Factoring Factorials
January 24, 2014
The trick is to build up the solution from the primes less than n instead of breaking down the solution starting from n!. Our solution is due to Will Ness. Consider 33!:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2
2 2
2
3 3 3 3 3 3 3 3 3 3 3
3 3 3
3
5 5 5 5 5 5
5
7 7 7 7
11 11 11
13 13
17 19 23 29 31
The pattern is obvious: For primes less than the square root of n, compute the number of times each prime and each prime power appears. For the remaining primes less than half of n, compute the number of times each prime appears (but not each prime power). Then each prime that still remains appears once. Like this:
33! = 2^( 33 div 2 + 33 div 4 + 33 div 8 + 33 div 16 + 33 div 32) *
3^( 33 div 3 + 33 div 9 + 33 div 27) *
5^( 33 div 5 + 33 div 25) *
----
7^( 33 div 7) * 11^( 33 div 11) * 13^( 33 div 13) *
----
17 * 19 * 23 * 29 * 31
It’s easy to reduce that calculation to Scheme:
(define (fact-fact n) ; prime factors of n factorial
(let loop ((ps (primes n)) (fs (list)))
(cond ((null? ps) (reverse fs))
((< (* (car ps) (car ps)) n)
(let ((p (car ps)))
(let ((k (let loop ((q p) (k 0))
(if (< n q) k
(loop (* q p) (+ k (quotient n q)))))))
(loop (cdr ps) (cons (cons (car ps) k) fs)))))
((< (+ (car ps) (car ps)) n)
(loop (cdr ps) (cons (cons (car ps) (quotient n (car ps))) fs)))
(else (loop (cdr ps) (cons (cons (car ps) 1) fs))))))
For example:
> (fact-fact 33)
((2 . 31) (3 . 15) (5 . 7) (7 . 4) (11 . 3) (13 . 2) (17 . 1)
(19 . 1) (23 . 1) (29 . 1) (31 . 1))
You can run the program at http://programmingpraxis.codepad.org/nFdoreYl, where there is also a demonstration that the calculation of the factors of 33! is correct.
Programming Praxis 
#include
#include
// reducing numbers from biggest to 2
// 16 -> 2*8 … 8 -> 2*4 … 4 -> 2*2 …
void main(int argc, char **argv) {
int n = atoi(argv[1]);
int *p = (int *) malloc(sizeof(int) * (n + 1));
int i, j, d;
for(i = 0; i 1; i–)
if(p[i]) {
for(j = i + i, d = 2; j <= n; j += i, d++) {
if(p[j]) {
p[i] += p[j];
p[d] += p[j];
p[j] = 0;
}
}
}
printf("1");
for(i = 2; i <= n; i++)
if(p[i])
printf(" * %i^%i", i, p[i]);
printf("\n");
}
// (sorry...) #include <stdio.h> #include <stdlib.h> // reducing numbers from biggest to 2 // 16 -> 2*8 ... 8 -> 2*4 ... 4 -> 2*2 ... void main(int argc, char **argv) { int n = atoi(argv[1]); int *p = (int *) malloc(sizeof(int) * (n + 1)); int i, j, d; for(i = 0; i <= n; i++) p[i] = 1; for(i = n; i > 1; i--) if(p[i]) { for(j = i + i, d = 2; j <= n; j += i, d++) { if(p[j]) { p[i] += p[j]; p[d] += p[j]; p[j] = 0; } } } printf("1"); for(i = 2; i <= n; i++) if(p[i]) printf(" * %i^%i", i, p[i]); printf("\n"); }Recursivelly adding factorizations
import Data.List import Data.Monoid import Data.Numbers.Primes facfac 1 = Factors [(1, 1)] facfac n = factorization n `mappend` facfac (n - 1) -- Factorizations Monoid data Factorization a = Factors [(a, a)] deriving Show factorization = Factors . map (\ps -> (head ps, length ps)) . group . primeFactors instance Integral a => Monoid (Factorization a) where mempty = Factors [] mappend (Factors []) fs = fs mappend fs (Factors []) = fs mappend xfs@(Factors (xf@(x, a):xs)) yfs@(Factors (yf@(y, b):ys)) = case x `compare` y of EQ -> Factors ((x, a + b):ms) LT -> Factors (xf:ns) GT -> Factors (yf:os) where Factors ms = mappend (Factors xs) (Factors ys) Factors ns = mappend (Factors xs) yfs Factors os = mappend xfs (Factors ys)(Excuse me, I can’t modify previous post)
In C version, we can reduce main loop from “i = n” to “i = n >> 1”.
That version has O(n log log n) and are not needed primes calculation nor mul nor div operations (space is O(n)).
Did in Go using my Project Euler library. Since it’s designed to solve PE problems everything is done with type uint64.
package main import ( . "fmt" . "github.com/dgottner/euler" "os" "strconv" ) func main() { var n, p uint64 if len(os.Args) < 2 { Println("Usage: factfact n") return } else { val, err := strconv.ParseInt(os.Args[1], 10, 64) if err != nil { Println(err.Error()) return } n = uint64(val) } Printf("%d! = 1", n) for p = 2; p <= n; p = NextPrime(p) { exp := n / p for fac := p * p; fac <= n; fac *= p { exp += n / fac } Printf(" * %d", p) if exp != 1 { Printf("^%d", exp) } } Println() }That’s a fun little problem. Here is my solution. It is quick and dirty, not efficient.
from itertools import count, takewhile def primes(): yield 2 primes = [2] n = 3 while True: if all(n % p for p in primes): yield n primes.append(n) n += 2 def pff(n): """prime factorization of n! returns a list of pairs (prime, power). """ up_to_n = lambda iter: takewhile(lambda p: p <= n, iter) return [(prime, sum(n/w for w in up_to_n(prime**power for power in count(1)))) for prime in up_to_n(primes())] from math import factorial from operator import mul print factorial(100) == reduce(mul, (a**b for (a, b) in pff(100)))Assuming `primes` is already implemented, returning a stream of prime numbers in order, in Haskell,
Another Python version. The optimization suggested by Will Ness did not work (in Python).
def facfac(n): for p in primes(n): nfacs = 0 f = n // p while f: nfacs += f f //= p yield p, nfacs