Skip to content
  • Categories
  • Recent
  • Tags
  • Popular
  • Users
  • Groups
Skins
  • Light
  • Brite
  • Cerulean
  • Cosmo
  • Flatly
  • Journal
  • Litera
  • Lumen
  • Lux
  • Materia
  • Minty
  • Morph
  • Pulse
  • Sandstone
  • Simplex
  • Sketchy
  • Spacelab
  • United
  • Yeti
  • Zephyr
  • Dark
  • Cyborg
  • Darkly
  • Quartz
  • Slate
  • Solar
  • Superhero
  • Vapor

  • Default (No Skin)
  • No Skin
Collapse

The New Coffee Room

  1. TNCR
  2. General Discussion
  3. Puzzle time - cutting the necklace

Puzzle time - cutting the necklace

Scheduled Pinned Locked Moved General Discussion
8 Posts 3 Posters 70 Views
  • Oldest to Newest
  • Newest to Oldest
  • Most Votes
Reply
  • Reply as topic
Log in to reply
This topic has been deleted. Only users with topic management privileges can see it.
  • jon-nycJ Online
    jon-nycJ Online
    jon-nyc
    wrote on last edited by
    #1

    Two thieves steal a necklace consisting of 10 rubies and 14 emeralds, fixed in some arbitrary order on a loop of golden string. Show that they can cut the necklace in two places so that when each thief takes one of the resulting pieces, he gets half the rubies and half the emeralds.

    If you don't take it, it can only good happen.

    1 Reply Last reply
    • HoraceH Offline
      HoraceH Offline
      Horace
      wrote on last edited by
      #2

      :::

      Starting with the emeralds, you have 14 different cut pairings available which splits them up 7 and 7. Let the cuts occur at the clockwise edge of the emerald, leaving any rubies between the adjacent emeralds on the clockwise side of the cuts.

      Then let the ruby imbalance be described by a number from -10 to 10. The objective is zero. Each of the 14 cut pairings gives an imbalance number. For every negative number, there will be an opposing positive number where the cut pairing is in the opposite order but otherwise identical. So if a given cut pairing has non-zero imbalance, you're guaranteed, as you move the cut pairing one position clockwise, that you'll eventually reach adjacent cut pairings where one has an imbalance to one side, and the next has an imbalance to the other side. In any such case, one of the cuts can be moved clockwise such that it captures exactly enough rubies to set the imbalance to zero.

      :::

      Education is extremely important.

      1 Reply Last reply
      • KlausK Offline
        KlausK Offline
        Klaus
        wrote on last edited by
        #3

        @Horace I was thinking along the same lines, but your last sentence is where I still scratch my head. For continuous functions, there's the "intermediate value theorem", but your imbalance is discrete - couldn't it "jump" over the 0, e.g., from -1 to +1?

        HoraceH 1 Reply Last reply
        • KlausK Klaus

          @Horace I was thinking along the same lines, but your last sentence is where I still scratch my head. For continuous functions, there's the "intermediate value theorem", but your imbalance is discrete - couldn't it "jump" over the 0, e.g., from -1 to +1?

          HoraceH Offline
          HoraceH Offline
          Horace
          wrote on last edited by Horace
          #4

          @klaus said in Puzzle time - cutting the necklace:

          @Horace I was thinking along the same lines, but your last sentence is where I still scratch my head. For continuous functions, there's the "intermediate value theorem", but your imbalance is discrete - couldn't it "jump" over the 0, e.g., from -1 to +1?

          Right, good point. So let the number be "rubies clockwise from first cut" minus "rubies clockwise from second cut". That number, for an even number of rubies, will always be even itself: in this case from the set [-10, -8, ..., 8, 10]. Any capturing of a ruby by moving the cut within the space between emeralds will modify that number by 2, up or down. If the sign of that number changes when the original pair of cuts is rotated clockwise, then you can go back to the previous pair and move one of the cuts clockwise to capture the appropriate number of rubies.

          Education is extremely important.

          1 Reply Last reply
          • jon-nycJ Online
            jon-nycJ Online
            jon-nyc
            wrote on last edited by jon-nyc
            #5

            That seems right to me. I hadn't solved it yet.

            If you don't take it, it can only good happen.

            1 Reply Last reply
            • KlausK Offline
              KlausK Offline
              Klaus
              wrote on last edited by
              #6

              Would the same technique work to partition a pearl necklace?

              1 Reply Last reply
              • jon-nycJ Online
                jon-nycJ Online
                jon-nyc
                wrote on last edited by
                #7

                I don’t know. Try as I might, I can never get them arranged in a ring.

                If you don't take it, it can only good happen.

                1 Reply Last reply
                • KlausK Offline
                  KlausK Offline
                  Klaus
                  wrote on last edited by
                  #8

                  Watch and learn!

                  :::

                  I can't believe you really clicked! Perv.
                  :::

                  1 Reply Last reply
                  Reply
                  • Reply as topic
                  Log in to reply
                  • Oldest to Newest
                  • Newest to Oldest
                  • Most Votes


                  • Login

                  • Don't have an account? Register

                  • Login or register to search.
                  • First post
                    Last post
                  0
                  • Categories
                  • Recent
                  • Tags
                  • Popular
                  • Users
                  • Groups