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

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太阳男子
太阳男子 2021-01-12 12:08

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

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  • 2021-01-12 12:32

    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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  • 2021-01-12 12:50

    I had this same question too (For itertools.permutations) and had a hard time tracing the complexities. This led me to visualize the code using matplotlib.pyplot;

    The code snippet is shown below

    result=[]
    import matplotlib.pyplot as plt
    import math
    x=list(range(1,11))
    def permutations(iterable, r=None): 
        count=0
        global result
        pool = tuple(iterable)
        n = len(pool)
        r = n if r is None else r
        if r > n:
            return
        indices = list(range(n))
        cycles = list(range(n, n-r, -1))
        yield tuple(pool[i] for i in indices[:r])
        while n:
            for i in reversed(range(r)):
                count+=1
                cycles[i] -= 1
                if cycles[i] == 0:
                    indices[i:] = indices[i+1:] + indices[i:i+1]
                    cycles[i] = n - i
                else:
                    j = cycles[i]
                    indices[i], indices[-j] = indices[-j], indices[i]
                    yield tuple(pool[i] for i in indices[:r])
                    break
            else:
                resulte.append(count)
                return
    for j in x:
        for i in permutations(range(j)):
            continue
    
    x=list(range(1,11))
    plt.plot(x,result)
    

    Time Complexity graph for itertools.permutation

    From the graph, it is observed that the time complexity is O(n!) where n=Input Size

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