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Posts Tagged ‘Three card brag’

Listener 4242 Killer Queen: A setter’s blog by gwizardry

Posted by gwizardry on 8 Jun 2013

As much as writing this is a pleasure in revisiting the subject following the thrill of the puzzle reaching publication, rather unfortunately I hadn’t at all thought about doing this at the time of creating the puzzle, and as such, much of the thought that went into it, as well as the luck (put that down to hard work, time and the fruitful results of trial and error if you will), has been forgotten.

So where to begin. Well a little about my solving history. I only came to the Listener on discovering Pentominos by Oyler, which in turn only came about from starting to read the Times on the basis of an excellent tongue-in-cheek opinion column on skiing from Giles Coren. Pentominos remains my favourite numerical Listener puzzle to date for its simple elegance. Since, I have at least completed the numericals that have followed and quite a few of the more interesting ones from the archives. On the literary side I fare rather less well, a half-cheated completion of Safe-Cracking by Kea is my best yet (stunning puzzle nonetheless). Still I’m making progress, so hopefully will be able to achieve a submission outside the 4 times per year I could currently manage.

So as many of you with a good memory will note, Pentominos shares the core mechanism with Killer Queen in that it is based on the unique factorisation of a number of products, as well as the use of the ‘0’ multiplier. However, that’s rather where the similarities end. I was browsing a couple of the excellent series on recreational mathematics by Martin Gardner which I had hoped to work into a puzzle, but had also been playing around with the idea of something based on games (games are covered a fair bit in those books). I don’t play cards much now, but I have a fair bit in the past and so somewhat accidently I came up with the idea of hands of cards which have to be deduced. It was a simple matter to work out a game which can meet the A*B<52 requirement (its nicer to use something closer to a full pack i thought) – so two players with 3 cards and we have a 6 by 8 grid. Incidently, although I myself had played ‘Brag’ (the three card variant of the poker class), I felt it might appear a little more accessible to readers of the preamble if called ‘poker’ as it is. This seems to have inevitably led to a mistake on my part, as a couple of solvers have pointed out to me – straights rank above flushes in brag, if not in 5 card variants I am more recently accustomed to – thank goodness there was no round with a straight vs a flush or it could have been catastrophic!

Anyway, back to the process, after selecting the game dynamic of 3-card Brag, I began to work around the idea of ‘a club sandwich’ either being the title or solution. incidentally that’s why the clubs appear on the same rows. After the first draft I had thought about reworking them to avoid this strange anomaly, but mostly laziness and lack of time but also an acknowledgment that the 1st draft wasn’t so bad after all kept me from doing so.

So as for the construction process. As Queen would be ranked 12 following a standard ordering, it was initially fairly clear that using it as a 12 would be heavy due to it having 2*2*3 as factors. I guessed this would make the solution either very easy or tedious (even more tedious than it is already!). So as J and K naturally fitted primes in 11 and 13, it seemed a good idea to use Q=0. Almost immediately the line She’s A Killer Queen hit me (or at least I could hear the song in my head and thought it fit well). At this point I dropped the club sandwich and thought of having a title answer which matched the Killer Queen theme. From this, the only way to work in an anagram was to use coordinates. I let this rest for a little, before working out a few of the hands roughly with the principle of using as many different types of deduction to be included in the solution. Such logical steps could use the following information:

  1. Remaining cards in the deck
  2. Remaining cards in each suit
  3. Remaining cards of certain face values
  4. Possible face value card positions based on the factors
  5. Possible suit positions based on suits listed
  6. Possible hands which fit the standard ordering rules
  7. Possible hands which fit the ‘winner’ result
  8. Possible entries which don’t invalidate what remains in other rows/columns (cross-matching)

The idea was that solvers would have to use different notions in the solution based on different combinations of information.

Early on I decided on the location of the 3 queens. This was necessary to avoid too many 0s and therefore a lack of information for solving. I didn’t want to use 4 as I wanted them to be different types of round and hands in each and also to save some of the Kings and Aces. I also had the loose idea of finding the other Queen (a hidden Queen of Hearts…), although nothing came of that.

I then proceeded to fill the face values for some of the rows (E, F and G particularly) though I forget the precise order. It was then a long process of trial and error and a strict checking of the logic to avoid missing any shortcuts. I didn’t mind the possibility of solvers being able to guess, as long as it was a precarious approach. I just had to make sure there was a single solution (this follows from the logical deduction…if you follow the logic correctly at each stage then there’s no risk in there being an alternative unless you’ve ‘guessed’ at some point).

So I’ll now review the first phase of solving (which matches well with this first phase of setting) [disclaimer….diagrams may contain errors as they were drawn up in haste for this blog and not during the setting process]

Solution 1st Phase


Keep a track of

a)      52 cards in the deck

b)      Number of Suit

c)       Number each Value

Factorise the rows and columns.

1)      GJ and GM are Q’s

2)      FJ or FM is a Q

3)      3 K’s else where so GI and GL are Aces

4)      E must contain JJJ1098 and Jacks are DHS so 10D 9D and 8D given 11s in columns I and K so  J J J 10 9 8 is hand order

5)      Row D are all clubs so must be 8 7 5 4 3 2 and so 8C in DL and so DI must be 7C

6)      3 7s remain all in rows B+C, so 2 must be in column I and as remaining factors

7)      Remaining J must be in FK and must be JC

8)      A must contain KK9553 and there must be KH, 9H, 5H, 3H

9)      AN must be 5 as no H and no K, and AK must be 3H so AJ is 5 and AL is K

10)   H contains K and no A’s as too many factors and so can’t be in middle columns, must be in HI and considering cross of suits in HI is H or S but must be S as H in and AN must be 5S


At this point the solver may reach somewhat of an impasse, in that a slight change in tack is required to continue the solving process. Although this pause does not coincide exactly with the second phase of setting there are key features which I fixed at an early stage. Whereas the the first phase used distinctive ideas for form the basis for the rounds (such as the location of the Jacks, Queens, Kings and 7s), the next phase of setting was more trial and error based. The solver must now consider in greater depth the possible combinations of each hand, particularly for rows B, C and D. The setting process involved continually taking a step back each time an element was added to ensure the placement of a new card didn’t reveal something to make things easier earlier on. I realised that if I could manage this and obtain a unique solution then the puzzle would remain challenging from start to finish. This isn’t always easy to achieve with the numericals, not due to any particular fault in the construction but simply as an artefact of the mechanism. So I was glad it seemed this puzzle would have a fairly constant or even increasing difficulty curve throughout.

The position of the remaining Ace was key here (the 4th being one of the 4 not played, in order to help maintain a viable solution) as considering it as a possible card greatly expands the combinations, this was also one reason to put 2 in front of the Queens in row (as 3 Aces in the other hands might have made it intractable).

Let’s now go through the second phase of the solution.

Solution 2nd Phase

11)   Row possibilities are 7-7-A-10-9-3, 7-7-9-A-10-3, 7-7-3-A-10-9, 7-7-9-A-6-5, 7-7-6-A-9-5 so 7 in column J

12)   Row includes either 8 8 4 2 2 or 8 4 4 4 2 so (as two 8’s placed and one 4 in row D) then this leaves one 8 and zero 4s or zero 8s and two 4s…and 7 is 7S

13)   Row must include two 10s and 6 otherwise need an 8 and a 4

14)   All remaining 2s 4s and 8s now fixed to B D H so has none, also only 10 left is 10C (others in which can’t be in C so 2 is in a 6, remaining values giving A 9 7 7 6 5 and 5 must be in CN and must be 5D (D or S but 5S taken) and AS (7s D+H taken) must be in CL (7s different suit)

15)   The 10s in H must be in J and L such that P1 wins (10H in L)

16)   FL must be 10

17)   BL BM must be 44 and then must be 8-2

18)   As in  must be H and D and so KH and 4H in column and also completes (no D for the K) and KC in FI

19)   7H JD also in I, 7D in CJ and 4D in BM

20)   Remaining and are spades


We’re now just a few elements from filling in the grid. There appeared to be some question over the uniqueness of the solution from some members of Answerbank, but as we’ll see this can be resolved using the right logic. Incidentally, this was a fault in the first draft, noted by Shane the co-editor. In order to resolve this it was necessary to move around the 6s, 3s and 2s to work in the possibility of a flush, only through avoiding this flush and therefore maintaining the ‘split’ round would the full solution be possible – as shown below.

Solution Phase 3

21)   Remaining Q must be in J

22)   GK must be S and only possibility is 3S (8S 4S 2S in B and 9S 6S in C

23)   3D in GN as 3C is in D

24)   Row  must be 742 448 for ordering

25)   8H not possible in column K so must be in and HK is D

26)   Remaining suits except Qs in and are now fixed

27)   Remaining clubs are 9 and 6 go into F

28)   No 3 factors in which fixes row and row D

29)   Row is then fixed

30)   Finally, QD must be in J to keep a split (otherwise flush)

Giving the coordinates AE GM CI AN AK OD resolving to A KIND OF MAGIC

Having completed the grid, it remains to work out the 4 word title. I was fairly sure that most could extract the ‘A _ _ _ _ of _ _ _ _ _’ or perhaps ‘A _ _ _ _ in _ _ _ _ _’ or ‘A _ _ _ _ if _ _ _ _ _’ and that seasoned solvers of the literery listeners would have no trouble at all.

In fact, working the title ‘A Kind of Magic’ into the puzzle had turned out to be quite troublesome. I had thought the song was called ‘It’s a Kind of Magic’ which had put me off it. Its of an odd number of letters and had a T and O so not fitting a simple way of defining the coordinates. This was a big disappointment as I liked how it related to my setter’s pseudonym (pronounced gee – wizardry as in ‘gwhizz’ and thus could equally be defined as ‘a kind of magic’), so I began to look at a list of Queen songs and albums but much to my delight spotted it was simply ‘A Kind of Magic’. All that remained was to arrange the rows and columns to fit the letters required. I played with quite a few names for the pack (yes pack or deck may be used depending on which side of the atlantic you’re from), in order to give the OD required.

I hope that the blog is of interest, though I fear it somewhat bursts the magic a little in that there was no ultra clever reverse logical engineering. Rather a classic exercise in the pragmatism of logical trial and error.


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