On 02/15/2012 01:20 PM, Franck Ditter wrote:
What is the cost of calling primes(n) below ? I'm mainly interested in
knowing if the call to append is O(1), even amortized.
Do lists in Python 3 behave like ArrayList in Java (if the capacity
is full, then the array grows by more than 1 element) ?
def sdiv(n) : # n>= 2
"""returns the smallest (prime) divisor of n"""
if n % 2 == 0 : return 2
for d in range(3,int(sqrt(n))+1,2) :
if n % d == 0 : return d
return n
def isPrime(n) :
"""Returns True iff n is prime"""
return n>= 2 and n == sdiv(n)
def primes(n) : # n>= 2
"""Returns the list of primes in [2,n]"""
res = []
for k in range(2,n+1) :
if isPrime(k) : res.append(k) # cost O(1) ?
return res
Thanks,
franck
Yes, lists behave the way you'd expect (see vector in C++), where when
they have to reallocate they do so exponentially.
However, realize that your algorithm is inefficient by a huge factor
more than any time spent expanding lists. The biggest single thing you
need to do is to memoize -- store the list of known primes, and add to
it as you encounter more. Then use that list instead of range(3, xxx,
2) for doing the trial divisions.
--
DaveA
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