问题
Possible Duplicate:
How to determine if a number is a prime with regex?
This page claims that this regular expression discovers non-prime numbers (and by counter-example: primes):
/^1?$|^(11+?)\1+$/
How does this find primes?
回答1:
I think the article explains it rather well, but I'll try my hand at it as well.
Input is in unary form. 1 is 1, 2 is 11, 3 is 111, etc. Zero is an empty string.
The first part of the regex matches 0 and 1 as non-prime. The second is where the magic kicks in.
(11+?) starts by finding divisors. It starts by being defined as 11, or 2. \1 is a variable referring to that previously captured match, so \1+ determines if the number is divisible by that divisor. (111111 starts by assigning the variable to 11, and then determines that the remaining 1111 is 11 repeated, so 6 is divisible by 2.)
If the number is not divisible by two, the regex engine increments the divisor. (11+?) becomes 111, and we try again. If at any point the regex matches, the number has a divisor that yields no remainder, and so the number cannot be prime.
回答2:
Took me a minute to realize this is intended for numbers in base-1 (unary?)
Several people in this ycombinator discussion explain this quite well. Actually those explanations are more succinct than I think I can get, so I'll leave it to the link.
来源：https://stackoverflow.com/questions/3296050/how-does-this-regex-find-primes