Before starting let me say: It's not homework, just plain, old, fun.
Now, I'm trying to come up with an algorithm that can answer this question 1/x + 1/y = 1/n!.
And as you can see by the link above, the author asked only for hints and not the actual answer, so I would kindly ask for the same.
I simplified the expression until (x - n!)(y - n!) = (n!)^2 as suggested by one of the answers, and by that time I understood that the number of combinations of (x,y) pairs is the same as the number of divisors of n!^2 (correct me if I'm wrong here).
So, as suggested by the accepted answer, I'm trying to get the multiplication of all the factors of each prime composing N!^2.
I've come up with some code in C using trial division to factorize N!^2 and the Sieve of Eratosthenes to get all the prime numbers up to sqrt(N!^2).
The problem now is memory, I have tried with N = 15 and my Mac (Quad Core 6GB of memory) almost died on me. The problem was memory. So I added some printf's and tried with N=11:
Sieve of Eratosthenes took 13339.910000 ms and used 152 mb of memory
n= 11; n!^2 = 1593350922240000; d = 6885
[2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,5,5,5,5,7,7,11,11]
The list is all the prime factors of N!^2 (besides 1 and N!^2 of course).
I would like some hints on how to minimize memory consumption and possible optimizations.
Code bellow, it was just a quick experiment so I'm sure it can be optimized.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <strings.h>
#include <sys/time.h>
#include <assert.h>
//Linked List
struct node {
struct node * next;
long val;
};
void addValue(struct node *list, long val) {
struct node *n = list;
if (n->val == -1) {
n->val = val;
return;
}
while (n->next) {
n = n->next;
}
struct node *newNode = malloc(sizeof(struct node));
newNode->val = val;
newNode->next = NULL;
n->next = newNode;
}
void freeLinkedList(struct node *list) {
struct node *c = list;
if (!c) return;
struct node *n = c->next;
free(c);
freeLinkedList(n);
}
void printList(struct node *list) {
struct node *n = list;
printf("[");
while (n) {
printf("%ld", n->val);
n = n->next;
if (n) {
printf(",");
}
}
printf("]\n");
}
//-----------
int fac(int n) {
if (n == 1) return 1;
return fac(n-1)*n;
}
//Sieve of Eratosthenes
int sieve_primes(long limit, long **list) {
struct timeval t1;
struct timeval t2;
double elapsedTime = 0;
gettimeofday(&t1, NULL);
assert(limit > 0);
//Create a list of consecutive integers from 2 to n: (2, 3, 4, ..., n).
long arrSize = limit-1;
long *arr = malloc(sizeof(long)*arrSize);
long c = 2;
for (long i = 0; i < arrSize; i++) {
arr[i] = c++;
}
assert(arr[arrSize-1] == limit);
for (long i = 0; i < arrSize; i++) {
//Let p be equal to the first number not crossed
long p = arr[i];
if (p == 0) continue;
//Starting from p, count up in increments of p and mark each of these numbers greater than p itself in the list.
for (long f = p+p; f < arrSize; f+=p) {
arr[f] = 0;
}
}
*list = arr;
gettimeofday(&t2, NULL);
elapsedTime = (t2.tv_sec - t1.tv_sec) * 1000.0; // sec to ms
elapsedTime += (t2.tv_usec - t1.tv_usec) / 1000.0; // us to ms
printf("Sieve of Eratosthenes took %f ms and used %lu mb of memory\n",elapsedTime, (arrSize * sizeof(int))/1024/1024);
return arrSize;
}
void trial_division(struct node* list, long n) { if (n == 1) {
addValue(list, 1);
return;
}
long *primes;
long primesSize = sieve_primes(sqrt(n), &primes);
struct timeval t1;
struct timeval t2;
double elapsedTime = 0;
gettimeofday(&t1, NULL);
for (long i = 0; i < primesSize; i++) {
long p = primes[i];
if (p == 0) continue;
if (p*p > n) break;
while (n % p == 0) {
addValue(list, p);
n/=p;
}
}
if (n > 1) {
addValue(list, n);
}
free(primes);
}
int main(int argc, char *argv[]) {
struct node *linkedList = malloc(sizeof(struct node));
linkedList->val = -1;
linkedList->next = NULL;
long n = 11;
long nF = fac(n);
long nF2 = nF*nF;
trial_division(linkedList, nF2);
long multOfAllPrimeFactors = 1;
struct node *c = linkedList;
while (c) {
long sumOfVal = 2;
long val = c->val;
c = c->next;
while(c) {
long val2 = c->val;
if (val == val2) {
sumOfVal++;
c = c->next;
} else break;
}
multOfAllPrimeFactors*=sumOfVal;
}
printf("n= %ld; n!^2 = %ld; d = %ld\n", n,nF2, multOfAllPrimeFactors);
printList(linkedList);
freeLinkedList(linkedList);
}
EDIT:
As an example I will show you the calculation for getting all the possible positive integer solutions to the initial equation:
3!^2 = 36 = (3^2*2^2*1^0)
So there are (1+2)(1+2)(1+0)=9 possible positive integer solutions to the diophantine equation. Double if you count negative integers. I'm using WolframAlpha to be sure.
EDIT 2:
I think I just found out "what a factorial is", I'm getting this very interesting output:
3! = [2,3]
3!^2 = [2,2,3,3]
3!^3 = [2,2,2,3,3,3]
3!^4 = [2,2,2,2,3,3,3,3]
Thanks :D