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    <p>Sorry, I don't understand LISP well enough to read your answer. But my first impression is that the time cost of the brute force solution should be:</p> <p>open bracket</p> <p>sqrt(k) to find the divisors of k (by trial division), square each one (constant time per factor), and sum them (constant time per factor). This is &sigma;<sub>2</sub>(k), which I will call x.</p> <p>plus</p> <p>not sure what the complexity of a good integer square root algorithm is, but certainly no worse than sqrt(x) (dumb trial multiplication). x might well be big-O larger than k, so I reserve judgement here, but x is obviously bounded above by k^3, because k has at most k divisors, each itself no bigger than k and hence its square no bigger than k^2. It's been so long since my maths degree that I have no idea how fast Newton-Raphson converges, but I suspect it's faster than sqrt(x), and if all else fails a binary chop is log(x).</p> <p>close bracket</p> <p>multiplied by n (as k ranges 1 .. n).</p> <p>So if your algorithm is worse than O(n * sqrt(n^3)) = O(n ^ (5/2)), in the dumb-sqrt case, or O(n * (sqrt(n) + log(n^3)) = O(n ^ 3/2) in the clever-sqrt case, I think something has gone wrong which should be identifiable in the algorithm. At this point I'm stuck because I can't debug your LISP.</p> <p>Oh, I've assumed that arithmetic is constant-time for the numbers in use. It darn well should be for numbers as small as 64 million, and the cube of that fits in a 64bit unsigned integer, barely. But even if your LISP implementation is making arithmetic worse than O(1), it shouldn't be worse than O(log n), so it won't have much affect on the complexity. Certainly won't make it super-polynomial.</p> <p>This is where someone comes along and tells me just how wrong I am.</p> <p>Oops, I just looked at your actual timing figures. They aren't worse than exponential. Ignoring the first and last values (because small times aren't accurately measurable and you haven't finished, respectively), multiplying n by 10 multiplies time by no more than 30-ish. 30 is about 10^1.5, which is about right for brute force as described above.</p>
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    1. POProject Euler #211 - efficiency issue
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    1. USSteve Jessop
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