Nearly-Square Divisors
August 5, 2016
The obvious answer uses brute force to find the divisors one-by-one until reaching the square root:
(define (nsq-div1 n)
(let loop ((d 1))
(if (and (zero? (modulo n d))
(<= n (* d d)))
(list d (/ n d))
(loop (+ d 1)))))
> (nsq-div1 36)
(6 6)
> (nsq-div1 60)
(10 6)
> (time (nsq-div1 224403121196654400))
(time (nsq-div1 224403121196654400))
no collections
55911 ms elapsed cpu time
55998 ms elapsed real time
598016 bytes allocated
(473753280 473670855)
A better answer factors n, computes the divisors as in a previous exercise, then takes the middle two divisors:
(define (factors n)
(if (even? n) (cons 2 (factors (/ n 2)))
(let loop ((n n) (f 3) (fs '()))
(cond ((< n (* f f)) (reverse (cons n fs)))
((zero? (modulo n f))
(loop (/ n f) f (cons f fs)))
(else (loop n (+ f 2) fs))))))
(define (divisors n)
(define (times x) (lambda (y) (* x y)))
(let divs ((fs (factors n)))
(unique = (sort <
(if (null? fs) '(1)
(let ((ds (divs (cdr fs))))
(append ds (map (times (car fs)) ds))))))))
(define (nsq-div2 n)
(let* ((divs (divisors n))
(len (length divs))
(b (list-ref divs (quotient (- len 1) 2))))
(list (/ n b) b)))
> (nsq-div2 36)
(6 6)
> (nsq-div2 60)
(10 6)
> (time (nsq-div2 224403121196654400))
(time (nsq-div2 224403121196654400))
5 collections
94 ms elapsed cpu time, including 16 ms collecting
104 ms elapsed real time, including 8 ms collecting
21878568 bytes allocated, including 19696976 bytes reclaimed
(473753280 473670855)
We’ve gone from a minute to a tenth of a second. Most of the time is spent computing the 80640 divisors of 224403121196654400; factorization takes a blink of an eye, as all the factors are less than 40. Of course, it may be necessary for large values of n to use some faster method of computing the factors.
We used unique from the Standard Prelude. You can run the program at http://ideone.com/VtTDGQ.
Programming Praxis 
Another method is to start with the square root of n, rounded down, then decrement by one until a divisor is found. On my computer this took 0.6 sec.
The pitfall with any method is to deal with a prime n.
In Python. The divisors are scanned for the maximum value less equal isqrt(n).
def nsd(n): s = isqrt(n) a = max(d for d in divisors(n) if d <= s) return a, n // a@Ernie had the same idea! Here’s a solution that does that in Racket.
(define (near-squares n) (let loop ((a (floor (sqrt n)))) (let ((b (/ n a))) (if (and (integer? b) (= n (* a b))) (list (inexact->exact a) (inexact->exact b)) (loop (sub1 a))))))I came up with the same idea as the previous commenters. For large n with a higher number of divisors this may become incredibly time consuming, e.g. n = 100#. So I guess there’s a quicker solution than simply testing all possible numbers from floor(sqrt(n)) downward.
Python:
import math def nsd(n): s = (int)(math.sqrt(n)) while n % s: s -= 1 return s, n // s print (nsd(224403121196654400))Output in approx. 0.012s: (473670855, 473753280)
Anyhow, this solution is based on the slow brute-force method. I’m almost certain there’s a much faster number-theoretical way to solve this for huge n.
@Ernie and @Milbrae. The number N=224403121196654400 has an incredibly large number of divisors and is the worst-case scenario for a solution using divisors and the best-case scenario for the solution walking down from the root. Now try to use both methods in the range [N-10, N+10). Timing is 90 ms for the divisor solution and the other solution takes forever.
Paul: That is point I was trying to make in my first post: if N has very few divisors, decrementing from the sqrt may take forever, and if N is prime, well don’t ask.
224403121196654400 is the 22nd “colossally abundant” number (and mentioned here: https://programmingpraxis.com/2016/04/22/gcd-sum/#comment-59608)
Finding nearly-square divisors seems about as hard as factoring (if we have a method for finding them we can apply it recursively to get a full factorization, and if we have a factorization, finding such divisors is straightforward).
Fermat factorization seems relevant too, but maybe only applies to odd numbers.
@Paul: You may have a point there. Anyhow, consider n having at least 30 distinct prime factors, each appearing once. n will then have 2**30 factors. You wouldn’t want to try almost all of them to find the correct NSD.
There’s a similar problem on Project Euler (https://projecteuler.net/problem=266). I don’t think the people you have already solved this did it through simply factoring.
@Milbrae: Good point, actually it’s a variant on the knapsack problem – find a maximal subset (or sub-multiset) of prime factors of n, the sum of whose logs is less than log n / 2 (and there are ways of solving the knapsack problem that are better than testing each alternative).
@Milbrae. Thanks for pointing me to Euler 266. I solved for the product of first 30 primes with divisors and it took about 4 hours. The Euler problem needs the first 42 primes. That will indeed need another approach.
@Matthew: A varint of knapsack? Could you elaborate on this?
@Milbrae: One form of knapsack problem is: given a set of real numbers S, and a limit k, find the maximal subset sum less (or equal) to k (ie. find a subset with sum less than or equal to k, and greater or equal to the sum of any other such subset). This is exactly what we want for the nearly-square root of a product of single primes (as in the Project Euler problem) – S is the set of logs of the primes, and k is the log of the square root.
Here’s a simple Haskell program that computes the log of the nearly square root of the product of the first 20 primes:
-- k target s: return largest subset sum of s <= target k target [] = 0 k target (a:s) = if a > target then k target s else max (a + k (target-a) s) (k target s) primes = [2,3,5,7,11,13,17,19,23,29, 31,37,41,43,47,53,59,61,67,71] s = map log(reverse primes) target = (sum s)/2 r = k target s main = print target >> print (k target s)Scan the tree of subsets, discarding a branch which cannot lead to a solution (here just subsets that would be too large – we could also keep track of the best solution found and discard branches that cannot lead to a better solution).
@Matthew: Wow!! Works fine and extremely fast. I assume correctly the second line of the output is the exponent to e to get the result? I know practically nothing about Haskell.
[…] a recent exercise, we discussed the problem of computing the nearly square divisors of a composite number n = a · b, […]
[…] nearly square divisors exercise has generated a considerable amount of interest, and several excellent solutions in the comments. […]
package test;
import java.util.*;
public class SquareDivisors {
public static int squareDivisor(int n){
int diff=10000;
if (n < 0){
return diff;
}
List pairs = new ArrayList();
//no need to look beyond sqrt hence check for (i*i )<=n
for(int i=1;(i*i)<=n;i++){
if (n%i==0){
System.out.println("Pairs : "+i +" "+ (n/i));
pairs.add(new Pairs(i,n/i));
}
}
for(Pairs pair: pairs){
diff= (diff < (Math.abs(pair.a-pair.b)))?diff:(Math.abs(pair.a-pair.b));
}
return diff;
}
public static void main(String args[]){
System.out.println("Diff : "+squareDivisor(36));
}
}
class Pairs{
public Pairs(int a,int b){
this.a=a;
this.b=b;
}
public int a,b;
}