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The New Coffee Room

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  3. Puzzle time - slicing the pie

Puzzle time - slicing the pie

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  • jon-nycJ Offline
    jon-nycJ Offline
    jon-nyc
    wrote on last edited by
    #7

    It’s not. Still the wrong series.

    "You never know what worse luck your bad luck has saved you from."
    -Cormac McCarthy

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    • KlausK Offline
      KlausK Offline
      Klaus
      wrote on last edited by Klaus
      #8

      OK, here's what all other sequences that start with 1,2,4,6,10,14,19 stored at that nice website say about it's 10th/11th element:

      A023115 - 45
      A076101 - 52
      A094281 - 51
      A323623 - 43
      A023536 - 46
      A047808 - 44
      A076268 - 42
      A024536 - 49
      A117237 - 52

      It must be one of those 😁

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

        Hint:

        ||Thinking of the cuts as lines on a circle, what’s the largest number of lines the nth cut can intersect?||

        "You never know what worse luck your bad luck has saved you from."
        -Cormac McCarthy

        KlausK 1 Reply Last reply
        • KlausK Klaus

          OK, here's what all other sequences that start with 1,2,4,6,10,14,19 stored at that nice website say about it's 10th/11th element:

          A023115 - 45
          A076101 - 52
          A094281 - 51
          A323623 - 43
          A023536 - 46
          A047808 - 44
          A076268 - 42
          A024536 - 49
          A117237 - 52

          It must be one of those 😁

          jon-nycJ Offline
          jon-nycJ Offline
          jon-nyc
          wrote on last edited by
          #10

          @Klaus said in Puzzle time - slicing the pie:

          OK, here's what all other sequences that start with 1,2,4,6,10,14,19 stored at that nice website say about it's 10th/11th element:

          A023115 - 45
          A076101 - 52
          A094281 - 51
          A323623 - 43
          A023536 - 46
          A047808 - 44
          A076268 - 42
          A024536 - 49
          A117237 - 52

          It must be one of those 😁

          And it’s none of those!

          "You never know what worse luck your bad luck has saved you from."
          -Cormac McCarthy

          1 Reply Last reply
          • jon-nycJ jon-nyc

            Hint:

            ||Thinking of the cuts as lines on a circle, what’s the largest number of lines the nth cut can intersect?||

            KlausK Offline
            KlausK Offline
            Klaus
            wrote on last edited by
            #11

            @jon-nyc said in Puzzle time - slicing the pie:

            Hint:

            ||Thinking of the cuts as lines on a circle, what’s the largest number of lines the nth cut can intersect?||

            Hey, I'm in a Zoom meeting right now. I can only do this as a low-priority background task that doesn't require me to think 🙂

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            • KlausK Offline
              KlausK Offline
              Klaus
              wrote on last edited by Klaus
              #12

              ||
              Oh, I see, my problem was that I screwed up on the start of the sequence. It's 1,2,4,7,11,16.

              The first answer I get from OEIS is A000124, which has the promising subtitle "maximal number of pieces formed when slicing a pancake with n cuts.".

              So the answer seems to be 56.

              I refer to the references given at OEIS for the rationale 😀
              ||

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              • jon-nycJ Offline
                jon-nycJ Offline
                jon-nyc
                wrote on last edited by jon-nyc
                #13

                Cheater.

                ||How about noticing, based on my hint, that the nth cut can, at most, intersect the other n-1 cuts just once. That forms n new line segments (considering the edges of the pizza as end points as well). Each new line segment corresponds to a new slice of pie.

                So f(n)=f(n-1)+n
                And we know f(0) is 1

                So:
                f(1) = 2
                f(2) = 4
                f(3) = 7
                f(4) = 11
                f(5) = 16
                f(6) = 22
                f(7) = 29
                f(8) = 37
                f(9) = 46
                f(10) = 56

                So to get the nth number:

                1+1+2+3+4+5+6..+n

                Note that's just 1 plus the sum of the natural numbers to n.

                So 1 + n(n+1)/2 or

                (n^2 + n + 2)/2||

                "You never know what worse luck your bad luck has saved you from."
                -Cormac McCarthy

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                • KlausK Offline
                  KlausK Offline
                  Klaus
                  wrote on last edited by Klaus
                  #14

                  Hey, I didn't google for the task or something; I merely tried out a few cuts and searched for known sequences based on that. For a 5% CPU usage background task I'm happy with my performance 🙂

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                  • KlausK Offline
                    KlausK Offline
                    Klaus
                    wrote on last edited by
                    #15

                    @jon-nyc said in Puzzle time - slicing the pie:

                    How about noticing, based on my hint, that the nth cut can, at most, intersect the other n-1 cuts just once.

                    That makes sense, but how do you know that this is not merely an upper bound but that you can always find a line that intersects n-1 cuts?

                    1 Reply Last reply
                    • jon-nycJ Offline
                      jon-nycJ Offline
                      jon-nyc
                      wrote on last edited by
                      #16

                      The short answer I can give you on my phone is “because there are infinite points on a circle”.

                      If you want a longer explanation I can do that later.

                      "You never know what worse luck your bad luck has saved you from."
                      -Cormac McCarthy

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