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Listener No 4319, Feature Film: A Setter’s Blog by Ferret

Posted by Listen With Others on 14 December 2014

When I started on the idea for this puzzle at the start of 2013, my first grid produced a very simple concertina-like staircase. When observed from above, the message FRED ASTAIRE & GINGER ROGERS appeared in six columns on the treads of the stairs. I was unhappy with grid for a number of reasons and I’m glad I abandoned the idea because, as it turned out, Homer’s excellent puzzle ‘Going Out in Style’ used the concertina concept in a much more thematic way than my puzzle and I’m sure the editors would not have accepted two such similar ideas.

In Swing Time, Fred and Ginger dance up their respective halves of the double staircase so I knuckled down and set about producing a structure that would permit that to happen in a 12×12 grid. I also wanted the film title, the name of the night club and ORIGAMI to appear. I eventually hit upon the cut and fold design for the staircase that you have seen but I wanted solvers to find the diagonal lines to make finding the names on the staircase more of a reveal. My first published puzzle appeared in the excellent Magpie magazine. It was called ‘Smith’s One’ and used the idea of diagonal lines to separate clashes so I thought I would try it again. I don’t know of any crossword producing software that is capable of handling this restriction so both that puzzle and this one had to be produced by hand on squared paper.

I produced a grid I was happy with sometime in February 2013 with jumbled ORIGAMI in a ‘two-step’ pattern (mentioned in my original preamble as a hint for dancing and stairs but cut due to lack of space). I then discovered that this type of cut and fold design is called KIRIGAMI so I made a small change to the grid in the top right to include the necessary K (in OKRAS) in case I decided to use that word (at the expense of the two-step hint) rather than ORIGAMI. Only one solver commented that the word should have been KIRIGAMI (in Collins and Wikipedia but not Chambers or ODE). Well, the preamble did say that only the instructions from across clues (the two types of fold) were related to the letters jumbled in ringed cells. It was the down clues that produced the instruction to cut.

Listener 4319

One solver also queried the staircase design. It is true that the double staircase appearing in the film is slightly more open at the bottom than the top but I needed the extra steps to have Fred and Ginger appear on the treads. The barred-off cells in conjunction with the ampersand sign were meant to be a help in determining the start and end point of the names, but also provided a useful way to ensure solvers had found the theme. The half-way platforms on each staircase were necessary to keep the grid together after the cuts were made. In fact, my first tiny model only had three rows in the centre giving a slightly better staircase and more room for the top platform where I was going to have Fred and Ginger swirl round one another, but the grid looked better with the dots, dashes and diagonals being symmetrical and I thought the symmetry would make what was going to be a tough puzzle easier to solve.

Many solvers mentioned ambiguities in the position of the diagonals, particularly in the NW corner of the grid. None of this was intentional. But all doubt was removed with the discovery of the symmetrical nature of the final grid and the need to locate the stars. A few solvers came up with another solution to the second clue. Girl comes back with fine vegetables (5) giving FLORA — [C]AROL< + F. But some admitted FLORA was not well defined by vegetables and the emerging message eventually discounted all but the correct OKRAS (SARA< after OK).

The puzzle went to the Listener editors in May 2013. Because of the necessarily large number of clues (to provide the instructions), the preamble was shortened and improved and I came up with some new clues to remove two ambiguities in the extra letters that I had missed and a few other refinements to avoid repeated use of words in clues. The puzzle just squeezed into the paper but needed a reduced font size for the clues. I must thank the editors Shane Shabankareh and Roger Phillips for their help and encouraging feedback, Roger for the beautiful 3D solution which appeared online and in the paper and John Green for sending me your comments.

I hope you felt the necessary cold-solving was worthwhile, even if the non-manipulation of the grid meant not everyone saw the final staircase or its relevance, which was mentioned in the wikipedia article (as was Silver Sandal) on Swing Time but really only confirmed through You Tube:

Swing Time 2min 20s in to see the stars rise!

Apologies for the delay in producing this requested blog but this really is the first opportunity I have had to produce it.

Ferret
11 December 2014

Posted in Setting Blogs | 1 Comment »

Solitaire II by Xanthippe

Posted by shirleycurran on 12 December 2014

DODOThere wasn’t going to be a Numpty blog this week as it can be no secret that one Numpty is hopeless with the numericals and the other one filled this one in in a rush as we were heading south to the Budock Vean crossword gathering (you can see what fun we had on Derek Harrison’s website)  via the Listener get-together that Quinapalus organised in Cambridge (where we just missed meeting Xanthippe who left minutes before we arrived!)

Aren’t friends great? Two have very generously given me their solving paths, which differ slightly from the one the other Numpty followed and intrinsically differ too. Both were full of admiration for this numerical which was an example of the pursuit of pure logic with no need for deep mathematical understanding.

My Numpty comment is that I could play the solitaire and reach a conclusion that produced three birds to complement the SOLITAIRE of the title and left the O of DODO in the centre cell, but couldn’t have created my solitaire set in the first place, so thumbs down for me – and Xanthippe didn’t give any clues that admitted him into the Oenophile Listener Setters Elite – however, he redeemed himself by leaving an empty glass on the table at the Castle in Cambridge – so all is well on that score. Many thanks Xanthippe and thanks, too, to both of the generous new LWO bloggers!

Here’s the first path by a speedy and gifted solver:

Solitaire II by Xanthippe

All answers are between 100 (256 dec) and FFE (4094); i.e. three hex digits with no leading 0’s.

I re-use x, y and ? to represent unknown digits throughout.

There are 10 properties:
P1: only numeric digits
P2: only letter digits
P3: only odd digits
P4: has repeated digit
P5:  > E00
P6: multiple of 6
P7: multiple of 7
P8: multiple of 10 (16 dec)
P9: multiple of 64 (100 dec)
P10: odd

10dn is a multiple of 7, 16 dec and 100 dec; so it is a multiple of 2800 decimal.  5600 is too big; so 10dn=2800=AF0.

By P8, 4ac, 11ac, 16ac all end in 0 and no other entries end in 0

By P5, 15dn starts with E or F.  If it starts with F, then so does the bigger 14ac.  But the 14a would be FFx and have a repeated digit (P4).  So 15dn starts with E.  14a = FEx for some x (and x must be a letter).  Whatever that x is, 16a is either xx0 or x00.  It cannot be the latter as the above shows that 9dn does not end in 0.  Also x must be even since 14a is not odd.  x cannot be F since 14ac has no repeats.  So x=A or C.  But A00 is not divisible by 6.  x must be C: 14ac = FEC, 16ac=CC0.

16dn = Cxy where x and y are both letters.  17ac has all odd digits, so x is odd.  The odd letters are B, D and F.  16dn<16ac=CC0 => x=B.  y is an even letter since 16dn is not odd and 16dn has no repeats so it is either CBA or CBE.   CBE (3262) is not divisible by 6.  So 16dn=CBA.

16dn < 9dn < 16ac.  CBA < 9dn=xyC < CC0 => 9dn=CBC

13a > 16a=CC0; 5dn is odd, so 13dn begins with an odd digit >= C.  So 13dn starts with D and 5dn = xyD.  5dn consists of all odd, letter digits. 5dn>16a, so x=D. 5dn cannot be DDD (the preamble states that no entry has all three digits the same.)  So 5dn = DBD or DFD.  But 11a<16dn=CBA so cannot begin with F.  5dn=DBD.

13ac is all letters, has no repeats and has an even digit so it is either DAF, DCF or DEF.  DCF=3535 is divisible by 7, so 13ac is either DAF or DEF.  16a=CC0 < 6dn < 5dn=DBD so 6dn starts with either C or D.  If 6dn starts with C then 5ac=DCx, but 5ac has a repeat and would then be DCC or DCD.  But it must have a numeric digit.  Hence 6dn starts with D.  11ac starts with B, ends in 0 and has a repeat so is either BB0 or B00.  If it were BB0, then 6dn would be all letters.  So 11ac=B00 and 6dn=D0A (3338) or D0E (3342).  6dn is divisible by 6, so 6dn=D0E and 13a=DEF.

8ac = xCA where x is numeric and odd (since 3dn is odd).  Of 1CA (458), 3CA (970), 5CA (1482), 7CA (1994) and 9CA (2506) only 5CA is divisible by 6.  8ac=5CA (and 8dn=58c).

11a=B00 < 3dn < 16dn=CBA.  3dn=x05, so x=B or C.  B05 (2821) is divisible by 7. So 3dn=C01.

5ac=DDx.  That x is the first digit of 7ac < 8ac=5CA.  It is also even since 1dn is not odd.  x=2 or 4.  But DD4 (3540) is divisible by 6.  So 5ac=DD2.  (That also fills in 7dn=20F).

1dn=x62 < 7dn=20F => 1dn=162.

1ac = 1xC has a repeat => 1ac = 11C or 1CC.  2dn>11a=B00 => 1ac=1CC.

2dn = Cxx < 3dn=C01 => 2dn = C01 (3073) or C03 (3075).  2dn is divisible by 7 => 2dn=C01, 7ac=215.

4ac = x00.  4ac > 8ac=5CA.  4ac must be one of 600 (1536), 700 (1792), 800 (2048), 900 (2304.  Of those only 600, 900 are divisible by 6.  12ac = xBF.  x must be an even, numeric digit since both B & F are odd letters.  12ac > 8ac = 5CA.  Hence x is 6 or 8.  4ac < 12ac so it cannot start with 9.  4ac = 600.

17ac is sandwiched between 4ac and 12ac and is all odd so it must start with 7, and 12ac must start with 8. 12a=8BF.

14dn=F7x is even, so x is even.  18a<8dn=58C.  So x=2 or 4.  F72 (3954) is divisible by 6, so 14dn=F74.

At this point the grid is completely filled except for 15dn = Exy (along with 17a and 18a).  There are twelve possibilities for 15dn, such that it along with 17ac and 18ac conform to the property restrictions.  However, note that neither 3 nor 9 appears in the grid.  Step b of the game requires that both of these digits appear in the grid.  Hence 15dn = E39 (3641) or E933 (3731).  The latter is divisible by 7 so 15dn=E39, 17a=73B and 18a=49A:

 

1 C C
6 0 0
D D 2 1 5 C A
B 0 0 8 B F
D E F E C C 0
7 3 B
4 9 A

 

Now for the game:
Step a) can only be 0 jumping 0:

1 C C
6 0 0
D D 2 1 5 C A
B 0 8 B F
D E F E C C 0
7 3 B
4 9 A

 

Step b) can be derived without too much trial and error using jumping pegs and jump direction in this order:
7 N, D in row 5 col 1 E, E in col 4 W, C in row 5 col 6 W, A in last row N, 8 S, 4 E  then N, 2 S, 1 in row 1 S, 0 in row 2 col 5 S then S then W then N then N resulting in (with I and O):

C C
O O
D D I C A
B O B F
E C O

 

Step C: Somehow (I used TEA) one determines that the three words are birds BECCAFICO, DOD and COB (SOLITAIRE is a bird as well).  BEC can be done using the D now found in col 1 (S then E then E).  Next jump A W, I E, O in col 7 N then W, C in row 1 col 5 W, then S:

O
D C O
O B
D

 

Next come COB using the D in col 2 moving E, E then S.  Using the same D pin jump W then N then jump the final O over that D to give DODO and leave that O in the center.

Here’s the second path to the creation of the solitaire set from the other brilliant solver:

Listener 4321: Solitaire II by Xanthippe 

From Solitaire game instructions, each digit 2-9 can only appear once in the grid

From clues, many cells can be identified as being either numbers or letters, odd or even, and if divisible by 16 must end in 0 (which can be entered immediately)

14ac contains only letter digits and >E00

13ac is odd and so its last digit (=first digit of 14ac) must be F(=15) rather than E(=14)

15dn>E00; its first digit (=second digit of 14ac) must be E as 14ac has no repeated digits

Since 14ac is not odd, last digit must be A or C (not E or else 14ac has repeated digit)

FEA (=4074) is a multiple of both 6 and 7

So 14ac = FEC

16ac (ending in 0) begins with C

16ac does have a repeated digit; since 9d only contains letters, must be the C

So 16ac = CC0 (=3264 which is a multiple of 6).

16dn<16ac; since 17ac is odd and ends in a letter, this must be B

So 16ac = CBA (=3258 which is a multiple of 6)

And 9dn = CBC as it is >16dn

6dn is >16ac but <E00 so must begin with D (middle digit is 0), as must 5ac/dn and 13ac

5dn (D?D) < 5ac (DD?) but 5dn contains only odd letters (B,D,F)

So 5dn = DBD, making 11ac = B00

13ac>5ac forces 13ac = DEF which makes 6dn = D0E

10dn is multiple of 7, 16 & 100; smallest possible answer is 2800 (=AF0), next 5600

5600>FFF and so 10dn = AF0

8ac is ?CA where ? is an odd number. By trial and error, 8ac = 5CA

3dn is ?05 where ? is either A or C; 3dn>10dn (=AF0) so 3dn is C05

2dn<3dn (=C05) so second digit must be 0; last digit (an odd number) must be 1 or 3

C03 (=3075) is not a multiple of 7, so 2dn = C01 (=3073 which is a multiple of 7)

4ac>8ac (=5CA) and all numbers, so first digit >5; and is a multiple of 6

By trial and error, can only be 600 or 900 but if 900 then no place in grid for a 6

So 4ac = 600

And the only place left in grid for an 8 is at start of 12ac, so 12ac = 8BF

17ac<12ac so first digit must be 7

14dn is F7? where ? is either 2 or 4

F72 is a multiple of 6 so 14dn = F74

1dn contains only numbers in format ?62 or ?64 with ? less than 5

? cannot be 4 otherwise 1dn>7dn

? cannot be 3 otherwise 1ac is 3CC (=972) which is a multiple of 6

So ? is 1 or 2 which leaves only spaces available in grid for 3 & 9 in 15dn

15dn = E39 (as E93 is a multiple of 7), making 17ac = 73B and 18ac = 49A

5ac is DD? with ? either 2 or 4 but DD4 (=3540) is a multiple of 6, so 5ac = DD2

So 7ac = 215 and 7dn = 20F

1dn<7dn so 1dn = 162 making 1ac = 1CC

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Listener 4321: Solitaire II by Xanthippe — Full Solution Path

Posted by Jaguar on 12 December 2014

Ah, the numerical puzzles! My highlights of the year… or at least they used to be, before I discovered the joys of the cryptic side of Listeners. Still, I look forward to them. A fresh challenge, a chance to have a break from leafing through the battered copy of Chambers. And the inevitable detailed solution blog to follow.

Xanthippe’s challenge this week involves the game of Solitaire and hexadecimal numbers. I never could solve Solitaire myself when younger, so now at last I might be walked through the solution. Well, that’s nice! But first a grid fill with not all that much to go on. Except for the interesting point that the numbers 2 to 9 will occur exactly once each in the filled grid, so that will give us something that may help constrain entries.

A while later, I’d stumbled my way to a final grid, and was left with following the instructions for playing the game. One or two hiccups later and, at last, I was staring at three final words. “Beccafico”, apparently some bird or other, and then of course the final words have to be “Cod” and … “Dobo”? Hmm… or I could go the other way round that final loop and recover the words “Cob” and “Dodo”, both of which are also birds, apparently. Along with Solitaire, which is also a bird. So far as I can tell, there’s no link beyond all being birds, although some species of solitaire are extinct, along with the dodo of course.

So that’s the destination sorted, and now it’s time to figure out the journey.

* * * * * * * *

 Solution Path

Obviously the starting point is understanding how to work with hexadecimal numbers. There are plenty of guides available. Still, I found it easier to interchange between normal numbers and hexadecimal ones, and excel is quite useful in this. Knowing that hex(100)=dec(256) and hex(FFE)=dec(4094), and then using Excel’s generous DEC2HEX function can generate all numbers in between.  Of this list we’ll only need 24 numbers in the end so it’s a bit of overkill, but still, nice to be able to have the full list as you can then easily trial possibilities. And anyway, arithmetic is easier in decimal for most people, and certainly it’s easier for me. One day, perhaps, I’ll get around to understanding how arithmetic works in different bases. That would have made Radix’s numerical offering in base 24 from last year one heck of a lot easier, for example.

Anyway…

The obvious entry point is 10dn, which we are told is a multiple of 7, 10 and 64 (all in base 16), or 7, 16 and 100 in decimal numbers. The only number in the range given that fits is 2800, or AF0. We can also enter final-digit 0’s in 11ac and 4ac. On the other hand, 9dn doesn’t end in 0, but 16ac contains a repeated digit and at least one letter. 16ac is also less than E00 but greater than AF0, so it can only be BB0, CC0 or DD0 (or 2992, 3264, 3536 in decimal). Of these, only 3264 divides by 6, so 16ac is CC0. (Alternatively, B and D are odd numbers, but 9dn is even, hence again only CC0 fits.)

14ac is greater than E00, has no repeated digits and contains only letters, while 15dn is also greater than E00, so both start E or F. So 14ac is FEC or EFC. But 7dn is odd, so 14ac is FEC.

After half of the grid has been filled.

After half of the grid has been filled.

The four possible starting moves in the game involve jumping from the middle cells of 11ac, 2dn, 12ac, 15dn to the central cell, but most also involve the same-value cells. Out of these four clues, only 11ac has a repeated digit, so 11ac must be x00 for x=B, C. Indeed, since 5dn is made of only odd digits, 11ac can be only B00. 5dn meanwhile is bigger than 16ac, less than E00, contains only odd digits and has a repeated digit. This means that it can be only DBB or DBD, but the first is ruled out because 13ac is also bigger than 16ac, making 5dn DBD.

5ac contains a repeated digit, but 7ac evidently starts with a number (contains only 0-9), whereas 6dn starts with a letter (greater than 16ac). Hence 5ac=DDx for even x=2,4,6,8. 13ac had only letters, but is also greater than 5dn, while 6dn is even, making 13ac=DCF or DEF. But DCF=3535 is divisible by 7, a property 13ac doesn’t have according to the list. Hence 6dn is D0E and 13ac is DEF.

16dn is a multiple of 6, hence it is even, and contains only letters without having a repeated digit. Hence 16dn ends in A or E. 17ac is odd so ends in B, D or F, making 16ac CBA, CBE, CDA, CDE, CFA or CFE. The last four are ruled out by noting that 16dn is

After the letters F to A have been removed. The 8 performed the final capture.

After the letters F to A have been removed. The 8 performed the final capture.

less than 16ac, while CBE=3262 is not divisible by 6, so than 16dn is CBA.

Just over half-way through the grid-fill now. Note that so far there has been very little actual arithmetic. A couple of divisions by 6 or 7, and noting that A, C, E are even numbers and B, D, F are odd, but that’s about it.

To continue… notice that 1ac is small, and starts with a small number therefore, but that 2dn and 3dn must both begin with B or C. Since 1ac has a repeated digit it must be xBB or xCC. But then 1ac is not odd, so it is xCC. 2dn is also less than 3dn, which is only possible if its second digit is 0. 9dn is sandwiched between 16dn and 16ac so must start with C, and as it cannot be CCC (no entry has the same three digits) it must be CBC.

8ac is a multiple of 6, so that its digits sum to a multiple of 3. This turns out to be possible only for 2CA, 5CA and 8CA. But 3dn is odd, so 8ac is 5CA.

At this point there are nine cells left to fill, but as we need at least one 1, and exactly one each of 2,3,4,6,7,8,9, we have eight of the nine cells almost sorted up to permutation.  This is part of what was meant by “paying attention to the removal instructions”.

After 9-6 have been removed, and showing the removal of the last 5 numbers. The D then jumps over the B to begin spelling out BECCAFICO.

After 9-6 have been removed, and showing the removal of the last 5 numbers. The D then jumps over the B to begin spelling out BECCAFICO.

7ac is less than 8ac while 5ac is even, hence 5ac becomes DD2 or DD4. But DD4 is divisible by 6, an option ruled out as 5ac was not listed as a multiple of 6, so it is DD2. 1ac therefore begins with 1 as it is less than 7dn and we can’t use another 2 in the grid. 18ac must begin with 4, as 14dn is even. 2dn is divisible by 7 and ends in 1 or 3. C03 isn’t divisible by 7, but C01 is. This means that there are two 1s (which was perhaps to be expected, as apparently in the final game the three words might use a 1, to be read as I), and the remaining empty cells are 3,6,7,8,9 in some order. Two out of 3, 7 and 9 will appear in 17ac (contains only odd digits),

and as 15dn is odd then in fact all three of the odd numbers are in the three empty cells at the bottom of the grid. Hence 4ac is 600 or 800 and 12ac is 6BF or 8BF, but 12ac is greater than 4ac so 4ac is 600 while 12ac is 8BF. 17ac is less than 12ac but greater than 8ac, so begins with 7. Therefore 15dn is E39 or E93. But E93 is divisible by seven! So 15dn is E39 and we have a completed final grid.

To play the game, following the instructions gives us little choice for the first few moves. The 0 must go first, then there’s only one way to remove an F; we need remove an E while making it possible to jump over a D next move.

Following the instructions leads to a workable first few moves. I’ll leave it to Dave Hennings, who might put together an animation of the game, to show how to play it, but I’ve included a few intermediate steps to the side.

Showing the final few moves. The D runs in a loop before being taken by the final 0, to give COB and DODO.

Showing the final few moves. The D runs in a loop before being taken by the final 0, to give COB and DODO.

 

So that marks the first time I’ve completed a game of Solitaire, I think! A very neat exercise in logic, so thanks Xanthippe.

With only five puzzles to go in the year, I wonder what is left in store for us…

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Listener No 4321: Solitaire II by Xanthippe

Posted by Dave Hennings on 12 December 2014

Well I was ready for the mathematical this week, and pleased to see Xanthippe in the chair again. Part of me thought that it would be Elap’s turn, and I was a bit nervous of what he would have on offer, but he would have to wait for another few months.

Listener 4321Here, we had a preamble that took up about twice the amount of space as the clues, and that couldn’t be good! We were to solve the clues, fill in the grid, and then play a game of solitaire. We were given hints as to what order the pegs would need removing: 0 then F down to 1 followed by three thematic words. I wondered if this game of solitaire would be needed at any point in the actual clue solving. In fact, I didn’t have to wait long to find out since the preamble told us that it would. (In hindsight, I think this was an unnecessary hint and should have been left for us to sort out.)

Anyway, the clues were of the type that I enjoy, whereby we were given a series of properties and told which clues fitted which property. Of course, this also told us which clues did not fit with any given property. This clue type certainly makes a pleasant change from the THIS * THAT = THE OTHER that is the norm. We were also given a list of the entries in ascending order. The entries themselves were in base-16.

There were a few properties which enabled me to pencil letters in the grid: only numeric digits were n, letter digits were l, only odd digits were o, as were the last digit of odd numbers. Multiples of 10 obviously had a 0 for their last digit. Care had to be taken with some of these properties as they were also in base-16, such that 64 (which is decimal 100) did not end in 00.

I liked the look of 14ac, 14dn and 15dn being greater than E00, and 16dn only letter digits. This made 14ac [EF][EF][ABCDEF], and since it didn’t have a repeating digit, it was either EF[ABCD] or FE[ABCD]. Moreover 16ac did have a repeated digit and was divisible by both 6 and 10. This meant that it had to be CC0. I was up and running.

From here on, it was a very enjoyable journey, deducing certain digits, eliminating others, and fitting the numbers into the ascending sequence that we had been given. Of course, I had the almost obligatory dead end… or at least I thought I did. My logic proved just a little faulty.

As is my wont, I’ll leave a full breakdown of the solving process to Shirley others (or, indeed, the Listener web site).

One of the features of the final grid was that, since we were told that O and I would be represented by 0 and 1 in spelling out the words, 2 to 9 could only appear once, namely in the initial countdown from F to 1. If S could represent a 5, then we would have been told. This enabled the entries at 17ac and 18ac to be resolved uniquely.

Thus I had a full grid in about three hours, and I was left wondering what three “thematic” words could be spelt out in the second half of the game. These words would have to consist only of the letters A–F, I and O, and the 9-letter word seemed as though it might be strange to say the least.

After a lot of umming and aahing (which spanned a couple of days), I wondered whether the words would be from one of the long lists in Mrs Bradford. I checked the 9-lettered fish… nothing. I also wondered whether the word ‘solitaire’ had an unusual meaning, but before I could reach for Chambers, I found myself checking Mrs B’s list of birds. BECCAFICO was there for the taking! Finally reaching for Chambers, I found that a solitaire was “an American or West Indian fly-catching thrush”.

Listener 4321 My EntryEliminating these letters from the grid, I was left with COB and DODO. However, it still took a bit of time to work out the exact sequence of the solitaire game. Commiserations to those solvers who got the mathematical bit, but failed with Xanthippe’s choice of birds. As with many Listener’s I had begun to think that the last step would stump me, but I got there in the end, I think.

Thanks for the enjoyment, Xanthippe.
 

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Listener 4320: Shrive by Llig

Posted by Jaguar on 5 December 2014

Only one week to go until the excitement of the year’s final numerical puzzle. In the meantime, Llig makes his first appearance in the Listener since 2011, apparently, although I’ve not met him before. Apparently Llig’s puzzles go back to before I was born!

Nothing too threatening up ahead, apparently, and the preamble strongly suggests a pangram with all 26 letters of the alphabet appearing in unchecked cells, in some pairs, and thirteen gimmicky clues hiding somewhere out there. So, to business!

It took only the third clue to reveal what might be going on, with the rather generous 11ac Force exercise in Egypt leading me straight to PE in Et. for E?PE?T, which I supposed could well be “Expect”. In which case the “pairs” would be something like AZ, BY, CX etc. So I suppose that’s the PDM sorted already. Still, got to fill the grid in now…

Not all that much to say, really, other than to assure Llig that I did enjoy this and appreciated the thought that had gone into it. Pangrams are hard enough to pull off as it it without the added challenge of including a set of words such as JacquardPrink and Humify. Can’t have been the easiest puzzle to put together, and I did like the cluing gimmick. Gentle but clever, and that’s just what the Listener Doctor ordered for a lazy Friday!

4320

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