Game on (13 June 2021)

In the last week I asked the following question: how many rows do we need to deal from the stock to be sure of procuring an empty column (assuming the worst possible permutation of unseen cards)?

First let us clear up the Captain-Obvious stuff: Column 2 is the only column with no unknown cards so we must focus on that. Also, there is not a single Nine anywhere until the second deal so the answer must be at least 2. By that time, two Sixes will appear in Column 2 so we need to find enough Sevens to take care of these Sixes.

Schistocerca Americana gave a correct answer of three rows. I say correct because the Grand Faster mucked up by not asking for the minimum number of deals. I should have asked what is the minimum number of rows we need to deal from the stock to be sure of procuring an empty column?

Bart gave another correct answer of two rows. Starting from the game state from last time:

The following moves do the trick:

Before deal: dj,aj
After deal 1 (J6254A388K) : db,ad,ba,ge
After deal 2 (562259AJ8Q): gc,bg,bf

Note that the first move dj is a typical tesuji (link) when playing with undo. This can only be explained by prior knowledge of cards in the stock – it is inconceivable an expert player can find some miniscule advantage of dj over “doing nothing” if playing without undo. Also observe that we got lucky with ba after deal 1: the Five and Six are the same suit, hence the move is indeed legal.

Of course, it will be desirable to achieve an empty column without dealing any rows from the stock. We can guarantee at least three turnovers in columns 1,6 and 7. On a good day, we will get an empty column without any of the shenanigans described above. The worst-case scenario says we are forced to deal two rows, take the empty column and proceed from there.

Our luck is in: the final hidden card is the Queen of Spades which can immediately go onto the King of Hearts in column 5. So now we know it is possible to get an empty column without dealing any of the shenanigans described above.

Our cheat sheet now looks like the following:

The power of an empty column should be pretty clear. For most of the columns it is easy to determine the next face-down card, then undo to recover the empty column.

It is time for a new question: Is it possible to determine the identity of the next unknown card for all columns containing at least one face-down card?

Assume we are allowed to restart from the very beginning, but cannot deal any cards from the stock.

Thanks to Bart Wright and Schistocerca Americana for once again reminding me of my lack of cultural knowledge (e.g. Kung Fu). There is only so much one can do with my favourite animal types from Phil Hellmuth’s book Play Poker Like The Pros 😊

4 thoughts on “Game on (13 June 2021)

    1. Correct. No doubt you already figured out rot13(gur Guerr bs Pyhof pna arire or fuvsgrq fvapr gur Gjb bs Qvnzbaqf zhfg or zbirq bagb gur Guerr bs Pyhof gb trg gur rzcgl pbyhza naq gurer ner ab hfrshy Guerrf be Sbhef naljurer ryfr)

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      1. I answered the question, right? But I was just having fun. You know I’m never terse or succinct. Here are my footnotes.
        —–
        Can we determine the identity of the next unknown card in each column? First, the easy cases (being sure to be specific and detailed).

        Columns 2 and 7 have no face-down cards and so are not part of the problem.

        Column 4. From final tableau, di,dj,dg,ed,ge “lemma”, creating happy in-suit king-queen pairs in columns d and e. Here we send the spade couple to space on vacation and win.
        5. Invoke the lemma and then send the hearts to space on vacation to win.
        6. Trivial fg
        9. ij,ig
        10. jg

        The remaining columns to work on are 1, 3, and 8. I cannot see how to do any of them. Starting out of order with column 3 helps a bit with the exposition.

        3. We have no receiving 9s, so the only way to get rid of that 8 is to put it in the space. The queen is easy to move — uncover the king in column 9 by moving the jack. What about the ace? I believe the ace cannot be moved while still making the space as done in the diagram. The only 2 is in column 7, That 2 is occupied by the ace from column 2. We had to find a home for that ace in order to expose the 8 at the bottom of the pile, which we need to be able to make our space. There is another way to make the space without digging in column 2 to find the 8 — we will find it in column 3 instead. This way of making the space is several moves shorter. Yet, looking in the cheat sheet, we see that in both columns 2 and 3 you have to move an ace and a queen to expose the 8, though in opposite order. So we can do that. But now we have a space, and an exposed 78 in column 3. If they were in-suit we could just put them in the space… but they are not, so there is no way to to get that 8 into the space. I have run out of things to try for exposing column 3.

        1. It’s easy to get that ten onto a jack. We then need homes for two 7s. One can be put in the space, but how to get rid of the other 7? We need an 8. There are 8s in column 2, 3, and 6. The 8 in column 6 is useless (it is “chemically inert” since the 7 on it needs a home. Any 8 we put it on means we have one fewer 8 in the total count. Plus one, followed by minus one equals zero.) We are counting on the empty column as home for one of our 7s, but that requires finding a home for the 7 that starts out in that column. As the board has developed, this was done by using the 8 in column 2 to absorb it. The only remaining 8 is in column 3. Can we access it and still have our vacant column? No. But in looking at column 3 I found another way to make the space by clearing out that column-3 8 rather than the one in column 2. It can be done, but the 7 from column 7 is sitting on that 8 and can’t receive a 7 from column 1. Looking just at the “home for 7s” problem, we could do it without every making a space, by putting those two 7s onto the 8s in column 2 and 3. The problem (or at least one problem) is that both columns have an ace that needs to be cleared, and the only receiving 2 is in column 7. Can’t move both of them. I think we’re out of things to try.

        8. There are no 4s anywhere, so the only way to move that 3 is into the space. But making that space requires moving a 2, and that is in fact the only 3 available. We can end up as in the final position shown in the blog entry. We have a 3/2 off-suit and no way to get the 3 into the space.

        Representatives of the 10 columns meet in a legislative session of the Republic of Spider Sophistry. The representatives of columns 2 and 7 are dismissed as not being part of the defining group, “columns containing at least one face-down card.” The question on the table is, “Can the player expose the next unknown card of each pile, with some line of play?” They decide to divide up the work so each representative finds out the answer for their column. When they are all done, 4, 5, 6, 9, and 10 are in favor (five votes) and 1, 3, and 8 are opposed (three votes). So the vote is 5 to 3 in favor, so the motion passes.

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  1. Master T and Esteemed Scholar B I hope that brevity is the virtue of the day as I don’t seem to have lots of time to do my normal twisted prose. People ask me, “ ’Hopper, I know you are retired, what do you do all day?” And my reply, “I don’t have the vaguest idea what I do all day, but it always takes me all day to do it”.

    I will not use my AncientExcel CheeterSheeter to move cards around, just to set up the opening position.

    Today’s assignment:
    “Is it possible to determine the identity of the next unknown card for all columns containing at least one face-down card?
    Assume we are allowed to restart from the very beginning, but cannot deal any cards from the stock.”

    No

    To turn over the card above the two 7’s in Col 1, we need a Jack, easy enough. We also need two 8’s. We have three 8’s. The 8 in Col 6 is already covered with a seven so is of no use here. In order to get to the other two 8’s we need to clear two Aces. but we only have one 2. Therefore we need to have a void for one of the Aces. We cannot create a void Col 2 because we have no 9. Col 7 also needs an 8 to clear its 7 in order to transform itself into a void and so we need three three usable 8’s and we only have 2 available.

    I do not believe you can turn over another card in Col 1. I hope the others have found a Magical Mystic Method to prove me wrong.

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