gcd (f(n), g(n))>1 ? - Page 3

LaurV · 2017-12-13, 05:49

We vote for deletion until you learn to use code tags...
(guess: nobody will waste the time to go through that pile of... ascii mixture to reply anything useful, this is a bit of an insult to us, like, in your eyes, you have no time to write it nice, your time is too valuable for you, but we have all the time in the world for free, to sort it up, understand what it does, come with a solution, reply to you in a nice format...)

CRGreathouse · 2017-12-13, 18:34

I think it's a nontrivial request, code tags or not. Probably the easiest way would be to ignore the subquadratic code entirely and trap it only at the Lehmer level.

bhelmes · 2018-02-20, 21:43

Quote:

Originally Posted by bhelmes

i choose two interesting quadratic function, for m=137 i can calculate up to 10^11 in 500 min on a single core and calculate the two function and make a gcd. I think there should be a speedup possible.

m=137 is the exponent for the mersenne number =
32032215596496435569.5439042183600204290159

Greetings from the primes
Bernhard

A peaceful and pretty calm night for everybody

I have made some speedtest for the mpz_gcd from gmp-library 6.1 :
m=77955737; 32,2 min / 10^6 gcds / single core
(m should be the exponent of the mersenne number)

Processor : 6x AMD FX(tm)-6300 Six-Core Processor,
L1 6* 16 KB 68939 MB/s,
L2 3*2048 KB 32858 MB/s,
L3 8 MB, 9554 MB/s,
4*8 GB Ram DDR3 1600 7813 MB/s,

Is this a good result or not ?

Greetings from the primes

Bernhard

bhelmes · 2018-03-07, 13:44

A pleasent day for all,

does anyone have a gmp gcd function with an additional minimum limit ?

The gcd function should stop the calculation, if gcd (n,m)<minimum.

I hope that this function will be faster than the pure gcd function.

Greetings from old Germany

Bernhard

bhelmes · 2018-04-05, 21:27

A peaceful night for all,

i search for some rules in order to calculate easier the gcd
of the values of two functions

f(n), g(m) and h(o) should be different quadratic irreducible polynomials,
n, m, o element of N
where f(n_o)=g(m_o)=h(o_o) for a special vulue n_o, m_o and o_o
and f(n_o)=a*b is not prime

I know that gcd [f(n_o-a), g(n_o-a)]=a

Can i speed up the calculation by reducing the three quadratic polynomials to two linear functions r(n), s(n) so that gcd [r(n_o-a), s(n_o-a)]=a

What i try to ask is, if the structure of the factors for the gcd remains ?

Greetings from the primes

Bernhard

P.S. I know that the mathematic description could be better formulated

bhelmes · 2018-05-02, 17:02

A peaceful and pleasant day for everyone,

if i have two linear functions f(x)=ax+b and g(x)=cx*d, x element N
how can i calculate the x for that the result of f(x)/g(x) is element N again ?

Thanks for your patience

Bernhard

2017-12-13, 05:49	#23
LaurV Romulan Interpreter Jun 2011 Thailand 7×1,373 Posts	We vote for deletion until you learn to use code tags... (guess: nobody will waste the time to go through that pile of... ascii mixture to reply anything useful, this is a bit of an insult to us, like, in your eyes, you have no time to write it nice, your time is too valuable for you, but we have all the time in the world for free, to sort it up, understand what it does, come with a solution, reply to you in a nice format...) Last fiddled with by LaurV on 2017-12-13 at 05:52

2017-12-13, 18:34	#24
CRGreathouse Aug 2006 3×1,993 Posts	I think it's a nontrivial request, code tags or not. Probably the easiest way would be to ignore the subquadratic code entirely and trap it only at the Lehmer level.

2018-03-07, 13:44	#26
bhelmes Mar 2016 3·5·23 Posts	A pleasent day for all, does anyone have a gmp gcd function with an additional minimum limit ? The gcd function should stop the calculation, if gcd (n,m)<minimum. I hope that this function will be faster than the pure gcd function. Greetings from old Germany Bernhard

2018-04-05, 21:27	#27
bhelmes Mar 2016 101011001₂ Posts	A peaceful night for all, i search for some rules in order to calculate easier the gcd of the values of two functions f(n), g(m) and h(o) should be different quadratic irreducible polynomials, n, m, o element of N where f(n_o)=g(m_o)=h(o_o) for a special vulue n_o, m_o and o_o and f(n_o)=a*b is not prime I know that gcd [f(n_o-a), g(n_o-a)]=a Can i speed up the calculation by reducing the three quadratic polynomials to two linear functions r(n), s(n) so that gcd [r(n_o-a), s(n_o-a)]=a What i try to ask is, if the structure of the factors for the gcd remains ? Greetings from the primes Bernhard P.S. I know that the mathematic description could be better formulated

2018-05-02, 17:02	#28
bhelmes Mar 2016 531₈ Posts	A peaceful and pleasant day for everyone, if i have two linear functions f(x)=ax+b and g(x)=cx*d, x element N how can i calculate the x for that the result of f(x)/g(x) is element N again ? Thanks for your patience Bernhard