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Listener 4242, Killer Queen by gwizardry: Solver’s Blog by Jaguar

Posted by Listen With Others on 7 June 2013

This blog provides my solution to 4242, and may deviate from the intended route. Towards the end of the initial solve things became very messy so at times I may have changed what I did to put things in a more logical order than actually happened, but then hopefully that makes it easier to follow.

Ground Work

It’s useful to explain the main techniques first, rather than as we go along:

  1. Unless it’s vital, until all or most cards are in place I found that it can save time to neglect trying to resolve suits. The only times it matters immediately are when one hand has to be a particular flush in order to win, or if other constraints also fix a card unambiguously quickly. In general it’s enough to know that there are only at most four of each card.

  2. Notation: squares are labelled by row letter then column letter, so that the top-left is AI and bottom-right cell is HN, etc.

  3. The preamble for this puzzle uses face value to mean two different things. Firstly, the face value can mean the numerical value of the card for this number puzzle, where Queens are counting as 0 and Aces as 1. Secondly, face value means the ordering of the cards as they appear in each hand. Cards are ranked, in descending order, as A, K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3, 2. This means that while the Queen and Ace have the lowest “number values”, they have high “ranking values”, so that a hand of a 2, Ace and Queen would appear as AQ2 not 2AQ. To avoid this confusion, this solution uses “face value” to mean the numerical values given in the puzzle; rank to mean the “ordering” of cards; and “strength” when comparing different hands such as flush and pair.

  4. We factorise all row and column numbers — which is the only real way to determine what card appears where.

  5. Finally, we will need to use the face values of all cards to factorise the entire used pack. As usual it’s best to do this in powers of prime numbers 2, 3, 5, 7, 11, 13. Note that an eight then counts as 2*2*2, so that an eight contributes three twos, and so on.

Rows:

A  114,075 = 3*3*3*5*5*13*13
B  7,168 = 2*2*2*2*2*2*2*2*2*2*7 (ten twos)
C  13,230 = 2*3*3*3*5*7*7
D  6,720 = 2*2*2*2*2*2*3*5*7 (six twos)
E  958,320 = 2*2*2*2*3*3*5*11*11*11
F  unknown
G  unknown
H  124,800 = 2*2*2*2*2*2*2*3*5*5*13 (seven twos)

Columns

I  5,738,733 = 3*3*7*7*7*11*13*13
J  unknown
K  156,816 = 2*2*2*2*3*3*3*3*11*11
L  416,000 = 2*2*2*2*2*2*2*2*5*5*5*13
M  unknown
N  230,400 = 2*2*2*2*2*2*2*2*2*2*3*3*5*5 (ten twos)

The pack, meanwhile, is missing an ace, a Queen, a 6 and a 2, but otherwise has the following:

three Aces (1)
three Queens (0)
four Kings (13 each)
four Jacks (11 each)
four 7s
thirty “2s” (one each from the three 2s, three 6es and four 10s, plus twelve from the 8s and eight from the 4s)
fifteen “3s” (four 3s, eight from the 9s and three from the 6es)
and eight “5s”, one each from the 5s and 10s.

After all this laborious counting we can start to solve the puzzle!

Solution

Step 1

Listener 4242 Jaguar Blog Fig 0First, notice that in rows B, C, D all of the four 7s appear. Moreover, three of these appear in column I, so that we can immediately put three 7s in squares BI, CI and DI. In particular, cell DI is the 7 of Clubs.

Now consider row D. Because it has six cards the same suit, all face values have to be different — and none of the cards can be an Ace, because this isn’t used in the main grid. The only way to split row D’s face-value product, 6,720 = 2*2*2*2*2*2*3*5*7, into six different cards without using an Ace is to use 2, 3, 4, 5, 7 and 8. Also, Player Two wins this hand. That would mean that he either has a higher-flush (8 of Clubs beats the 7) or a straight flush, which would be 4, 3, 2, say. This option, though, is ruled out because we also know that whenever possible cards are ordered in descending order of rank. The 7 comes first in Player One’s hand, so he cannot have the 8, so Player Two has it. Therefore cell DL is the 8 of Clubs.

Now turn to row B. This was given as 7,168= 7*(ten 2s), so that now we have accounted for the 7, the remaining five cards must be 2s, 4s, 8s or (possibly) Aces. Counting powers of two if we used only 4s and 2s shows that we would fall one power of 2 short, so that there is at least one 8 in row B. Player Two must have this, for the only way Player One could have the 8 is if he had a pair of 7s (such a hand would be entered 7 7 8 in the grid). This is not possible — not enough factors of 7 in row B to use — so Player Two has the 8. How then can Player One win? He must have a flush, all Spades, so that his hand is 7S, 4S, 2S, as the remaining cards must be powers of 2 that are less than 7. Anything else would lose or would not fit — so we have Player One’s hand in row B. The 7 in CI must be the 7 of Hearts, as column I contains no Diamonds.

The next place to turn our attention to is row E. This has the factorisation:

958,320 = 2*2*2*2*3*3*5*11*11*11

Which tells us that row E has three Jacks and three other cards. The only way to absorb all of 2*2*2*2*3*3*5 into just three cards is to use 10, 9, 8. We know that columns L and N contain no Jacks so that if Player Two has any Jacks he has one in Column N. But this would result either in a pair of Jacks (imposibel) or hand that is not in descending order of rank. So Player One holds all three Jacks. How then can he lose? Note the rules that tell us that three of a kind beats a straight, but that a “straight flush” beats three of a kind. So Player Two must hold the 10, 9, 8 of the same suit — specifically, Diamonds. The remaining three Jacks are the JD, JH and JS, but we don’t yet know their order.

Step 2

Listener 4242 Jaguar Blog Fig 1The final Jack is hiding somewhere in column K, so must be in either row F or row G. What do we know about these hidden rows? Firstly, the fact that row G is a tie means that the two hands must match exactly. In particular, if there is a Jack in GK, then there would have to be a second Jack appearing in GN. This is not the case, as there is only one Jack unaccounted for. So the final Jack (which is the Jack of Clubs) is in FK.

We can also start to place the Queens. They are hiding in rows F and G, and Columns J and M. That places all Queens in three of the four cells FJ, GJ, FM, GM. In particular there must be at least one Queen in row G. And, because the hands tie, there must be two Queens in row G — specifically, in GJ and GM. What cards can go to the left of these Queens? The only possibilities, bearing in mind that hands are entered in descending order of rank or as pairs first, are:

AQx ; KQx; QQx — where x is the third (unknown) card. We discount the two Queens possibility as that would mean column I would also be “killed”. Also, there can only be one King in row G as there are two in row A and one in row H, so that only one King is missing. So the only option is AQx, so that cells GI and GL contain Aces.

There are three cards left to place in column I. These must include two Kings (the 13’s) and a 9 from the factorisation. The 9 counts as two 3s, though, and row H can only contain a single 3, so HI can’t contain the 9 and must have a King.

Returning to the hidden rows, note that in particular exactly six Clubs are hidden in these rows — only row D contains Clubs, and the Ace is not used — and these six are the K, Q, J, 10, 9, 6. That King is the only King in these two rows, so it cannot be in a pair, nor can it be in row G as it should appear to the left of a Queen. So the King of Clubs is in either cell FI, or FL (not FN as column N has no Kings). Also missing in the hidden rows is precisely one of the eight “5s” in the pack. That stops the 10 from appearing in row G, as this would force the other empty cell in row G to have a 10 so that the hands tie. Nor can it appear in FI or FJ, as that would mean a pair of 10s (the hand would be 10, 10, J). So the 10 appears in FL, FM or FN. Finally, there are also only two missing “2s” in rows F and G (you can count 28 2s in all other rows, out of 30). So the 6 cannot appear in row G either. The only place for the 6 of Clubs is then cell FN.

Recalling that cell FI is either a 9 or a King, we can try placing a 9 in FI, but find that this cannot work as then the only possible fit is 99JKQ6 with nowhere for the ten to go, or 99JK106 when the remaining Queen can’t fit anywhere either. So cell FI is the King of Clubs, and FJ must be the final Queen, so that FL and FM are the 10C and 9C. But then the only way for Player One to win in row F is if the Queen were the Queen of Clubs — “straight flush” beats “normal flush”, but “flush beats straight”. So row F is KC QC JC 10C 9C 6C. Finally, there are two missing factors of 3 in the hidden rows, so that the final two cells in row G are both 3s; and cell AI is the missing 9.

Step 3

Listener 4242 Jaguar Blog Fig 2The remaining two Kings appear in row A. With Player One holding a 9 as his first card, it must be Player Two that has the pair of Kings. That accounts for all but the factors 5*5*3 in row A; splitting these between three cards can only be done using two 5s and a 3. Again, Player One can’t hold the pair of 5s as these would appear on the left of his hand, so Player Two has one 5 and Player One the other 5 and 3.

At first glance Player One now has some junk — but we are told he wins this hand over a pair of Kings. He does not have a straight; he must have a flush, of three Hearts. (Player One’s having a lot of luck so far — two flushes to open and a straight flush in hand F!)

We’re down now to just thirteen totally empty cells. Let’s turn to row B. We know that the remaining cards contain seven factors of 2. This can be done by 448 or 882. But we can be even more specific than this, as the only 2 unaccounted for is the 2 of Diamonds — we haven’t fixed the 2 of Clubs yet, but it’s in row D somewhere, and the 2 of Hearts is unused. The two 8’s missing are the 8 of Hearts and 8 of Spades. Suppose we did finish off row B as 882. Then look at column L. On putting an 8 in BL the only missing factor in row L is 5, so that the last two cards would be a 5 and an Ace, one of which goes in HL.

Note that Player One has to win in row H. But his hand is heavily constrained already by the lack of available cards. The only hands better than “King high” are all ruled out:

i) A pair of Kings means five Kings appearing in the grid;
ii) A straight would mean KQJ using a fifth Jack (and also “killing” information about row H due to the Queen);
iii) A flush is impossible as there are only two cards in any one suit appearing in row H.

Therefore Player One has only got King high. Any hand with an Ace automatically beats this. If cell HL had a five instead, then possible hands are:

55x ; 543 ; 542 ; 532.

55x and 543 both would beat King high, while 542 and 532 both use a fourth 2 when there are only three in the grid — one appearing in BK, one in BN and the third in row D somewhere. So cell HL can have neither an Ace nor a 5, so cell BL cannot have an 8. Therefore Player Two’s hand in row B is 448, with the 8 being the 8 of Spades as column N has no Hearts. Now cell HL can be either an Ace (ruled out already as above), or a 2, or a 10. If it were a 2, then Player Two’s hand in row H could only be 22x which again beats King high. So HL is a 10, and CL is the final Ace.

Step 4

Listener 4242 Jaguar Blog Fig 3Now’s the time when it definitely helps to have a pack of cards handy. Laying aside all the cards that are accounted for, we can see that the eight cards remaining are, ignoring suits:

10, 9, 8, 7, 6, 6, 5, 2.

While we’ve not yet placed them, the 2, 3, 4, 5 of Clubs all go in row D, and there are still suits to be sorted out for a few cards yet.

Return to row H. The factorisation was:

124,800 = 2*2*2*2*2*2*2*3*5*5*13

We have so far accounted for the 13 and 5*2 = 10, leaving six 2s , a 5 and a 3. That factor of 3 has to appear as a 6 because all of the 3s are accounted for. This leaves five 2s and a 5 in three cards. The only way we can do this using the remaining card values is by using 10, 8, 2 (8, 5, 4 would use a fifth 4). The 10 cannot appear in HM, as this would give Player Two a winning pair, so Player One has that 10 in cell HJ.

Meanwhile row C contains the 9, 7, other 6 and the 5. Player One wins this hand too, and can do so with either the hand 776 or the 765 straight. In this second case, that would give Player Two a hand of A97. But row N does not contain a factor 7, so we exclude this, giving Player One 776 and Player Two A95.

To complete the number fill, we start with column K. Dividing its product 156,816 by the face values of all cards in column K leaves 156,816/(11*11*2*3*3*6) = 12. So the remaining cards are either a 6 and a 2 or a 4 and a 3. Cell HK can only contain 8, 6, or 2. It cannot be 8 (which now goes in HM) but cannot be 2 either, as DK would then be the already-used 6 of Clubs. So HK is a 6 and DK is the 2 of Clubs.

This leaves a 2 to go in HN. Now looking at column N we see that the remaining card must be 230,400/(5*8*5*8*6*3*2) = 4, or the 4 of Clubs. So DM is the 5 of Clubs, and DJ is the 3, meaning that all card ranks are known for each cell.

Step 5

Listener 4242 Jaguar Blog Fig 4

All that remains is to determine the suits of each card. For the most part this is obvious as several times only one suit is left for that number. The information given quickly resolves the remaining ambiguities. If Player Two holds all three Spades that appear in row C, for example, then he would have a winning flush, but Player One wins, so that CK is the 6 of Spades not the 6 of Diamonds; the 6 of Diamonds is then in HK, which is the only Diamond appearing in columns K or I — making EJ the Jack of Diamonds. Column K also contains two Hearts which cannot both be 3s, so that GK is the 3 of Spades, EK the Jack of Hearts and EI the Jack of Spades.

We can also use simple counting. Allowing for the unused cards — a Club, two Hearts and a Spade — then we count twelve Spades appearing in rows A to E and row H. The thirteenth Spade, we just saw, was the 3, so that no more Spades can appear in row G. This fixes CL as the Ace of Spades — making CN the 5 of Diamonds, in turn giving AN and HI as Spades and GI as the Ace of Hearts and GL as the Ace of Diamonds. Then we see that the remaining cards in column L (AL, BL, HL) are all Hearts.

 
Finishing off gives us:

Listener 4242 Jaguar Blog Fig 5What of the two Queens? These can be suited, because they have to be a Heart and a Diamond, but if GM held the Queen of Diamonds then that would give Player Two a winning flush. So he has the Heart and Player One the Diamond Queen.

We can complete the puzzle by reading off the coordinates of the cards:

KC, QH, 7H, 5S, 3H, 2H.

Note that the Two of Hearts is given as in the “Outstanding Deck” = OD. The other five cards are seen to be in FI, GM, CI, AN, AK respectively, giving FIGMCIANAKOD or, rearranging, “A Kind of Magic”, which, along with “Killer Queen”, is a song by the band Queen.

 

Final solution:

Listener 4242 Jaguar Blog Fig 6

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One Response to “Listener 4242, Killer Queen by gwizardry: Solver’s Blog by Jaguar”

  1. […] with a worked solution for last year’s card puzzle by gwizardry (Killer Queen, Listener 4242) and then a couple of weeks later I covered for Shirley when her own fine puzzle At Spes Non […]

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