Project Euler Problem 3
September 20, 2011
Here’s our solution:
(define (factors n) ; trial division
(let loop ((n n) (fs '()))
(if (even? n) (loop (/ n 2) (cons 2 fs))
(let loop ((n n) (f 3) (fs fs))
(if (< n (* f f)) (reverse (cons n fs))
(if (zero? (modulo n f))
(loop (/ n f) f (cons f fs))
(loop n (+ f 2) fs)))))))
And here’s a sample run:
> (factors 13195)
(5 7 13 29)
You can run the program at http://programmingpraxis.codepad.org/9NbfghJq.
Programming Praxis 
It looks like this may not behave well for powers of two, though it handles other evens fine:
Python:
def factors(n): fs = [] while (n & 1 == 0): fs.append(2) n //= 2 f = 3 while n >= f * f: if (n % f == 0): fs.append(f) n //= f else: f += 1 if n != 1: fs.append(n) return fsHuzzah for native big integers!
How embarrassing! I need a step 2.5 that reads “If n = 1, output the list of factors and stop.” and make the corresponding change in the function.
Trying to learn Google Go, first time using big.Int
Have to say that I don’t find it very intuitive to use, but then again, I’m only just starting..
func Factors(n *big.Int) (answer vector.Vector) { var f, q, r, incr, zero, tmp big.Int f.SetInt64(2) incr.SetInt64(2) for n.Bit(0) == 0 { answer.Push(new(big.Int).Set(&f)) n.Div(n,&f) } f.SetInt64(3) for tmp.Mul(&f,&f).Cmp(n) < 1 { q.DivMod(n, &f, &r) if r.Cmp(&zero) > 0 { f.Add(&f, &incr) } else { n.Set(&q) answer.Push(new(big.Int).Set(&f)) } } if n.Cmp(new(big.Int).SetInt64(1)) != 0 { answer.Push(new(big.Int).Set(n)) } return }Oh, sorry for the cruft.. used an online syntax highlighter, but doesn’t appear that it worked very well.
In ruby (pretty much exactly the same as Graham’s python version) …
def prime_factors(n) fs = [] while n % 2 == 0 fs << 2 n /= 2 end f = 3 while n >= f**2 if n % f == 0 fs << f n /= f else f += 1 end end fs << n if n != 1 fs endYou could also monkey patch Integer to provide this in a more ruby like way.
In Python. Slightly more verbose, and with the steps interspersed through the code as comments. I’m gradually getting the hang of Python.
Source
Same problem solved in C# language
https://sites.google.com/site/eulerproblemsincsharp/home/problem-3
Here’s a slight twist on the described function. Rather than try all
odd numbers >= 3, I used a 30-wheel to generate the possible factors.
Also, the function is a generator which returns successive factors of its
argument. Use list() to get the last factor.
from itertools import accumulate, chain, cycle def genfactors(n): for f in accumulate(chain((2,1,2,2), cycle((4,2,4,2,4,6,2,6)))): if f*f > n: break while not n % f: yield f n //= f if n > 1: yield n #some tests print(list(genfactors(440190069936))) # -> [2, 2, 2, 2, 3, 317, 3803, 7607]@graham/@slabounty – I think line 12/14 should be f += 2, so that f skips even numbers.
Thanks for catching that, @Mike; I could have sworn I put in += 2. Anyway, nice wheel! I’ll have to read up on wheel factorization and Python3’s accumulate.
Graham, you can read about wheel factorization at the earlier exercise at Programming Praxis.
C# version
public class Euler3 { public static long[] PrimeFactors(long n) { // 1. Create a list of factors, initially empty. var factors = new List<long>(); // 2. If n is even, add 2 to the end of the list of factors, divide n by 2, and go to Step 2. while( (n & 1) == 0) n /= 2; // 3. Set f ← 3. long f = 3; while(true) { // 4. Calculate the quotient q and remainder r when dividing n by f, so that n = q f + r with 0 ≤ r < f. long q = n / f; long r = n - q * f; // 5. If r > 0, set f ← f + 2 and go to Step 4. if ( r > 0 ) { f += 2; } else { // 6. Otherwise, r = 0. Add f to the end of the list of factors. Set n ← q. factors.Add(f); n = q; // 7. If n < f2, add n to the end of the list of factors, output the list of factors, and stop. if(n < f * f) { factors.Add(n); break; } } } return factors.ToArray(); } public static void Main(string[] args) { var x1 = 13195; var fx1 = PrimeFactors(x1); Console.WriteLine("Prime factors of '{0}' = [{1}]", x1, String.Join(", ", fx1)); var x2 = 600851475143; var fx2 = PrimeFactors(x2); Console.WriteLine("Prime factors of '{0}' = [{1}]", x2, String.Join(", ", fx2)); Console.WriteLine("Largest prime factor = {0}", fx2[fx2.Length - 1]); Console.ReadLine(); } }in c language
#include
int main()
{
long long int n = 600851475143;
long long int a[100];
long long int i,f,q,r,j;
i=0;
while(n%2==0)
{
a[i] = 2;
i++;
n=n/2;
}
f=3;
while(n > (f*f))
{
q = n/f;
r = n%f;
if(r>0)
{
f = f+2;
}
else
{
a[i] = f;
i++;
n = q;
if(n < (f*f))
{
a[i] = n;
i++;
}
}
}
for(j=0; j<i; ++j)
printf("%lld ",a[j]);
printf("\n");
return 0;
}
FORTH Version to find greatest prime factor. Rather than factor all even numbers, the first step factors all even numbers > 2. When n is equal to 2, the square test fails, so the odd factoring loop does not execute. Therefore there is no special case to check for n = 1 at the end, but more importantly the invariant is preserved that when the function ends, the greatest prime factor is on the stack as a 64 bit integer. (If the final “d.” and printing is removed, the FORTH word would become a function to return the largest prime factor.)
: deven? ( d -- ? ) drop 1 and 0= ; : d/mod ( d n -- d/n d%n ) locals| n | \ FORTH does not have remainder function for double division \ so we multiply the quotient by the divisor and subtract... \ 2dup 1 n M*/ 2dup 2>r \ get quotient & save a copy n 1 M*/ d- drop \ compute remainder 2r> ; \ restore quotient : .factors ( d -- ) BEGIN 2dup deven? >r 2. 2over d< r> and WHILE 2 . d2/ REPEAT 3 locals| f | BEGIN 2dup f dup m* d< not WHILE 2dup f d/mod rot 0= IF f . 2nip \ drop n; quotient on stack replaces it ELSE 2drop \ drop quotient f 2 + TO f THEN REPEAT d. ;Testing:
C#:
long num = 600851475143;
long f = 1;
long z = 1;
while (z != num)
{
if (num % f == 0)
{
Console.WriteLine(f);
z = z * f;
}
f++;
}
Console.WriteLine(“DONE!”);
Console.ReadLine();
In Java:
public class LargestPrimeFactor {
private static double primefactor(double n) {
double i=2;
for(;itemp)? i:temp;
}
}
return i;
}
public static void main(String[] args) {
System.out.println(primefactor(600851475143.0));
}
}
HMM the above code was cut off
public class LargestPrimeFactor {
private static double primefactor(double n) {
double i=2;
for(;itemp)? i:temp;
}
}
return i;
}
public static void main(String[] args) {
System.out.println(primefactor(600851475143.0));
}
}
I have done its very easy and convenient solution. Check this out here @ http://crispylogs.com/project-euler-problem-3-solution/
[…] See: https://programmingpraxis.com/2011/09/20/project-euler-problem-3/ […]