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Release calc version 2.12.5.5

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Landon Curt Noll
2017-05-20 19:24:06 -07:00
parent 2726ae9d23
commit 8dd7a3cd2a
23 changed files with 1126 additions and 422 deletions
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cal/lucas.cal
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@@ -1,7 +1,7 @@
/*
* lucas - perform a Lucas primality test on h*2^n-1
*
* Copyright (C) 1999 Landon Curt Noll
* Copyright (C) 1999,2017 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
@@ -17,9 +17,9 @@
* received a copy with calc; if not, write to Free Software Foundation, Inc.
* 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA.
*
* @(#) $Revision: 30.2 $
* @(#) $Id: lucas.cal,v 30.2 2013/09/27 08:58:46 chongo Exp $
* @(#) $Source: /usr/local/src/bin/calc/cal/RCS/lucas.cal,v $
* @(#) $Revision: 30.4 $
* @(#) $Id: lucas.cal,v 30.4 2017/05/20 21:54:16 chongo Exp $
* @(#) $Source: /usr/local/src/bin/calc-RHEL7/cal/RCS/lucas.cal,v $
*
* Under source code control: 1990/05/03 16:49:51
* File existed as early as: 1990
@@ -28,6 +28,12 @@
* Share and enjoy! :-) http://www.isthe.com/chongo/tech/comp/calc/
*/
/*
* For a general tutorial on how to find a new largest known prime, see:
*
* http://www.isthe.com/chongo/tech/math/prime/prime-tutorial.pdf
*/
/*
* NOTE: This is a standard calc resource file. For information on calc see:
*
@@ -71,10 +77,15 @@
* NOTE: Both largest known and largest known twin prime records have been
* 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 after all!
*/
/* ON GAINING A WORLD RECORD:
/*
* ON GAINING A WORLD RECORD:
*
* For a general tutorial on how to find a new largest known prime, see:
*
* http://www.isthe.com/chongo/tech/math/prime/prime-tutorial.pdf
*
* The routines in calc were designed to be portable, and to work on
* numbers of 'sane' size. The Amdahl 6 team used a 'ultra-high speed
@@ -83,6 +94,13 @@
* The heart of the package was a multiplication and square routine that
* was based on the PFA Fast Fourier Transform and on Winograd's radix FFTs.
*
* NOTE: While the PFA Fast Fourier Transform and Winograd's radix FFTs
* might have been optimal for the Amdahl 6 team at the time,
* they might not be optimal for your CPU architecture. See
* the above mentioned tutorial for information on better
* methods of performing multiplications and squares of very
* large numbers.
*
* Having a fast computer, and a good multi-precision package are
* critical, but one also needs to know where to look in order to have
* a good chance at a record. Knowing what to test is beyond the scope
@@ -139,12 +157,10 @@
* be the factors of another candidate.
*
* Finally, one should eliminate all values of 'h*2^n-1' where
* 'h*2^n+1' is divisible by a small primes. The ideas behind this
* point is beyond the scope of this program.
* 'h*2^n+1' is divisible by a small primes.
*/
global pprod256; /* product of "primes up to 256" / "primes up to 46" */
pprod256 = 0; /* product of "primes up to 256" / "primes up to 46" */
/*
* lucas - lucas primality test on h*2^n-1
@@ -171,15 +187,33 @@ global pprod256; /* product of "primes up to 256" / "primes up to 46" */
* "Introduction to Analytic Number Theory", by Tom A. Apostol,
* Springer-Verlag, 1984, p 188.
*
* An excellent 5-page paper by Oystein J. Rodseth (we apologize that the
* ASCII character set does not allow us to spell his name with the
* umlaut marks on the O's):
*
* NOTE: The original Amdahl 6 method predates the publication of Ref4.
* The gen_v1() function used by lucas() uses the Ref4 method.
* See the 'Amdahl 6 legacy code' section below for the original
* method of generating v(1).
*
* Ref4:
*
* "A note on primality tests for N = h*2^n-1", by Oystein J. Rodseth,
* Department of Mathematics, University of Bergen, BIT Numerical
* Mathematics. 34 (3): pp 451-454.
*
* http://folk.uib.no/nmaoy/papers/luc.pdf
*
* This test is performed as follows: (see Ref1, Theorem 5)
*
* a) generate u(0) (see the function gen_u0() below)
* a) generate u(2) (see the function gen_u2() below)
* (NOTE: some call this u(0))
*
* b) generate u(n-2) according to the rule:
* b) generate u(n) according to the rule:
*
* u(i+1) = u(i)^2-2 mod h*2^n-1
*
* c) h*2^n-1 is prime if and only if u(n-2) == 0 Q.E.D. :-)
* c) h*2^n-1 is prime if and only if u(n) == 0 Q.E.D. :-)
*
* Now the following conditions must be true for the test to work:
*
@@ -188,7 +222,7 @@ global pprod256; /* product of "primes up to 256" / "primes up to 46" */
* h < 2^n
* h mod 2 == 1
*
* A few misc notes:
* A few miscellaneous notes:
*
* In order to reduce the number of tests, as attempt to eliminate
* any number that is divisible by a prime less than 257. Valid prime
@@ -222,7 +256,7 @@ lucas(h, n)
local testval; /* h*2^n-1 */
local shiftdown; /* the power of 2 that divides h */
local u; /* the u(i) sequence value */
local v1; /* the v(1) generator of u(0) */
local v1; /* the v(1) generator of u(2) */
local i; /* u sequence cycle number */
local oldh; /* pre-reduced h */
local oldn; /* pre-reduced n */
@@ -364,18 +398,17 @@ lucas(h, n)
}
/*
* try to compute u(0)
* try to compute u(2) (NOTE: some call this u(0))
*
* We will use gen_v1() to give us a v(1) using the values
* of 'h' and 'n'. We will then use gen_u0() to convert
* the v(1) into u(0).
* of 'h' and 'n'. We will then use gen_u2() to convert
* the v(1) into u(2).
*
* 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
* 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
* table, we might have been able to show that it is prime.
* generate a test for h*2^n-1. The legacy function,
* legacy_gen_v1() used by the Amdahl 6 could have returned
* -1. The new gen_v1() based on the method outlined in Ref4
* will never return -1.
*/
v1 = gen_v1(h, n);
if (v1 < 0) {
@@ -384,10 +417,10 @@ lucas(h, n)
ldebug("lucas", "unknown: no v(1)");
return -1;
}
u = gen_u0(h, n, v1);
u = gen_u2(h, n, v1);
/*
* compute u(n-2)
* compute u(n) (NOTE: some call this u(n-2))
*/
for (i=3; i <= n; ++i) {
/* u = (u^2 - 2) % testval; */
@@ -407,11 +440,19 @@ lucas(h, n)
}
/*
* gen_u0 - determine the initial Lucas sequence for h*2^n-1
* gen_u2 - determine the initial Lucas sequence for h*2^n-1
*
* Historically many start the Lucas sequence with u(0).
* Some, like the author of this code, prefer to start
* with U(2). This is so one may say:
*
* 2^p-1 is prime if u(p) = 0 mod 2^p-1
* or:
* h*2^p-1 is prime if u(p) = 0 mod h*2^p-1
*
* According to Ref1, Theorem 5:
*
* u(0) = alpha^h + alpha^(-h)
* u(2) = alpha^h + alpha^(-h) (NOTE: Ref1 calls it u(0))
*
* Now:
*
@@ -419,7 +460,7 @@ lucas(h, n)
*
* Therefore:
*
* u(0) = v(h)
* u(2) = v(h) (NOTE: Ref1 calls it u(0))
*
* We calculate v(h) as follows: (Ref1, top of page 873)
*
@@ -447,11 +488,11 @@ lucas(h, n)
* v1 - gen_v1(h,n) (see function below)
*
* returns:
* u(0) - initial value for Lucas test on h*2^n-1
* -1 - failed to generate u(0)
* u(2) - initial value for Lucas test on h*2^n-1
* -1 - failed to generate u(2)
*/
define
gen_u0(h, n, v1)
gen_u2(h, n, v1)
{
local shiftdown; /* the power of 2 that divides h */
local r; /* low value: v(n) */
@@ -500,7 +541,7 @@ gen_u0(h, n, v1)
* at least 2 bits long for the loop below to work.
*/
if (h == 1) {
ldebug("gen_u0", "quick h == 1 case");
ldebug("gen_u2", "quick h == 1 case");
/* return r%(h*2^n-1); */
return hnrmod(r, h, n, -1);
}
@@ -540,21 +581,502 @@ gen_u0(h, n, v1)
return r;
}
/*
* gen_u0 - determine the initial Lucas sequence for h*2^n-1
*
* Historically many start the Lucas sequence with u(0).
* Some, like the author of this code, prefer to start
* with u(2). This is so one may say:
*
* 2^p-1 is prime if u(p) = 0 mod 2^p-1
* or:
* h*2^n-1 is prime if U(n) = 0 mod h*2^n-1
*
* For those using the old code with gen_u0(), we
* simply call gen_u2() instead.
*
* See the function gen_u2() for details.
*
* input:
* h - h as in h*2^n-1
* n - n as in h*2^n-1
* v1 - gen_v1(h,n) (see function below)
*
* returns:
* u(2) - initial value for Lucas test on h*2^n-1
* -1 - failed to generate u(2)
*/
define
gen_u0(h, n, v1)
{
return gen_u2(h, n, v1);
}
/*
* rodseth_xhn - determine if v(1) == x for h*2^n-1
*
* For a given h*2^n-1, v(1) == x if:
*
* jacobi(x-2, h*2^n-1) == 1 (Ref4, condition 1) part 1
* jacobi(x+2, h*2^n-1) == -1 (Ref4, condition 1) part 2
*
* Now when x-2 <= 0:
*
* jacobi(x-2, h*2^n-1) == 0
*
* because:
*
* jacobi(x,y) == 0 if x <= 0
*
* So for (Ref4, condition 1) part 1 to be true:
*
* x-2 > 0
*
* And therefore:
*
* x > 2
*
* input:
* x - potential v(1) value
* h - h as in h*2^n-1
* n - n as in h*2^n-1
*
* returns:
* 1 if v(1) == x for h*2^n-1
* 0 otherwise
*/
define
rodseth_xhn(x, h, n)
{
local testval; /* h*2^n-1 */
/*
* check arg types
*/
if (!isint(h)) {
quit "bad args: h must be an integer";
}
if (!isint(n)) {
quit "bad args: n must be an integer";
}
if (!isint(x)) {
quit "bad args: x must be an integer";
}
/*
* firewall
*/
if (x <= 2) {
return 0;
}
/*
* Check for jacobi(x-2, h*2^n-1) == 1 (Ref4, condition 1) part 1
*/
testval = h*2^n-1;
if (jacobi(x-2, testval) != 1) {
return 0;
}
/*
* Check for jacobi(x+2, h*2^n-1) == -1 (Ref4, condition 1) part 2
*/
if (jacobi(x+2, testval) != -1) {
return 0;
}
/*
* v(1) == x for this h*2^n-1
*/
return 1;
}
/*
* Trial tables used by gen_v1()
*
* When h mod 3 == 0, one needs particular values of D, a and b (see gen_v1
* documentation) in order to find a value of v(1).
* When h mod 3 == 0, according to Ref4 we need to find the first value X where:
*
* This table defines 'quickmax' possible tests to be taken in ascending
* order. The v1_qval[x] refers to a v(1) value from Ref1, Table 1. A
* related D value is found in d_qval[x]. All D values expect d_qval[1]
* are also taken from Ref1, Table 1. The case of D == 21 as listed in
* Ref1, Table 1 can be changed to D == 7 for the sake of the test because
* of {note 6}.
* jacobi(X-2, h*2^n-1) == 1 (Ref4, condition 1) part 1
* jacobi(X+2, h*2^n-1) == -1 (Ref4, condition 1) part 2
*
* We can show that X > 2. See the comments in the rodseth_xhn(x,h,n) above.
*
* Some values of X satisfy more often than others. For example a large sample
* of odd h, h multiple of 3 and large n (some around 1e4, some near 1e6, others
* near 3e7) where the sample size was 66 973 365, here is the count of the
* smallest value of X that satisfies conditions in Ref4, condition 1:
*
* count X
* ----------
* 26791345 3
* 17223016 5
* 7829600 9
* 6988774 11
* 3301093 15
* 1517149 17
* 910346 21
* 711791 29
* 573403 20
* 390395 27
* 288637 35
* 149751 36
* 107733 39
* 58743 41
* 35619 45
* 25052 32
* 17775 51
* 13031 44
* 7563 56
* 7540 49
* 7060 59
* 4407 57
* 2948 65
* 2502 55
* 2388 69
* 2094 71
* 689 77
* 626 81
* 491 66
* 426 95
* 219 80
* 203 67
* 185 84
* 152 99
* 127 72
* 102 74
* 98 87
* 67 90
* 55 104
* 48 101
* 32 105
* 17 109
* 16 116
* 15 111
* 13 92
* 12 125
* 7 129
* 3 146
* 2 140
* 2 120
* 1 165
* 1 161
* 1 155
*
* The above distribution was found to hold fairly well over many values of
* odd h that are a multiple of 3 and for many values of n where h < 2^n.
*
* Given this information, when odd h is a multiple of 3 we try, in order,
* these values of X:
*
* 3, 5, 9, 11, 15, 17, 21, 29, 20, 27, 35, 36, 39, 41, 45, 32, 51, 44,
* 56, 49, 59, 57, 65, 55, 69, 71, 77, 81, 66, 95, 80, 67, 84, 99, 72,
* 74, 87, 90, 104, 101, 105, 109, 116, 111, 92
*
* And stop on the first value of X where:
*
* jacobi(X-2, h*2^n-1) == 1
* jacobi(X+2, h*2^n-1) == -1
*
* If no value in that list works, we start simple search starting with X = 120
* and incrementing by 1 until a value of X is found.
*
* The x_tbl[] matrix contains those common values of X to try in order.
* If all x_tbl_len fail to satisfy Ref4 condition 1, then we begin a
* linear search at next_x until we find a proper X value.
*
* IMPORTANT NOTE: Using this table will not find the smallest possible v(1)
* for a given h and n. This is not a problem because using
* a larger value of v(1) does not impact the primality test.
* Furthermore after lucas(h, n) generates a few u(n) terms,
* the values will wrap (due to computing mod h*2^n-1).
* Finally on average, about 1/4 of the values of X work as
* v(1) for a given n when h is a multiple of 3. Skipping
* rarely used v(1) will not doom gen_v1() to a long search.
*/
x_tbl_len = 45;
mat x_tbl[x_tbl_len];
x_tbl = {
3, 5, 9, 11, 15, 17, 21, 29, 20, 27, 35, 36, 39, 41, 45, 32, 51, 44,
56, 49, 59, 57, 65, 55, 69, 71, 77, 81, 66, 95, 80, 67, 84, 99, 72,
74, 87, 90, 104, 101, 105, 109, 116, 111, 92
};
next_x = 120;
/*
* gen_v1 - compute the v(1) for a given h*2^n-1 if we can
*
* This function assumes:
*
* n > 2 (n==2 has already been eliminated)
* h mod 2 == 1
* h < 2^n
* h*2^n-1 mod 3 != 0 (h*2^n-1 has no small factors, such as 3)
*
* The generation of v(1) depends on the value of h. There are two cases
* to consider, h mod 3 != 0, and h mod 3 == 0.
*
***
*
* Case 1: (h mod 3 != 0)
*
* This case is easy.
*
* In Ref1, page 869, one finds that if: (or see Ref2, page 131-132)
*
* h mod 6 == +/-1
* h*2^n-1 mod 3 != 0
*
* which translates, gives the functions assumptions, into the condition:
*
* h mod 3 != 0
*
* If this case condition is true, then:
*
* u(2) = (2+sqrt(3))^h + (2-sqrt(3))^h (see Ref1, page 869)
* = (2+sqrt(3))^h + (2+sqrt(3))^(-h) (NOTE: some call this u(2))
*
* and since Ref1, Theorem 5 states:
*
* u(2) = alpha^h + alpha^(-h) (NOTE: some call this u(2))
* r = abs(2^2 - 1^2*3) = 1
*
* and the bottom of Ref1, page 872 states:
*
* v(x) = alpha^x + alpha^(-x)
*
* If we let:
*
* alpha = (2+sqrt(3))
*
* then
*
* u(2) = v(h) (NOTE: some call this u(2))
*
* so we simply return
*
* v(1) = alpha^1 + alpha^(-1)
* = (2+sqrt(3)) + (2-sqrt(3))
* = 4
*
***
*
* Case 2: (h mod 3 == 0)
*
* For the case where h is a multiple of 3, we turn to Ref4.
*
* The central theorem on page 3 of that paper states that
* we may set v(1) to the first value X that satisfies:
*
* jacobi(X-2, h*2^n-1) == 1 (Ref4, condition 1)
* jacobi(X+2, h*2^n-1) == -1 (Ref4, condition 1)
*
* NOTE: Ref4 uses P, which we shall refer to as X.
* Ref4 uses N, which we shall refer to as h*2^n-1.
*
* NOTE: Ref4 uses the term Legendre-Jacobi symbol, which
* we shall refer to as the Jacobi symbol.
*
* 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
* the following definitions of jacobi(a,b) and L(a,p):
*
* The Legendre symbol L(a,p) takes the value:
*
* L(a,p) == 1 => a is a quadratic residue of p
* L(a,p) == -1 => a is NOT a quadratic residue of p
*
* when:
*
* p is prime
* p mod 2 == 1
* gcd(a,p) == 1
*
* The value a is a quadratic residue of b if there exists some integer z
* such that:
*
* z^2 mod b == a
*
* The Jacobi symbol jacobi(a,b) takes the value:
*
* jacobi(a,b) == 1 => b is not prime,
* or a is a quadratic residue of b
* jacobi(a,b) == -1 => a is NOT a quadratic residue of b
*
* when
*
* b mod 2 == 1
* gcd(a,b) == 1
*
* It is worth noting for the Legendre symbol, in order for L(X+/-2,
* h*2^n-1) to be defined, we must ensure that neither X-2 nor X+2 are
* factors of h*2^n-1. This is done by pre-screening h*2^n-1 to not
* have small factors and keeping X+2 less than that small factor
* limit. It is worth noting that in lucas(h, n), we first verify
* that h*2^n-1 does not have a factor < 257 before performing the
* primality test. So while X+/-2 < 257, we know that
* gcd(X+/-2, h*2^n-1) == 1.
*
* Returning to the testing of conditions in Ref4, condition 1:
*
* jacobi(X-2, h*2^n-1) == 1
* jacobi(X+2, h*2^n-1) == -1
*
* When such an X is found, we set:
*
* v(1) = X
*
***
*
* In conclusion, we can compute v,(1) by attempting to do the following:
*
* h mod 3 != 0
*
* we return:
*
* v(1) == 4
*
* h mod 3 == 0
*
* we return:
*
* v(1) = X
*
* where X > 2 in a integer such that:
*
* jacobi(X-2, h*2^n-1) == 1
* jacobi(X+2, h*2^n-1) == -1
*
***
*
* input:
* h h as in h*2^n-1
* n n as in h*2^n-1
*
* output:
* returns v(1), or -1 is there is no quick way
*/
define
gen_v1(h, n)
{
local x; /* potential v(1) to test */
local i; /* x_tbl index */
/*
* check arg types
*/
if (!isint(h)) {
quit "bad args: h must be an integer";
}
if (!isint(n)) {
quit "bad args: n must be an integer";
}
/*
* check for Case 1: (h mod 3 != 0)
*/
if (h % 3 != 0) {
/* v(1) is easy to compute */
return 4;
}
/*
* What follow is Case 2: (h mod 3 == 0)
*/
/*
* We will look for x that satisfies conditions in Ref4, condition 1:
*
* jacobi(X-2, h*2^n-1) == 1 part 1
* jacobi(X+2, h*2^n-1) == -1 part 2
*/
for (i=0; i < x_tbl_len; ++i) {
/*
* test Ref4 condition 1:
*/
x = x_tbl[i];
if (rodseth_xhn(x, h, n) == 1) {
/*
* found a x that satisfies Ref4 condition 1
*/
ldebug("gen_v1", "h= " + str(h) + " n= " + str(n) +
" v1= " + str(x) + " using tbl[ " +
str(i) + " ]");
return x;
}
}
/*
* We are in that rare case (about 1 in 2 300 000) where none of the
* common X values satisfy Ref4 condition 1. We start a linear search
* at next_x from here on.
*
* However, we also need to keep in mind that when x+2 >= 257, we
* need to verify that gcd(x-2, h*2^n-1) == 1 and
* and to verify that gcd(x+2, h*2^n-1) == 1.
*/
x = next_x;
while (rodseth_xhn(x, h, n) != 1) {
++x;
}
/* finally found a v(1) value */
ldebug("gen_v1", "h= " + str(h) + " n= " + str(n) +
" v1= " + str(x) + " beyond tbl");
return x;
}
/*
* ldebug - print a debug statement
*
* input:
* funct name of calling function
* str string to print
*/
define
ldebug(funct, str)
{
if (config("resource_debug") & 8) {
print "DEBUG:", funct:":", str;
}
return;
}
/*
************************
* Amdahl 6 legacy code *
************************
*
* NOTE: What follows is legacy code based on the method used by the
* Amdahl 6 group:
*
* John Brown, Landon Curt Noll, Bodo Parady, Gene Smith,
* Joel Smith and Sergio Zarantonello
*
* This method generated v(1) for nearly all values, except for a
* few rare cases when h mod 3 == 0. The code is NOT used by lucas.cal
* above. The gen_v1() function above is based on an improved method
* outlined in Ref4. That method generated v(1) for all h.
*
* The code below is kept for historical purposes only. The functions
* and global variables of the Amdahl 6 legacy code all begin with legacy_.
*/
/*
* Trial tables used by legacy_gen_v1()
*
* When h mod 3 == 0, one needs particular values of D, a and b (see
* legacy_gen_v1 documentation) in order to find a value of v(1).
*
* This table defines 'legacy_quickmax' possible tests to be taken in ascending
* order. The legacy_v1_qval[x] refers to a v(1) value from Ref1, Table 1. A
* related D value is found in legacy_d_qval[x]. All D values expect
* legacy_d_qval[1] are also taken from Ref1, Table 1. The case of D == 21 as
* listed in Ref1, Table 1 can be changed to D == 7 for the sake of the test
* because of {note 6}.
*
* It should be noted that the D values all satisfy the selection values
* as outlined in the gen_v1() function comments. That is:
* as outlined in the legacy_gen_v1() function comments. That is:
*
* D == P*(2^f)*(3^g)
*
@@ -571,20 +1093,20 @@ gen_u0(h, n, v1)
* where Q == 1. No further processing is needed to compute v(1) when r
* is of this form.
*/
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 */
legacy_quickmax = 8;
mat legacy_d_qval[legacy_quickmax];
mat legacy_v1_qval[legacy_quickmax];
legacy_d_qval[0] = 5; legacy_v1_qval[0] = 3; /* a=1 b=1 r=4 */
legacy_d_qval[1] = 7; legacy_v1_qval[1] = 5; /* a=3 b=1 r=12 D=21 */
legacy_d_qval[2] = 13; legacy_v1_qval[2] = 11; /* a=3 b=1 r=4 */
legacy_d_qval[3] = 11; legacy_v1_qval[3] = 20; /* a=3 b=1 r=2 */
legacy_d_qval[4] = 29; legacy_v1_qval[4] = 27; /* a=5 b=1 r=4 */
legacy_d_qval[5] = 53; legacy_v1_qval[5] = 51; /* a=53 b=1 r=4 */
legacy_d_qval[6] = 17; legacy_v1_qval[6] = 66; /* a=17 b=1 r=1 */
legacy_d_qval[7] = 19; legacy_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
* legacy_gen_v1 - compute the v(1) for a given h*2^n-1 if we can
*
* This function assumes:
*
@@ -613,12 +1135,12 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
*
* If this case condition is true, then:
*
* u(0) = (2+sqrt(3))^h + (2-sqrt(3))^h (see Ref1, page 869)
* = (2+sqrt(3))^h + (2+sqrt(3))^(-h)
* u(2) = (2+sqrt(3))^h + (2-sqrt(3))^h (see Ref1, page 869)
* = (2+sqrt(3))^h + (2+sqrt(3))^(-h) (some call this u(0))
*
* and since Ref1, Theorem 5 states:
*
* u(0) = alpha^h + alpha^(-h)
* u(2) = alpha^h + alpha^(-h)
* r = abs(2^2 - 1^2*3) = 1
*
* and the bottom of Ref1, page 872 states:
@@ -631,7 +1153,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
*
* then
*
* u(0) = v(h)
* u(2) = v(h)
*
* so we simply return
*
@@ -666,7 +1188,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
*
* where L(x,y) is the Legendre symbol (see below), then:
*
* u(0) = alpha^h + alpha^(-h)
* u(2) = alpha^h + alpha^(-h)
*
* The bottom of Ref1, page 872 states:
*
@@ -674,7 +1196,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
*
* thus since:
*
* u(0) = v(h)
* u(2) = v(h)
*
* so we want to return:
*
@@ -929,7 +1451,7 @@ d_qval[7] = 19; v1_qval[7] = 74; /* a=38 b=1 r=2 */
* returns v(1), or -1 is there is no quick way
*/
define
gen_v1(h, n)
legacy_gen_v1(h, n)
{
local d; /* the 'D' value to try */
local val_mod; /* h*2^n-1 mod 'D' */
@@ -947,10 +1469,10 @@ gen_v1(h, n)
* We will try all 'D' values until we find a proper v(1)
* or run out of 'D' values.
*/
for (i=0; i < quickmax; ++i) {
for (i=0; i < legacy_quickmax; ++i) {
/* grab our 'D' value */
d = d_qval[i];
d = legacy_d_qval[i];
/* compute h*2^n-1 mod 'D' quickly */
val_mod = (h*pmod(2,n%(d-1),d)-1) % d;
@@ -965,13 +1487,13 @@ gen_v1(h, n)
/* D mod 4 == 1, so check for J(D, h*2^n-1) == -1 */
if (jacobi(val_mod, d) == -1) {
/* it worked, return the related v(1) value */
return v1_qval[i];
return legacy_v1_qval[i];
}
} else {
/* D mod 4 == -1, so check for J(D, h*2^n-1) == 1 */
if (jacobi(val_mod, d) == 1) {
/* it worked, return the related v(1) value */
return v1_qval[i];
return legacy_v1_qval[i];
}
}
}
@@ -1029,19 +1551,3 @@ gen_v1(h, n)
/* no quick and dirty v(1), so return -1 */
return -1;
}
/*
* ldebug - print a debug statement
*
* input:
* funct name of calling function
* str string to print
*/
define
ldebug(funct, str)
{
if (config("resource_debug") & 8) {
print "DEBUG:", funct:":", str;
}
return;
}