Could someone walk me line-by line through this prime sieve? I'm attempting to develop my own sieve but most of the faster ones contain syntax that I don't understand (like the lines i've made comments on). Sorry if this is sort of a newbie question -- also if you have any links that could help me with sieve writing they'd be greatly appreciated.
def rwh_primes1(n):
""" Returns a list of primes < n """
sieve = [True] * (n/2) # What does the '[True]' mean?
for i in xrange(3,int(n**0.5)+1,2):
if sieve[i/2]: # The same complaint as in line 3, I don't understand the variable 'sieve'
sieve[i*i/2::i] = [False] * ((n-i*i-1)/(2*i)+1) # how is sieve used here - it seems to be as a list but i'm unsure
return [2] + [2*i+1 for i in xrange(1,n/2) if sieve[i]] # 'sieve' is confusing me here yet again.
I'll put the meaning of each line of code in italics, and add my comments in normal-face text.
sieve = [True] * (n/2)
Declare a new local variable sieve and initialise it to the value [True] * (n/2). What does that expression mean? [True] is a single-element list containing only the Boolean value True. Multiplying a list by a number repeats the list, so sieve is now an n/2-element list of all-True values.
for i in xrange(3, int(n**0.5)+1, 2):
Start counting in steps of 2, starting at 3 and ending at sqrt(n). This particular choice of range is an optimisation of the Sieve algorithm: we could count all the way from 1 to n in steps of 1, but that'd be less efficient.
if sieve[i/2]:
Look at the i/2th element of the sieve list. If it's True, then continue down the if branch. The author is using i/2 to compensate for the fact that the code is counting in steps of 2. This way you can work with a shorter list and use less memory.
sieve[i*i/2::i] = [False] * ((n-i*i-1)/(2*i)+1)
Update every i-th element of sieve to False, starting at i*i/2. sieve[i*i/2::i] is called slice notation - because it appears on the left-hand side of an assignment this code will update that slice of sieve. The right-hand side is a repeat of the list-times-number trick. It computes an all-False list of the correct length.
return [2] + [2*i+1 for i in xrange(1,n/2) if sieve[i]]
This code just turns the True/False values in sieve into a list of prime numbers. The calculation is compensating for the fact that sieve doesn't contain any even numbers by multiplying each True value's index by 2.