**2**

# Performance issues in simple program

###
#1
Members - Reputation: **863**

Posted 25 October 2012 - 10:12 AM

[source lang="cpp"]//Searches for number prime factors and return biggest prime factor 2012-10-23(24)#include <stdio.h>#include <stdlib.h>int isPrime(int num); // checks if number is prime: 1 - prime 0 - non-primevoid setPrimeArray(int * arr, int num); // composes array of primes up to num to test factorsint * primeFactors(int * arr, int num);//finds prime factors of numberint biggestPrimeFactor(int * factors,int size);//finds biggest prime factorint setArray (int * arr); // deletes 0 from array and returns setted array(using dynamic reallocation)int main(){ int num = 0; int * numFactors = 0; int numOfFac = 0; int i = 0; printf("Enter the number whose prime factors you want to find: \n"); scanf("%i", &num); int * primeArray = calloc(num,sizeof(int)); setPrimeArray(primeArray,num);//optimize this function int s = setArray(primeArray); printf("Prime numbers up to %i:\n",num); for(;i < s; i++)printf("%i ",primeArray[i]); i = 0; // primeArray -=3; numFactors = primeFactors(primeArray, num); numOfFac = setArray(numFactors); printf("\nPrime Factors of %i is:\n",num); for(;i < numOfFac; i++)printf("%i ",numFactors[i]); printf("\nThe biggest prime factor of %i is: %i\n",num, biggestPrimeFactor(numFactors,numOfFac)); system("PAUSE"); free(primeArray); free(numFactors); return 0;}int isPrime(int num){ int factor = 0; if (num == 1) return 1; int i = 1; for (; i <= num; i ++) { if( num % i == 0) factor++; } if (factor <= 2) return 1; else return 0;}void setPrimeArray(int * arr, int num){ int i = 1; int j = 0; for(; i <= num; i++) { if(isPrime(i)) { arr[j] = i; j++; } }}int * primeFactors(int * arr, int num){ int i = 1;//not 0 - do not include prime factor 1, starts from 2 int j = 0;//for indexing primeFactors array; int * factors = calloc(num, sizeof(int));//HERE WAS ERROR - MUST USE DYNAMIC MEMORY WITH FACTORS for(;i < num;) { if (num % arr[i] == 0) // { factors[j] = arr[i]; //BIG, BIG //*factors = arr[i]; // //factors++; // j++; // MISTAKE... num /= arr[i]; // i = 1; // } else i++; if (num == 1) return factors; } return factors;}int biggestPrimeFactor(int * factors,int size){ int i = 0; int max = factors[0]; for(;i < size; i ++) { if (factors[i] > max) max = factors[i]; } return max;}int setArray (int * arr){ int i = 0; while(*arr != 0) { arr++; i++; } arr = realloc(arr, sizeof(int)*i); return i;}[/source]

And I think I need optimize this function:

[source lang="cpp"]void setPrimeArray(int * arr, int num){ int i = 1; int j = 0; for(; i <= num; i++) { if(isPrime(i)) { arr[j] = i; j++; } }}[/source]

Because it takes a lot of time with bigger numbers in this function. It makes an array of primes up to numbers, so I can later use it to find prime factors of number. So if anyone have ideas how I can optimize my code, please share them with me.

Honore de Balzac

###
#2
Members - Reputation: **510**

Posted 25 October 2012 - 10:59 AM

While your solution certainly works, it's not terribly efficient overall in terms of memory (as you've seen). My suggestion is to go back to the drawing board and see if you can figure out how to calculate the primes without requiring so much memory. You're almost there, you're just being a little overzealous with calloc...

###
#3
Crossbones+ - Reputation: **2901**

Posted 25 October 2012 - 11:03 AM

The problem is that there is so far between the big primes, and you test way too many numbers.

You could cut it in half if you only test odd numbers... (i+=2) but there is probably smarter ways...

But even better, since prime numbers never change, why not precalculate them once, save to file, and then load the prime table from file instead?

**Edited by Olof Hedman, 25 October 2012 - 11:08 AM.**

###
#6
Members - Reputation: **510**

Posted 25 October 2012 - 11:55 AM

Each number has a pair of factors that are, themselves, numbers that are either primes or smaller numbers with pairs of factors. If you keep splitting your number this way, eventually you'll have a particular type of data structure containing a bunch of numbers at the very "bottom" (my word, not the actual term typically used) that are all prime. All you have to do at that point is pick the right one...

You don't necessarily need to store factors in a file, though there's no reason not too if you really want to do it.

###
#7
Crossbones+ - Reputation: **13645**

Posted 25 October 2012 - 12:07 PM

long largest_prime_factor(long n) { // Take out all the factors of 2 while (n%2==0) n/=2; if (n==1) return 2; // Take out all the factors of 3 while (n%3==0) n/=3; if (n==1) return 3; for (long p=5; p*p<=n; p+=6) { // Take out all the factors of p while (n%p==0) n/=p; if (n==1) return p; long q = p+2; // Take out all the factors of q while (n%q==0) n/=q; if (n==1) return q; } return n; // If there's something left that doesn't have a divisor <= sqrt(n), then n is prime and we can return it. }

###
#8
Members - Reputation: **137**

Posted 25 October 2012 - 12:28 PM

I have written something quickly and it seems to be working.

[source lang="cpp"]#include <stdio.h>int main() { int number; printf("Enter number: "); scanf("%d", &number); for (int i = 2; i < number; ++i) { while (number % i == 0) number /= i; } printf("Biggest prime: %d\n", number); return 0;}[/source]

**Edited by aavci, 25 October 2012 - 12:32 PM.**

###
#9
Crossbones+ - Reputation: **2123**

Posted 25 October 2012 - 07:15 PM

The main algorithmic improvement here is diving only by numbers up to the square root of the number.

...code snippet...

Nice. Here's my version:

int largestPrimeFactor = 2; for (int i=2; i<=n; ++i) { while (n % i == 0) { n/=i; largestPrimeFactor = i; } }

###
#10
Crossbones+ - Reputation: **13645**

Posted 27 October 2012 - 08:28 PM

alnite's code is correct, but it will take a very long time to run when n is a large prime, which might make it unsuitably slow for the Project Euler challenge.

###
#11
Crossbones+ - Reputation: **9055**

Posted 27 October 2012 - 08:49 PM

I would expect Project Euler to specifically choose a large number to prevent you from using trial division, forcing you to implement a better algorithm. Look into Pollard's rho algorithm, or Pollard's p + 1. There's also a continued fraction algorithm (SQUFOF) which is somewhat arcane, but works. Quadratic sieve is the next step up but is considerably harder to implement. Number Field Sieve (NFS) is state-of-the-art but good luck implementing that, seriously. You can also look into elliptic curve factorization (elliptic curve method or ECM for short) which is pretty good for average-size factors but is a bit tricky to get right and difficult to understand - you need to understand elliptic curves.

For reference, quadratic sieve can factorize a 100-digit semiprime into two 50-digit primes in four hours or so. Trial division would have done the same in, say, a few trillion years.

All of this is probably overkill for 64-bit integers though.

**Edited by Bacterius, 27 October 2012 - 08:53 PM.**

The slowsort algorithm is a perfect illustration of the multiply and surrender paradigm, which is perhaps the single most important paradigm in the development of reluctant algorithms. The basic multiply and surrender strategy consists in replacing the problem at hand by two or more subproblems, each slightly simpler than the original, and continue multiplying subproblems and subsubproblems recursively in this fashion as long as possible. At some point the subproblems will all become so simple that their solution can no longer be postponed, and we will have to surrender. Experience shows that, in most cases, by the time this point is reached the total work will be substantially higher than what could have been wasted by a more direct approach.

- *Pessimal Algorithms and Simplexity Analysis*