What is the computational complexity of `itertools.combinations` in python?
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0
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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)
python time-complexity complexity-theory
add a comment |
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)
python time-complexity complexity-theory
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.
– eukaryota
Nov 22 at 1:32
@eukaryota makes sense
– juanpa.arrivillaga
Nov 22 at 3:06
add a comment |
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)
python time-complexity complexity-theory
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
python time-complexity complexity-theory
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.
– eukaryota
Nov 22 at 1:32
@eukaryota makes sense
– juanpa.arrivillaga
Nov 22 at 3:06
add a comment |
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.
– eukaryota
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.
– eukaryota
Nov 22 at 1:32
@juanpa.arrivillaga There is going to be at least another factor r for the tuple generation.
– eukaryota
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
add a comment |
1 Answer
1
active
oldest
votes
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.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
add a comment |
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.
add a comment |
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.
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.
edited Nov 22 at 1:33
answered Nov 22 at 1:20
eukaryota
1,154115
1,154115
add a comment |
add a comment |
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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.
– eukaryota
Nov 22 at 1:32
@eukaryota makes sense
– juanpa.arrivillaga
Nov 22 at 3:06