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The New Coffee Room

  1. TNCR
  2. General Discussion
  3. Puzzle Time

Puzzle Time

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  • jon-nycJ Online
    jon-nycJ Online
    jon-nyc
    wrote on last edited by
    #6

    Yes

    Only non-witches get due process.

    • Cotton Mather, Salem Massachusetts, 1692
    1 Reply Last reply
    • jon-nycJ Online
      jon-nycJ Online
      jon-nyc
      wrote on last edited by
      #7

      So yes to all 4 of your questions. The ‘face up’ point just means the 4! combinations doesn’t become 2^4 * 4! by varying whether each card is face up or down.

      Only non-witches get due process.

      • Cotton Mather, Salem Massachusetts, 1692
      1 Reply Last reply
      • AxtremusA Away
        AxtremusA Away
        Axtremus
        wrote on last edited by
        #8

        This is probably a cheat ... but if I can exploit the “orientation” of how a card is placed (e.g., the angle the card is placed against the edge of the table), one card can encode arbitrarily many possibilities, limited only by George’s ability to place a card at the desired orientation precisely and Cats’ ability to tell apart one orientation from another (think of it like a Shannon limit of sort).

        Thinking about encoding many possibilities with four cards more conventionally, I don’t think we need to encode 52 possibilities, but rather at most 48 for any given hand since the four cards laid out can be eliminated, so we just need one more “dimension” to double that 4! permutations. I have not work it out yet.

        1 Reply Last reply
        • jon-nycJ Online
          jon-nycJ Online
          jon-nyc
          wrote on last edited by
          #9

          Yeah that’s a cheat. You have the order of the cards. That’s all.

          Only non-witches get due process.

          • Cotton Mather, Salem Massachusetts, 1692
          1 Reply Last reply
          • KlausK Offline
            KlausK Offline
            Klaus
            wrote on last edited by Klaus
            #10

            I'm not quite there yet, but I think I'm on the right track. That said, maybe I'm making this too complicated.

            :::

            Let's give the card names from 1 to 52.

            The selected cards are c-1 to c-5, whereby c-1 < c-2 < c-3 < c-4 < c-5.

            He removes c3.

            Now George proceeds as follows:

            Let's call the 24 different orders in which four cards can be arranged s-1 to s-24.

            Case 1: c-2 and c-4 < 26.

            In this case, c-3 must be between 1 and 26, but not c-2 or c-4. This means, at most 24 possibilities remain. If c-3 was the n'th card (not counting c-2 and c-4), choose order s-n.

            Case 2: c-2 and c-4 >= 26.

            This case is symmetrical to case 1.

            Case 3: c-2 < 26 and c-4 >= 26.

            Here I'm stuck.

            :::

            It would all be easy if somehow one additional bit of information could be transmitted, which would bump up the 24 sorting orders to 48, which would be enough (since only 48 cards remain).

            1 Reply Last reply
            • KlausK Offline
              KlausK Offline
              Klaus
              wrote on last edited by
              #11

              Ah, got it!

              1 Reply Last reply
              • jon-nycJ Online
                jon-nycJ Online
                jon-nyc
                wrote on last edited by
                #12

                I will say this much - it’s definitely the case that the selection within the five by George has to be strategic as well as the ordering of the other four.

                Only non-witches get due process.

                • Cotton Mather, Salem Massachusetts, 1692
                AxtremusA 1 Reply Last reply
                • jon-nycJ jon-nyc

                  I will say this much - it’s definitely the case that the selection within the five by George has to be strategic as well as the ordering of the other four.

                  AxtremusA Away
                  AxtremusA Away
                  Axtremus
                  wrote on last edited by
                  #13

                  @jon-nyc said in Puzzle Time:

                  I will say this much - it’s definitely the case that the selection within the five by George has to be strategic as well as the ordering of the other four.

                  :::

                  On that point, I’m thinking the selection of that card has to be suite based. Since there are only four suites and George gets five cards, it is guaranteed that at least two cards will come from the same suite.

                  :::

                  1 Reply Last reply
                  • KlausK Offline
                    KlausK Offline
                    Klaus
                    wrote on last edited by Klaus
                    #14

                    Got it now, I think.

                    Here's how to get that extra bit of information out of the card choice.

                    :::

                    Now consider the colors (spades, clubs etc.) of the cards.

                    The notation 4.1 means that there is one color that shows up four times and 1 color that shows up once.

                    For the five cards there are these possibilities.

                    a) 4.1
                    b) 3.2
                    c) 3.1.1
                    d) 2.2.1
                    e) 2.1.1.1

                    For four cards, there are these possibilities:

                    i) 4.0
                    ii) 3.1
                    iii) 2.2
                    iv) 2.1.1
                    v) 1.1.1.1

                    From each of a) to e) you can choose a card to reach either one of i), iii) or v), or one of ii) and iv).

                    Let's call i), iii) and v) the "zero" case and ii) and iv) the "one" case.

                    There's the extra bit. Together with the 24 sorting orders I can then encode a number between 1 and 48.

                    :::

                    1 Reply Last reply
                    • jon-nycJ Online
                      jon-nycJ Online
                      jon-nyc
                      wrote on last edited by jon-nyc
                      #15

                      How would you do the encoding of the 48? What kind of prima facie coding scheme would you use when you don’t know in advance which 48 cards you’ll be distinguishing among?

                      Ax - you’re on the right track. Or a right track at least.

                      Only non-witches get due process.

                      • Cotton Mather, Salem Massachusetts, 1692
                      KlausK 1 Reply Last reply
                      • jon-nycJ jon-nyc

                        How would you do the encoding of the 48? What kind of prima facie coding scheme would you use when you don’t know in advance which 48 cards you’ll be distinguishing among?

                        Ax - you’re on the right track. Or a right track at least.

                        KlausK Offline
                        KlausK Offline
                        Klaus
                        wrote on last edited by Klaus
                        #16

                        @jon-nyc said in Puzzle Time:

                        What kind of prima facie coding scheme would you use when you don’t know in advance which 48 cards you’ll be distinguishing among?

                        That part's easy.

                        From the 52 cards, subtract the 4 cards that were handed to Cats. Sort them. Name them 1 to 48.

                        The number encoding is: Name the 24 sorting orders s1 to s24. If you transmit 0 via the method I described above, sorting s-n stands for card n. If you transmit 1, it stands for 24+n.

                        1 Reply Last reply
                        • jon-nycJ Online
                          jon-nycJ Online
                          jon-nyc
                          wrote on last edited by
                          #17

                          Right but the need to choose a particular card to implement your ‘bit’ could conflict with your ability to represent the number, couldn't it?

                          Only non-witches get due process.

                          • Cotton Mather, Salem Massachusetts, 1692
                          1 Reply Last reply
                          • KlausK Offline
                            KlausK Offline
                            Klaus
                            wrote on last edited by
                            #18

                            Yes, but I was kind of hoping you wouldn't notice.

                            1 Reply Last reply
                            • jon-nycJ Online
                              jon-nycJ Online
                              jon-nyc
                              wrote on last edited by
                              #19

                              Lol.

                              There’s a much easier way which people could easily do in their head in real time to make the trick work.

                              Only non-witches get due process.

                              • Cotton Mather, Salem Massachusetts, 1692
                              1 Reply Last reply
                              • HoraceH Offline
                                HoraceH Offline
                                Horace
                                wrote on last edited by Horace
                                #20

                                :::

                                Considering 5 random distinct numbers from 1 to 52, the magician could choose either the highest or the lowest number to keep secret, while ordering the remaining four to identify a number between 1 and 24, since there are 24 orderings of four distinct things A,B,C, and D where A is the lowest number, B the second lowest, etc. The magician would choose either the highest or lowest of the 5 numbers, whichever one was within 24 of the remaining highest number, considering 52 to roll over to 1 as numbers increase. This should always be possible. The assistant would identify the hidden number as the indicated offset from the highest of their four cards, rolling over from 52 to 1 if necessary.

                                :::

                                Education is extremely important.

                                KlausK 1 Reply Last reply
                                • jon-nycJ Online
                                  jon-nycJ Online
                                  jon-nyc
                                  wrote on last edited by jon-nyc
                                  #21

                                  :::

                                  Yes that should work.

                                  A conceptually similar solution that is easier to compute in your head in ‘magic trick time’:

                                  Based in Ax’s observation that there is always at least one suit with two cards within the 5.

                                  George chooses one of the two cards of the same suit, puts the other on top of the four he hands to Cats.

                                  With the other three cards he can communicate a number between 1-6. He use that number to indicate how much higher the reserved card is than the top card, allowing for the cycle from K->A->2.

                                  Example. He puts the queen of clubs on top with the reserved card being the 5. He indicates 6 with the other three cards. Cats knows that from Q you count K,A,2,3,4,5 so the reserved card is a 5.

                                  What if the reserved card was a six? In that case George would have reserved the Q and handed Cats the 6 on top of the deck, still indicating 6 with the other three cards. (7,8,9,10,J,Q). In other words, by strategically choosing between the two cards of that suit, the additional 3 cards can convey enough information.

                                  :::

                                  Only non-witches get due process.

                                  • Cotton Mather, Salem Massachusetts, 1692
                                  1 Reply Last reply
                                  • HoraceH Offline
                                    HoraceH Offline
                                    Horace
                                    wrote on last edited by Horace
                                    #22

                                    Yeah that's conceptually identical but for the range being 1-13 rather than 1-52. But with some practice I don't think either is particularly difficult to compute in magic trick time. To identify the number between 0 and 23 based on the order of 4 cards, let hte first card indicate which quartile the number is in, then the second card indicate which tri-cile of that, and the third card which of the two remaining numbers it is.

                                    Education is extremely important.

                                    1 Reply Last reply
                                    • HoraceH Horace

                                      :::

                                      Considering 5 random distinct numbers from 1 to 52, the magician could choose either the highest or the lowest number to keep secret, while ordering the remaining four to identify a number between 1 and 24, since there are 24 orderings of four distinct things A,B,C, and D where A is the lowest number, B the second lowest, etc. The magician would choose either the highest or lowest of the 5 numbers, whichever one was within 24 of the remaining highest number, considering 52 to roll over to 1 as numbers increase. This should always be possible. The assistant would identify the hidden number as the indicated offset from the highest of their four cards, rolling over from 52 to 1 if necessary.

                                      :::

                                      KlausK Offline
                                      KlausK Offline
                                      Klaus
                                      wrote on last edited by
                                      #23

                                      @Horace said in Puzzle Time:

                                      The magician would choose either the highest or lowest of the 5 numbers, whichever one was within 24 of the remaining highest number

                                      Ah yes, that was the puzzle piece I was missing.

                                      1 Reply Last reply
                                      • HoraceH Offline
                                        HoraceH Offline
                                        Horace
                                        wrote on last edited by
                                        #24

                                        jon where do you get these puzzles? Or do you invent them?

                                        Education is extremely important.

                                        1 Reply Last reply
                                        • jon-nycJ Online
                                          jon-nycJ Online
                                          jon-nyc
                                          wrote on last edited by
                                          #25

                                          Mostly from a mailing list from a university ‘math circle‘ we were involved with in NC

                                          Only non-witches get due process.

                                          • Cotton Mather, Salem Massachusetts, 1692
                                          1 Reply Last reply
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