/usr/share/calc/mfactor.cal is in apcalc-common 2.12.5.0-1.
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* mfactor - return the lowest factor of 2^n-1, for n > 0
*
* Copyright (C) 1999 Landon Curt Noll
*
* Calc is open software; you can redistribute it and/or modify it under
* the terms of the version 2.1 of the GNU Lesser General Public License
* as published by the Free Software Foundation.
*
* Calc is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
* or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Lesser General
* Public License for more details.
*
* A copy of version 2.1 of the GNU Lesser General Public License is
* distributed with calc under the filename COPYING-LGPL. You should have
* received a copy with calc; if not, write to Free Software Foundation, Inc.
* 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA.
*
* @(#) $Revision: 30.1 $
* @(#) $Id: mfactor.cal,v 30.1 2007/03/16 11:09:54 chongo Exp $
* @(#) $Source: /usr/local/src/bin/calc/cal/RCS/mfactor.cal,v $
*
* Under source code control: 1996/07/06 06:09:40
* File existed as early as: 1996
*
* chongo <was here> /\oo/\ http://www.isthe.com/chongo/
* Share and enjoy! :-) http://www.isthe.com/chongo/tech/comp/calc/
*/
/*
* hset method
*
* We will assume that mfactor is called with p_elim == 17.
*
* n = (the Mersenne exponent we are testing)
* Q = 4*2*3*5*7*11*13*17 (4 * pfact(of some reasonable integer))
*
* We first determine all values of h mod Q such that:
*
* gcd(h*n+1, Q) == 1 and h*n+1 == +/-1 mod 8
*
* There will be 2*1*2*4*6*10*12*16 such values of h.
*
* For efficiency, we keep the difference between consecutive h values
* in the hset[] difference array with hset[0] being the first h value.
* Last, we multiply the hset[] values by n so that we only need
* to add sequential values of hset[] to get factor candidates.
*
* We need only test factors of the form:
*
* (Q*g*n + hx) + 1
*
* where:
*
* g is an integer >= 0
* hx is computed from hset[] difference value described above
*
* Note that (Q*g*n + hx) is always even and that hx is a multiple
* of n. Thus the typical factor form:
*
* 2*k*n + 1
*
* implies that:
*
* k = (Q*g + hx/n)/2
*
* This allows us to quickly eliminate factor values that are divisible
* by 2, 3, 5, 7, 11, 13 or 17. (well <= p value found below)
*
* The following loop shows how test_factor is advanced to higher test
* values using hset[]. Here, hcount is the number of elements in hset[].
* It can be shown that hset[0] == 0. We add hset[hcount] to the hset[]
* array for looping control convenience.
*
* (* increase test_factor thru other possible test values *)
* test_factor = 0;
* hindx = 0;
* do {
* while (++hindx <= hcount) {
* test_factor += hset[hindx];
* }
* hindx = 0;
* } while (test_factor < some_limit);
*
* The test, mfactor(67, 1, 10000) took on an 200 Mhz r4k (user CPU seconds):
*
* 210.83 (prior to use of hset[])
* 78.35 (hset[] for p_elim = 7)
* 73.87 (hset[] for p_elim = 11)
* 73.92 (hset[] for p_elim = 13)
* 234.16 (hset[] for p_elim = 17)
* p_elim == 19 requires over 190 Megs of memory
*
* Over a long period of time, the call to load_hset() becomes insignificant.
* If we look at the user CPU seconds from the first 10000 cycle to the
* end of the test we find:
*
* 205.00 (prior to use of hset[])
* 75.89 (hset[] for p_elim = 7)
* 73.74 (hset[] for p_elim = 11)
* 70.61 (hset[] for p_elim = 13)
* 57.78 (hset[] for p_elim = 17)
* p_elim == 19 rejected because of memory size
*
* The p_elim == 17 overhead takes ~3 minutes on an 200 Mhz r4k CPU and
* requires about ~13 Megs of memory. The p_elim == 13 overhead
* takes about 3 seconds and requires ~1.5 Megs of memory.
*
* The value p_elim == 17 is best for long factorizations. It is the
* fastest even thought the initial startup overhead is larger than
* for p_elim == 13.
*
* NOTE: The values above are prior to optimizations where hset[] was
* multiplied by n plus other optimizations. Thus, the CPU
* times you may get will not likely match the above values.
*/
/*
* mfactor - find a factor of a Mersenne Number
*
* Mersenne numbers are numbers of the form:
*
* 2^n-1 for integer n > 0
*
* We know that factors of a Mersenne number are of the form:
*
* 2*k*n+1 and +/- 1 mod 8
*
* We make use of the hset[] difference array to eliminate factor
* candidates that would otherwise be divisible by 2, 3, 5, 7 ... p_elim.
*
* given:
* n attempt to factor M(n) = 2^n-1
* start_k the value k in 2*k*n+1 to start the search (def: 1)
* rept_loop loop cycle reporting (def: 10000)
* p_elim largest prime to eliminate from test factors (def: 17)
*
* returns:
* factor of (2^n)-1
*
* NOTE: The p_elim argument is optional and defaults to 17. A p_elim value
* of 17 is faster than 13 for even medium length runs. However 13
* uses less memory and has a shorter startup time.
*/
define mfactor(n, start_k, rept_loop, p_elim)
{
local Q; /* 4*pfact(p_elim), hset[] cycle size */
local hcount; /* elements in the hset[] difference array */
local loop; /* report loop count */
local q; /* test factor of 2^n-1 */
local g; /* g as in test candidate form: (Q*g*hset[i])*n + 1 */
local hindx; /* hset[] index */
local i;
local tmp;
local tmp2;
/*
* firewall
*/
if (!isint(n) || n <= 0) {
quit "n must be an integer > 0";
}
if (!isint(start_k)) {
start_k = 1;
} else if (!isint(start_k) || start_k <= 0) {
quit "start_k must be an integer > 0";
}
if (!isint(rept_loop)) {
rept_loop = 10000;
}
if (rept_loop < 1) {
quit "rept_loop must be an integer > 0";
}
if (!isint(p_elim)) {
p_elim = 17;
}
if (p_elim < 3) {
quit "p_elim must be an integer > 2 (try 13 or 17)";
}
/*
* declare our global values
*/
Q = 4*pfact(p_elim);
hcount = 2;
/* allocate the h difference array */
for (i=2; i <= p_elim; i = nextcand(i)) {
hcount *= (i-1);
}
local mat hset[hcount+1];
/*
* load the hset[] difference array
*/
{
local x; /* h*n+1 mod 8 */
local h; /* potential h value */
local last_h; /* previous valid h value */
last_h = 0;
for (i=0,h=0; h < Q; ++h) {
if (gcd(h*n+1,Q) == 1) {
x = (h*n+1) % 8;
if (x == 1 || x == 7) {
hset[i++] = (h-last_h) * n;
last_h = h;
}
}
}
hset[hcount] = Q*n - last_h*n;
}
/*
* setup
*
* determine the next g and hset[] index (hindx) values such that:
*
* 2*start_k <= (Q*g + hset[hindx])
*
* and (Q*g + hset[hindx]) is a minimum and where:
*
* Q = (4 * pfact(of some reasonable integer))
* g = (some integer) (hset[] cycle number)
*
* We also compute 'q', the next test candidate.
*/
g = (2*start_k) // Q;
tmp = 2*start_k - Q*g;
for (tmp2=0, hindx=0;
hindx < hcount && (tmp2 += hset[hindx]/n) < tmp;
++hindx) {
}
if (hindx == hcount) {
/* we are beyond the end of a hset[] cycle, start at the next */
++g;
hindx = 0;
tmp2 = hset[0]/n;
}
q = (Q*g + tmp2)*n + 1;
/*
* look for a factor
*
* We ignore factors that themselves are divisible by a prime <=
* some small prime p.
*
* This process is guaranteed to find the smallest factor
* of 2^n-1. A smallest factor of 2^n-1 must be prime, otherwise
* the divisors of that factor would also be factors of 2^n-1.
* Thus we know that if a test factor itself is not prime, it
* cannot be the smallest factor of 2^n-1.
*
* Eliminating all non-prime test factors would take too long.
* However we can eliminate 80.81% of the test factors
* by not using test factors that are divisible by a prime <= 17.
*/
if (pmod(2,n,q) == 1) {
return q;
} else {
/* report this loop */
printf("at 2*%d*%d+1, cpu: %f\n",
(q-1)/(2*n), n, usertime());
fflush(files(1));
loop = 0;
}
do {
/*
* determine if we need to report
*
* NOTE: (q-1)/(2*n) is the k value from 2*k*n + 1.
*/
if (rept_loop <= ++loop) {
/* report this loop */
printf("at 2*%d*%d+1, cpu: %f\n",
(q-1)/(2*n), n, usertime());
fflush(files(1));
loop = 0;
}
/*
* skip if divisable by a prime <= 449
*
* The value 281 was determined by timing loops
* which found that 281 was at or near the
* minimum time to factor 2^(2^127-1)-1.
*
* The addition of the do { ... } while (factor(q, 449)>1);
* loop reduced the factoring loop time (36504 k values with
* the hset[] initialization time removed) from 25.69 sec to
* 15.62 sec of CPU time on a 200Mhz r4k.
*/
do {
/*
* determine the next factor candidate
*/
q += hset[++hindx];
if (hindx >= hcount) {
hindx = 0;
/*
* if we cared about g,
* then we wound ++g here too
*/
}
} while (factor(q, 449) > 1);
} while (pmod(2,n,q) != 1);
/*
* return the factor found
*
* q is a factor of (2^n)-1
*/
return q;
}
if (config("resource_debug") & 3) {
print "mfactor(n [, start_k=1 [, rept_loop=10000 [, p_elim=17]]])"
}
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