

All hope abandon ye who enter here!

Just to be sure, I checked with my seventh edition of the ODQ and CONSTERNATION! There, in black and white, was a new translation that didn’t fit the bill at all. ‘ABANDON ALL HOPE YOU WHO ENTER!’ The preamble was clear that we had to rearrange a string of letters, to show in a thematically appropriate position, a quotation (using the word order found in the Oxford Dictionary of Quotations) (my underlining). If I use the current ODQ, that is impossible as there is no U in that string of letters.

Fortunately we have some very old versions of the ODQ on the shelves, and the second edition, published in 1953 and a 1972 concise edition have the desired version of the quotation, but I do feel concern for any new, young solver who has a shiny latest edition and a dilemma. I wonder whether this was an editorial oversight.

Apart from that, the solving was great fun, so many thanks Flying Tortoise.

## Listener No. 4333: In the Event of Fire by Flying Tortoise

Posted by Dave Hennings on 6 March 2015

Listener number two from this setter, number one having the stage show Wicked and Defying Gravity as its theme. This week, we were faced with a small 10×10 asymmetrical grid and a quotation to be found with which we “solvers of Listener crosswords might identify”. The only other thing to look out for were some “grid treatments that affect some entries”.

Starting on the across clues, I failed on 1, but got 6 WANE, 9 AMPLY and 10 BELIE. This last was followed by another clue 10, and that had a length of (6)! I proceeded to check the given lengths of other clues and found thirteen more that didn’t match their entry lengths. Of these, 21ac and 30dn consisted of just one square. That made 14 out of 35 that needed “treatment”.

I failed with the second clue 10, which was a shame as it might have helped me with the north-east corner a bit more quickly — since BELIE didn’t appear in the finished grid at all! 12ac LEAP YEAR was the first entry that I could see would definitely need treating, but exactly how would have to wait.

17 Universe’s underlying principle introducing destiny and origin (3) looked like it should be DAO, and we were told it was in the ODE. Well it wasn’t in mine — mine went from Danzig to dap. It had Tao, but then so did Chambers and with an etymology of “[Chin dào way, path]”.

I reached 21 Murphy’s father drowned in the river (7) which effectively had six unchecked letters. (Even getting the checked E later on, it would stump me for ages.) 23 was another that needed treatment, and SUMERIAN also had to lose half its letters.

Reaching the down clues, it was a relief to be able to enter BALLADEER and EYE-OPENER without the need for any jiggery-pokery, and 2/18 gave OMERTA and IMAGINE and it looked as though A and IM would be dropped.

Thus, with the two 9-letter entries in place, the rest of the grid gradually came together, and I wrote down eight words formed from the bits of words which had been dropped from various entries: BELIEF (a delightfully sneaky use of the two 10acs), YEARN, DESIRE, ASSUME, AIM, EXPECT, CONFIDENCE and LONG. 21ac DESIREE was the last clue I solved, even though it didn’t need to be, since EYE-OPENER provided its solitary E.

And so to the endgame.

The eight words thus revealed seemed to have a similar sort of meaning. I decided to use Chambers Crossword Dictionary and found that aim, expect, confidence and long all had ‘hope’ in their list of synonyms, and hope itself had all eight of the dropped words.

A quick hop, skip and jump led me to hope in my ODQ index, and from there to the quotation from Dante’s Inferno “ABANDON ALL HOPE, YE WHO ENTER HERE”, except my ODQ has “…YOU WHO ENTER”, not YE and not HERE. It didn’t take too long to find all the letters around the outside of the grid if YE and HERE are used: hence the unusual “using the word order” given by the preamble, although I didn’t feel it was a strong enough indication of a missing word. The original Italian is “Lasciate ogne speranza, voi ch’intrate”, and that doesn’t include ‘qui’ (‘here’) either.

And what about the letters being arranged to show “in a thematically appropriate position”? Well, I’m sorry, just around the outside of the grid didn’t seem right somehow, especially since nearly every picture of the words are written above the entrance to Hell.

And so to my dilemma: “in a thematically appropriate position” didn’t resolve the ambiguity as I currently saw it:

• Did “Abandon all hope ye who enter” go clockwise round the grid starting near the bottom of column 1?
• Did “Abandon all hope ye who enter here” go clockwise round the grid starting in the bottom left square?
• Did “Abandon all hope ye who enter [here]” go anti-clockwise round the grid; after all that wouldn’t be against Listener rules, would it? (Well, that did seem unlikely.)
• Did “Abandon all hope ye who enter” go somewhere else in the grid?
• Where was Hell in all this?
• Why were we told to use the ODQ when “Abandon all hope ye who enter” is in the Quotations section at the back of Chambers?
• What exactly was that “thematically appropriate position”?

I spent time on and off (mainly off, thank goodness) over the next week trying to find the correct solution and see if I was missing something. I decided that around the outside of the grid just wasn’t the appropriate position, although I could see that it could be the appropriate shape for the entrance to hell. I had found DIS diagonally in the main SW–NE diagonal early on, although it looked somewhat innocuous.

At 3pm on the Tuesday before the deadline, I was resigned to putting ABANDON ALL HOPE, YE WHO ENTER HERE around the outside of the grid when, with a bit more jiggery-pokery, I had ABANDON ALL HOPE, YE WHO ENTER replacing POEYE / ENAW N H TOOB / BALLADEER HN anti-clockwise in the perimeter. The final HN would go up the SW–NE diagonal to meet DIS and thus disappear into Hell.

I was home and dry, wasn’t I? Well, no…

Fast forward four days to the Listener Dinner in Harrogate. Pretty much everyone I spoke to there had the full quotation going round the grid from bottom left to bottom right. No dilemmas for them, just rearrange all the letters and plonk them in the grid.

Far from resolving the ambiguity, the preamble had actually forced me down what was the wrong (or at least, unintended) path. Thanks, but apologies to Flying Tortoise; it was a good idea, and I loved the disappearing hope… which somewhat echoes my own feelings for an all-correct run this year.

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## Listener No. 4332, Two’s Company: A Setter’s Blog by Augeas

Posted by Listen With Others on 1 March 2015

Two’s Company started life almost exactly three years ago following the discovery in ODQ of the quotation. Clearly the length of the quotation, as well as its meaning, required two letters in each cell, and the double bed called for mirror symmetry. Filling the grid, done with pencil and eraser as always, was much easier than many commentators have suggested – very early on in the process that it was decided to allow the letter pairs to appear in either order, thus allowing much more latitude in finding checking words, although all had to have an even number of letters. This would necessitate the entering of letters in a precise manner, with the two letters in each cell appearing in the correct order both across and down. Many solvers found this a fertile source of trips, but it was not designed so. Indeed, a lot of effort was spent in agreeing a form of words in the Preamble which made it perfectly clear that “letters [were] encountered in the correct order in all entries and the perimeter”. In a small number of cells where the same letter appears twice any orientation would be acceptable.

By the time it had reached the point of serious consideration by the Editors (last summer) the sesquicentenary of Mrs Patrick Campbell’s birth suggested that the first week in February 2015 would be a good time for it to be published. In that version solvers were required to highlight the C and L at the start and finish of her surname to indicate a full grasp of what was going on, but this was later removed as “unnecessary after the solver has done the hard slog of filling the grid and finding the quotation”. One of Augeas’s earlier puzzles had involved a quotation by Nancy Mitford which did not appear in the then current Edition of ODQ. Since then he has taken the view that if a quotation appears and no indication is made then it must appear in the current ODQ. A recent puzzle shows that not all setters (or Editors) share this view, as we shall see next week. Where a quotation is given elsewhere then this must be specified (eg. in ODQ Second and Third Editions only). Augeas is one of those odd people who, of necessity, buys each new Edition of Chambers – like many solvers – and of ODQ. One solver, in her kind letter commenting, said that it wasn’t in her ODQ (Second Edition). Maybe a Christmas present idea for her to suggest to someone?

Anyway, back at the editing face things were moving. The ur-version had a different arrangement in what George calls the New England corner, and removing a couple of bars (and allowing the non-Chambers SHON) overcame the perceived problem there. It was pointed out that once the grid was filled there wasn’t much help from the letters in the perimeter to finding the quotation (a view with which quite a few solvers disagreed), so the idea of an anagram of the unches emerged. The serendipitous discovery that CHESTERFIELD could be obtained added weight to the idea. There was originally no indication of where the quotation started, and it was felt that, given the difficulty of getting a handle on the quotation itself, giving the initial cell would make the task less troublesome.

Clearly there is an urgent requirement for the BBC to show that fine film THEM!, the existence of which perplexed many solvers. Odd that one may regard oneself as too young to know about a 1950s movie, but not too young to be familiar with Haydn or Descartes. The general reaction from solvers has been positive, and Augeas thanks those who have ventured a response, in writing and, before all the fuss, on line. Augeas also owes a debt of gratitude to JEG, whose task was made a lot more difficult by the requirement to check orientation as well as content, thus slowing his work-rate significantly. Sorry, John.

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