20181227, 16:52  #45 
Sep 2002
Database er0rr
2^{2}·5·181 Posts 
Here is 2selfridge BailliePSW code:
Last fiddled with by paulunderwood on 20181227 at 17:17 
20181228, 15:36  #46 
Sep 2002
Database er0rr
7044_{8} Posts 
I have tidied all programs. I also fixed a problem with factorial primes/PRPs where a suitable jacobi symbols could not be found.
s2: (x+2)^(n+1) == 5+2*a (mod n, x^2a*x+1) where a minimal such that jacobi(a^24,n) == 1. lucasPRP: x^(n+1) == 1 (mod n, x^2a*x+1) where minimal a>2 such that jacobi(a^24,n) == 1. hybrid: as "s2" for a=0 or a=1 and otherwise (x+1)^(n+1) = 2+a (mod n, x^2a*x+1) where minimal a>2 such that jacobi(a^24,n) == 1, fusion: as "hybrid" except tests (2*x)^((n+1)/2) == jacobi(2*(a+2),n)*2 (mod m, x^2a*x+1) for a>2 such that jacobi(a^24,n) == 1 bpsw2: (2*x)^((n+1)/2) == jacobi(2*(a+2),n)*2 (mod n, x^2a*x+1) for minimal a>2 such that jacobi(a^24,n) == 1 Last fiddled with by paulunderwood on 20181228 at 15:56 
20181230, 06:58  #47 
Sep 2002
Database er0rr
111000100100_{2} Posts 
Below is the result of not using threads for this size of number. Obviously, I should test the input size to set the number of threads.
4 threads: Code:
time ./pfgw64 k f0 od q'39924*10^30000+3*2^99654+455'  ../../coding/gwnum/prp2  Is 2PRP! real 0m17.745s user 0m33.124s sys 0m13.808s Code:
time ./pfgw64 k f0 od q'39924*10^30000+3*2^99654+455'  ../../coding/gwnum/prp2  Is 2PRP! real 0m11.615s user 0m11.776s sys 0m0.004s Code:
if (LEN > 1000000) gwset_num_threads ( gwdata, 4 ); Last fiddled with by paulunderwood on 20181230 at 07:39 
20190204, 20:10  #48 
Sep 2002
Database er0rr
2^{2}·5·181 Posts 
Array Arithmetic
Although what I present here may not be applicable to GWNUM, it is useful where one has to construct one's own arbitrary arithmetic using arrays.
Let w be 2^64 in 64bit machines, be 2^54 in JavaScript etc. Find minimal r>0 such that jacobiSymbol(w^(2*r)4,n)==1 where n is the prime candidate. Then test: (w*x)^((n+1)/2)==w*kronecker(w^r+2,n) (mod n, x^2w^r*x+1) Intermediate calculations in the left to right binary exponentiation are then: Squaring: (s*x + t)^2 > s*(w^r*s + 2*t)*x + (t  s)*(t + s) (mod n) Multiplying by the base: (s*x + t)*w*x > (w^(r + 1)*s + w*t)*x  w*s (mod n) With some simple manipulation of array indices, this should be quite quick. I would be impressed by someone if they implemented this. Last fiddled with by paulunderwood on 20190204 at 20:44 
20190205, 02:18  #49 
Sep 2002
Database er0rr
3620_{10} Posts 
On second thoughts, a large "r" could really balloon the modular reductions in the above. However if w=2 and the test is:
(2*x)^((n+1)/2) == 2*jacobi(2*(2^r+2)) (mod n, x^22^r*x+1) then the squaring part would be: (s*x+t)^2 > s*(s<<r+t<<1)*x + (ts)*(t+s) mod n, thus saving a multiplication (by "a") in favour of shifts by "r". I don't think that there would be too much excess to modularly reduce most of the time. A similar situation arises for multiplication by the base, 2*x: (s*x+t)*2*x > (s<<(r+1)+t<<1)*x  s<<1 mod n. Last fiddled with by paulunderwood on 20190205 at 02:27 
20190508, 15:43  #50 
Sep 2002
Database er0rr
2^{2}×5×181 Posts 
During left to right binary exponentiation of S*x+T to a power over x^2+1, the square of the intermediate value s*x+t is 2*s*t*x+t^2s^2.
My question is about FFT. Is it quicker to do 2 forward FFTs, 3 convolutions, and 3 reverse FFTs than it is to do 4 forward FFTs, 2 convolutions and 2 reverse FFTs? Or, to put it another way, is it better to compute the difference of squares in this case? Last fiddled with by paulunderwood on 20190508 at 16:30 
20190508, 19:38  #51  
"Robert Gerbicz"
Oct 2005
Hungary
2·3^{6} Posts 
Quote:
t^2s^2=(ts)*(t+s) and also note that (ts)+(t+s)=2*t, but you'd loss in precision if you calculate one transform with this. 

20190508, 20:01  #52 
"Robert Gerbicz"
Oct 2005
Hungary
2×3^{6} Posts 

20190508, 20:21  #53 
Sep 2002
Database er0rr
7044_{8} Posts 
Answering my own question: It is quicker to go down the (ts)*(t+s) route I conclude after running my own tests. It seems that forward and inverse FFTs need less computation than pointwise multiplication.
Last fiddled with by paulunderwood on 20190508 at 20:50 
20200731, 11:03  #54 
Sep 2002
Database er0rr
2^{2}·5·181 Posts 
Here is a gwnum program that tests Euler+Frobenius as per https://mersenneforum.org/showpost.p...6&postcount=74.
power_quadratic.c tests over x^22*x2^s which arguably safer than quadratic.c (x^22*xb). Last fiddled with by paulunderwood on 20200804 at 20:10 
20210210, 12:46  #55 
Sep 2002
Database er0rr
2^{2}·5·181 Posts 
xp1_euler_frobenius
Here is my GWNUM code for the test given in this thread:
I will review the Frobenius section and would welcome any improvements to it. Rather than calculating 2*s*t and (a*s+t)*(s+t)  (a+1)*s*t, by precalculating forward FFTs the latter has fewer additions and mults by small numbers if it is calculated as a*s^2+t^2.. Done! 15% faster. One thing that will be done is to use gwswap instead of gwcopy in the Frobenius "if" clause. Done! A test shows this program to be a good 1+2.5 selfridges Branched Euler PRP  if 2 then we can use gwstartnextfft (illegally)  I have also uploaded "quickest.c" that does the following (with code using forward FFTs) Code:
condition base quadratic ~selfridges n == 3 mod 4 (x+2) [x^2+1] 2.2 n == 5 mod 8 (x+1) [x^2+2] 2.8 n == 5 mod 6 (x+2) [x^2x+1] 2.2 default (x+1) [x^2a*x+1] 2.2 (Fixed compile instructions corrupted by global remove of "quick" ) EDIT: I removed illegal instructions from both programs . Sorry for the inconvenience. Last fiddled with by paulunderwood on 20210306 at 14:35 
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