Square form factorization: Difference between revisions
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m Minor code improvement |
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Line 449:
for ( uint32_t i = 0; i < 20; ++i ) {
uint64_t test = distribution(random);
uint64_t factor = squfof(test);
if ( factor == 0 ) {
std::cout << test << "
} else {
std::cout << test << " = " << factor << " * " << test / factor << std::endl;
}
std::cout << std::endl;
Line 463 ⟶ 462:
{{ out }}
<pre>
</pre>
|
Revision as of 10:36, 28 September 2023
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
You are encouraged to solve this task according to the task description, using any language you may know.
- Task.
Daniel Shanks's Square Form Factorization (SquFoF).
Invented around 1975, ‘On a 32-bit computer, SquFoF is the clear champion factoring algorithm for numbers between 1010 and 1018, and will likely remain so.’
An integral binary quadratic form is a polynomial
f(x,y) = ax2 + bxy + cy2
with integer coefficients and discriminant D = b2 – 4ac.
For each positive discriminant there are multiple forms (a, b, c).
The next form in a periodic sequence (cycle) of adjacent forms is found by applying a reduction operator rho, essentially a variant of Euclid's algorithm for finding the continued fraction of a square root. Using floor(√N), rho constructs a principal form (1, b, c) with D = 4N.
SquFoF is based on the existence of cycles containing ambiguous forms, with the property that a divides b. They come in pairs of associated forms (a, b, c) and (c, b, a) called symmetry points. If an ambiguous form is found (there is one for each divisor of D), write the discriminant as (ak)2 – 4ac = a(a·k2 – 4c) = 4N and (if a is not equal to 1 or 2) N is split.
Shanks used square forms to jump to a random ambiguous cycle. Fact: if any form in an ambiguous cycle is squared, that square form will always land in the principal cycle. Conversely, the square root of any form in the principal cycle lies in an ambiguous cycle. (Possibly the principal cycle itself).
A square form is easy to find: the last coefficient c is a perfect square. This happens about once every ∜N-th cycle step and for even indices only. Let rho compute the inverse square root form and track the ambiguous cycle backward until the symmetry point is reached. (Taking the inverse reverses the cycle). Then a or a/2 divides D and therefore N.
To avoid trivial factorizations, Shanks created a list (queue) to hold small coefficients appearing early in the principal cycle, that may be roots of square forms found later on. If these forms are skipped, no roots land in the principal cycle itself and cases a = 1 or a = 2 do not happen.
Sometimes the cycle length is too short to find a proper square form. This is fixed by running five instances of SquFoF in parallel, with input N and 3, 5, 7, 11 times N; the discriminants then will have different periods. If N is prime or the cube of a prime, there are improper squares only and the program will duly report failure.
- Reference.
[1] A detailed analysis of SquFoF (2007)
C
Based on Wp
From the Wikipedia entry for Shanks's square forms factorization [[2]]
#include <math.h>
#include <stdio.h>
#define nelems(x) (sizeof(x) / sizeof((x)[0]))
const unsigned long multiplier[] = {1, 3, 5, 7, 11, 3*5, 3*7, 3*11, 5*7, 5*11, 7*11, 3*5*7, 3*5*11, 3*7*11, 5*7*11, 3*5*7*11};
unsigned long long gcd(unsigned long long a, unsigned long long b)
{
while (b != 0)
{
a %= b;
a ^= b;
b ^= a;
a ^= b;
}
return a;
}
unsigned long long SQUFOF( unsigned long long N )
{
unsigned long long D, Po, P, Pprev, Q, Qprev, q, b, r, s;
unsigned long L, B, i;
s = (unsigned long long)(sqrtl(N)+0.5);
if (s*s == N) return s;
for (int k = 0; k < nelems(multiplier) && N <= 0xffffffffffffffff/multiplier[k]; k++) {
D = multiplier[k]*N;
Po = Pprev = P = sqrtl(D);
Qprev = 1;
Q = D - Po*Po;
L = 2 * sqrtl( 2*s );
B = 3 * L;
for (i = 2 ; i < B ; i++) {
b = (unsigned long long)((Po + P)/Q);
P = b*Q - P;
q = Q;
Q = Qprev + b*(Pprev - P);
r = (unsigned long long)(sqrtl(Q)+0.5);
if (!(i & 1) && r*r == Q) break;
Qprev = q;
Pprev = P;
};
if (i >= B) continue;
b = (unsigned long long)((Po - P)/r);
Pprev = P = b*r + P;
Qprev = r;
Q = (D - Pprev*Pprev)/Qprev;
i = 0;
do {
b = (unsigned long long)((Po + P)/Q);
Pprev = P;
P = b*Q - P;
q = Q;
Q = Qprev + b*(Pprev - P);
Qprev = q;
i++;
} while (P != Pprev);
r = gcd(N, Qprev);
if (r != 1 && r != N) return r;
}
return 0;
}
int main(int argc, char *argv[]) {
int i;
const unsigned long long data[] = {
2501,
12851,
13289,
75301,
120787,
967009,
997417,
7091569,
13290059,
42854447,
223553581,
2027651281,
11111111111,
100895598169,
1002742628021,
60012462237239,
287129523414791,
9007199254740931,
11111111111111111,
314159265358979323,
384307168202281507,
419244183493398773,
658812288346769681,
922337203685477563,
1000000000000000127,
1152921505680588799,
1537228672809128917,
4611686018427387877};
for(int i = 0; i < nelems(data); i++) {
unsigned long long example, factor, quotient;
example = data[i];
factor = SQUFOF(example);
if(factor == 0) {
printf("%llu was not factored.\n", example);
}
else {
quotient = example / factor;
printf("Integer %llu has factors %llu and %llu\n",
example, factor, quotient);
}
}
}
- Output:
Integer 2501 has factors 61 and 41 Integer 12851 has factors 71 and 181 Integer 13289 has factors 137 and 97 Integer 75301 has factors 293 and 257 Integer 120787 has factors 43 and 2809 Integer 967009 has factors 1609 and 601 Integer 997417 has factors 257 and 3881 Integer 7091569 has factors 2663 and 2663 Integer 13290059 has factors 3119 and 4261 Integer 42854447 has factors 9689 and 4423 Integer 223553581 has factors 11213 and 19937 Integer 2027651281 has factors 46061 and 44021 Integer 11111111111 has factors 21649 and 513239 Integer 100895598169 has factors 112303 and 898423 1002742628021 was not factored. Integer 60012462237239 has factors 6862753 and 8744663 Integer 287129523414791 has factors 6059887 and 47381993 Integer 9007199254740931 has factors 10624181 and 847801751 Integer 11111111111111111 has factors 2071723 and 5363222357 Integer 314159265358979323 has factors 317213509 and 990371647 Integer 384307168202281507 has factors 415718707 and 924440401 Integer 419244183493398773 has factors 48009977 and 8732438749 Integer 658812288346769681 has factors 62222119 and 10588072199 Integer 922337203685477563 has factors 110075821 and 8379108103 1000000000000000127 was not factored. Integer 1152921505680588799 has factors 139001459 and 8294312261 1537228672809128917 was not factored. Integer 4611686018427387877 has factors 343242169 and 13435662733
Classical heuristic
See Discussion.
//SquFoF: minimalistic version without queue.
//Classical heuristic. Tested: tcc 0.9.27
#include <math.h>
#include <stdio.h>
//input maximum
#define MxN ((unsigned long long) 1 << 62)
//reduce indefinite form
#define rho(a, b, c) { \
t = c; c = a; a = t; t = b; \
q = (rN + b) / a; \
b = q * a - b; \
c += q * (t - b); }
//initialize
#define rhoin(a, b, c) { \
rho(a, b, c) h = b; \
c = (mN - h * h) / a; }
#define gcd(a, b) while (b) { \
t = a % b; a = b; b = t; }
//multipliers
const unsigned long m[] = {1, 3, 5, 7, 11, 0};
//square form factorization
unsigned long squfof( unsigned long long N ) {
unsigned long a, b, c, u, v, w, rN, q, t, r;
unsigned long long mN, h;
int i, ix, k = 0;
if ((N & 1)==0) return 2;
h = floor(sqrt(N)+ 0.5);
if (h * h == N) return h;
while (m[k]) {
if (k && N % m[k]==0) return m[k];
//check overflow m * N
if (N > MxN / m[k]) break;
mN = N * m[k++];
r = floor(sqrt(mN));
h = r; //float64 fix
if (h * h > mN) r -= 1;
rN = r;
//principal form
b = r; c = 1;
rhoin(a, b, c)
//iteration bound
ix = floor(sqrt(2*r)) * 4;
//search principal cycle
for (i = 2; i < ix; i += 2) {
rho(a, b, c)
//even step
r = floor(sqrt(c)+ 0.5);
if (r * r == c) {
//square form found
//inverse square root
v = -b; w = r;
rhoin(u, v, w)
//search ambiguous cycle
do { r = v;
rho(u, v, w)
} while (v != r);
//symmetry point
h = N; gcd(h, u)
if (h != 1) return h;
}
rho(a, b, c)
//odd step
}
}
return 1;
}
void main(void) {
const unsigned long long data[] = {
2501,
12851,
13289,
75301,
120787,
967009,
997417,
7091569,
5214317,
20834839,
23515517,
33409583,
44524219,
13290059,
223553581,
2027651281,
11111111111,
100895598169,
1002742628021,
60012462237239,
287129523414791,
9007199254740931,
11111111111111111,
314159265358979323,
384307168202281507,
419244183493398773,
658812288346769681,
922337203685477563,
1000000000000000127,
1152921505680588799,
1537228672809128917,
4611686018427387877,
0};
unsigned long long N, f;
int i = 0;
while (1) {
N = data[i++];
//scanf("%llu", &N);
if (N < 2) break;
printf("N = %llu\n", N);
f = squfof(N);
if (N % f) f = 1;
if (f == 1) printf("fail\n\n");
else printf("f = %llu N/f = %llu\n\n", f, N/f);
}
}
- Showing problem cases only:
... N = 5214317 f = 73 N/f = 71429 N = 20834839 f = 3361 N/f = 6199 N = 23515517 f = 53 N/f = 443689 N = 33409583 f = 991 N/f = 33713 N = 44524219 f = 593 N/f = 75083 ... N = 1000000000000000127 f = 111756107 N/f = 8948056861 ... N = 1537228672809128917 f = 26675843 N/f = 57626245319 ...
C++
#include <cmath>
#include <cstdint>
#include <iostream>
#include <numeric>
#include <random>
uint64_t test_value = 0;
uint64_t sqrt_test_value = 0;
class BQF { // Binary quadratic form
public:
BQF(const uint64_t& a, const uint64_t& b, const uint64_t& c) : a(a), b(b), c(c) {
q = ( sqrt_test_value + b ) / c;
bb = q * c - b;
}
BQF rho() {
return BQF(c, bb, a + q * ( b - bb ));
}
BQF rho_inverse() {
return BQF(c, bb, ( test_value - bb * bb ) / c);
}
uint64_t a, b, c;
private:
uint64_t q, bb;
};
uint64_t squfof(const uint64_t& number) {
const uint32_t sqrt = std::sqrt(number);
if ( sqrt * sqrt == number ) {
return sqrt;
}
test_value = number;
sqrt_test_value = std::sqrt(test_value);
// Principal form
BQF form(0, sqrt_test_value, 1);
form = form.rho_inverse();
// Search principal cycle
for ( uint32_t i = 0; i < 4 * std::sqrt(2 * sqrt_test_value); i += 2 ) {
// Even step
form = form.rho();
uint64_t sqrt_c = std::sqrt(form.c);
if ( sqrt_c * sqrt_c == form.c ) { // Square form found
// Inverse square root
BQF form_inverse(0, -form.b, sqrt_c);
form_inverse = form_inverse.rho_inverse();
// Search ambiguous cycle
uint64_t previous_b = 0;
do {
previous_b = form_inverse.b;
form_inverse = form_inverse.rho();
} while ( form_inverse.b != previous_b );
// Symmetry point
const uint64_t g = std::gcd(number, form_inverse.a);
if ( g != 1 ) {
return g;
}
}
// Odd step
form = form.rho();
}
if ( number % 2 == 0 ) {
return 2;
}
return 0; // Failed to factorise, possibly a prime number
}
int main() {
std::random_device random;
std::mt19937 generator(random());
const uint64_t lower_limit = 100'000'000'000'000'000;
std::uniform_int_distribution<uint64_t> distribution(lower_limit, 10 * lower_limit);
for ( uint32_t i = 0; i < 20; ++i ) {
uint64_t test = distribution(random);
uint64_t factor = squfof(test);
if ( factor == 0 ) {
std::cout << test << " - failed to factorise" << std::endl;
} else {
std::cout << test << " = " << factor << " * " << test / factor << std::endl;
}
std::cout << std::endl;
}
}
- Output:
822140815871714649 = 141 * 5830785928168189 473377979025428817 = 3 * 157792659675142939 482452941918160803 = 4410431 * 109389069213 165380937127655630 = 65438 * 2527292049385 191677853606692475 = 7589219 * 25256598025 480551815975206727 = 2843 * 169029833265989 178710207362206205 = 5 * 35742041472441241 484660189375949842 = 1094 * 443016626486243 758704390319635770 = 1605 * 472713015775474 820453356193182720 = 97280 * 8433936638499 706982627912630220 = 121273 * 5829678724140 614913973550671312 = 437204432 * 1406467841 601482456081568543 = 131 * 4591469130393653 610533314488947626 = 14 * 43609522463496259 336343281182924332 = 70108 * 4797502156429 308127213282933401 = 7 * 44018173326133343 582455924775519843 = 3 * 194151974925173281 694215100094443276 = 32070628 * 21646445467 398821795604697523 = 181 * 2203435334832583 477964959783291032 = 517029608 * 924444079
FreeBASIC
' ***********************************************
'subject: Shanks's square form factorization:
' ambiguous forms of discriminant 4N
' give factors of N.
'tested : FreeBasic 1.08.1
'------------------------------------------------
const MxN = culngint(1) shl 62
'input maximum
const qx = (1 shl 5) - 1
'queue size
type arg
'squfof arguments
as ulong m, f
as integer vb
end type
type bqf
declare sub rho ()
'reduce indefinite form
declare function issq (byref r as ulong) as integer
'return -1 if c is square, set r:= sqrt(c)
declare sub qform (byref g as string, byval t as integer)
'print binary quadratic form #t (a, 2b, c)
as ulong rN, a, b, c
as integer vb
end type
type queue
declare sub enq (byref P as bqf)
'enqueue P.c, P.b if appropriate
declare function pro (byref P as bqf, byval r as ulong) as integer
'return -1 if a proper square form is found
as ulong a(qx), L, m
as integer k, t
end type
'global variables
dim shared N as ulongint
'the number to split
dim shared flag as integer
'signal to end all threads
dim shared as ubyte q1024(1023), q3465(3464)
'quadratic residue tables
'------------------------------------------------
sub bqf.rho ()
dim as ulong q, t
swap a, c
'residue
q = culng(rN + b) \ a
t = b: b = q * a - b
'pseudo-square
c += q * (t - b)
end sub
'initialize form
#macro rhoin(F)
F.rho : h = F.b
F.c = (mN - h * h) \ F.a
#endmacro
function bqf.issq (byref r as ulong) as integer
if q1024(c and 1023) andalso q3465(c mod 3465) then
'98.6% non-squares filtered
r = culng(sqr(c))
if r * r = c then return -1
end if
issq = 0
end function
sub bqf.qform (byref g as string, byval t as integer)
if vb = 0 then exit sub
dim as longint u = a, v = b, w = c
if t and 1 then
w = -w
else
u = -u
end if
v shl= 1
print g;str(t);" = (";u;",";v;",";w;")"
end sub
'------------------------------------------------
#macro red(r, a)
r = iif(a and 1, a, a shr 1)
if m > 2 then
r = iif(r mod m, r, r \ m)
end if
#endmacro
sub queue.enq (byref P as bqf)
dim s as ulong
red(s, P.c)
if s < L then
'circular queue
k = (k + 2) and qx
if k > t then t = k
'enqueue P.b, P.c
a(k) = P.b mod s
a(k + 1) = s
end if
end sub
function queue.pro (byref P as bqf, byval r as ulong) as integer
dim as integer i, sw
'skip improper square forms
for i = 0 to t step 2
sw = (P.b - a(i)) mod r = 0
sw and= a(i + 1) = r
if sw then return 0
next i
pro = -1
end function
'------------------------------------------------
sub squfof (byval ap as any ptr)
dim as arg ptr rp = cptr(arg ptr, ap)
dim as ulong L2, m, r, t, f = 1
dim as integer ix, i, j
dim as ulongint mN, h
'principal and ambiguous cycles
dim as bqf P, A
dim Q as queue
if (N and 1) = 0 then
rp->f = 2 ' even N
flag =-1: exit sub
end if
h = culngint(sqr(N))
if h * h = N then
'N is square
rp->f = culng(h)
flag =-1: exit sub
end if
rp->f = 1
'multiplier
m = rp->m
if m > 1 then
if (N mod m) = 0 then
rp->f = m ' m | N
flag =-1: exit sub
end if
'check overflow m * N
if N > (MxN \ m) then exit sub
end if
mN = N * m
r = int(sqr(mN))
'float64 fix
if culngint(r) * r > mN then r -= 1
P.rN = r
A.rN = r
P.vb = rp->vb
A.vb = rp->vb
'verbosity switch
if P.vb then print "r = "; r
Q.k = -2: Q.t = -1: Q.m = m
'Queue entry bounds
Q.L = int(sqr(r * 2))
L2 = Q.L * m shl 1
'principal form
P.b = r: P.c = 1
rhoin(P)
P.qform("P", 1)
ix = Q.L shl 2
for i = 2 to ix
'search principal cycle
if P.c < L2 then Q.enq(P)
P.rho
if (i and 1) = 0 andalso P.issq(r) then
'square form found
if Q.pro(P, r) then
P.qform("P", i)
'inverse square root
A.b =-P.b: A.c = r
rhoin(A): j = 1
A.qform("A", j)
do
'search ambiguous cycle
t = A.b
A.rho: j += 1
if A.b = t then
'symmetry point
A.qform("A", j)
red(f, A.a)
if f = 1 then exit do
flag = -1
'factor found
end if
loop until flag
end if ' proper square
end if ' square form
if flag then exit for
next i
rp->f = f
end sub
'------------------------------------------------
data 2501
data 12851
data 13289
data 75301
data 120787
data 967009
data 997417
data 7091569
data 13290059
data 23515517
data 42854447
data 223553581
data 2027651281
data 11111111111
data 100895598169
data 1002742628021
data 60012462237239
data 287129523414791
data 9007199254740931
data 11111111111111111
data 314159265358979323
data 384307168202281507
data 419244183493398773
data 658812288346769681
data 922337203685477563
data 1000000000000000127
data 1152921505680588799
data 1537228672809128917
data 4611686018427387877
data 0
'main
'------------------------------------------------
const tx = 4
dim as double tim = timer
dim h(4) as any ptr
dim a(4) as arg
dim as ulongint f
dim as integer s, t
width 64, 30
cls
'tabulate quadratic residues
for t = 0 to 1540
s = t * t
q1024(s and 1023) =-1
q3465(s mod 3465) =-1
next t
a(0).vb = 0
'set one verbosity switch only
a(0).m = 1
'multipliers
a(1).m = 3
a(2).m = 5
a(3).m = 7
a(4).m = 11
do
print
do : read N
loop until N < MxN
if N < 2 then exit do
print "N = "; N
flag = 0
for t = 1 to tx + 1 step 2
if t < tx then
h(t) = threadcreate(@squfof, @a(t))
end if
squfof(@a(t - 1))
f = a(t - 1).f
if t < tx then
threadwait(h(t))
if f = 1 then f = a(t).f
end if
if f > 1 then exit for
next t
'assert
if N mod f then f = 1
if f = 1 then
print "fail"
else
print "f = ";f;" N/f = ";N \ f
end if
loop
print "total time:"; csng(timer - tim); " s"
end
- Examples:
N = 2501 f = 61 N/f = 41 N = 12851 f = 71 N/f = 181 N = 13289 f = 97 N/f = 137 N = 75301 f = 293 N/f = 257 N = 120787 f = 43 N/f = 2809 N = 967009 f = 1609 N/f = 601 N = 997417 f = 257 N/f = 3881 N = 7091569 f = 2663 N/f = 2663 N = 13290059 f = 3119 N/f = 4261 N = 23515517 f = 53 N/f = 443689 N = 42854447 f = 4423 N/f = 9689 N = 223553581 f = 11213 N/f = 19937 N = 2027651281 f = 44021 N/f = 46061 N = 11111111111 f = 21649 N/f = 513239 N = 100895598169 f = 112303 N/f = 898423 N = 1002742628021 fail N = 60012462237239 f = 6862753 N/f = 8744663 N = 287129523414791 f = 6059887 N/f = 47381993 N = 9007199254740931 f = 10624181 N/f = 847801751 N = 11111111111111111 f = 2071723 N/f = 5363222357 N = 314159265358979323 f = 317213509 N/f = 990371647 N = 384307168202281507 f = 415718707 N/f = 924440401 N = 419244183493398773 f = 48009977 N/f = 8732438749 N = 658812288346769681 f = 62222119 N/f = 10588072199 N = 922337203685477563 f = 110075821 N/f = 8379108103 N = 1000000000000000127 f = 111756107 N/f = 8948056861 N = 1152921505680588799 f = 139001459 N/f = 8294312261 N = 1537228672809128917 f = 26675843 N/f = 57626245319 N = 4611686018427387877 f = 343242169 N/f = 13435662733 total time: 0.0170462 s
Go
Yet another solution based on the Wikipedia C code.
Rather than juggling with big.Int, I've just allowed the two 'awkward' cases to fail.
package main
import (
"fmt"
"math"
)
func isqrt(x uint64) uint64 {
x0 := x >> 1
x1 := (x0 + x/x0) >> 1
for x1 < x0 {
x0 = x1
x1 = (x0 + x/x0) >> 1
}
return x0
}
func gcd(x, y uint64) uint64 {
for y != 0 {
x, y = y, x%y
}
return x
}
var multiplier = []uint64{
1, 3, 5, 7, 11, 3 * 5, 3 * 7, 3 * 11, 5 * 7, 5 * 11, 7 * 11, 3 * 5 * 7, 3 * 5 * 11, 3 * 7 * 11, 5 * 7 * 11, 3 * 5 * 7 * 11,
}
func squfof(N uint64) uint64 {
s := uint64(math.Sqrt(float64(N)) + 0.5)
if s*s == N {
return s
}
for k := 0; k < len(multiplier) && N <= math.MaxUint64/multiplier[k]; k++ {
D := multiplier[k] * N
P := isqrt(D)
Pprev := P
Po := Pprev
Qprev := uint64(1)
Q := D - Po*Po
L := uint32(isqrt(8 * s))
B := 3 * L
i := uint32(2)
var b, q, r uint64
for ; i < B; i++ {
b = uint64((Po + P) / Q)
P = b*Q - P
q = Q
Q = Qprev + b*(Pprev-P)
r = uint64(math.Sqrt(float64(Q)) + 0.5)
if (i&1) == 0 && r*r == Q {
break
}
Qprev = q
Pprev = P
}
if i >= B {
continue
}
b = uint64((Po - P) / r)
P = b*r + P
Pprev = P
Qprev = r
Q = (D - Pprev*Pprev) / Qprev
i = 0
for {
b = uint64((Po + P) / Q)
Pprev = P
P = b*Q - P
q = Q
Q = Qprev + b*(Pprev-P)
Qprev = q
i++
if P == Pprev {
break
}
}
r = gcd(N, Qprev)
if r != 1 && r != N {
return r
}
}
return 0
}
func main() {
examples := []uint64{
2501,
12851,
13289,
75301,
120787,
967009,
997417,
7091569,
13290059,
42854447,
223553581,
2027651281,
11111111111,
100895598169,
1002742628021,
60012462237239,
287129523414791,
9007199254740931,
11111111111111111,
314159265358979323,
384307168202281507,
419244183493398773,
658812288346769681,
922337203685477563,
1000000000000000127,
1152921505680588799,
1537228672809128917,
4611686018427387877,
}
fmt.Println("Integer Factor Quotient")
fmt.Println("------------------------------------------")
for _, N := range examples {
fact := squfof(N)
quot := "fail"
if fact > 0 {
quot = fmt.Sprintf("%d", N/fact)
}
fmt.Printf("%-20d %-10d %s\n", N, fact, quot)
}
}
- Output:
Integer Factor Quotient ------------------------------------------ 2501 61 41 12851 71 181 13289 137 97 75301 293 257 120787 43 2809 967009 1609 601 997417 257 3881 7091569 2663 2663 13290059 3119 4261 42854447 9689 4423 223553581 11213 19937 2027651281 46061 44021 11111111111 21649 513239 100895598169 112303 898423 1002742628021 0 fail 60012462237239 6862753 8744663 287129523414791 6059887 47381993 9007199254740931 10624181 847801751 11111111111111111 2071723 5363222357 314159265358979323 317213509 990371647 384307168202281507 415718707 924440401 419244183493398773 48009977 8732438749 658812288346769681 62222119 10588072199 922337203685477563 110075821 8379108103 1000000000000000127 0 fail 1152921505680588799 139001459 8294312261 1537228672809128917 0 fail 4611686018427387877 343242169 13435662733
J
J does not have an unsigned fixed width integer type, which is one of the reasons that (in J) this algorithm is less optimal than advertised:
sqff=: {{
s=. <.%:y
if. y=*:s do. s return. end.
for_D. (x:y)*/:~*/@>,{1,each}.p:i.5 do.
if. -.'integer'-:datatype D=. x:inv D do. break. end.
P=. <.%:D
Q=. 1, D-P*P
lim=. <:6*<.%:2*s
for_i. }.i.lim do.
b=. <.(+/0 _1{P)%{:Q
P=. P,|(b*{:Q)-{:P
Q=. Q,|(_2{Q)+b*-/_2{.P
if. 2|i do. if. (=<.&.%:){:Q do. break. end. end.
end.
if. i>:lim do. continue. end.
Q=. <.%:{:Q
b=. <.(-/0 _1{P)%Q
P=. ,(b*Q)+{:P
Q=. Q, <.|(D-*:P)%Q
whilst. ~:/_2{.P do.
b=. <.(+/0 _1{P)%{:Q
P=. P,|(b*{:Q)-{:P
Q=. Q,|(_2{Q)+b*-/_2{.P
end.
f=. y+.x:_2{Q
if. -. f e. 1,y do. f return. end.
end.
1
}}
Task examples:
task ''
2501: 61 * 41
12851: 71 * 181
13289: 137 * 97
75301: 293 * 257
120787: 43 * 2809
967009: 601 * 1609
997417: 257 * 3881
7091569: 2663 * 2663
13290059: 3119 * 4261
42854447: 9689 * 4423
223553581: 11213 * 19937
2027651281: 46061 * 44021
11111111111: 21649 * 513239
100895598169: 112303 * 898423
1002742628021 was not factored
60012462237239: 6862753 * 8744663
287129523414791: 6059887 * 47381993
9007199254740931: 10624181 * 847801751
11111111111111111: 2071723 * 5363222357
314159265358979323: 317213509 * 990371647
384307168202281507: 415718707 * 924440401
419244183493398773: 48009977 * 8732438749
658812288346769681: 62222119 * 10588072199
922337203685477563: 110075821 * 8379108103
1000000000000000127 was not factored
1152921505680588799: 139001459 * 8294312261
1537228672809128917 was not factored
4611686018427387877 was not factored
where
task=: {{
for_num. nums do.
factor=. x:sqff num
if. 1=factor do. echo num,&":' was not factored'
else. echo num,&":': ',factor,&":' * ',":x:num%factor
end.
end.
}}
nums=: ".{{)n
2501
12851
13289
75301
120787
967009
997417
7091569
13290059
42854447
223553581
2027651281
11111111111
100895598169
1002742628021
60012462237239
287129523414791
9007199254740931
11111111111111111
314159265358979323
384307168202281507
419244183493398773
658812288346769681
922337203685477563
1000000000000000127
1152921505680588799
1537228672809128917
4611686018427387877x
}}-.LF
Java
import java.math.BigInteger;
import java.util.List;
import java.util.concurrent.ThreadLocalRandom;
public final class SquareFormFactorization {
public static void main(String[] args) {
ThreadLocalRandom random = ThreadLocalRandom.current();
final long lowerLimit = 10_000_000_000_000_000L;
final List<Long> tests = random.longs(20, lowerLimit, 10 * lowerLimit).boxed().toList();
for ( long test : tests ) {
long factor = squfof(test);
if ( factor == 0 ) {
System.out.println(test + " - failed to factorise");
} else if ( factor == 1 ) {
System.out.println(test + " is a prime number");
} else {
System.out.println(test + " = " + factor + " * " + test / factor);
}
System.out.println();
}
}
private static long squfof(long number) {
if ( BigInteger.valueOf(number).isProbablePrime(15) ) {
return 1; // Prime number
}
final int sqrt = (int) Math.sqrt(number);
if ( sqrt * sqrt == number ) {
return sqrt;
}
testValue = number;
sqrtTestValue = (long) Math.sqrt(testValue);
// Principal form
BQF form = new BQF(0, sqrtTestValue, 1);
form = form.rhoInverse();
// Search principal cycle
for ( int i = 0; i < 4 * (long) Math.sqrt(2 * sqrtTestValue); i += 2 ) {
// Even step
form = form.rho();
long sqrtC = (long) Math.sqrt(form.c);
if ( sqrtC * sqrtC == form.c ) { // Square form found
// Inverse square root
BQF formInverse = new BQF(0, -form.b, sqrtC);
formInverse = formInverse.rhoInverse();
// Search ambiguous cycle
long previousB = 0;
do {
previousB = formInverse.b;
formInverse = formInverse.rho();
} while ( formInverse.b != previousB );
// Symmetry point
final long gcd = gcd(number, formInverse.a);
if ( gcd != 1 ) {
return gcd;
}
}
// Odd step
form = form.rho();
}
if ( number % 2 == 0 ) {
return 2;
}
return 0; // Failed to factorise
}
private static long gcd(long a, long b) {
while ( b != 0 ) {
long temp = a; a = b; b = temp % b;
}
return a;
}
private static class BQF { // Binary quadratic form
public BQF(long aA, long aB, long aC) {
a = aA; b = aB; c = aC;
q = ( sqrtTestValue + b ) / c;
bb = q * c - b;
}
public BQF rho() {
return new BQF(c, bb, a + q * ( b - bb ));
}
public BQF rhoInverse() {
return new BQF(c, bb, ( testValue - bb * bb ) / c);
}
private long a, b, c;
private long q, bb;
}
private static long testValue, sqrtTestValue;
}
- Output:
20096060843736547 = 433 * 46411225967059 24628423963378844 = 7 * 3518346280482692 68276045265502398 = 37 * 1845298520689254 61072103663732497 = 8477 * 7204447760261 63462639942509072 = 16 * 3966414996406817 60313009405143787 = 89288189 * 675486983 76093594148871700 = 377900 * 201359074223 31796652636180617 is a prime number 87047981623879461 = 243 * 358222146600327 71567116631895554 = 73 * 980371460710898 50852012325831410 = 2 * 25426006162915705 65816967116185802 = 131280559 * 501345878 89627452852493643 = 31 * 2891208156532053 41735751565855318 = 10004047 * 4171886794 97291513005945602 = 2 * 48645756502972801 88974788272758998 = 59 * 1508047258860322 53903340306287681 = 21727 * 2480938017503 10811459482792395 = 546427 * 19785734385 95115727966103864 = 26105228 * 3643550938 11340988571009785 = 5 * 2268197714201957
jq
Adapted from Wren
Works with gojq, the Go implementation of jq
gojq has support for unbounded-precision integer arithmetic and accordingly the output shown below is from a run thereof; the C implementation of jq produces correct results up to and including [287129523414791,6059887,47381993].
Preliminaries
def gcd(a; b):
# subfunction expects [a,b] as input
# i.e. a ~ .[0] and b ~ .[1]
def rgcd: if .[1] == 0 then .[0]
else [.[1], .[0] % .[1]] | rgcd
end;
[a,b] | rgcd;
# for infinite precision integer-arithmetic
def idivide($p; $q): ($p - ($p % $q)) / $q ;
def idivide($q): (. - (. % $q)) / $q ;
def isqrt:
def irt:
. as $x
| 1 | until(. > $x; . * 4) as $q
| {$q, $x, r: 0}
| until( .q <= 1;
.q |= idivide(4)
| .t = .x - .r - .q
| .r |= idivide(2)
| if .t >= 0
then .x = .t
| .r += .q
else .
end)
| .r ;
if type == "number" and (isinfinite|not) and (isnan|not) and . >= 0
then irt
else "isqrt requires a non-negative integer for accuracy" | error
end ;
The Tasks
def multipliers:
[
1, 3, 5, 7, 11, 3*5, 3*7, 3*11, 5*7, 5*11, 7*11, 3*5*7, 3*5*11, 3*7*11, 5*7*11, 3*5*7*11
];
# input should be a number
def squfof:
def toi : floor | tostring | tonumber;
. as $N
| (($N|sqrt + 0.5)|toi) as $s
| if ($s*$s == $N) then $s
else label $out
| {}
| multipliers[] as $multiplier
| ($N * $multiplier) as $D
| .P = ($D|isqrt)
| .Pprev = .P
| .Pprev as $Po
| .Qprev = 1
| .Q = $D - $Po*$Po
| (($s * 8)|isqrt) as $L
| (3 * $L) as $B
| .i = 2
| .b = 0
| .q = 0
| .r = 0
| .stop = false
| until( (.i >= $B) or .stop;
.b = idivide($Po + .P; .Q)
| .P = .b * .Q - .P
| .q = .Q
| .Q = .Qprev + .b * (.Pprev - .P)
| .r = (((.Q|isqrt) + 0.5)|toi)
| if ((.i % 2) == 0 and (.r*.r) == .Q) then .stop = true
else
.Qprev = .q
| .Pprev = .P
| .i += 1
end )
| if .i < $B
then
.b = idivide($Po - .P; .r)
| .P = .b*.r + .P
| .Pprev = .P
| .Qprev = .r
| .Q = idivide($D - .Pprev*.Pprev; .Qprev)
| .i = 0
| .stop = false
| until (.stop;
.b = idivide($Po + .P; .Q)
| .Pprev = .P
| .P = .b * .Q - .P
| .q = .Q
| .Q = .Qprev + .b * (.Pprev - .P)
| .Qprev = .q
| .i += 1
| if (.P == .Pprev) then .stop = true else . end )
| .r = gcd($N; .Qprev)
| if .r != 1 and .r != $N then .r, break $out else empty end
else empty
end
end
// 0 ;
def examples: [
"2501",
"12851",
"13289",
"75301",
"120787",
"967009",
"997417",
"7091569",
"13290059",
"42854447",
"223553581",
"2027651281",
"11111111111",
"100895598169",
"1002742628021",
"60012462237239",
"287129523414791",
"9007199254740931",
"11111111111111111",
"314159265358979323",
"384307168202281507",
"419244183493398773",
"658812288346769681",
"922337203685477563",
"1000000000000000127",
"1152921505680588799",
"1537228672809128917",
"4611686018427387877"
];
"[Integer, Factor, Quotient]"
"---------------------------",
(examples[] as $example
| ($example|tonumber) as $N
| ($N | squfof) as $fact
| if $fact == 0 then "fail"
else idivide($N; $fact) as $quot
| [$N, $fact, $quot]
end
)
- Output:
[Integer, Factor, Quotient] --------------------------- [2501,61,41] [12851,71,181] [13289,137,97] [75301,293,257] [120787,43,2809] [967009,1609,601] [997417,257,3881] [7091569,2663,2663] [13290059,3119,4261] [42854447,9689,4423] [223553581,11213,19937] [2027651281,46061,44021] [11111111111,21649,513239] [100895598169,112303,898423] fail [60012462237239,6862753,8744663] [287129523414791,6059887,47381993] [9007199254740931,10624181,847801751] [11111111111111111,2071723,5363222357] [314159265358979323,317213509,990371647] [384307168202281507,415718707,924440401] [419244183493398773,48009977,8732438749] [658812288346769681,62222119,10588072199] [922337203685477563,110075821,8379108103] [1000000000000000127,111756107,8948056861] [1152921505680588799,139001459,8294312261] [1537228672809128917,26675843,57626245319] [4611686018427387877,343242169,13435662733]
Julia
Modified from Wikipedia's article at [[3]]
function square_form_factor(n::T)::T where T <: Integer
multiplier = T.([1, 3, 5, 7, 11, 3*5, 3*7, 3*11, 5*7, 5*11, 7*11, 3*5*7, 3*5*11, 3*7*11, 5*7*11, 3*5*7*11])
s = T(round(sqrt(n)))
s * s == n && return s
for k in multiplier
T != BigInt && n > typemax(T) ÷ k && break
d = k * n
p0 = pprev = p = isqrt(d)
qprev = one(T)
Q = d - p0 * p0
l = T(floor(2 * sqrt(2 * s)))
B, i = 3 * l, 2
while i < B
b = (p0 + p) ÷ Q
p = b * Q - p
q = Q
Q = qprev + b * (pprev - p)
r = T(round(sqrt(Q)))
iseven(i) && r * r == Q && break
qprev, pprev = q, p
i += 1
end
i >= B && continue
b = (p0 - p) ÷ r
pprev = p = b * r + p
qprev = r
Q = (d - pprev * pprev) ÷ qprev
i = 0
while true
b = (p0 + p) ÷ Q
pprev = p
p = b * Q - p
q = Q
Q = qprev + b * (pprev - p)
qprev = q
i += 1
p == pprev && break
end
r = gcd(n, qprev)
r != 1 && r != n && return r
end
return zero(T)
end
println("Integer Factor Quotient\n", "-"^45)
@time for n in Int128.([
2501, 12851, 13289, 75301, 120787, 967009, 997417, 7091569, 13290059, 42854447, 223553581,
2027651281, 11111111111, 100895598169, 1002742628021, 60012462237239, 287129523414791,
9007199254740931, 11111111111111111, 314159265358979323, 384307168202281507, 419244183493398773,
658812288346769681, 922337203685477563, 1000000000000000127, 1152921505680588799,
1537228672809128917, 4611686018427387877])
print(rpad(n, 22))
factr = square_form_factor(n)
print(rpad(factr, 10))
println(factr == 0 ? "fail" : n ÷ factr)
end
- Output:
Integer Factor Quotient --------------------------------------------- 2501 61 41 12851 71 181 13289 137 97 75301 293 257 120787 43 2809 967009 1609 601 997417 257 3881 7091569 2663 2663 13290059 3119 4261 42854447 9689 4423 223553581 11213 19937 2027651281 46061 44021 11111111111 21649 513239 100895598169 112303 898423 1002742628021 0 fail 60012462237239 6862753 8744663 287129523414791 6059887 47381993 9007199254740931 10624181 847801751 11111111111111111 2071723 5363222357 314159265358979323 317213509 990371647 384307168202281507 415718707 924440401 419244183493398773 48009977 8732438749 658812288346769681 62222119 10588072199 922337203685477563 110075821 8379108103 1000000000000000127 111756107 8948056861 1152921505680588799 139001459 8294312261 1537228672809128917 26675843 57626245319 4611686018427387877 343242169 13435662733 0.039027 seconds (698 allocations: 38.312 KiB)
Nim
import math, strformat
const M = [uint64 1, 3, 5, 7, 11]
template isqrt(n: uint64): uint64 = uint64(sqrt(float(n)))
template isEven(n: uint64): bool = (n and 1) == 0
proc squfof(n: uint64): uint64 =
if n.isEven: return 2
var h = uint64(sqrt(float(n)) + 0.5)
if h * h == n: return h
for m in M:
if m > 1 and (n mod m == 0): return m
# Check overflow m * n.
if n > uint64.high div m: break
let mn = m * n
var r = isqrt(mn)
if r * r > mn: dec r
let rn = r
# Principal form.
var b = r
var a = 1u64
h = (rn + b) div a * a - b
var c = (mn - h * h) div a
for i in 2..<(4 * isqrt(2 * r)):
# Search principal cycle.
swap a, c
var q = (rn + b) div a
let t = b
b = q * a - b
c += q * (t - b)
if i.isEven:
r = uint64(sqrt(float(c)) + 0.5)
if r * r == c: # Square form found?
# Inverse square root.
q = (rn - b) div r
var v = q * r + b
var w = (mn - v * v) div r
# Search ambiguous cycle.
var u = r
while true:
swap w, u
r = v
q = (rn + v) div u
v = q * u - v
if v == r: break
w += q * (r - v)
# Symmetry point.
h = gcd(n, u)
if h != 1: return h
result = 1
const Data = [2501u64,
12851u64,
13289u64,
75301u64,
120787u64,
967009u64,
997417u64,
7091569u64,
13290059u64,
42854447u64,
223553581u64,
2027651281u64,
11111111111u64,
100895598169u64,
1002742628021u64,
60012462237239u64,
287129523414791u64,
9007199254740931u64,
11111111111111111u64,
314159265358979323u64,
384307168202281507u64,
419244183493398773u64,
658812288346769681u64,
922337203685477563u64,
1000000000000000127u64,
1152921505680588799u64,
1537228672809128917u64,
4611686018427387877u64]
echo "N f N/f"
echo "======================================"
for n in Data:
let f = squfof(n)
let res = if f == 1: "fail" else: &"{f:<10} {n div f}"
echo &"{n:<22} {res}"
- Output:
N f N/f ====================================== 2501 61 41 12851 71 181 13289 97 137 75301 293 257 120787 43 2809 967009 1609 601 997417 257 3881 7091569 2663 2663 13290059 3119 4261 42854447 4423 9689 223553581 11213 19937 2027651281 44021 46061 11111111111 21649 513239 100895598169 112303 898423 1002742628021 fail 60012462237239 6862753 8744663 287129523414791 6059887 47381993 9007199254740931 10624181 847801751 11111111111111111 2071723 5363222357 314159265358979323 317213509 990371647 384307168202281507 415718707 924440401 419244183493398773 48009977 8732438749 658812288346769681 62222119 10588072199 922337203685477563 110075821 8379108103 1000000000000000127 111756107 8948056861 1152921505680588799 139001459 8294312261 1537228672809128917 26675843 57626245319 4611686018427387877 343242169 13435662733
Perl
use strict;
use warnings;
use feature 'say';
use ntheory <is_prime gcd forcomb vecprod>;
my @multiplier;
my @p = <3 5 7 11>;
forcomb { push @multiplier, vecprod @p[@_] } scalar @p;
sub sff {
my($N) = shift;
return 1 if is_prime $N; # if n is prime
return sqrt $N if sqrt($N) == int sqrt $N; # if n is a perfect square
for my $k (@multiplier) {
my $P0 = int sqrt($k*$N); # P[0]=floor(sqrt(N)
my $Q0 = 1; # Q[0]=1
my $Q = $k*$N - $P0**2; # Q[1]=N-P[0]^2 & Q[i]
my $P1 = $P0; # P[i-1] = P[0]
my $Q1 = $Q0; # Q[i-1] = Q[0]
my $P = 0; # P[i]
my $Qn = 0; # $P[$i+1];
my $b = 0; # b[i]
until (sqrt($Q) == int(sqrt($Q))) { # until Q[i] is a perfect square
$b = int( int(sqrt($k*$N) + $P1 ) / $Q); # floor(floor(sqrt(N+P[i-1])/Q[i])
$P = $b*$Q - $P1; # P[i]=b*Q[i]-P[i-1]
$Qn = $Q1 + $b*($P1 - $P); # Q[i+1]=Q[i-1]+b(P[i-1]-P[i])
($Q1, $Q, $P1) = ($Q, $Qn, $P);
}
$b = int( int( sqrt($k*$N)+$P ) / $Q ); # b=floor((floor(sqrt(N)+P[i])/Q[0])
$P1 = $b*$Q0 - $P; # P[i-1]=b*Q[0]-P[i]
$Q = ( $k*$N - $P1**2 )/$Q0; # Q[1]=(N-P[0]^2)/Q[0] & Q[i]
$Q1 = $Q0; # Q[i-1] = Q[0]
while () {
$b = int( int(sqrt($k*$N)+$P1 ) / $Q ); # b=floor(floor(sqrt(N)+P[i-1])/Q[i])
$P = $b*$Q - $P1; # P[i]=b*Q[i]-P[i-1]
$Qn = $Q1 + $b*($P1 - $P); # Q[i+1]=Q[i-1]+b(P[i-1]-P[i])
last if $P == $P1; # until P[i+1]=P[i]
($Q1, $Q, $P1) = ($Q, $Qn, $P);
}
for (gcd $N, $P) { return $_ if $_ != 1 and $_ != $N }
}
return 0
}
for my $data (
11111, 2501, 12851, 13289, 75301, 120787, 967009, 997417, 4558849, 7091569, 13290059,
42854447, 223553581, 2027651281, 11111111111, 100895598169, 1002742628021, 60012462237239,
287129523414791, 11111111111111111, 384307168202281507, 1000000000000000127, 9007199254740931,
922337203685477563, 314159265358979323, 1152921505680588799, 658812288346769681,
419244183493398773, 1537228672809128917) {
my $v = sff($data);
if ($v == 0) { say 'The number ' . $data . ' is not factored.' }
elsif ($v == 1) { say 'The number ' . $data . ' is a prime.' }
else { say "$data = " . join ' * ', sort {$a <=> $b} $v, int $data/int($v) }
}
- Output:
11111 = 41 * 271 2501 = 41 * 61 12851 = 71 * 181 13289 = 97 * 137 75301 = 257 * 293 120787 = 43 * 2809 967009 = 601 * 1609 997417 = 257 * 3881 4558849 = 383 * 11903 7091569 = 2663 * 2663 13290059 = 3119 * 4261 42854447 = 4423 * 9689 223553581 = 11213 * 19937 2027651281 = 44021 * 46061 11111111111 = 21649 * 513239 100895598169 = 112303 * 898423 The number 1002742628021 is a prime. 60012462237239 = 6862753 * 8744663 287129523414791 = 6059887 * 47381993 11111111111111111 = 2071723 * 5363222357 384307168202281507 = 415718707 * 924440401 1000000000000000127 = 111756107 * 8948056861 9007199254740931 = 10624181 * 847801751 922337203685477563 = 110075821 * 8379108103 314159265358979323 = 317213509 * 990371647 1152921505680588799 = 139001459 * 8294312261 658812288346769681 = 62222119 * 10588072199 419244183493398773 = 48009977 * 8732438749 1537228672809128917 = 26675843 * 57626245319
Phix
--requires(64) -- (decided to limit 32-bit explicitly instead) constant MxN = power(2,iff(machine_bits()=32?53:63)), m = {1, 3, 5, 7, 11} function squfof(atom N) -- square form factorization integer h, a=0, b, c, u=0, v, w, rN, q, r, t if remainder(N,2)==0 then return 2 end if h = floor(sqrt(N) + 0.5) if h*h==N then return h end if for k=1 to length(m) do integer mk = m[k] if mk>1 and remainder(N,mk)==0 then return mk end if //check overflow m * N if N>MxN/mk then exit end if atom mN = N*mk r = floor(sqrt(mN)) if r*r>mN then r -= 1 end if rN = r //principal form {b,a} = {r,1} h = floor((rN+b)/a)*a-b c = floor((mN-h*h)/a) for i=2 to floor(sqrt(2*r)) * 4-1 do //search principal cycle {a,c,t} = {c,a,b} q = floor((rN+b)/a) b = q*a-b c += q*(t-b) if remainder(i,2)==0 then r = floor(sqrt(c)+0.5) if r*r==c then //square form found //inverse square root q = floor((rN-b)/r) v = q*r+b w = floor((mN-v*v)/r) //search ambiguous cycle u = r while true do {u,w,r} = {w,u,v} q = floor((rN+v)/u) v = q*u-v if v==r then exit end if w += q*(r-v) end while //symmetry point h = gcd(N,u) if h