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 Showing results 1 to 25 of 1000 Search took 0.31 seconds. Search: Posts Made By: R. Gerbicz
 Forum: Math 2020-08-13, 10:24 Replies: 3 Views: 186 Posted By R. Gerbicz Yes, a numeric evaluation, integration is quite... Yes, a numeric evaluation, integration is quite easy. But an analytic formula for a fixed [n-1,n] (or subinterval) is still possible, I've done exactly the same way to get the above rho(x) when done...
 Forum: Math 2020-08-12, 12:06 Replies: 3 Views: 186 Posted By R. Gerbicz You can get the rho function's Taylor polynom for... You can get the rho function's Taylor polynom for each n in the [n-1,n] interval, if you want the first n=maxn polynom, then just call ff(maxn,maxe) if you want the max degree=e in the polynoms, I've...
 Forum: GpuOwl 2020-08-07, 16:10 Replies: 2,393 Views: 135,743 Posted By R. Gerbicz Basically that prp-cf test it the same as the prp... Basically that prp-cf test it the same as the prp test, just in the end there is a big division. Even if the server already accepted such runs from your gpuowl (is it/was it possible?), then it is...
 Forum: Puzzles 2020-08-06, 23:05 Replies: 27 Views: 936 Posted By R. Gerbicz Checked those 480780, 100622 numbers. Turned out... Checked those 480780, 100622 numbers. Turned out that it is using more multiplications, I counted every mults, and to get the final product from the y[] array you also need mults. I mean this is not...
 Forum: Puzzles 2020-08-06, 20:58 Replies: 27 Views: 936 Posted By R. Gerbicz What addition chains here is now obsolete in the... What addition chains here is now obsolete in the method. See one of my post, the required final task for E=2 to get: R=y[1]*y[3]^3*y[5]^5*y[7]^7 mod M.
 Forum: Puzzles 2020-08-06, 20:18 Replies: 27 Views: 936 Posted By R. Gerbicz Yes, I've used the popular sliding window name,... Yes, I've used the popular sliding window name, but here there is no such precomputation here. Yeah it is hard to believe there is such an algorithm, because in that sliding window you are actually...
 Forum: Puzzles 2020-08-06, 19:42 Replies: 27 Views: 936 Posted By R. Gerbicz Searched today exactly that crypto paper, thanks.... Searched today exactly that crypto paper, thanks. I'll check it.
 Forum: Puzzles 2020-08-06, 17:54 Replies: 27 Views: 936 Posted By R. Gerbicz Really don't understand, could you explain it? ... Really don't understand, could you explain it? But I've just revisited my pari code and forget to subtract one, so my updated code gives only 104341 multiplications for E=13 (and that is still the...
 Forum: Puzzles 2020-08-05, 22:43 Replies: 27 Views: 936 Posted By R. Gerbicz Yes, so used the "number of set bits"="number of... Yes, so used the "number of set bits"="number of multiplications" to describe the solution's goodness. Written a small Pari code to get that number for E=13 for the given N.
 Forum: Puzzles 2020-08-05, 22:20 Replies: 27 Views: 936 Posted By R. Gerbicz See: the original task for P-1 method is to get... See: the original task for P-1 method is to get x^N mod M, but with multiplying N with a "small" number is still good for the method, so when you compute r(k)=x^(k*N) mod M [and then gcd(r(k)-1,M)]....
 Forum: Puzzles 2020-08-05, 19:12 Replies: 27 Views: 936 Posted By R. Gerbicz OK, here it is: Using an improved method... OK, here it is: Using an improved method Houston, reached 111.181 set bits (!), and with lower limits of the solution this could work easily on GPU also. [note that we started from 720737 ones of N...
 Forum: Puzzles 2020-08-05, 16:26 Replies: 27 Views: 936 Posted By R. Gerbicz https://www.youtube.com/watch?v=EY27lgnPKWI&featur... https://www.youtube.com/watch?v=EY27lgnPKWI&feature=youtu.be&t=9 I'm at set bits=644353 for an equivalent defined problem AND still counting.
 Forum: Software 2020-08-02, 20:38 Replies: 581 Views: 74,222 Posted By R. Gerbicz With the new version we could lower the trial... With the new version we could lower the trial factoring limit for all p by roughly one bit, as you'd do only one prp test with a quick proof instead of two tests. Any thoughts?
 Forum: PARI/GP 2020-08-01, 10:03 Replies: 12 Views: 430 Posted By R. Gerbicz Ok, but for f(x)=x^3 it is weaker than the binary... Ok, but for f(x)=x^3 it is weaker than the binary search. Just try this: cnt=0;solve(x=-1,2,cnt+=1;print(cnt" "x);x^3) So it is doing at most 259 iterations.
 Forum: PARI/GP 2020-07-30, 14:45 Replies: 12 Views: 430 Posted By R. Gerbicz Good question, it is not binary search, it should... Good question, it is not binary search, it should be another root finding algorithm. Btw in some really trivial cases the solve breaks: ? solve(x=-1,2,x^3) *** at top-level: solve(x=-1,2,x^3)...
 Forum: PARI/GP 2020-07-30, 12:04 Replies: 12 Views: 430 Posted By R. Gerbicz What is missing here is that you need a... What is missing here is that you need a continuous function. Otherwise the result is crap: ? solve(x=0,3,if(x<2,-1,1)) %3 = 1.9999999999999999999999999999999999999 ?
 Forum: Math 2020-07-28, 17:32 Replies: 22 Views: 720 Posted By R. Gerbicz You have a really good chance to halve the number... You have a really good chance to halve the number of ones from the expected L/2 to L/4, assuming log2(n)=L: binomial(10*L,L/4)>39^(L/4)=n^(log(39)/log(2)/4)=n^1.321 >> n
 2020-07-18, 20:28 Replies: 4 Views: 258 Posted By R. Gerbicz See... See https://en.wikipedia.org/wiki/Zsigmondy%27s_theorem and you don't need that p,q are primes, check me, but with few exception this should be working, since there is a prime r for that r | a^n-b^n...
 2020-07-05, 17:56 Replies: 10 Views: 722 Posted By R. Gerbicz Yes, that is a divisor: ?... Yes, that is a divisor: ? d=886407410000361345663448535540258622490179142922169401; ? Mod(2,d)^(2^127-1)+1 %2 = Mod(0, 886407410000361345663448535540258622490179142922169401) ? ## *** last...
 Forum: Math 2020-06-23, 09:12 Replies: 225 Views: 10,171 Posted By R. Gerbicz With those recursive calls it will use power... With those recursive calls it will use power residues in stack memory. Is it intended? You could do this also in disk (using the same size).
 Forum: Math 2020-06-23, 01:43 Replies: 225 Views: 10,171 Posted By R. Gerbicz You could use sliding window technique in void... You could use sliding window technique in void exponentiate (gwhandle *gwdata, gwnum x, uint64_t power) .
 Forum: Puzzles 2020-06-22, 09:50 Replies: 7 Views: 363 Posted By R. Gerbicz I've also that in my mind. Generated 1e8 random... I've also that in my mind. Generated 1e8 random boards, and counted 8499941 configurations that has no 5 or more same piece in a line. So roughly we are expecting 0.085*binomial(80,40) for the whole...
 Forum: Puzzles 2020-06-22, 06:33 Replies: 7 Views: 363 Posted By R. Gerbicz So there is no at least 5 in a line? So there is no at least 5 in a line?
 Forum: Puzzles 2020-06-21, 22:56 Replies: 7 Views: 370 Posted By R. Gerbicz A faster approach: if the discriminant is not a... A faster approach: if the discriminant is not a square then you can eliminate that triplet. With that you don't even need to calculate the coefficients, just check for ~50 smallest primes if the...
 Forum: Math 2020-06-20, 10:11 Replies: 225 Views: 10,171 Posted By R. Gerbicz You can easily avoid this just set hash=2*hash+1,... You can easily avoid this just set hash=2*hash+1, even if you'd this for every computed hash then no hash value will be zero among h_i.
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