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June 6, 2014 17:28
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Source code for Segmented Sieve of Zakiya (SSoZ) for P7.
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/* | |
This C++ source file will compile to an executable program to | |
perform the Segmented Sieve of Zakiya (SSoZ) to find primes <= N. | |
It is based on the P7 Strictly Prime (SP) Prime Generator. | |
Prime Genrators have the form: mod*k + ri; ri -> {1,r1..mod-1} | |
The residues ri are integers coprime to mod, i.e. gcd(ri,mod) = 1 | |
For P7, mod = 2*3*5*7 = 210 and the number of residues are | |
rescnt = (2-1)(3-1)(5-1)(7-1) = 48, which are {1,11,13,17..209}. | |
Adjust segment byte length parameter B (usually L1|l2 cache size) | |
for optimum operational performance for cpu being used. | |
On Linux use -O compiler flag to compile for optimum speed: | |
$ g++ -O ssozp7cpp.cpp -o ssozp7cpp | |
Then run executable: $ ./ssozp7cpp <cr>, and enter value for N. | |
As coded, input values cover the range: 11 -- 2^64-1 | |
Related code, papers, and tutorials, are downloadable here: | |
http://www.4shared.com/folder/TcMrUvTB/_online.html | |
Use of this code is free subject to acknowledgment of copyright. | |
Copyright (c) 2014 Jabari Zakiya -- jzakiya at gmail dot com | |
Version Date: 2014/05/22 | |
This code is provided under the terms of the | |
GNU General Public License Version 3, GPLv3, or greater. | |
License copy/terms are here: http://www.gnu.org/licenses/ | |
*/ | |
#include <cmath> | |
#include <vector> | |
#include <cstdlib> | |
#include <iostream> | |
#include <stdint.h> | |
using namespace std; | |
typedef uint64_t uint64; | |
char pbits[256] = { | |
8,7,7,6,7,6,6,5,7,6,6,5,6,5,5,4,7,6,6,5,6,5,5,4,6,5,5,4,5,4,4,3 | |
,7,6,6,5,6,5,5,4,6,5,5,4,5,4,4,3,6,5,5,4,5,4,4,3,5,4,4,3,4,3,3,2 | |
,7,6,6,5,6,5,5,4,6,5,5,4,5,4,4,3,6,5,5,4,5,4,4,3,5,4,4,3,4,3,3,2 | |
,6,5,5,4,5,4,4,3,5,4,4,3,4,3,3,2,5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1 | |
,7,6,6,5,6,5,5,4,6,5,5,4,5,4,4,3,6,5,5,4,5,4,4,3,5,4,4,3,4,3,3,2 | |
,6,5,5,4,5,4,4,3,5,4,4,3,4,3,3,2,5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1 | |
,6,5,5,4,5,4,4,3,5,4,4,3,4,3,3,2,5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1 | |
,5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1,4,3,3,2,3,2,2,1,3,2,2,1,2,1,1,0 | |
}; | |
int residues[49] = { | |
1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67 | |
, 71, 73, 79, 83, 89, 97,101,103,107,109,113,121,127,131,137,139 | |
,143,149,151,157,163,167,169,173,179,181,187,191,193,197,199,209,211 | |
}; | |
// Global parameters | |
uint B; // segment byte size | |
uint KB; // segment resgroup size | |
uint mod = 210; // prime generator mod value | |
uint bprg = 6; // number of bytes per resgroups | |
uint rescnt = 48; // number of residues for prime generator | |
uint pcnt; // number of primes from r1..sqrt(N) | |
uint64 primecnt; // number of primes <= N | |
uint64 *next; // pointer to array of primes first nonprimes | |
uint *primes; // pointer to array of primes <= sqrt(N) | |
char *seg; // pointer to seg[B] segment byte array | |
void sozP7(uint val) | |
{ | |
int posn[210]; | |
for (int i=0; i < rescnt; i++) posn[residues[i]] = i-1; | |
uint i, k, r, modk; | |
uint num = val-1 | 1; // if value even number then subtract 1 | |
k=num/mod; modk = mod*k; r=1; // kth residue group, base num value | |
while (num >= modk+residues[r]) r++; // find last pc position <= num | |
uint maxprms = k*rescnt + r-1; // maximum number of prime candidates | |
vector<char> prms(maxprms); // array of prime candidates set False | |
uint sqrtN = (uint) sqrt((double) num); | |
modk=0; r=0; k=0; | |
// sieve to eliminate nonprimes from small primes prms array | |
for (i=0; i < maxprms; i++){ | |
r++; if (r > rescnt) {r=1; modk += mod; k++;} | |
if (prms[i]) continue; | |
uint res_r = residues[r]; | |
uint prime = modk + res_r; | |
if (prime > sqrtN) break; | |
uint prmstep = prime * rescnt; | |
for (int ri=1; ri < (rescnt+1); ri++){ | |
uint product = res_r * residues[ri]; // residues cross-product | |
uint nonprm = (k*(prime+residues[ri])+product/mod)*rescnt; | |
nonprm += posn[product % mod]; // residue track value | |
for (; nonprm < maxprms; nonprm += prmstep) prms[nonprm]=true; | |
} | |
} | |
// the prms array now has all the positions for primes r1..N | |
// approximate primes array size; make greater than N/ln(N) | |
uint max = (double) ((num/log( (double) num) * 1.13)+8); | |
primes = new uint[max]; // allocate mem at runtime | |
pcnt = 0; | |
// extract prime numbers and count from prms into prims array | |
modk=0; r=0; | |
for (i=0; i < maxprms; i++){ | |
r++; if (r > rescnt) {r=1; modk +=mod;} | |
if (!prms[i]) primes[pcnt++] = modk + residues[r]; | |
} | |
} | |
void nextinit() | |
{ | |
int pos[mod]; | |
for (int i=1; i < rescnt; ++i) pos[residues[i]] = i-1; | |
pos[1] = rescnt-1; | |
// for each prime store resgroup on each restrack for prime*(modk+ri) | |
for (uint j = 0; j < pcnt; ++j) { // for the pcnt primes r1..sqrt(N) | |
uint prime = primes[j]; // for each prime | |
uint64 k = (prime-2)/mod; // find the resgroup it's in | |
uint r = (prime-2)%mod + 2; // its base residue value | |
for (uint i=1; i <= rescnt; ++i){ // for each residue value | |
uint prod = r * residues[i]; // create residues cross-product r*ri | |
uint row = pos[prod % mod]; // find residue track its on | |
next[row*pcnt + j] = k*(prime + residues[i]) + (prod-2)/mod; | |
} | |
} | |
} | |
void segsieve(uint Kn) | |
{ // for Kn resgroups in segment | |
for (uint b=0; b < Kn*bprg; ++b) // for every byte in the segment | |
seg[b] = 0; // set every byte bit to prime (0) | |
for (uint r=0; r < rescnt; ++r) { // for each residue track | |
uint byti = r >> 3; // for each residues group byte[i] | |
uint biti = 1 << (r&7); // set the ith residue track bit mask | |
uint row = (byti*8 + (r&7))*pcnt; // set address to ith row in next[] | |
for (uint j=0; j < pcnt; ++j){ // for each prime <= sqrt(N) for restrack | |
if (next[row+j] < Kn) { // if 1st mult resgroup index <= seg size | |
uint k = next[row+j]; // get its resgroup value | |
uint ki = k*bprg + byti; // convert it to a byte address in seg[] | |
uint prime = primes[j]; // get prime, convert it to number of | |
uint prmstep = prime * bprg; // bytes to next primenth resgroup byte | |
for (; k < Kn; k += prime){ // for each primeth resgroup in segment | |
seg[ki] |= biti; // set ith residue in byte as nonprime | |
ki += prmstep; // create next nonprime byte address | |
} | |
next[row+j] = k - Kn; // 1st resgroup in next eligible segment | |
} | |
else next[row+j] -= Kn; // if 1st mult resgroup index > seg size | |
} | |
} | |
// count the primes in the segment | |
for (uint s = 0; s < Kn*bprg; ++s) // for the Kn resgroup bytes | |
primecnt += pbits[seg[s] & 0xff]; // count the '0' bits as primes | |
} | |
void printprms(uint Kn, uint64 Ki) | |
{ | |
// Extract and print the primes in each segment: | |
// recommend piping output to less: ./ssozxxx | less | |
// can then use Home, End, Pgup, Pgdn keys to see primes | |
for (uint k = 0; k < Kn; ++k) // for Kn residues groups | |
for (uint r = 0; r < rescnt; ++r) // for each residue bit | |
if (!(seg[k*bprg + r/8] & (1 << r%8))) // if its prime '0', show value | |
cout << mod*(Ki+k) + residues[r+1] << " "; | |
cout << "\n"; | |
} | |
main() | |
{ | |
cout << "Enter number value: "; | |
uint64 val; // find primes <= val (11..2^64-1) | |
cin >> val; | |
B = 262144; // L2D_CACHE_SIZE 256*1024 bytes, I5 cpu | |
bprg = (rescnt-1)/8 + 1; // number of bytes per resgroups | |
KB = B/bprg; // number of resgroups per segment | |
B = KB*bprg; // number of bytes per segment | |
seg = new char[B]; // create segment array of B bytesize | |
cout << "segment has "<< B << " bytes and " << KB << " residues groups\n"; | |
int r; | |
uint64 k, modk; | |
uint64 num = val-1 | 1; // if val even subtract 1 | |
k=num/mod; modk = mod*k; r=1; // kth residue group, base num value | |
while (num >= modk+residues[r]) r++; // find last pc position <= num | |
uint64 maxpcs =k*rescnt + r-1; // maximum number of prime candidates | |
uint64 Kmax = (num-2)/mod + 1; // maximum number of resgroups for val | |
cout <<"prime candidates = "<< maxpcs <<"; resgroups = "<< Kmax << endl; | |
uint sqrtN = (uint) sqrt((double) num); | |
sozP7(sqrtN); // get pcnt and primes <= sqrt(nun) | |
cout << "create next["<< rescnt << "x" << pcnt << "] array\n"; | |
next = new uint64[rescnt*pcnt]; // create the next[] array | |
nextinit(); // load with first nonprimes resgroups | |
primecnt = 4; // 2,3,5,7 the P7 excluded primes count | |
uint Kn = KB; // set sieve resgroups size to segment size | |
uint64 Ki = 0; // starting resgroup index for each segment | |
cout << "perform segmented SoZ\n"; | |
for (; Ki < Kmax; Ki += KB) { // for KB size resgroup slices up to Kmax | |
if (Kmax-Ki < KB) Kn=Kmax-Ki; // set sieve resgroups size for last segment | |
segsieve(Kn); // sieve primes for current segment | |
//printprms(Kn,Ki); // print primes for the segment (optional) | |
} | |
uint64 lprime=0; // get last prime and primecnt <= val | |
modk = mod*(Kmax-1); // mod for last resgroup in last segment | |
uint b = (Kn-1)*bprg; // num bytes to last resgroup in segment | |
r = rescnt-1; // from last restrack in resgroup | |
while (true) { // repeat until last prime determined | |
if (!(seg[b + r/8] & 1 << (r&7))) { // if restrack in byte[i] is prime | |
lprime = modk+residues[r+1]; // determine the prime value | |
if (lprime <= num) break; // if <= num exit from loop with lprime | |
else primecnt--; // else reduce total primecnt | |
} // reduce restrack, setup next resgroup | |
r--; if (r < 0) {r=rescnt-1; modk -= mod; b -= bprg;} // if necessary | |
} | |
cout << "last segment = " << Kn << " resgroups; segment iterations = " << Ki/KB << "\n"; | |
cout << "last prime (of " << primecnt << ") is " << lprime << endl; | |
delete[] next; delete[] primes; | |
return 0; | |
} |
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