Simple Sieve of Eratosthenes in Python 3
$begingroup$
So yet another Sieve of Eratosthenes in Python 3.
The function returns a list of all primes smaller but not equal max_n.
The motivation is, as a practice, a simple implementation of the algorithm that is faithful, short, readable and transparent, while still getting a reasonable performance.
def primes(max_n):
"""Return a list of primes smaller than max_n."""
sieve = [True] * max_n
# p contains the largest prime yet found.
p = 2
# Only for p < sqrt(max_n) we check,
# i.e. p ** 2 < max_n, to avoid float issues.
while p ** 2 < max_n:
# Cross-out all true multiples of p:
for z in range(2 * p, max_n, p):
sieve[z] = False
# Find the next prime:
for z in range(p + 1, max_n):
if sieve[z]:
p = z
break
# 0 and 1 are not returned:
return [z for z in range(2, max_n) if sieve[z]]
IMHO it would be preferable to avoid p ** 2 < max_n and instead use p < max_n ** 0.5. Can we do this? It surprisingly seems to work as long as max_n ** 0.5 fits into the float mantissa, even if max_n doesn’t.
The second for-loop doesn’t look very nice with the break but I don’t have any idea how to do it otherwise…
Do you have any suggestions?
Are there still any simplifications possible? Or non-hackish ways to increase performance?
beginner python-3.x sieve-of-eratosthenes
asked 10 mins ago
wolf-revo-catswolf-revo-cats
1386
$endgroup$
add a comment |
$begingroup$
So yet another Sieve of Eratosthenes in Python 3.
The function returns a list of all primes smaller but not equal max_n.
The motivation is, as a practice, a simple implementation of the algorithm that is faithful, short, readable and transparent, while still getting a reasonable performance.
def primes(max_n):
"""Return a list of primes smaller than max_n."""
sieve = [True] * max_n
# p contains the largest prime yet found.
p = 2
# Only for p < sqrt(max_n) we check,
# i.e. p ** 2 < max_n, to avoid float issues.
while p ** 2 < max_n:
# Cross-out all true multiples of p:
for z in range(2 * p, max_n, p):
sieve[z] = False
# Find the next prime:
for z in range(p + 1, max_n):
if sieve[z]:
p = z
break
# 0 and 1 are not returned:
return [z for z in range(2, max_n) if sieve[z]]
IMHO it would be preferable to avoid p ** 2 < max_n and instead use p < max_n ** 0.5. Can we do this? It surprisingly seems to work as long as max_n ** 0.5 fits into the float mantissa, even if max_n doesn’t.
The second for-loop doesn’t look very nice with the break but I don’t have any idea how to do it otherwise…
Do you have any suggestions?
Are there still any simplifications possible? Or non-hackish ways to increase performance?
beginner python-3.x sieve-of-eratosthenes
asked 10 mins ago
wolf-revo-catswolf-revo-cats
1386
$endgroup$
add a comment |
$begingroup$
So yet another Sieve of Eratosthenes in Python 3.
The function returns a list of all primes smaller but not equal max_n.
The motivation is, as a practice, a simple implementation of the algorithm that is faithful, short, readable and transparent, while still getting a reasonable performance.
def primes(max_n):
"""Return a list of primes smaller than max_n."""
sieve = [True] * max_n
# p contains the largest prime yet found.
p = 2
# Only for p < sqrt(max_n) we check,
# i.e. p ** 2 < max_n, to avoid float issues.
while p ** 2 < max_n:
# Cross-out all true multiples of p:
for z in range(2 * p, max_n, p):
sieve[z] = False
# Find the next prime:
for z in range(p + 1, max_n):
if sieve[z]:
p = z
break
# 0 and 1 are not returned:
return [z for z in range(2, max_n) if sieve[z]]
IMHO it would be preferable to avoid p ** 2 < max_n and instead use p < max_n ** 0.5. Can we do this? It surprisingly seems to work as long as max_n ** 0.5 fits into the float mantissa, even if max_n doesn’t.
The second for-loop doesn’t look very nice with the break but I don’t have any idea how to do it otherwise…
Do you have any suggestions?
Are there still any simplifications possible? Or non-hackish ways to increase performance?
beginner python-3.x sieve-of-eratosthenes
asked 10 mins ago
wolf-revo-catswolf-revo-cats
1386
$endgroup$
So yet another Sieve of Eratosthenes in Python 3.
The function returns a list of all primes smaller but not equal max_n.
The motivation is, as a practice, a simple implementation of the algorithm that is faithful, short, readable and transparent, while still getting a reasonable performance.
def primes(max_n):
"""Return a list of primes smaller than max_n."""
sieve = [True] * max_n
# p contains the largest prime yet found.
p = 2
# Only for p < sqrt(max_n) we check,
# i.e. p ** 2 < max_n, to avoid float issues.
while p ** 2 < max_n:
# Cross-out all true multiples of p:
for z in range(2 * p, max_n, p):
sieve[z] = False
# Find the next prime:
for z in range(p + 1, max_n):
if sieve[z]:
p = z
break
# 0 and 1 are not returned:
return [z for z in range(2, max_n) if sieve[z]]
IMHO it would be preferable to avoid p ** 2 < max_n and instead use p < max_n ** 0.5. Can we do this? It surprisingly seems to work as long as max_n ** 0.5 fits into the float mantissa, even if max_n doesn’t.
The second for-loop doesn’t look very nice with the break but I don’t have any idea how to do it otherwise…
Do you have any suggestions?
Are there still any simplifications possible? Or non-hackish ways to increase performance?
beginner python-3.x sieve-of-eratosthenes
beginner python-3.x sieve-of-eratosthenes
asked 10 mins ago
wolf-revo-catswolf-revo-cats
1386
asked 10 mins ago
wolf-revo-catswolf-revo-cats
1386
asked 10 mins ago
wolf-revo-catswolf-revo-cats
1386
asked 10 mins ago
wolf-revo-catswolf-revo-cats
1386
asked 10 mins ago
wolf-revo-catswolf-revo-cats
1386
1386
add a comment |
add a comment |
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