问题
I've written a program that attempts to find Amicable Pairs. This requires finding the sums of the proper divisors of numbers.
Here is my current sumOfDivisors() method:
int sumOfDivisors(int n)
{
int sum = 1;
int bound = (int) sqrt(n);
for(int i = 2; i <= 1 + bound; i++)
{
if (n % i == 0)
sum = sum + i + n / i;
}
return sum;
}
So I need to do lots of factorization and that is starting to become the real bottleneck in my application. I typed a huge number into MAPLE and it factored it insanely fast.
What is one of the faster factorization algorithms?
回答1:
Pulled directly from my answer to this other question.
The method will work, but will be slow. "How big are your numbers?" determines the method to use:
- Less than 2^16 or so: Lookup table.
- Less than 2^70 or so: Richard Brent's modification of Pollard's rho algorithm.
- Less than 10^50: Lenstra elliptic curve factorization
- Less than 10^100: Quadratic Sieve
- More than 10^100: General Number Field Sieve
回答2:
The question in the title (and the last line) seems to have little to do with the actual body of the question. If you're trying to find amicable pairs, or computing the sum of divisors for many numbers, then separately factorising each number (even with the fastest possible algorithm) is absolutely an inefficient way to do it.
The sum-of-divisors function, σ(n) = (sum of divisors of n), is a multiplicative function: for relatively prime m and n, we have σ(mn) = σ(m)σ(n), so
σ(p1k1…prkr) = [(p1k1+1-1)/(p1-1)]…[(prkr+1-1)/(pr-1)].
So you would use any simple sieve (e.g. an augmented version of the Sieve of Eratosthenes) to find the primes up to n, and, in the process, the factorisation of all numbers up to n. (For example, as you do your sieve, store the smallest prime factor of each n. Then you can later factorize any number n by iterating.) This would be faster (overall) than using any separate factorization algorithm several times.
BTW: several known lists of amicable pairs already exist (see e.g. here and the links at MathWorld) – so are you trying to extend the record, or doing it just for fun?
回答3:
Shor's Algorithm: http://en.wikipedia.org/wiki/Shor%27s_algorithm
Of course you need a quantum computer though :D
回答4:
I would suggest starting from the same algorithm used in Maple, the Quadratic Sieve.
- Choose your odd number n to factorize,
- Choose a natural number k,
- Search all p <= k so that k^2 is not congruent to (n mod p) to obtain a factor base B = p1, p2, ..., pt,
- Starting from r > floor(n) search at least t+1 values so that y^2 = r^2 - n all have just factors in B,
- For every y1, y2, ..., y(t+1) just calculated you generate a vector v(yi) = (e1, e2, ..., et) where ei is calculated by reducing over modulo 2 the exponent pi in yi,
- Use Gaussian Elimination to find some of the vectors that added together give a null vector
- Set x as the product of ri related to yi found in the previous step and set y as p1^a * p2^b * p3^c * .. * pt^z where exponents are the half of the exponents found in the factorization of yi
- Calculate d = mcd(x-y, n), if 1 < d < n then d is a non-trivial factor of n, otherwise start from step 2 choosing a bigger k.
The problem about these algorithms is that they really imply a lot of theory in numerical calculus..
回答5:
This is a paper of the Integer Factorization in Maple.
"Starting from some very simple instructions—“make integer factorization faster in Maple” — we have implemented the Quadratic Sieve factoring algorithm in a combination of Maple and C..."
http://www.cecm.sfu.ca/~pborwein/MITACS/papers/percival.pdf
回答6:
Depends how big your numbers are. If you're searching for amicable pairs you're doing a lot of factorisations, so the key may not be to factor as quickly as possible, but to share as much work as possible between different calls. To speed up trial division you could look at memoization, and/or precalculating primes up to the square root of the biggest number you care about. It's quicker to get the prime factorisation, then calculate the sum of all factors from that, than it is to loop all the way up to sqrt(n) for every number.
If you're looking for really big amicable pairs, say bigger than 2^64, then on a small number of machines you can't do it by factorising every single number no matter how fast your factorisation is. The short-cuts which you're using to find candidates might help you factor them.
回答7:
A more 2015 C++ version 227 lookup table implementation for 1GB memory:
#include <iostream.h> // cerr, cout, and NULL
#include <string.h> // memcpy()
#define uint unsigned __int32
uint *factors;
const uint MAX_F=134217728; // 2^27
void buildFactors(){
factors=new (nothrow) uint [(MAX_F+1)*2]; // 4 * 2 * 2^27 = 2^30 = 1GB
if(factors==NULL)return; // not able to allocate enough free memory
int i;
for(i=0;i<(MAX_F+1)*2;i++)factors[i]=0;
//Sieve of Eratosthenese
factors[1*2]=1;
factors[1*2+1]=1;
for(i=2;i*i<=MAX_F;i++){
for(;factors[i*2] && i*i<=MAX_F;i++);
factors[i*2]=1;
factors[i*2+1]=i;
for(int j=2;i*j<=MAX_F;j++){
factors[i*j*2]=i;
factors[i*j*2+1]=j;
}
}
for(;i<=MAX_F;i++){
for(;i<=MAX_F && factors[i*2];i++);
if(i>MAX_F)return;
factors[i*2]=1;
factors[i*2+1]=i;
}
}
uint * factor(uint x, int &factorCount){
if(x > MAX_F){factorCount=-1;return NULL;}
uint tmp[70], at=x; int i=0;
while(factors[at*2]>1){
tmp[i++]=factors[at*2];
cout<<"at:"<<at<<" tmp:"<<tmp[i-1]<<endl;
at=factors[at*2+1];
}
if(i==0){
cout<<"at:"<<x<<" tmp:1"<<endl;
tmp[i++]=1;
tmp[i++]=x;
}else{
cout<<"at:"<<at<<" tmp:1"<<endl;
tmp[i++]=at;
}
factorCount=i;
uint *ret=new (nothrow) uint [factorCount];
if(ret!=NULL)
memcpy(ret, tmp, sizeof(uint)*factorCount);
return ret;
}
void main(){
cout<<"Loading factors lookup table"<<endl;
buildFactors(); if(factors==NULL){cerr<<"Need 1GB block of free memory"<<endl;return;}
int size;
uint x=30030;
cout<<"\nFactoring: "<<x<<endl;
uint *f=factor(x,size);
if(size<0){cerr<<x<<" is too big to factor. Choose a number between 1 and "<<MAX_F<<endl;return;}
else if(f==NULL){cerr<<"ran out of memory trying to factor "<<x<<endl;return;}
cout<<"\nThe factors of: "<<x<<" {"<<f[0];
for(int i=1;i<size;i++)
cout<<", "<<f[i];
cout<<"}"<<endl;
delete [] f;
x=30637;
cout<<"\nFactoring: "<<x<<endl;
f=factor(x,size);
cout<<"\nThe factors of: "<<x<<" {"<<f[0];
for(int i=1;i<size;i++)
cout<<", "<<f[i];
cout<<"}"<<endl;
delete [] f;
delete [] factors;
}
Update: or sacrificing some simplicity for a bit more range just past 228
#include <iostream.h> // cerr, cout, and NULL
#include <string.h> // memcpy(), memset()
//#define dbg(A) A
#ifndef dbg
#define dbg(A)
#endif
#define uint unsigned __int32
#define uint8 unsigned __int8
#define uint16 unsigned __int16
uint * factors;
uint8 *factors08;
uint16 *factors16;
uint *factors32;
const uint LIMIT_16 = 514; // First 16-bit factor, 514 = 2*257
const uint LIMIT_32 = 131074;// First 32-bit factor, 131074 = 2*65537
const uint MAX_FACTOR = 268501119;
//const uint64 LIMIT_64 = 8,589,934,594; // First 64-bit factor, 2^33+1
const uint TABLE_SIZE = 268435456; // 2^28 => 4 * 2^28 = 2^30 = 1GB 32-bit table
const uint o08=1, o16=257 ,o32=65665; //o64=4294934465
// TableSize = 2^37 => 8 * 2^37 = 2^40 1TB 64-bit table
// => MaxFactor = 141,733,953,600
/* Layout of factors[] array
* Indicies(32-bit) i Value Size AFactorOf(i)
* ---------------- ------ ---------- ----------------
* factors[0..128] [1..513] 8-bit factors08[i-o08]
* factors[129..65408] [514..131073] 16-bit factors16[i-o16]
* factors[65409..268435455] [131074..268501119] 32-bit factors32[i-o32]
*
* Note: stopping at i*i causes AFactorOf(i) to not always be LargestFactor(i)
*/
void buildFactors(){
dbg(cout<<"Allocating RAM"<<endl;)
factors=new (nothrow) uint [TABLE_SIZE]; // 4 * 2^28 = 2^30 = 1GB
if(factors==NULL)return; // not able to allocate enough free memory
uint i,j;
factors08 = (uint8 *)factors;
factors16 = (uint16 *)factors;
factors32 = factors;
dbg(cout<<"Zeroing RAM"<<endl;)
memset(factors,0,sizeof(uint)*TABLE_SIZE);
//for(i=0;i<TABLE_SIZE;i++)factors[i]=0;
//Sieve of Eratosthenese
//8-bit values
dbg(cout<<"Setting: 8-Bit Values"<<endl;)
factors08[1-o08]=1;
for(i=2;i*i<LIMIT_16;i++){
for(;factors08[i-o08] && i*i<LIMIT_16;i++);
dbg(cout<<"Filtering: "<<i<<endl;)
factors08[i-o08]=1;
for(j=2;i*j<LIMIT_16;j++)factors08[i*j-o08]=i;
for(;i*j<LIMIT_32;j++)factors16[i*j-o16]=i;
for(;i*j<=MAX_FACTOR;j++)factors32[i*j-o32]=i;
}
for(;i<LIMIT_16;i++){
for(;i<LIMIT_16 && factors08[i-o08];i++);
dbg(cout<<"Filtering: "<<i<<endl;)
if(i<LIMIT_16){
factors08[i-o08]=1;
j=LIMIT_16/i+(LIMIT_16%i>0);
for(;i*j<LIMIT_32;j++)factors16[i*j-o16]=i;
for(;i*j<=MAX_FACTOR;j++)factors32[i*j-o32]=i;
}
}i--;
dbg(cout<<"Setting: 16-Bit Values"<<endl;)
//16-bit values
for(;i*i<LIMIT_32;i++){
for(;factors16[i-o16] && i*i<LIMIT_32;i++);
factors16[i-o16]=1;
for(j=2;i*j<LIMIT_32;j++)factors16[i*j-o16]=i;
for(;i*j<=MAX_FACTOR;j++)factors32[i*j-o32]=i;
}
for(;i<LIMIT_32;i++){
for(;i<LIMIT_32 && factors16[i-o16];i++);
if(i<LIMIT_32){
factors16[i-o16]=1;
j=LIMIT_32/i+(LIMIT_32%i>0);
for(;i*j<=MAX_FACTOR;j++)factors32[i*j-o32]=i;
}
}i--;
dbg(cout<<"Setting: 32-Bit Values"<<endl;)
//32-bit values
for(;i*i<=MAX_FACTOR;i++){
for(;factors32[i-o32] && i*i<=MAX_FACTOR;i++);
factors32[i-o32]=1;
for(j=2;i*j<=MAX_FACTOR;j++)factors32[i*j-o32]=i;
}
for(;i<=MAX_FACTOR;i++){
for(;i<=MAX_FACTOR && factors32[i-o32];i++);
if(i>MAX_FACTOR)return;
factors32[i-o32]=1;
}
}
uint * factor(uint x, int &factorCount){
if(x > MAX_FACTOR){factorCount=-1;return NULL;}
uint tmp[70], at=x; int i=0;
while(at>=LIMIT_32 && factors32[at-o32]>1){
tmp[i++]=factors32[at-o32];
dbg(cout<<"at32:"<<at<<" tmp:"<<tmp[i-1]<<endl;)
at/=tmp[i-1];
}
if(at<LIMIT_32){
while(at>=LIMIT_16 && factors16[at-o16]>1){
tmp[i++]=factors16[at-o16];
dbg(cout<<"at16:"<<at<<" tmp:"<<tmp[i-1]<<endl;)
at/=tmp[i-1];
}
if(at<LIMIT_16){
while(factors08[at-o08]>1){
tmp[i++]=factors08[at-o08];
dbg(cout<<"at08:"<<at<<" tmp:"<<tmp[i-1]<<endl;)
at/=tmp[i-1];
}
}
}
if(i==0){
dbg(cout<<"at:"<<x<<" tmp:1"<<endl;)
tmp[i++]=1;
tmp[i++]=x;
}else{
dbg(cout<<"at:"<<at<<" tmp:1"<<endl;)
tmp[i++]=at;
}
factorCount=i;
uint *ret=new (nothrow) uint [factorCount];
if(ret!=NULL)
memcpy(ret, tmp, sizeof(uint)*factorCount);
return ret;
}
uint AFactorOf(uint x){
if(x > MAX_FACTOR)return -1;
if(x < LIMIT_16) return factors08[x-o08];
if(x < LIMIT_32) return factors16[x-o16];
return factors32[x-o32];
}
void main(){
cout<<"Loading factors lookup table"<<endl;
buildFactors(); if(factors==NULL){cerr<<"Need 1GB block of free memory"<<endl;return;}
int size;
uint x=13855127;//25255230;//30030;
cout<<"\nFactoring: "<<x<<endl;
uint *f=factor(x,size);
if(size<0){cerr<<x<<" is too big to factor. Choose a number between 1 and "<<MAX_FACTOR<<endl;return;}
else if(f==NULL){cerr<<"ran out of memory trying to factor "<<x<<endl;return;}
cout<<"\nThe factors of: "<<x<<" {"<<f[0];
for(int i=1;i<size;i++)
cout<<", "<<f[i];
cout<<"}"<<endl;
delete [] f;
x=30637;
cout<<"\nFactoring: "<<x<<endl;
f=factor(x,size);
cout<<"\nThe factors of: "<<x<<" {"<<f[0];
for(int i=1;i<size;i++)
cout<<", "<<f[i];
cout<<"}"<<endl;
delete [] f;
delete [] factors;
}
来源：https://stackoverflow.com/questions/2267146/what-is-the-fastest-factorization-algorithm