On Chameleon Island exists a peculiar sort of chameleon. At any given time any given chameleon is either red, blue or green. When two chameleons of unlike color meet, both immediately change to the remaining possible color.

I scientist has collected 36 of these animals, 12 of each color, and kept them in 36 separate containers to prevent color change, but he wants to keep them in two terraria.

When kept together in small numbers, there's a danger that all the lizards will ultimately go to one color, as exemplified by the following scenario starting out with 1 red, 4 blue and 13 green chameleons. The two letters at the left of each line specify the meeting that changed the count to the one on the given line:

      r  b  g
      1  4 13
rg    0  6 12
bg    2  5 11
rg    1  7 10
rg    0  9  9
bg    2  8  8
bg    4  7  7
bg    6  6  6
bg    8  5  5
bg   10  4  4
bg   12  3  3
bg   14  2  2
bg   16  1  1
bg   18  0  0

From then on, this scenario has all 18 of its chameleons red.

How can the scientist divide his 36 chameleons between the two terraria without posing the possibility of all becoming one color in either terrarium? Assume that no births or deaths occur. There's more than one way.

For any one terrarium, the number of each of the three colors must be mod(x,3)=0, mod(y,3)=1, and mod(z,3)=2, for these and only these prevent unicolorization; adding two to one color, and subtracting one from each of the others, preserves this parity.  If we divide the chamelea evenly, with 18 in each terrarium, there are 14 ways (and permutations) to populate each terrarium, with the 12s complements in the other.

Hence the simplest solution would be to put 18 into each terrarium.  For red, blue, and green these could then be 5 red, 6 blue, 7 green in one, and 7, 6, 5 respectively in the other.  At the other extreme we could have 0,1,2 in one and 12,11,10 in the other.

Posted by ed bottemiller on 2009-09-25 17:56:59