David Wilson wrote:
Suppose we tile the plane randomly with black and white unit square tiles in a grid pattern, black and white tiles occuring with equal probablity (sort of like the static you used to see on your old black and white TV).
This general setup is what's called a "percolation model" in the biz (where the biz in question is statistical mechanics). Well, not quite: they usually think of one of the two colors as "open" and the other as "closed", and only ask questions about the connectivity of open sites. Rearranging your mail a bit:
I am thinking that if black squares occur with probability 1/2, the probability of an infinite polyomino is 0. Certainly if they occur with probability 1, the probability of an infinite polyomino is 1. What about choosing black squares with probability 3/4? Other probabilities?
The standard big results here are: (1) there is a critical probability, usually denoted p_c, such that if you declare squares open with probability p < p_c, the probability of an infinite open component is zero, while with p > p_c it is 1. In dimension two, it's even known that p <= p_c is the probability zero condition; what happens at p_c exactly is known for some other lattices, but I don't know the full story.
What is the probability that a randomly chosen square is in a unomino (I find only one page on the web with that word!)? What is the probability that a randomly chosen polyomino is a unomino? What is the most frequent polyomino in both senses?
(I think people use "monomino.") That's just the probability that you're surrounded by squares whose color is not your own, so (1/2)^4 = 1/16, right? Likewise for a domino, there are four neighbors you could domin with (now *there's* a non-sactioned noun-to-verb conversion), and six border squares that would have to be of the opposite color. So the odds that a random square is in a domino is 4 * (1/2)^7 = 1/32. And picking random squares makes dominos twice as findable as monominos, so both come up one sixteeth of the time if you pick a random square and look at its connected component. Right? Unless I'm missing something, it would be easy to do this calculation for any specific polyomino. Eg, for an L-triomino, there are (I think) 12 ways for a square to lie in an L-triomino, and it needs the cooperation of two same-color and seven opposite-color squares, so 12*(1/2)^9 = 3/128 of all squares are in L's, and 9/128 of the time a random square's polyomino is an L. (That's bigger than 1/16, so it's the new record-holder.) I have to go do real work, so I'll let someone else (write the computer program to) expand this table (and tell me if I'm doing something dumb)... --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.