What is the best time complexity possible to find a solution to x^2 + y^2 = m where m can be any natural number?

Question

I have to find a solution to the equation x² + y² = m where m can be any natural number and x,y are integers. This can be done in O(m^1/2) by using brute force easily. Is there a way I can do this in constant time?

Answer 1

I'm not sure you can do it any better than O(sqrt(m)) for an arbitrary m but that's pretty damn good , much better than linear time :-)

The approach is to start x and y at opposite ends of the solution space and either increment the low end or decrement the high end, depending on whether the result is too low or too high (obviously if the result is perfect, return the values):

def solveForM(m):
    set lo to 0
    set hi to sqrt(m), rounded up.
    while lo < hi:
        testVal = lo * lo + hi * hi
        if testVal == m:
            return (lo, hi)
        if testVal > m:
            hi = hi - 1
        else lo = lo + 1
    return "No solution"

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If m can be limited somehow, you could achieve O(1) by use of a lookup table (many optimisations come down to trading space for time), such as:

 0 ->  0, 0
 1 ->  0, 1
 2 ->  1, 1
       No solution for 3
 4 ->  0, 2
 5 ->  1, 2
       No solution for 6, 7
 8 ->  2, 2
 9 ->  0, 3
10 ->  1, 3
       No solution for 11, 12
13 ->  2, 3
... and so on.

A table like this can be generated with a small program along the lines of (Python 3):

for hi in range(1001):
    for lo in range(1001):
        m = lo * lo + hi * hi
        print("%5d -> %d, %d"%(m, lo, hi))

You have to sort (and possibly remove duplicates) afterwards to create a fast look-up table but the performance is okay with the generation of an unsorted list taking fifteen seconds for all m up to two million.

In any case, this is only run once, after which you would place the table into code where the time expense is incurred at compile time.