.
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).
@Ernie had the same idea! Here’s a solution that does that in Racket.
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:
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:)
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 (). 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:
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;
}