
chameleons
we can sort of reduce the island, as far as what were
concerned in, to a state machine.
the state is described by 3 numbers, which represent
the difference in population between R G B chameleons.
ie at the start
RG = 2, GB = 2, RB = 4
any meeting of chameleons is a transition in our state
machine. in terms of state, we can generalize a
transition t to the following transformation
t
{
Num A = Num A
Num B = Num B  3
Num C = Num C  3
}
where the mapping of NumX to RG GB RB is
determined by who meets who (try it)
whatever the transitions you choose, the effect
will be that of adding a multiple of 3 to the
three state variables.
so after any number of meetings
RG = 2 + 3x, GB = 2 + 3y RB = 4 +3z
where x y z are determined by what meetings occur.
the island can be populated entirely by one color when
any of the three numbers is zero, since at that point
the two colors involved can cancel each other out and
yield only the third color. in the language of the state
machine this occurs when
RG = 0 or GB = 0 o or RB = 0
so in conclusion
RG = 0 and RG = 2 + 3x or
BG = 0 and RG = 2 + 3y or
RB = 0 and RB = 4 + 3z or
all which are unsatisfiable, so the answer is no.
the other conclusion is that it would be possible
if any of the differences between populations
was a multiple of 3.
hmm now that i read it, it looks like
utter bullshit! hahahahah
Cheradenine Zakalwe
Wednesday, June 05, 2002
You can get all one color.
Here's an example:
Start with {13, 15, 17}
A red meets a green, they both change to blue.
Now we have {12, 14, 19}
Same thing happens again and again . . .
{11, 13, 21}
{10, 12, 23}
. . . .
{0, 2, 43}
Now, we have a blue meet a green, and get:
{1, 1, 42}
The red and green meet, and we're at:
{0, 0, 44}
So that's just proof by example. Someone want to provide a more general proof?
Andrew Hogue
Tuesday, July 16, 2002
Where did you put the 45th lizard?
Malachi Brown
Friday, September 06, 2002
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