Adam R. Wood (zotmeister) wrote,
Adam R. Wood

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Puzzle 52: Heartbreaker

"I couldn't follow suit," Tom said brokenheartedly.

- Adam R. Wood, original Tom Swifty

I love playing cards. I love collecting them; I love handling them; I love playing games with them; I love giving (and receiving) decks as gifts. I had to make Puzzle 52 based on playing cards. The problem was that there weren't any appropriate types of playing-card-based puzzles that I was aware of and liked... so I invented my own. Took awhile, but I'm quite happy with what resulted. Inspired by various card games and based on a fairly awful pun, I submit for your entertainment Heartbreaker:

To the left is an unsolved sample puzzle; to the right is the same puzzle with all the fun pre-removed. In addition to the solution (in blue), I also highlighted (in green) the heartbreaker in order to make explaining the rules (and the sample solution) that much easier.

The concise instructions: ...actually, I'm not going to bother this time - this is one of those puzzles, the kind that are relatively simple in execution but for which a rigorous explanation of the rules is, well, a real heartbreaker. Hopefully this will do:

  1. Each cell of the grid is to hold a playing card; the deck in use will be defined in the text accompanying the puzzle. Each card of the deck is to be placed in exactly one cell; the object of the puzzle is to determine the position for every card in the deck.

  2. Some cells may have a rank, or suit, or both depicted in it; the card placed in such a cell must match what's already there.

  3. Loops (in the style of Perfect Fit) may appear around some cells. I call these 'cardtouches', as Egyptologists could use a clever pun after all the obvious 'mummy' jokes and I'm going for broke on pun overload already anyhow. The cards to be placed into the cells inside a cardtouche must make a valid Rummy meld - that is to say, they must either be (a) [a 'set'] all of matching rank, but each a unique suit; or (b) [a 'run'] all of matching suit, but in sequential ranks (although the physical positioning of the cards doesn't have to match that sequence). Remember that aces are only sequential to deuces for this purpose - aces are always LOW in Rummy!

  4. Some rows and/or columns may have "signs" (a labeled arrow) pointing at them. The labels are all Poker hand ranks; the labels for the standard ranks, from weakest to strongest, are: N, 1P, 2P, 3K, S, F, FH, 4K, SF. [I'm not beyond using freak hands if they will make for an interesting puzzle; I will define any label other than these in the text accompanying a puzzle that uses one.] The strongest Poker hand that could be made with any combination of up to five cards from the row/column a sign points to must match that sign's label.

  5. Every Hearts penalty card - cards of the heart suit and Queens of Spades - must be part of a snake. That's a technical term puzzle veterans will recognize: the cells of a snake must form a chain, each orthogonally adjacent to the next. Snakes cannot touch themselves - that is to say, cells in the snake can only be orthogonally adjacent if they are consecutive in the chain, and can only be diagonally adjacent if they are two links in the chain apart. [Alternatively, they cannot comprise any two-by-two blocks of cells, nor can they enclose an area on all sides (that is, not using a grid edge to do so) leaving that area orthogonally disconnected from the rest of the grid.] If multiple snakes are to appear (as would be the case if a double deck were prescribed), they can't touch each other, either - cells belonging to different snakes cannot be adjacent, not even diagonally.

  6. Each snake must have exactly one of every rank of heart, which must all be physically in order in the snake, from ace down to whatever the lowest-ranked heart in the deck is. Remember - Ace is always HIGH in Hearts!

  7. Each snake must also have exactly one Queen of Spades as well. However, it can appear ANYWHERE in the chain - before the ace, after the lowest card, or anywhere in between, interrupting the otherwise perfect sequence of hearts. You guessed it - just as some play the card game, the Queen of Spades breaks hearts. I told you it was awful.

The deck for the sample puzzle is a mere twenty cards, just Ace through Ten in each of the four standard suits. This means it must have exactly one snake of exactly six cards, basically a heart royal flush in order with a Queen of Spades tagging along or butting in. Look at those six cards in the solution above and you'll find they're wrapped around the top of the central gap; the Spade Queen is between the Heart Jack and Ten, but the hearts are otherwise in order - Ace, King, Queen, Jack, Ten. This is what needs to happen.

That '3F' is for "three-card flush" - that is, the three cards in that column need to be the same suit. (Technically, they can't be sequential ranks either, as then it would be a three-card straight flush, but that isn't important for this puzzle.) It's actually a fairly illuminating sample puzzle - one of my better ones in my opinion - and should prove an interesting solve if you want to tackle it yourself. Of course, if you would just like a walkthrough - and after that rules definition, I wouldn't blame you - that can be arranged:

(Start of sample puzzle solution)
The Ace of Hearts is given, so let's start there. As the highest Heart in the snake, it needs to have either the Queen of Spades or the King of Hearts (or both) next to it. Neither can be below it - there's a club there - so it will have to be in the diagonal cardtouche that includes both the cell above it and the cell to its left. This cardtouche also has the King of Diamonds in it already. We can't put the Queen of Spades in there with it, since it doesn't match suit OR rank, and all cards in a cardtouche will need to have one or the other in common (a very useful thing to remember). Therefore, the King of Hearts will have to go in to extend the snake.

But which way - left or up? Remember the snake can't touch itself. If we go left, then we'll need to go either up or down from there, both of which are traps. Even if that club weren't there, down would be just as bad as up - we'll be forced to step back right next, making the fourth card in the snake touch the first (the Ace) from above or below. [Alternatively, those four cards would make a two-by-two square.] So the King of Hearts needs to be above the Ace.

We now have two kings in that cardtouche. This tells us it must be a set (matching ranks) as opposed to a run (matching suits), so that third card (to the left of the Ace of Hearts) must also be a king. We don't know if it's the Club or Spade King yet, but we don't need to know for the next step: the other cardtouche the King of Diamonds is in, the horizontal one, can no longer be a set. If it were, we'd need five kings in the deck, and there are only four! So it must be a run, its two other cards being the Queen and Jack of Diamonds in some order. (Remember, the Ace is low in Rummy - Ace, King, Queen is NOT a run! Also remember the cards don't have to be in order inside the loop, so having the King in the center like that is still fine for a run.)

Let us now return to the snake. We're at an interesting point: the next card in the snake must be a queen - we just don't know if it's hearts or spades! What we do know is where it has to be, and that is to the right of the King of Hearts (left of course is a dead end). What happens next is tricky but key, so follow closely: we now have partial information on two cards in the four-card cardtouche. One is a diamond, and either a jack or queen; another is a queen, and either a heart or spade. Those two cards need to be in a set or run together, so either their ranks must match or their suits must match. The latter is impossible, so they must both be queens, and the entire cardtouche must be a set of queens. This means every queen in the grid must be in that cardtouche! If the Queen of Spades were the card to the right of the King of Hearts, the Queen of Hearts would need to be next in the snake... but that would be impossible, since the snake would have to step diagonally to get to it! So the Queen of Hearts is to the right of the King, and with the knowledge that the heartbreaker can't be outside that loop, we can place the Jack of Hearts to the right of that.

You may want to get an actual deck of cards at this point if you haven't already - it can help keep track of what cards you've placed and which you still have left. For example, we can resolve the three-of-a-kind the second row needs now. We have a king, a queen, and a jack in that row already - all different ranks - so the last two cards in that row need to be a pair, matching one of those three ranks. But we're out of queens, and we've only one king left, so they'll have to be jacks. This results in all four jacks now being assigned, although between the Club and Spade Jacks we still don't know which is which.

Time to finish the snake. The Jack of Hearts is now surrounded on all four sides, none of which can be the Ten of Hearts, so that Queen below it will have to be the Queen of Spades, giving us the sequence break we need. That last Queen at the bottom right has to be the Queen of Clubs - it's the only suit left - which means the Ten of Hearts can't be to the right of the Queen of Spades (we'd again have two cards in the same 'touche with different ranks AND different suits, always a no-no). It'll have to be below it instead, and with that we've placed all the hearts.

From here, we just need to mop up. That last heart landed in a cardtouche, so that clearly can't be a run anymore - we're out of that suit! (Not to mention that a diagonal cardtouche couldn't be a run in Hearts anyway, unless the rules of Rummy were changed to allow two-card runs...) Placing tens into the other two cells in that loop, that vertical cardtouche is now locked into a club run, Q-J-10. That means the Spade Jack is the one in the leftmost column, and in filling in that three-card flush we finally know it's the Spade King and not the Club King in that first loop! To complete the two pair in the fourth column, that diamond at the bottom will have to be the last ten, forcing the ten to its left to be a spade. We now have only four cards left - three aces and a king - and a cardtouche still untouched. We don't even have the cards for ANY run left, so the aces will have to form a set in there. Lastly, the leftmost cell in the bottom row has to be the King of Clubs, by the virtue of it being the only card left in the deck! We're finished.
(End of sample puzzle solution)

This first full-size Heartbreaker uses a standard 52-card poker deck. In some ways it's actually easier than the sample - I tried to keep it cute as opposed to educational - but I did manage to highlight a few curiosities. Certain possibilities came to mind as I composed it, so I'll have some tricks up my sleeve - uh, no pun intended that time - for the next one. Actually, maybe the one after next, since I'm probably using a Pinochle deck next time... - ZM

Tags: breaker, puzzles
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