Complexity question on Python 3 lists

[Dave Angel]

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