What is the computational complexity of `itertools.combinations` in python?











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itertools.combinations in python is a powerful tool for finding all combination of r terms, however, I want to know about its computational complexity.



Let's say I want to know the complexity in terms of n and r, and certainly it will give me all the r terms combination from a list of n terms.



According to the Official document, this is the rough implementation.



def combinations(iterable, r):
# combinations('ABCD', 2) --> AB AC AD BC BD CD
# combinations(range(4), 3) --> 012 013 023 123
pool = tuple(iterable)
n = len(pool)
if r > n:
return
indices = list(range(r))
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != i + n - r:
break
else:
return
indices[i] += 1
for j in range(i+1, r):
indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in indices)









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  • 3




    It's gonna be nCr, i.e O(n choose r)
    – juanpa.arrivillaga
    Nov 21 at 19:51








  • 1




    @juanpa.arrivillaga There is going to be at least another factor r for the tuple generation.
    – user10605163
    Nov 22 at 1:32










  • @eukaryota makes sense
    – juanpa.arrivillaga
    Nov 22 at 3:06















up vote
0
down vote

favorite












itertools.combinations in python is a powerful tool for finding all combination of r terms, however, I want to know about its computational complexity.



Let's say I want to know the complexity in terms of n and r, and certainly it will give me all the r terms combination from a list of n terms.



According to the Official document, this is the rough implementation.



def combinations(iterable, r):
# combinations('ABCD', 2) --> AB AC AD BC BD CD
# combinations(range(4), 3) --> 012 013 023 123
pool = tuple(iterable)
n = len(pool)
if r > n:
return
indices = list(range(r))
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != i + n - r:
break
else:
return
indices[i] += 1
for j in range(i+1, r):
indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in indices)









share|improve this question


















  • 3




    It's gonna be nCr, i.e O(n choose r)
    – juanpa.arrivillaga
    Nov 21 at 19:51








  • 1




    @juanpa.arrivillaga There is going to be at least another factor r for the tuple generation.
    – user10605163
    Nov 22 at 1:32










  • @eukaryota makes sense
    – juanpa.arrivillaga
    Nov 22 at 3:06













up vote
0
down vote

favorite









up vote
0
down vote

favorite











itertools.combinations in python is a powerful tool for finding all combination of r terms, however, I want to know about its computational complexity.



Let's say I want to know the complexity in terms of n and r, and certainly it will give me all the r terms combination from a list of n terms.



According to the Official document, this is the rough implementation.



def combinations(iterable, r):
# combinations('ABCD', 2) --> AB AC AD BC BD CD
# combinations(range(4), 3) --> 012 013 023 123
pool = tuple(iterable)
n = len(pool)
if r > n:
return
indices = list(range(r))
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != i + n - r:
break
else:
return
indices[i] += 1
for j in range(i+1, r):
indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in indices)









share|improve this question













itertools.combinations in python is a powerful tool for finding all combination of r terms, however, I want to know about its computational complexity.



Let's say I want to know the complexity in terms of n and r, and certainly it will give me all the r terms combination from a list of n terms.



According to the Official document, this is the rough implementation.



def combinations(iterable, r):
# combinations('ABCD', 2) --> AB AC AD BC BD CD
# combinations(range(4), 3) --> 012 013 023 123
pool = tuple(iterable)
n = len(pool)
if r > n:
return
indices = list(range(r))
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != i + n - r:
break
else:
return
indices[i] += 1
for j in range(i+1, r):
indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in indices)






python time-complexity complexity-theory






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asked Nov 21 at 19:49









ArtificiallyIntelligence

326312




326312








  • 3




    It's gonna be nCr, i.e O(n choose r)
    – juanpa.arrivillaga
    Nov 21 at 19:51








  • 1




    @juanpa.arrivillaga There is going to be at least another factor r for the tuple generation.
    – user10605163
    Nov 22 at 1:32










  • @eukaryota makes sense
    – juanpa.arrivillaga
    Nov 22 at 3:06














  • 3




    It's gonna be nCr, i.e O(n choose r)
    – juanpa.arrivillaga
    Nov 21 at 19:51








  • 1




    @juanpa.arrivillaga There is going to be at least another factor r for the tuple generation.
    – user10605163
    Nov 22 at 1:32










  • @eukaryota makes sense
    – juanpa.arrivillaga
    Nov 22 at 3:06








3




3




It's gonna be nCr, i.e O(n choose r)
– juanpa.arrivillaga
Nov 21 at 19:51






It's gonna be nCr, i.e O(n choose r)
– juanpa.arrivillaga
Nov 21 at 19:51






1




1




@juanpa.arrivillaga There is going to be at least another factor r for the tuple generation.
– user10605163
Nov 22 at 1:32




@juanpa.arrivillaga There is going to be at least another factor r for the tuple generation.
– user10605163
Nov 22 at 1:32












@eukaryota makes sense
– juanpa.arrivillaga
Nov 22 at 3:06




@eukaryota makes sense
– juanpa.arrivillaga
Nov 22 at 3:06












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I would say it is θ[r (n choose r)], the n choose r part is the number of times the generator has to yield and also the number of times the outer while iterates.



In each iteration at least the output tuple of length r needs to be generated, which gives the additional factor r. The other inner loops will be O(r) per outer iteration as well.



This is assuming that the tuple generation is actually O(r) and that the list get/set are indeed O(1) at least on average given the particular access pattern in the algorithm. If this is not the case, then still Ω[r (n choose r)] though.



As usual in this kind of analysis I assumed all integer operations to be O(1) even if their size is not bounded.






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    up vote
    1
    down vote



    accepted










    I would say it is θ[r (n choose r)], the n choose r part is the number of times the generator has to yield and also the number of times the outer while iterates.



    In each iteration at least the output tuple of length r needs to be generated, which gives the additional factor r. The other inner loops will be O(r) per outer iteration as well.



    This is assuming that the tuple generation is actually O(r) and that the list get/set are indeed O(1) at least on average given the particular access pattern in the algorithm. If this is not the case, then still Ω[r (n choose r)] though.



    As usual in this kind of analysis I assumed all integer operations to be O(1) even if their size is not bounded.






    share|improve this answer



























      up vote
      1
      down vote



      accepted










      I would say it is θ[r (n choose r)], the n choose r part is the number of times the generator has to yield and also the number of times the outer while iterates.



      In each iteration at least the output tuple of length r needs to be generated, which gives the additional factor r. The other inner loops will be O(r) per outer iteration as well.



      This is assuming that the tuple generation is actually O(r) and that the list get/set are indeed O(1) at least on average given the particular access pattern in the algorithm. If this is not the case, then still Ω[r (n choose r)] though.



      As usual in this kind of analysis I assumed all integer operations to be O(1) even if their size is not bounded.






      share|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        I would say it is θ[r (n choose r)], the n choose r part is the number of times the generator has to yield and also the number of times the outer while iterates.



        In each iteration at least the output tuple of length r needs to be generated, which gives the additional factor r. The other inner loops will be O(r) per outer iteration as well.



        This is assuming that the tuple generation is actually O(r) and that the list get/set are indeed O(1) at least on average given the particular access pattern in the algorithm. If this is not the case, then still Ω[r (n choose r)] though.



        As usual in this kind of analysis I assumed all integer operations to be O(1) even if their size is not bounded.






        share|improve this answer














        I would say it is θ[r (n choose r)], the n choose r part is the number of times the generator has to yield and also the number of times the outer while iterates.



        In each iteration at least the output tuple of length r needs to be generated, which gives the additional factor r. The other inner loops will be O(r) per outer iteration as well.



        This is assuming that the tuple generation is actually O(r) and that the list get/set are indeed O(1) at least on average given the particular access pattern in the algorithm. If this is not the case, then still Ω[r (n choose r)] though.



        As usual in this kind of analysis I assumed all integer operations to be O(1) even if their size is not bounded.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 22 at 1:33

























        answered Nov 22 at 1:20









        user10605163

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