Factor an integer to something as close to a square as possible

问题

I have a function that reads a file byte by byte and converts it to a floating point array. It also returns the number of elements in said array. Now I want to reshape the array into a 2D array with the shape being as close to a square as possible.

As an example let's look at the number 800:

sqrt(800) = 28.427...

Now by I can figure out by trial and error that 25*32 would be the solution I am looking for. I do this by decrementing the sqrt (rounded to nearest integer) if the result of multiplying the integers is to high, or incrementing them if the result is too low.

I know about algorithms that do this for primes, but this is not a requirement for me. My problem is that even the brute force method I implemented will sometimes get stuck and never finish (which is the reason for my arbitrary limit of iterations):

import math

def factor_int(n):
    nsqrt = math.ceil(math.sqrt(n))

    factors = [nsqrt, nsqrt]
    cd = 0
    result = factors[0] * factors[1]
    ii = 0
    while (result != n or ii > 10000):
        if(result > n):
            factors[cd] -= 1
        else:
            factors[cd] += 1
        result = factors[0] * factors[1]
        print factors, result
        cd = 1 - cd
        ii += 1

    return "resulting factors: {0}".format(factors)

input = 80000
factors = factor_int(input)

using this script above the output will get stuck in a loop printing

[273.0, 292.0] 79716.0
[273.0, 293.0] 79989.0
[274.0, 293.0] 80282.0
[274.0, 292.0] 80008.0
[273.0, 292.0] 79716.0
[273.0, 293.0] 79989.0
[274.0, 293.0] 80282.0
[274.0, 292.0] 80008.0
[273.0, 292.0] 79716.0
[273.0, 293.0] 79989.0
[274.0, 293.0] 80282.0
[274.0, 292.0] 80008.0
[273.0, 292.0] 79716.0
[273.0, 293.0] 79989.0
[274.0, 293.0] 80282.0
[274.0, 292.0] 80008.0
[273.0, 292.0] 79716.0
[273.0, 293.0] 79989.0
[274.0, 293.0] 80282.0

But I wonder if there are more efficient solutions for this? Certainly I can't be the first to want to do something like this.

回答1:

def factor_int(n):
    nsqrt = math.ceil(math.sqrt(n))
    solution = False
    val = nsqrt
    while not solution:
        val2 = int(n/val)
        if val2 * val == float(n):
            solution = True
        else:
            val-=1
    return val, val2, n

try it with:

for x in xrange(10, 20):
      print factor_int(x)

回答2:

Interesting question, here's a possible solution to your problem:

import math


def min_dist(a, b):
    dist = []
    for Pa in a:
        for Pb in b:
            d = math.sqrt(
                math.pow(Pa[0] - Pb[0], 2) + math.pow(Pa[1] - Pb[1], 2))
            dist.append([d, Pa])

    return sorted(dist, key=lambda x: x[0])


def get_factors(N):
    if N < 1:
        return N

    N2 = N / 2
    NN = math.sqrt(N)

    result = []
    for a in range(1, N2 + 1):
        for b in range(1, N2 + 1):
            if N == (a * b):
                result.append([a, b])

    result = min_dist(result, [[NN, NN]])
    if result:
        return result[0][1]
    else:
        return [N, 1]


for i in range(801):
    print i, get_factors(i)

The key of this method is finding the minimum distance to the cartesian point of [math.sqrt(N), math.sqrt(N)] which meets the requirements N=a*b, a&b integers.

回答3:

I think the modulus operator is a good fit for this problem:

import math 

def factint(n):
    pos_n = abs(n)
    max_candidate = int(math.sqrt(pos_n))
    for candidate in xrange(max_candidate, 0, -1):
        if pos_n % candidate == 0:
            break
    return candidate, n / candidate

回答4:

Here's a direct method that finds the smallest, closest integers a, b, such that a * b >= n, and a <= b, where n is any number:

def factor_int(n):
    a = math.floor(math.sqrt(n))
    b = math.ceil(n/float(a))
    return a, b

try it with:

for x in xrange(10, 20):
    print factor_int(x)

回答5:

Here's a shorter code based on the currently accepted answer that is shorter and takes about 25%-75% less time to run than their code (from basic timeit tests):

from math import sqrt, ceil
def factor_int_2(n):
    val = ceil(sqrt(n))
    while True:
        if not n%val:
            val2 = n//val
            break
        val -= 1
return val, val2

And here's a small and messy test I made to test the efficiency of the method:

print("Method 2 is shorter and about {}% quicker".format(100*(1 - timeit(lambda: factor_int_2(12345))/timeit(lambda: factor_int(12345)))))
#returns 'Method 2 is shorter and about 75.03810670186826% quicker'

来源：https://stackoverflow.com/questions/39248245/factor-an-integer-to-something-as-close-to-a-square-as-possible