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Release calc version 2.11.0t10

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Landon Curt Noll
1999-11-11 05:15:39 -08:00
parent 86c8e6dcf1
commit 96c34adee3
283 changed files with 2380 additions and 3032 deletions
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lib/lucas.cal
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@@ -19,10 +19,10 @@
* OTHER TORTIOUS ACTION, ARISING OUT OF OR IN CONNECTION WITH THE USE OR
* PERFORMANCE OF THIS SOFTWARE.
*
* Landon Curt Noll
* http://reality.sgi.com/chongo/
* Landon Curt Noll
* http://reality.sgi.com/chongo/
*
* chongo <was here> /\../\
* chongo <was here> /\../\
*/
/*
* lucas - perform a Lucas primality test on h*2^n-1
@@ -34,26 +34,26 @@
* Sergio Zarantonello proved the following 65087 digit number to be prime:
*
* 216193
* 391581 * 2 -1
* 391581 * 2 -1
*
* At the time of discovery, this number was the largest known prime.
* The primality was demonstrated by a program implementing the test
* found in these routines. An Amdahl 1200 takes 1987 seconds to test
* the primality of this number. A Cray 2 took several hours to
* confirm this prime. As of 31 Dec 1995, this prime was the 3rd
* confirm this prime. As of 31 Dec 1995, this prime was the 3rd
* largest known prime and the largest known non-Mersenne prime.
*
* The same team also discovered the following twin prime pair:
*
* 11235 11235
* 1706595 * 2 -1 1706595 * 2 +1
* 1706595 * 2 -1 1706595 * 2 +1
*
* At the time of discovery, this was the largest known twin prime pair.
*
* NOTE: Both largest known and largest known twin prime records have been
* broken. Rather than update this file each time, I'll just
* broken. Rather than update this file each time, I'll just
* congratulate the finders and encourage others to try for
* larger finds. Records were made to be broken afterall!
* larger finds. Records were made to be broken afterall!
*
* ON GAINING A WORLD RECORD:
*
@@ -89,7 +89,7 @@
* 'h', the time to test each number remains relatively constant.
*
* It is clearly a win to eliminate potential test candidates by
* rejecting numbers that that are divisible by 'small' primes. We
* rejecting numbers that that are divisible by 'small' primes. We
* (the "Amdahl 6") rejected all numbers that were divisible by primes
* less than '2^40'. We stopped looking for small factors at '2^40'
* when the rate of candidates being eliminated was slowed down to
@@ -132,7 +132,7 @@ global pprod256; /* product of "primes up to 256" / "primes up to 46" */
* ABOUT THE TEST:
*
* This routine will perform a primality test on h*2^n-1 based on
* the mathematics of Lucas, Lehmer and Riesel. One should read
* the mathematics of Lucas, Lehmer and Riesel. One should read
* the following article:
*
* Ref1:
@@ -153,7 +153,7 @@ global pprod256; /* product of "primes up to 256" / "primes up to 46" */
*
* This test is performed as follows: (see Ref1, Theorem 5)
*
* a) generate u(0) (see the function gen_u0() below)
* a) generate u(0) (see the function gen_u0() below)
*
* b) generate u(n-2) according to the rule:
*
@@ -163,9 +163,9 @@ global pprod256; /* product of "primes up to 256" / "primes up to 46" */
*
* Now the following conditions must be true for the test to work:
*
* n >= 2
* n >= 2
* h >= 1
* h < 2^n
* h < 2^n
* h mod 2 == 1
*
* A few misc notes:
@@ -179,9 +179,9 @@ global pprod256; /* product of "primes up to 256" / "primes up to 46" */
* as 2^39).
*
* The condition 'h mod 2 == 1' is not a problem. Say one is testing
* 'j*2^m-1', where j is even. If we note that:
* 'j*2^m-1', where j is even. If we note that:
*
* j mod 2^x == 0 for x>0 implies j*2^m-1 == ((j/2^x)*2^(m+x))-1,
* j mod 2^x == 0 for x>0 implies j*2^m-1 == ((j/2^x)*2^(m+x))-1,
*
* then we can let h=j/2^x and n=m+x and test 'h*2^n-1' which is the value.
* We need only consider odd values of h because we can rewrite our numbers
@@ -227,7 +227,7 @@ lucas(h, n)
*/
oldh = h;
oldn = n;
shiftdown = fcnt(h,2); /* h % 2^shiftdown == 0, max shiftdown */
shiftdown = fcnt(h,2); /* h % 2^shiftdown == 0, max shiftdown */
if (shiftdown > 0) {
h >>= shiftdown;
n += shiftdown;
@@ -238,13 +238,13 @@ lucas(h, n)
*/
if (h <= 0 || n <= 0) {
print "ERROR: reduced args violate the rule: 0 < h < 2^n";
print " ERROR: h=":oldh, "n=":oldn, "reduced h=":h, "n=":n;
print " ERROR: h=":oldh, "n=":oldn, "reduced h=":h, "n=":n;
ldebug("lucas", "unknown: h <= 0 || n <= 0");
return -1;
}
if (highbit(h) >= n) {
print "ERROR: reduced args violate the rule: h < 2^n";
print " ERROR: h=":oldh, "n=":oldn, "reduced h=":h, "n=":n;
print " ERROR: h=":oldh, "n=":oldn, "reduced h=":h, "n=":n;
ldebug("lucas", "unknown: highbit(h) >= n");
return -1;
}
@@ -351,7 +351,7 @@ lucas(h, n)
* the v(1) into u(0).
*
* If gen_v1() returns a negative value, then we failed to
* generate a test for h*2^n-1. This is because h mod 3 == 0
* generate a test for h*2^n-1. This is because h mod 3 == 0
* is hard to do, and in rare cases, exceed the tables found
* in this program. We will generate an message and assume
* the number is not prime, even though if we had a larger
@@ -561,14 +561,14 @@ gen_u0(h, n, v1)
quickmax = 8;
mat d_qval[quickmax];
mat v1_qval[quickmax];
d_qval[0] = 5; v1_qval[0] = 3; /* a=1 b=1 r=4 */
d_qval[1] = 7; v1_qval[1] = 5; /* a=3 b=1 r=12 D=21 */
d_qval[2] = 13; v1_qval[2] = 11; /* a=3 b=1 r=4 */
d_qval[3] = 11; v1_qval[3] = 20; /* a=3 b=1 r=2 */
d_qval[4] = 29; v1_qval[4] = 27; /* a=5 b=1 r=4 */
d_qval[5] = 53; v1_qval[5] = 51; /* a=53 b=1 r=4 */
d_qval[6] = 17; v1_qval[6] = 66; /* a=17 b=1 r=1 */
d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
d_qval[0] = 5; v1_qval[0] = 3; /* a=1 b=1 r=4 */
d_qval[1] = 7; v1_qval[1] = 5; /* a=3 b=1 r=12 D=21 */
d_qval[2] = 13; v1_qval[2] = 11; /* a=3 b=1 r=4 */
d_qval[3] = 11; v1_qval[3] = 20; /* a=3 b=1 r=2 */
d_qval[4] = 29; v1_qval[4] = 27; /* a=5 b=1 r=4 */
d_qval[5] = 53; v1_qval[5] = 51; /* a=53 b=1 r=4 */
d_qval[6] = 17; v1_qval[6] = 66; /* a=17 b=1 r=1 */
d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
/*
* gen_v1 - compute the v(1) for a given h*2^n-1 if we can
@@ -671,7 +671,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
* are true, and return the related v(1).
*
* Before we address the two conditions, we need some background information
* on two symbols, Legendre and Jacobi. In Ref 2, pp 278, 284-285, we find
* on two symbols, Legendre and Jacobi. In Ref 2, pp 278, 284-285, we find
* the following definitions of J(a,p) and L(a,n):
*
* The Legendre symbol L(a,p) takes the value:
@@ -724,7 +724,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
*
* From Ref2, table 32:
*
* p mod 8 == +/-1 implies L(2,p) == 1 {note 3}
* p mod 8 == +/-1 implies L(2,p) == 1 {note 3}
* p mod 12 == +/-1 implies L(3,p) == 1 {note 4}
*
* Since h*2^n-1 mod 8 == -1, for n>2, note 3 implies:
@@ -737,14 +737,14 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
*
* By use of {A3.5}, {note 2}, {note 5} and {note 6}, one can show:
*
* L((2^g)*(3^l)*(z^2), h*2^n-1) == 1 (g>=0,l>=0,z>0,n>2) {note 7}
* L((2^g)*(3^l)*(z^2), h*2^n-1) == 1 (g>=0,l>=0,z>0,n>2) {note 7}
*
* Returning to the testing of conditions, take condition 1:
*
* L(D, h*2^n-1) == -1 [condition 1]
*
* In order for J(D, h*2^n-1) to be defined, we must ensure that D
* is not a factor of h*2^n-1. This is done by pre-screening h*2^n-1 to
* is not a factor of h*2^n-1. This is done by pre-screening h*2^n-1 to
* not have small factors and selecting D less than that factor check limit.
*
* By use of {note 7}, we can show that when we choose D to be:
@@ -765,7 +765,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
* == J(h*2^n-1 mod P, P)*(-1)^((h*2^n-2)*(P-1)/4) {note 0}
*
* When does J(h*2^n-1 mod P, P)*(-1)^((h*2^n-2)*(P-1)/4) take the value of -1,
* thus satisfy [condition 1]? The answer depends on P. Now P is a prime>2,
* thus satisfy [condition 1]? The answer depends on P. Now P is a prime>2,
* thus P mod 4 == 1 or -1.
*
* Take P mod 4 == 1:
@@ -780,7 +780,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
*
* Take P mod 4 == -1:
*
* P mod 4 == -1 implies (-1)^((h*2^n-2)*(P-1)/4) == -1
* P mod 4 == -1 implies (-1)^((h*2^n-2)*(P-1)/4) == -1
*
* Thus:
*
@@ -832,7 +832,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
* == J(h*2^n-1 mod Q, Q)*(-1)^((h*2^n-2)*(Q-1)/4) {note 0}
*
* When does J(h*2^n-1 mod Q, Q)*(-1)^((h*2^n-2)*(Q-1)/4) take the value of 1,
* thus satisfy [condition 2]? The answer depends on Q. Now Q is a prime>2,
* thus satisfy [condition 2]? The answer depends on Q. Now Q is a prime>2,
* thus Q mod 4 == 1 or -1.
*
* Take Q mod 4 == 1:
@@ -847,7 +847,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
*
* Take Q mod 4 == -1:
*
* Q mod 4 == -1 implies (-1)^((h*2^n-2)*(Q-1)/4) == -1
* Q mod 4 == -1 implies (-1)^((h*2^n-2)*(Q-1)/4) == -1
*
* Thus:
*
@@ -855,7 +855,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
* == L(h*2^n-1 mod Q, Q) * -1
* == -J(h*2^n-1 mod Q, Q)
*
* Therefore [condition 2] is met by selecting D = Q*(2^j)*(3^k)*(z^2),
* Therefore [condition 2] is met by selecting D = Q*(2^j)*(3^k)*(z^2),
* where Q is prime>2, j>=0, k>=0, z>0; if and only if one of the following
* to cases are true:
*
@@ -894,7 +894,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
*
* r == Q*(2^j)*(3^k)*(z^2) (Q==1 or Q is prime>2, j>=0, k>=0, z>0)
*
* one of the following is true:
* one of the following is true:
* P mod 4 == 1 and J(h*2^n-1 mod P, P) == -1
* P mod 4 == -1 and J(h*2^n-1 mod P, P) == 1
*
@@ -903,7 +903,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
* Q mod 4 == -1 and J(h*2^n-1 mod Q, Q) == -1
*
* If we cannot find a v(1) quickly enough, then we will give up
* testing h*2^n-1. This does not happen too often, so this hack
* testing h*2^n-1. This does not happen too often, so this hack
* is not too bad.
*
***
@@ -971,28 +971,28 @@ gen_v1(h, n)
*
* We will check with:
*
* v(1)=81 D=6557 a=79 b=1 r=316
* v(1)=81 D=6557 a=79 b=1 r=316
*
* Now, D==79*83 and r=79*2^2. If we show that:
* Now, D==79*83 and r=79*2^2. If we show that:
*
* J(h*2^n-1 mod 79, 79) == -1
* J(h*2^n-1 mod 83, 83) == 1
*
* then we will satisfy [condition 1]. Observe:
* then we will satisfy [condition 1]. Observe:
*
* 79 mod 4 == -1 implies (-1)^((h*2^n-2)*(79-1)/4) == -1
* 83 mod 4 == -1 implies (-1)^((h*2^n-2)*(83-1)/4) == -1
* 79 mod 4 == -1 implies (-1)^((h*2^n-2)*(79-1)/4) == -1
* 83 mod 4 == -1 implies (-1)^((h*2^n-2)*(83-1)/4) == -1
*
* J(D, h*2^n-1) == J(83, h*2^n-1) * J(79, h*2^n-1)
* == J(h*2^n-1, 83) * (-1)^((h*2^n-2)*(83-1)/4) *
* J(h*2^n-1, 79) * (-1)^((h*2^n-2)*(79-1)/4)
* J(h*2^n-1, 79) * (-1)^((h*2^n-2)*(79-1)/4)
* == J(h*2^n-1 mod 83, 83) * -1 *
* J(h*2^n-1 mod 79, 79) * -1
* J(h*2^n-1 mod 79, 79) * -1
* == 1 * -1 *
* -1 * -1
* == -1
*
* We will also satisfy [condition 2]. Observe:
* We will also satisfy [condition 2]. Observe:
*
* (a^2 - b^2*D)/r == (79^2 - 1^1*6557)/316
* == -1