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More chameleons (Posted on 2002-07-30) Difficulty: 3 of 5
Long ago, there existed a species of fighting chameleons. These chameleons were divided into six types of matching color and strength:
  • Black were the strongest, followed by
  • blue,
  • green,
  • orange,
  • yellow and
  • white which were the weakest.

    Whenever two chameleons of the same color met, they would fight to the death and the victor would become stronger and change color (eg white to yellow). Black chameleons would fight eternally.

    The small island of Ula was initially populated by a group of fighting chameleons. For this group

    a) the colors present each had an equal number of chameleons (for example, group = 3 black, 3 green and 3 yellow)

    b) it was not made up entirely of white chameleons

    After all the possible fighting was done, there remained one black and green and no blue or orange chameleons.

    How many white chameleons remained in the island? Prove it.

  • See The Solution Submitted by Cheradenine    
    Rating: 3.5000 (14 votes)

    Comments: ( Back to comment list | You must be logged in to post comments.)
    Some Thoughts Numbers | Comment 10 of 27 |
    Well, my initial thought to this now is that there must be no white chameleons left at the end.

    Given the two extra conditions of "all possible fights have happened" and "only chameleons of the same colour fight", we can conclude that there can only ever end up with zero or one of every colour. Any more and they would pair up and continue to fight - this wouldn't be an end state.

    So, for each colour we have 0 or 1 of each type left at the end. So, our answer must be one or the other, obviously...

    Also, given a possible start state of 2 Blues and 2 Oranges, it is easy to see here this would end up with our given end scenario of 1 Black, 1 Green and no other chameleons. So, here, we have 0 Whites.

    So, if we're required to find a single solution for all start states that end up at the given end solution, I assume the answer is "zero" whites will be left.

    These are just some initial thoughts over lunch - anyone any ideas on a quick way of proving this is the case for all working start states? Only extra thing I can think of is we have to end up proving that the 'equal number' stated in case a) must be even somehow (an even number of whites would result in none being left after the fighting)...
      Posted by Nick Reed on 2002-07-30 02:07:32
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