Getting time limit exceeded on submitting answer. I am also facing same problem with 2-3 more questions that I have submitted on spoj.com.
http://www.spoj.com/problems/PRIME1/
Peter wants to generate some prime numbers for his cryptosystem. Help him! Your task is to generate all prime numbers between two given numbers!
Input
The input begins with the number t of test cases in a single line (t<=10). In each of the next t lines there are two numbers m and n (1 <= m <= n <= 1000000000, n-m<=100000) separated by a space.
Output
For every test case print all prime numbers p such that m <= p <= n, one number per line, test cases separated by an empty line.
Example
Input:
2
1 10
3 5
Output:
2
3
5
7
3
5
Warning: large Input/Output data, be careful with certain languages (though most should be OK if the algorithm is well designed)
Here is my code in C.
#include <stdio.h>
int main()
{
int t,i,k,count;
long long int j=0,m=0,n=0;
scanf("%d",&t);
for(i=1;i<=t;i++)
{
scanf("%lld%lld",&m,&n);
for(j=m;j<=n;j++)
{
count=0;
for(k=1;k<=j/2;k++)
{
if(j%k==0)
count++;
if(count>1)
break;
}
if(count==1)
printf("%lld\n",j);
}
printf("\n");
}
return 0;
}
Your algorithm is O((m-n)*n) which of course won't run within the allocated time limit. Let's go over your code:
count=0;
for(k=1;k<=j/2;k++)
{
if(j%k==0)
count++;
if(count>1)
break;
}
if(count==1)
printf("%lld\n",j);
Micro optimization: Why do you need a counter? You could get away with a bool.
Optimization: Why are you testing primes j/2? If j has a divisor greater than 1 than it's guaranteed that j has a divisor that's at most sqrt(j).
Micro Optimization: Don't consider even numbers at all, except for 2.
bool prime = j==2 || j%2==1 ;
for(k=2;prime && k*k<=j;k++)
{
if(j%k==0) prime = false;
}
}
if(prime) printf("%lld\n",j);
Now this is O((m-n)*sqrt(n)) which is a lot faster.
I suppose this won't make the limit. You could extend the second micro-optimization to skip numbers divisible by 3 very easy.
Optimization: If this is still not enough then you have to do a pseudo-primality test. One test that's very easy to implement in O(log(n)) is https://en.wikipedia.org/wiki/Fermat_primality_test. With this the complexity is down to O((m-n)*log(n)) which should be run in the available time limit.
I am unable to understand why i am getting runtime error with this code. Problem is every number >=6 can be represented as sum of two prime numbers.
My code is ...... Thanks in advance problem link is http://poj.org/problem?id=2262
#include "stdio.h"
#include "stdlib.h"
#define N 1000000
int main()
{
long int i,j,k;
long int *cp = malloc(1000000*sizeof(long int));
long int *isprime = malloc(1000000*sizeof(long int));
//long int *isprime;
long int num,flag;
//isprime = malloc(2*sizeof(long int));
for(i=0;i<N;i++)
{
isprime[i]=1;
}
j=0;
for(i=2;i<N;i++)
{
if(isprime[i])
{
cp[j] = i;
j++;
for(k=i*i;k<N;k+=i)
{
isprime[k] = 0;
}
}
}
//for(i=0;i<j;i++)
//{
// printf("%d ",cp[i]);
//}
//printf("\n");
while(1)
{
scanf("%ld",&num);
if(num==0) break;
flag = 0;
for(i=0;i<j&&num>cp[i];i++)
{
//printf("%d ",cp[i]);
if(isprime[num-cp[i]])
{
printf("%ld = %ld + %ld\n",num,cp[i],num-cp[i]);
flag = 1;
break;
}
}
if(flag==0)
{
printf("Goldbach's conjecture is wrong.\n");
}
}
free(cp);
free(isprime);
return 0;
}
Two possibilities immediately spring to mind. The first is that the user input may be failing if whatever test harness is being used does not provide any input. Without knowing more detail on the harness, this is a guess at best.
You could check that by hard-coding a value rather than accepting one from standard input.
The other possibility is the rather large memory allocations being done. It may be that you're in a constrained environment which doesn't allow that.
A simple test for that is to drop the value of N (and, by the way, use it rather than the multiple hardcoded 1000000 figures in your malloc calls). A better way would be to check the return value from malloc to ensure it's not NULL. That should be done anyway.
And, aside from that, you may want to check your Eratosthenes Sieve code. The first item that should be marked non-prime for the prime i is i + i rather than i * i as you have. I think it should be:
for (k = i + i; k < N; k += i)
The mathematical algorithm is actually okay since any multiple of N less than N * N will already have been marked non-prime by virtue of the fact it's a multiple of one of the primes previously checked.
Your problem lies with integer overflow. At the point where N becomes 46_349, N * N is 2_148_229_801 which, if you have a 32-bit two's complement integer (maximum value of 2_147_483_647), will wrap around to -2_146_737_495.
When that happens, the loop keeps going since that negative number is still less than your limit, but using it as an array index is, shall we say, inadvisable :-)
The reason it works with i + i is because your limit is well short of INT_MAX / 2 so no overflow happens there.
If you want to make sure that this won't be a problem if you get up near INT_MAX / 2, you can use something like:
for (k = i + i; (k < N) && (k > i); k += i)
That extra check on k should catch the wraparound event, provided your wrapping follows the "normal" behaviour - technically, I think it's undefined behaviour to wrap but most implementations simply wrap two positives back to a negative due to the two's complement nature. Be aware then that this is actually non-portable, but what that means in practice is that it will only work on 99.999% of machines out there :-)
But, if you're a stickler for portability, there are better ways to prevent overflow in the first place. I won't go into them here but to say they involve subtracting one of the terms being summed from MAX_INT and comparing it to the other term being summed.
The only way I can get this to give an error is if I enter a value greater than 1000000 or less than 1 to the scanf().
Like this:
ubuntu#amrith:/tmp$ ./x
183475666
Segmentation fault (core dumped)
ubuntu#amrith:/tmp$
But the reason for that should be obvious. Other than that, this code looks good.
Just trying to find what went wrong!
If the sizeof(long int) is 4 bytes for the OS that you are using, then it makes this problem.
In the code:
for(k=i*i;k<N;k+=i)
{
isprime[k] = 0;
}
Here, when you do k = i*i, for large values if i, the value of k goes beyond 4 bytesand get truncated which may result in negative numbers and so, the condition k<N is satisfied but with a negative number :). So you get a segmentation fault there.
It's good that you need only i+i, but if you need to increase the limit, take care of this problem.
I don't know where I am doing wrong in trying to calculate prime factorizations using Pollard's rho algorithm.
#include<stdio.h>
#define f(x) x*x-1
int pollard( int );
int gcd( int, int);
int main( void ) {
int n;
scanf( "%d",&n );
pollard( n );
return 0;
}
int pollard( int n ) {
int i=1,x,y,k=2,d;
x = rand()%n;
y = x;
while(1) {
i++;
x = f( x ) % n;
d = gcd( y-x, n);
if(d!=1 && d!=n)
printf( "%d\n", d);
if(i == k) {
y = x;
k = 2 * k;
}
}
}
int gcd( int a, int b ) {
if( b == 0)
return a;
else
return gcd( b, a % b);
}
One immediate problem is, as Peter de Rivaz suspected the
#define f(x) x*x-1
Thus the line
x = f(x)%n;
becomes
x = x*x-1%n;
and the precedence of % is higher than that of -, hence the expression is implicitly parenthesised as
x = (x*x) - (1%n);
which is equivalent to x = x*x - 1; (I assume n > 1, anyway it's x = x*x - constant;) and if you start with a value x >= 2, you have overflow before you had a realistic chance of finding a factor:
2 -> 2*2-1 = 3 -> 3*3 - 1 = 8 -> 8*8 - 1 = 63 -> 3968 -> 15745023 -> overflow if int is 32 bits
That doesn't immediately make it impossible that gcd(y-x,n) is a factor, though. It just makes it likely that at a stage where theoretically, you would have found a factor, the overflow destroys the common factor that mathematically would exist - more likely than a common factor introduced by overflow.
Overflow of signed integers is undefined behaviour, so there are no guarantees how the programme behaves, but usually it behaves consistently so the iteration of f still produces a well-defined sequence for which the algorithm in principle works.
Another problem is that y-x will frequently be negative, and then the computed gcd can also be negative - often -1. In that case, you print -1.
And then, it is a not too rare occurrence that iterating f from a starting value doesn't detect a common factor because the cycles modulo both prime factors (for the example of n a product of two distinct primes) have equal length and are entered at the same time. You make no attempt at detecting such a case; whenever gcd(|y-x|, n) == n, any further work in that sequence is pointless, so you should break out of the loop when d == n.
Also, you never check whether n is a prime, in which case trying to find a factor is a futile undertaking from the start.
Furthermore, after fixing f(x) so that the % n applies to the complete result of f(x), you have the problem that x*x still overflows for relatively small x (with the standard signed 32-bit ints, for x >= 46341), so factoring larger n may fail due to overflow. At least, you should use unsigned long long for the computations, so that overflow is avoided for n < 2^32. However, factorising such small numbers is typically done more efficiently with trial division. Pollard's Rho method and other advanced factoring algorithms are meant for larger numbers, where trial division is no longer efficient or even feasible.
I'm just a novice at C++, and I am new to Stack Overflow, so some of what I have written is going to look sloppy, but this should get you going in the right direction. The program posted here should generally find and return one non-trivial factor of the number you enter at the prompt, or it will apologize if it cannot find such a factor.
I tested it with a few semiprime numbers, and it worked for me. For 371156167103, it finds 607619 without any detectable delay after I hit the enter key. I didn't check it with larger numbers than this. I used unsigned long long variables, but if possible, you should get and use a library that provides even larger integer types.
Editing to add, the single call to the method f for X and 2 such calls for Y is intentional and is in accordance with the way the algorithm works. I thought to nest the call for Y inside another such call to keep it on one line, but I decided to do it this way so it's easier to follow.
#include "stdafx.h"
#include <stdio.h>
#include <iostream>
typedef unsigned long long ULL;
ULL pollard(ULL numberToFactor);
ULL gcd(ULL differenceBetweenCongruentFunctions, ULL numberToFactor);
ULL f(ULL x, ULL numberToFactor);
int main(void)
{
ULL factor;
ULL n;
std::cout<<"Enter the number for which you want a prime factor: ";
std::cin>>n;
factor = pollard(n);
if (factor == 0) std::cout<<"No factor found. Your number may be prime, but it is not certain.\n\n";
else std::cout<<"One factor is: "<<factor<<"\n\n";
}
ULL pollard(ULL n)
{
ULL x = 2ULL;
ULL y = 2ULL;
ULL d = 1ULL;
while(d==1||d==n)
{
x = f(x,n);
y = f(y,n);
y = f(y,n);
if (y>x)
{
d = gcd(y-x, n);
}
else
{
d = gcd(x-y, n);
}
}
return d;
}
ULL gcd(ULL a, ULL b)
{
if (a==b||a==0)
return 0; // If x==y or if the absolute value of (x-y) == the number to be factored, then we have failed to find
// a factor. I think this is not proof of primality, so the process could be repeated with a new function.
// For example, by replacing x*x+1 with x*x+2, and so on. If many such functions fail, primality is likely.
ULL currentGCD = 1;
while (currentGCD!=0) // This while loop is based on Euclid's algorithm
{
currentGCD = b % a;
b=a;
a=currentGCD;
}
return b;
}
ULL f(ULL x, ULL n)
{
return (x * x + 1) % n;
}
Sorry for the long delay getting back to this. As I mentioned in my first answer, I am a novice at C++, which will be evident in my excessive use of global variables, excessive use of BigIntegers and BigUnsigned where other types might be better, lack of error checking, and other programming habits on display which a more skilled person might not exhibit. That being said, let me explain what I did, then will post the code.
I am doing this in a second answer because the first answer is useful as a very simple demo of how a Pollard's Rho algorithm is to implement once you understand what it does. And what it does is to first take 2 variables, call them x and y, assign them the starting values of 2. Then it runs x through a function, usually (x^2+1)%n, where n is the number you want to factor. And it runs y through the same function twice each cycle. Then the difference between x and y is calculated, and finally the greatest common divisor is found for this difference and n. If that number is 1, then you run x and y through the function again.
Continue this process until the GCD is not 1 or until x and y are equal again. If the GCD is found which is not 1, then that GCD is a non-trivial factor of n. If x and y become equal, then the (x^2+1)%n function has failed. In that case, you should try again with another function, maybe (x^2+2)%n, and so on.
Here is an example. Take 35, for which we know the prime factors are 5 and 7. I'll walk through Pollard Rho and show you how it finds a non-trivial factor.
Cycle #1: X starts at 2. Then using the function (x^2+1)%n, (2^2+1)%35, we get 5 for x. Y starts at 2 also, and after one run through the function, it also has a value of 5. But y always goes through the function twice, so the second run is (5^2+1)%35, or 26. The difference between x and y is 21. The GCD of 21 (the difference) and 35 (n) is 7. We have already found a prime factor of 35! Note that the GCD for any 2 numbers, even extremely large exponents, can be found very quickly by formula using Euclid's algorithm, and that's what the program I will post here does.
On the subject of the GCD function, I am using one library I downloaded for this program, a library that allows me to use BigIntegers and BigUnsigned. That library also has a GCD function built in, and I could have used it. But I decided to stay with the hand-written GCD function for instructional purposes. If you want to improve the program's execution time, it might be a good idea to use the library's GCD function because there are faster methods than Euclid, and the library may be written to use one of those faster methods.
Another side note. The .Net 4.5 library supports the use of BigIntegers and BigUnsigned also. I decided not to use that for this program because I wanted to write the whole thing in C++, not C++/CLI. You could get better performance from the .Net library, or you might not. I don't know, but I wanted to share that that is also an option.
I am jumping around a bit here, so let me start now by explaining in broad strokes what the program does, and lastly I will explain how to set it up on your computer if you use Visual Studio 11 (also called Visual Studio 2012).
The program allocates 3 arrays for storing the factors of any number you give it to process. These arrays are 1000 elements wide, which is excessive, maybe, but it ensures any number with 1000 prime factors or less will fit.
When you enter the number at the prompt, it assumes the number is composite and puts it in the first element of the compositeFactors array. Then it goes through some admittedly inefficient while loops, which use Miller-Rabin to check if the number is composite. Note this test can either say a number is composite with 100% confidence, or it can say the number is prime with extremely high (but not 100%) confidence. The confidence is adjustable by a variable confidenceFactor in the program. The program will make one check for every value between 2 and confidenceFactor, inclusive, so one less total check than the value of confidenceFactor itself.
The setting I have for confidenceFactor is 101, which does 100 checks. If it says a number is prime, the odds that it is really composite are 1 in 4^100, or the same as the odds of correctly calling the flip of a fair coin 200 consecutive times. In short, if it says the number is prime, it probably is, but the confidenceFactor number can be increased to get greater confidence at the cost of speed.
Here might be as good a place as any to mention that, while Pollard's Rho algorithm can be pretty effective factoring smaller numbers of type long long, the Miller-Rabin test to see if a number is composite would be more or less useless without the BigInteger and BigUnsigned types. A BigInteger library is pretty much a requirement to be able to reliably factor large numbers all the way to their prime factors like this.
When Miller Rabin says the factor is composite, it is factored, the factor stored in a temp array, and the original factor in the composites array divided by the same factor. When numbers are identified as likely prime, they are moved into the prime factors array and output to screen. This process continues until there are no composite factors left. The factors tend to be found in ascending order, but this is coincidental. The program makes no effort to list them in ascending order, but only lists them as they are found.
Note that I could not find any function (x^2+c)%n which will factor the number 4, no matter what value I gave c. Pollard Rho seems to have a very hard time with all perfect squares, but 4 is the only composite number I found which is totally impervious to it using functions in the format described. Therefore I added a check for an n of 4 inside the pollard method, returning 2 instantly if so.
So to set this program up, here is what you should do. Go to https://mattmccutchen.net/bigint/ and download bigint-2010.04.30.zip. Unzip this and put all of the .hh files and all of the C++ source files in your ~\Program Files\Microsoft Visual Studio 11.0\VC\include directory, excluding the Sample and C++ Testsuite source files. Then in Visual Studio, create an empty project. In the solution explorer, right click on the resource files folder and select Add...existing item. Add all of the C++ source files in the directory I just mentioned. Then also in solution expolorer, right click the Source Files folder and add a new item, select C++ file, name it, and paste the below source code into it, and it should work for you.
Not to flatter overly much, but there are folks here on Stack Overflow who know a great deal more about C++ than I do, and if they modify my code below to make it better, that's fantastic. But even if not, the code is functional as-is, and it should help illustrate the principles involved in programmatically finding prime factors of medium sized numbers. It will not threaten the general number field sieve, but it can factor numbers with 12 - 14 digit prime factors in a reasonably short time, even on an old Core2 Duo computer like the one I am using.
The code follows. Good luck.
#include <string>
#include <stdio.h>
#include <iostream>
#include "BigIntegerLibrary.hh"
typedef BigInteger BI;
typedef BigUnsigned BU;
using std::string;
using std::cin;
using std::cout;
BU pollard(BU numberToFactor);
BU gcda(BU differenceBetweenCongruentFunctions, BU numberToFactor);
BU f(BU x, BU numberToFactor, int increment);
void initializeArrays();
BU getNumberToFactor ();
void factorComposites();
bool testForComposite (BU num);
BU primeFactors[1000];
BU compositeFactors[1000];
BU tempFactors [1000];
int primeIndex;
int compositeIndex;
int tempIndex;
int numberOfCompositeFactors;
bool allJTestsShowComposite;
int main ()
{
while(1)
{
primeIndex=0;
compositeIndex=0;
tempIndex=0;
initializeArrays();
compositeFactors[0] = getNumberToFactor();
cout<<"\n\n";
if (compositeFactors[0] == 0) return 0;
numberOfCompositeFactors = 1;
factorComposites();
}
}
void initializeArrays()
{
for (int i = 0; i<1000;i++)
{
primeFactors[i] = 0;
compositeFactors[i]=0;
tempFactors[i]=0;
}
}
BU getNumberToFactor ()
{
std::string s;
std::cout<<"Enter the number for which you want a prime factor, or 0 to quit: ";
std::cin>>s;
return stringToBigUnsigned(s);
}
void factorComposites()
{
while (numberOfCompositeFactors!=0)
{
compositeIndex = 0;
tempIndex = 0;
// This while loop finds non-zero values in compositeFactors.
// If they are composite, it factors them and puts one factor in tempFactors,
// then divides the element in compositeFactors by the same amount.
// If the element is prime, it moves it into tempFactors (zeros the element in compositeFactors)
while (compositeIndex < 1000)
{
if(compositeFactors[compositeIndex] == 0)
{
compositeIndex++;
continue;
}
if(testForComposite(compositeFactors[compositeIndex]) == false)
{
tempFactors[tempIndex] = compositeFactors[compositeIndex];
compositeFactors[compositeIndex] = 0;
tempIndex++;
compositeIndex++;
}
else
{
tempFactors[tempIndex] = pollard (compositeFactors[compositeIndex]);
compositeFactors[compositeIndex] /= tempFactors[tempIndex];
tempIndex++;
compositeIndex++;
}
}
compositeIndex = 0;
// This while loop moves all remaining non-zero values from compositeFactors into tempFactors
// When it is done, compositeFactors should be all 0 value elements
while (compositeIndex < 1000)
{
if (compositeFactors[compositeIndex] != 0)
{
tempFactors[tempIndex] = compositeFactors[compositeIndex];
compositeFactors[compositeIndex] = 0;
tempIndex++;
compositeIndex++;
}
else compositeIndex++;
}
compositeIndex = 0;
tempIndex = 0;
// This while loop checks all non-zero elements in tempIndex.
// Those that are prime are shown on screen and moved to primeFactors
// Those that are composite are moved to compositeFactors
// When this is done, all elements in tempFactors should be 0
while (tempIndex<1000)
{
if(tempFactors[tempIndex] == 0)
{
tempIndex++;
continue;
}
if(testForComposite(tempFactors[tempIndex]) == false)
{
primeFactors[primeIndex] = tempFactors[tempIndex];
cout<<primeFactors[primeIndex]<<"\n";
tempFactors[tempIndex]=0;
primeIndex++;
tempIndex++;
}
else
{
compositeFactors[compositeIndex] = tempFactors[tempIndex];
tempFactors[tempIndex]=0;
compositeIndex++;
tempIndex++;
}
}
compositeIndex=0;
numberOfCompositeFactors=0;
// This while loop just checks to be sure there are still one or more composite factors.
// As long as there are, the outer while loop will repeat
while(compositeIndex<1000)
{
if(compositeFactors[compositeIndex]!=0) numberOfCompositeFactors++;
compositeIndex ++;
}
}
return;
}
// The following method uses the Miller-Rabin primality test to prove with 100% confidence a given number is composite,
// or to establish with a high level of confidence -- but not 100% -- that it is prime
bool testForComposite (BU num)
{
BU confidenceFactor = 101;
if (confidenceFactor >= num) confidenceFactor = num-1;
BU a,d,s, nMinusOne;
nMinusOne=num-1;
d=nMinusOne;
s=0;
while(modexp(d,1,2)==0)
{
d /= 2;
s++;
}
allJTestsShowComposite = true; // assume composite here until we can prove otherwise
for (BI i = 2 ; i<=confidenceFactor;i++)
{
if (modexp(i,d,num) == 1)
continue; // if this modulus is 1, then we cannot prove that num is composite with this value of i, so continue
if (modexp(i,d,num) == nMinusOne)
{
allJTestsShowComposite = false;
continue;
}
BU exponent(1);
for (BU j(0); j.toInt()<=s.toInt()-1;j++)
{
exponent *= 2;
if (modexp(i,exponent*d,num) == nMinusOne)
{
// if the modulus is not right for even a single j, then break and increment i.
allJTestsShowComposite = false;
continue;
}
}
if (allJTestsShowComposite == true) return true; // proven composite with 100% certainty, no need to continue testing
}
return false;
/* not proven composite in any test, so assume prime with a possibility of error =
(1/4)^(number of different values of i tested). This will be equal to the value of the
confidenceFactor variable, and the "witnesses" to the primality of the number being tested will be all integers from
2 through the value of confidenceFactor.
Note that this makes this primality test cryptographically less secure than it could be. It is theoretically possible,
if difficult, for a malicious party to pass a known composite number for which all of the lowest n integers fail to
detect that it is composite. A safer way is to generate random integers in the outer "for" loop and use those in place of
the variable i. Better still if those random numbers are checked to ensure no duplicates are generated.
*/
}
BU pollard(BU n)
{
if (n == 4) return 2;
BU x = 2;
BU y = 2;
BU d = 1;
int increment = 1;
while(d==1||d==n||d==0)
{
x = f(x,n, increment);
y = f(y,n, increment);
y = f(y,n, increment);
if (y>x)
{
d = gcda(y-x, n);
}
else
{
d = gcda(x-y, n);
}
if (d==0)
{
x = 2;
y = 2;
d = 1;
increment++; // This changes the pseudorandom function we use to increment x and y
}
}
return d;
}
BU gcda(BU a, BU b)
{
if (a==b||a==0)
return 0; // If x==y or if the absolute value of (x-y) == the number to be factored, then we have failed to find
// a factor. I think this is not proof of primality, so the process could be repeated with a new function.
// For example, by replacing x*x+1 with x*x+2, and so on. If many such functions fail, primality is likely.
BU currentGCD = 1;
while (currentGCD!=0) // This while loop is based on Euclid's algorithm
{
currentGCD = b % a;
b=a;
a=currentGCD;
}
return b;
}
BU f(BU x, BU n, int increment)
{
return (x * x + increment) % n;
}
As far as I can see, Pollard Rho normally uses f(x) as (x*x+1) (e.g. in these lecture notes ).
Your choice of x*x-1 appears not as good as it often seems to get stuck in a loop:
x=0
f(x)=-1
f(f(x))=0
have wrote the code for what i see to be a good algorithm for finding the greatest prime factor for a large number using recursion. My program crashes with any number greater than 4 assigned to the variable huge_number though. I am not good with recursion and the assignment does not allow any sort of loop.
#include <stdio.h>
long long prime_factor(int n, long long huge_number);
int main (void)
{
int n = 2;
long long huge_number = 60085147514;
long long largest_prime = 0;
largest_prime = prime_factor(n, huge_number);
printf("%ld\n", largest_prime);
return 0;
}
long long prime_factor (int n, long long huge_number)
{
if (huge_number / n == 1)
return huge_number;
else if (huge_number % n == 0)
return prime_factor (n, huge_number / n);
else
return prime_factor (n++, huge_number);
}
any info as to why it is crashing and how i could improve it would be greatly appreciated.
Even fixing the problem of using post-increment so that the recursion continues forever, this is not a good fit for a recursive solution - see here for why, but it boils down to how fast you can reduce the search space.
While your division of huge_number whittles it down pretty fast, the vast majority of recursive calls are done by simply incrementing n. That means you're going to use a lot of stack space.
You would be better off either:
using an iterative solution where you won't blow out the stack (if you just want to solve the problem) (a); or
finding a more suitable problem for recursion if you're just trying to learn recursion.
(a) An example of such a beast, modeled on your recursive solution, is:
#include <stdio.h>
long long prime_factor_i (int n, long long huge_number) {
while (n < huge_number) {
if (huge_number % n == 0) {
huge_number /= n;
continue;
}
n++;
}
return huge_number;
}
int main (void) {
int n = 2;
long long huge_number = 60085147514LL;
long long largest_prime = 0;
largest_prime = prime_factor_i (n, huge_number);
printf ("%lld\n", largest_prime);
return 0;
}
As can be seen from the output of that iterative solution, the largest factor is 10976461. That means the final batch of recursions in your recursive solution would require a stack depth of ten million stack frames, not something most environments will contend with easily.
If you really must use a recursive solution, you can reduce the stack space to the square root of that by using the fact that you don't have to check all the way up to the number, but only up to its square root.
In addition, other than 2, every other prime number is odd, so you can further halve the search space by only checking two plus the odd numbers.
A recursive solution taking those two things into consideration would be:
long long prime_factor_r (int n, long long huge_number) {
// Debug code for level checking.
// static int i = 0;
// printf ("recursion level = %d\n", ++i);
// Only check up to square root.
if (n * n >= huge_number)
return huge_number;
// If it's a factor, reduce the number and try again.
if (huge_number % n == 0)
return prime_factor_r (n, huge_number / n);
// Select next "candidate" prime to check against, 2 -> 3,
// 2n+1 -> 2n+3 for all n >= 1.
if (n == 2)
return prime_factor_r (3, huge_number);
return prime_factor_r (n + 2, huge_number);
}
You can see I've also removed the (awkward, in my opinion) construct:
if something then
return something
else
return something else
I much prefer the less massively indented code that comes from:
if something then
return something
return something else
But that's just personal preference. In any case, that gets your recursion level down to 1662 (uncomment the debug code to verify) rather than ten million, a rather sizable reduction but still not perfect. That runs okay in my environment.
You meant n+1 instead of n++. n++ increments n after using it, so the recursive call gets the original value of n.
You are overflowing stack, because n++ post-increments the value, making a recursive call with the same values as in the current invocation.
the crash reason is stack overflow. I add a counter to your program and execute it(on ubuntu 10.04 gcc 4.4.3) the counter stop at "218287" before core dump. the better solution is using loop instead of recursion.