## Efficient Prime Factorization for large numbers

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number of prime factors

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how to find prime factors of a large number in python

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I've been working on a little problem where I need to compute 18-digit numbers into their respective prime factorization. Everything compiles and it runs just fine, considering that it actually works, but I am looking to reduce the run time of the prime factorization. I have implemented recursion and threading but I think I might need some help in understanding possible algorithms for large number computation.

Every time I run this on the 4 numbers I have pre-made, it takes about 10 seconds. I would like to reduce this to possibly 0.06 seconds if there are any ideas out there.

I noticed a few algorithms like Sieve of Eratosthenes and producing a list of all the prime numbers prior to computing. I'm just wondering if someone could elaborate on it. For instance, I'm having issues understanding how to implement Sieve of Eratosthenes into my program or if it would even be a good idea. Any and all pointers on how to approach this better would be really helpful!

Here is my code:

#include <iostream> #include <thread> #include <vector> #include <chrono> using namespace std; using namespace std::chrono; vector<thread> threads; vector<long long> inputVector; bool developer = false; vector<unsigned long long> factor_base; vector<long long> primeVector; class PrimeNumber { long long initValue; // the number being prime factored vector<long long> factors; // all of the factor values public: void setInitValue(long long n) { initValue = n; } void addToVector(long long m) { factors.push_back(m); } void setVector(vector<long long> m) { factors = m; } long long getInitValue() { return initValue; } vector<long long> getVector() { return factors; } }; vector<PrimeNumber> primes; // find primes recursively and have them returned in vectors vector<long long> getPrimes(long long n, vector<long long> vec) { double sqrt_of_n = sqrt(n); for (int i = 2; i <= sqrt_of_n; i++) { if (n % i == 0) { return vec.push_back(i), getPrimes(n / i, vec); //cause recursion } } // pick up the last prime factorization number vec.push_back(n); //return the finished vector return vec; } void getUserInput() { long long input = -1; cout << "Enter all of the numbers to find their prime factors. Enter 0 to compute" << endl; do { cin >> input; if (input == 0) { break; } inputVector.push_back(input); } while (input != 0); } int main() { vector<long long> temp1; // empty vector vector<long long> result1; // temp vector if (developer == false) { getUserInput(); } else { cout << "developer mode active" << endl; long long a1 = 771895004973090566; long long b1 = 788380500764597944; long long a2 = 100020000004324000; long long b2 = 200023423420000000; inputVector.push_back(a1); inputVector.push_back(b2); inputVector.push_back(b1); inputVector.push_back(a2); } high_resolution_clock::time_point time1 = high_resolution_clock::now(); // give each thread a number to comput within the recursive function for (int i = 0; i < inputVector.size(); i++) { PrimeNumber prime; prime.setInitValue(inputVector.at(i)); threads.push_back(thread([&]{ prime.setVector(result1 = getPrimes(inputVector.at(i), temp1)); primes.push_back(prime); })); } // allow all of the threads to join back together. for (auto& th : threads) { cout << th.get_id() << endl; th.join(); } high_resolution_clock::time_point time2 = high_resolution_clock::now(); // print all of the information for (int i = 0; i < primes.size(); i++) { vector<long long> temp = primes.at(i).getVector(); for (int m = 0; m < temp.size(); m++) { cout << temp.at(m) << " "; } cout << endl; } cout << endl; // so the running time auto duration = duration_cast<microseconds>(time2 - time1).count(); cout << "Duration: " << (duration / 1000000.0) << endl; return 0; }

Trial division is only suitable for factoring small numbers. For *n* up to 2^64, you'll need a better algorithm: I recommend starting with wheel factorization to get the small factors, followed by Pollard's rho algorithm to get the rest. Where trial division is O(sqrt(n)), rho is O(sqrt(sqrt(n))), so it's much faster. For 2^64, sqrt(n) = 2^32, but sqrt(sqrt(n)) = 2^16, which is a huge improvement. You should expect to factor your numbers in a few milliseconds, at most.

I don't have C++ code for factoring, but I do have readable Python code. Let me know if you want me to post it. If you want to know more about wheel factorization and the rho algorithm, I have lots of prime number stuff at my blog.

**Integer factorization,** In number theory, integer factorization is the decomposition of a composite number into a product of smaller integers. If these factors are further restricted to prime numbers, the process is called prime factorization. When the numbers are sufficiently large, no efficient, non-quantum integer Prime factors of a big number. Given a number N, print all the prime factors and their powers. Examples : Input : 250 Output : 2 1 5 3 Explanation: The prime factors of 250 are 2 and 5. 2 appears once in the prime factorization of and 5 is thrice in it.

for(int i = 2; i * i <= n; ++i) //no sqrt, please { while(n%i == 0) //while, not if { factors.push_back(i); n/=i; } } if(n != 1) { factors.push_back(n); }

This is basically a neater implementation of your algorithm. Its complexity is sqrt of N. It will work pretty quickly even for a 18-digit number, but only if the prime factors are all small. If it's a product of two large prime numbers, or worse, is prime itself, this will run for approximately 10 seconds.

**FAST Prime Factorization (5-digit number),** Trial division is only suitable for factoring small numbers. For n up to 2^64, you'll need a better algorithm: I recommend starting with wheel Efficient program to print all prime factors of a given number. Given a number n, write an efficient function to print all prime factors of n. For example, if the input number is 12, then output should be “2 2 3”. And if the input number is 315, then output should be “3 3 5 7”.

I found that the Sieve of Eratosthenes on your modern processor thrashes the cache, so that main memory bandwidth is the limiting factor. I found this when trying to run multiple threads and failing to speed things up by as much as I was hoping for.

So, I recommend breaking the sieve into segments which will fit in the L3 cache. Also, if you exclude multiples of 2, 3 and 5 from the bit vector, then an 8 bit byte can represent 30 numbers on the number line, with 1 bit for each number which is 1, 7, 11, 13, 17, 19, 23 or 29 modulo 30 -- so that a bit map for primes up to 10^9 takes ~32MB -- 10^9 / (30 * 1024 * 1024). This is almost half the size of a bit map which just excludes multiples of 2, which is ~60MB -- 10^9 / (2 * 8 * 1024 * 1024).

Obviously, to run the sieve up to 10^9 you need the primes up to sqrt(10^9) -- which requires some 1,055 bytes, from which you can generate any part of the full sieve up to 10^9.

FWIW, the results I get on a modest AMD Phenom II x6 1090T (8MB L3 cache), for primes up to 10^9 are:

1. 1 core, 1 segment 3.260 seconds elapsed 2. 5 cores, 1 segment 1.830 seconds elapsed 3. 1 core, 8 segments 1.800 seconds elapsed 4. 5 cores, 40 segments 0.370 seconds elapsed

where by "segment" I mean a part of the sieve. In this case the sieve is ~32MB, so where there are multiple segments they are using about 4MB of L3 cache at any one time.

Those times include the time required to scan the completed sieve and generate all the primes as an array of integers. That takes about 0.5 secs of CPU ! So, to run the sieve without actually extracting the primes from it, takes 0.270 seconds elapsed in case (4) above.

FWIW, I get a small improvement -- to 0.240 seconds in case (4) -- by initialising each segment using a precalculated pattern that removes multiples of 7, 11, 13 and 17. That pattern is 17,017 bytes.

Clearly, to do a single factorization in 0.06 secs... you need the sieve to be pre-computed !

**Efficient Prime Factorization for large numbers,** There are many ways to factor numbers. One approach I like is the Lehmer sieve which is a mechanical device that tries to find a pair of numbers x,y such that The Prime Number factors are: WolframAlpha also provides accurate and efficient prime-number factorizations for large numbers. SOCR Resource Visitor number , since Jan. 01, 2002

A simple speedup of two can easily be achieved by changing your loop:

if (n % 2) { return vec.push_back(i), getPrimes(n / i, vec); } for (int i = 3; i <= sqrt_of_n; i += 2) { if (n % i == 0) { return vec.push_back(i), getPrimes(n / i, vec); //cause recursion } }

You first should test the number by two. Then, starting from 3 you test again incrementing your loop by two at a time. You already know thay 4, 6, 8, ... are even numbers and have 2 as a factor. Testing against even numbers you're reducing your complexity by half.

To factor a number `N`

you only need the prime numbers <= sqrt(N). For a 18 digit number you only need to test against all primes less than `1e9`

, and since there are 98 millon primes less than `2e9`

you can easily store 100 millon numbers on today's computers and run the factoring in parallel. If each number takes 8 bytes of RAM (`int64_t`

), 100 millon primes would take 800 MB of memory. This algorithm is the classic solution to SPOJ problem #2, Prime Generator.

The best way to list all the small primes that can fit on a 32-bit int is to build a Sieve of Eratostenes. I told you that we need the primes less than sqrt(N) to factor any N, so to factor 64 bit integers you need all the primes that fit as a 32-bit number.

**How to find prime factors of big numbers made up of big prime ,** As you find factors you can divide them out. If there are no factors smaller than √n the number is prime, which is much smaller (for large n) than To factor a number N you only need the prime numbers <= sqrt(N). For a 18 digit number you only need to test against all primes less than 1e9, and since there are 98 millon primes less than 2e9 you can easily store 100 millon numbers on today's computers and run the factoring in parallel.

Algorithm :

- Divide number by 2 until it is not divisible by it ,store the result and display it .
- Divide number by 3 until it is not divisible by 3 and display the result,
- Repeat same process for 5,7... etc till square root of n.
If at the end resultant number is prime , display its count as 1.

int main() { long long n; cin >> n; int count =0 ; while(!(n%2)){ n = n / 2; count++; } if(count > 0) { cout<<"2^"<<count<<" "; } for(long long i=3;i<=sqrt(n) ; i+=2){ count=0; while(n%i == 0){ count++; n = n/i; } if(count){ cout << i <<"^" <<count<<" "; } } if(n>2){ cout<<n <<"^1"; } return 0; }

Input : 100000000 Output 2^8 5^8

**Fastest way to find all the prime factors of a very large number ,** The first few prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19 and 23, and we have a prime "Prime Factorization" is finding which prime numbers multiply together to make That is because factoring very large numbers is very hard, and can take The first few prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19 and 23, and we have a prime number chart if you need more. If we can make it by multiplying other whole numbers it is a Composite Number. Here we see it in action: 2 is Prime, 3 is Prime, 4 is Composite (=2×2), 5 is Prime, and so on

**Prime Factorization,** Mathematicians are working all the time trying to find faster ways of factoring, because we don't know of fast ways of factoring such large numbers. For a number, N You have not made much of a distinction about large numbers or small numbers; for the moment small numbers are good enough. At some point I realized I wanted a C++ command to factor any 32 bit signed integer, so absolute value up to $2^{31} - 1$ or 2,147,483,647.

**Which is the fastest prime factorization algorithm to date?,** Is there a way to factor really big numbers into primes other than the usual It is very effective for numbers up to about 1000 when doing the A prime number is a natural number (greater than 1) that has exactly two factors, 1 and itself. In order to check if a number is prime or not, we can count the number of factors. If it is 2, then we say that the number is prime, else it is a composite number. Side note, non-prime numbers are called composite numbers. Table of Contents

**Factoring Very Large Numbers into Primes - Math Forum,** These two JavaScript calculators compute the prime factorization for large also provides accurate and efficient prime-number factorizations for large numbers. Few additional thoughts to Fast Number Factorization in Python answer. is_prime() In case if you have multiple consequent calls you should use something like Sieve_of_Eratosthenes. If you will, time to generate sieve will depend on maximum value of number to factorize but total time will be reduces. prime_factors()

##### Comments

- This question is better suited for Code Review
- oh shoot haha didnt even know they had a section for this. thanks!
- possible duplicate of Problems with prime numbers
- It's good that you stop looking for factors when you reach sqrt(n), but it's not so good that when you recurse, you start again at 2. If no number smaller than
`i`

is a factor of`n`

, then no number smaller than`i`

is a factor of`n/i`

either, and there's no need checking all of those again. In particular, if`i`

is the smallest factor of`n`

and`i`

is greater than the*cube root*of`n`

, then`i`

and`n/i`

are the only factors of`n`

. It's not necessary to do that specific check, though; it will be automatic if you start the next search at`i`

. - Hmm. Your first test number,
`a1 = 771895004973090566`

, can be factored in less than 1/2000 second (or better), because it is 2 x 385947502486545283. The factor 2 is of course found instantly. Then, 385947502486545283 is easily determined to be prime using Miller–Rabin. Similarly,`a2 = 788380500764597944`

can be factored almost instantly to 2 x 2 x 2 x 7 x 14078223227939249. The challenge is actually to factor hard semiprimes like 18436839306515468081 = 2988873347 x 6168491323, and for that you want Shanks's Square Forms Factorization, Hart's One-Line Factorization, or Brent–Pollard Rho. - While I agree with this answer (+1), Pollard rho works in expected O(sqrt(p)) time where p is the prime factor. So why not just use Pollard rho the whole way through to make things simpler? It will find the small prime factors very fast, so probably no noticeable slow down. Also, if someone wants guaranteed running time, Pollard rho only gives expected running time, and worse, the "expected" time relies on the assumption that a pseudo-random polynomial mod n will behave enough like a random sequence to make the random assumption valid to compute expected running time.
- To use Pollard rho, you have to at least remove factors of 2. Pollard rho may yield composite factors, so you have to test primality, which slows things down, and trial division lets you get small prime factors without need of a primality test. I agree with you that the crossover point from trial division to rho is low; I typically use a 2,3,5-wheel to 10000, then cut over to rho.
- It can be faster to compute
`sqrt`

once than to compute`i*i`

repeatedly. - @MarkRansom It is better to take i*i than to take sqrt as multiplication is faster than taking square root.
- @ABcDexter it depends. If you only need to do
`sqrt`

once, but you need to do`i*i`

1000 times, the`sqrt`

can be faster. - @MarkRansom Can you explain this comment please ? How are they differ ? If you take sqrt(n) and check with i - or if you take i^2 and check with n ... Mathematically these are equivalent am I wrong ?
- @BeTa yes they're equivalent, that's the point. The question is which is
*faster*. If you look at the construction of the loop,`i*i`

must be recomputed each time`i`

changes, which is each iteration of the loop. On the other hand`sqrt(n)`

must be computed only once. Even though the computation of`sqrt(n)`

is slower than`i*i`

, once you factor in all the repetitions`sqrt`

is faster overall. - Packing the primes mod 30 (sans 2,3,5) into a single byte is brilliant. Did you think that up? Very nice.
- This algorithm is capable of producing correct results, but it's akin to suggesting someone use a Bubble Sort for sorting a list. Outside of certain learning scenarios, it's just not the right tool for the job.