Optimize python algorithm

Question

I have three sorted lists, examplewise

a, b, c = [10,9,8], [9,8,7], [13,5,1]

I want to get all the combinations x, y, z where x in a, y in b and z in c and 1/x + 1/y + 1/z < 1 in the fastest time possible. I've been trying some different approaches,

for x, y, z in product(a, b, c):
    if predicative(x,y,z):
        yield (x, y, z)

Obviously, this takes too long time considering I'm checking everything, and the lists a, b, c are already sorted. I have tried sorting product(a,b,c) on the sum, but that is reaaally slow, as it uses all the products. My initial plan with having a, b and c sorted is so I could break out of the loop as soon as one fails. Any ideas?

Thanks.

Have you considered itertools.takewhile? The pure Python equivalent implementation is effectively your current approach plus two lines (else: break). — jonrsharpe
– jonrsharpe, Commented Jan 18, 2015 at 12:58
I don't think takewhile will work to be honest, as it goes linearly through. The way product is sorted, I could easily miss combinations. Please correct me if I'm wrong. — Martol1ni
– Martol1ni, Commented Jan 18, 2015 at 13:08
@Martol1ni Since the data is already sorted, this is the best you can get I guess. — thefourtheye
– thefourtheye, Commented Jan 18, 2015 at 13:24
The use of continue isn't right though, is it? If the first one is not valid, the second one will definately not be valid. — Martol1ni
– Martol1ni, Commented Jan 18, 2015 at 13:40

amit · Accepted Answer · 2015-01-18 12:59:35Z

2

A simple solution that could speed it up a bit would be to store 1/z for each z in c in a list, and for each pair of x,y in a,b - use binary search for the highest 1/z (in the auxillary list) such that 1/x + 1/y + 1/z < 1 - this will effectively trim many search from the 3rd lists and get you some speed up.
This will reduce time complexity to O(log(n)*n^2 + Y), where Y is the output size (number of triplets produced).

Note however, that since the output size itself is O(n^3) (consider 3 lists with all elements > 3.333) - you cannot avoid the worst case slow time, since you might be needing to generate n^3 triplets.

If you want only the count however, the approach suggested can easily find it in O(n^2*logn).

answered Jan 18, 2015 at 12:59

amit

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Martol1ni Over a year ago

That is very interesting. I will end up with O(n^3) whatever I do I suppose.

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