python - sieve of Eratosthenes when number is prime print "1" when not print "0"

Question

I must create a function f(n) which value is 1 when the number is prime or value is 0 when isn't prime. Code is working but prints in revers order. For example: f(6)= 0 0 1 1 0 1 0

def prime(n):
    if n<2: return False
    for i in range(2,n):
        if n%i == 0:
            return False
    return True

def f(n):
    print('0', end=' ')
    a=1
    while a<=n:
        n=n-1
        print('1' if prime(n) else '0', end=' ')

f(6)

That is _not_ a [sieve](http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes). It's just a loop counting down numbers and naively testing whether they are prime. But if you want to reverse the order of the output, just reverse the loop. — tobias_k, Apr 22 '15 at 15:01
Check [this post](http://stackoverflow.com/a/3941967/764322) for a sieve example. What you have is just a very inefficient prime check. — Imanol Luengo, Apr 22 '15 at 15:08
I would also switch out n with sqrt(n). It'll save some time :) — Benjy Kessler, Apr 22 '15 at 15:08

paddyg · Accepted Answer · 2015-04-22T15:16:32.757

1

Reverse the loop like:

def f(n):
  print('0')
  for a in range(1, n+1):
    print('1' if prime(a) else '0')

PS I have seen good examples of how to actually implement a sieve of Eratosthenes in python on Stack Overflow, worth searching for a better solution.

edited Apr 22 '15 at 15:16

answered Apr 22 '15 at 15:05

paddyg

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python - sieve of Eratosthenes when number is prime print "1" when not print "0"

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