On 19 January 2015 at 23:03, mmarco mma...@unizar.es wrote:
It is much faster to work with absolute fields instead of towers of
extensions:
sage: K.sqrt3=QuadraticField(3)
sage: F.sqrt5=K.extension(x^2-5)
sage: R.a1,a2,a3,a4,a5 = F[]
sage: %time _=(a1+a2+a3+sqrt5*a4+sqrt3*a5)^25
CPU times:
On 2015-01-20 10:14, John Cremona wrote:
? Mod(x,x^2-3) + Mod(x,x^2-5)
%1 = 0
That's a PARI feature: the result would lie in the ring
QQ[x]/(x^2-3, x^2-5) = QQ[x]/(1)
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any idea how to get the number of
terms in a better way in Sage?):
sage: K.sqrt3 = QuadraticField(3)
sage: L.sqrt5 = K.extension(x^2-5)
sage: R.a1,a2,a3,a4,a5 = L[]
sage: time f = (a1+a2+a3+sqrt5*a4+sqrt3*a5)^18
CPU times: user 2.43 s, sys: 3.94 ms, total: 2.44 s
Wall time: 2.44
It is much faster to work with absolute fields instead of towers of
extensions:
sage: K.sqrt3=QuadraticField(3)
sage: F.sqrt5=K.extension(x^2-5)
sage: R.a1,a2,a3,a4,a5 = F[]
sage: %time _=(a1+a2+a3+sqrt5*a4+sqrt3*a5)^25
CPU times: user 27.4 s, sys: 12 ms, total: 27.4 s
Wall time: 27.5 s
sage:
Hi Miguel,
On Mon, Jan 19, 2015 at 4:03 PM, mmarco mma...@unizar.es wrote:
It is much faster to work with absolute fields instead of towers of
extensions:
sage: K.sqrt3=QuadraticField(3)
sage: F.sqrt5=K.extension(x^2-5)
sage: R.a1,a2,a3,a4,a5 = F[]
sage: %time
On Monday, January 19, 2015 at 9:27:44 PM UTC-8, Ondřej Čertík wrote:
and your approach returns a wrong number of terms, so something is
wrong. But it is quite fast.
The term count doesn't tell you that. The representation of sqrt3 and sqrt5
doesn't consist of single term expressions:
What is here?
AMD Phenom 3GHz, 8GB RAM, no other big jobs
Since that expression is large, the cache size of the CPU might
significantly impact performance.
Wouldn't that affect any of the following?
│ Sage Version 6.5.beta5, Release Date: 2015-01-05 │
│ Type
On Sunday, January 18, 2015 at 9:18:53 AM UTC+1, vdelecroix wrote:
Your example can be reduced to polynomials
sage: K.sqrt3 = QuadraticField(3)
sage: R.a1,a2,a3,a4,a5 = K[]
sage: timeit((a1+a2+a3+a4+sqrt3*a5)^25)
5 loops, best of 3: 81 ms per loop
How do you get this speed? Here
On Mon, Jan 19, 2015 at 8:55 AM, Ralf Stephan gtrw...@gmail.com wrote:
On Sunday, January 18, 2015 at 9:18:53 AM UTC+1, vdelecroix wrote:
Your example can be reduced to polynomials
sage: K.sqrt3 = QuadraticField(3)
sage: R.a1,a2,a3,a4,a5 = K[]
sage: timeit((a1+a2+a3+a4+sqrt3*a5)^25)
5
On Mon, Jan 19, 2015 at 9:46 AM, Ralf Stephan gtrw...@gmail.com wrote:
What is here?
AMD Phenom 3GHz, 8GB RAM, no other big jobs
Since that expression is large, the cache size of the CPU might
significantly impact performance.
Wouldn't that affect any of the following?
│ Sage Version
On Mon, Jan 19, 2015 at 11:19 AM, Ondřej Čertík ondrej.cer...@gmail.com wrote:
Hi Vincent,
On Sun, Jan 18, 2015 at 10:06 AM, Vincent Delecroix
20100.delecr...@gmail.com wrote:
Hi,
2015-01-18 18:03 UTC+01:00, Ondřej Čertík ondrej.cer...@gmail.com:
Can you invent an example, that can't be
Hello Ondrej,
For such questions of Sage usage, it is better to discuss on
ask.sagemath.org or sage-support.
You can also deal with all algebraic numbers at once with QQbar
sage: sqrt3 = QQbar(sqrt(3))
sage: sqrt5 = QQbar(sqrt(5))
But then polynomials over QQbar are much slower.
Vincent
Hi Vincent,
On Mon, Jan 19, 2015 at 11:30 AM, Vincent Delecroix
20100.delecr...@gmail.com wrote:
Hello Ondrej,
For such questions of Sage usage, it is better to discuss on
ask.sagemath.org or sage-support.
You can also deal with all algebraic numbers at once with QQbar
sage: sqrt3 =
Hi Vincent,
On Sun, Jan 18, 2015 at 10:06 AM, Vincent Delecroix
20100.delecr...@gmail.com wrote:
Hi,
2015-01-18 18:03 UTC+01:00, Ondřej Čertík ondrej.cer...@gmail.com:
Can you invent an example, that can't be converted to polynomials?
Perhaps (a1+a2+a3+sqrt(5)*a4+sqrt(3)*a5)^25?
Still
On Monday, January 19, 2015 at 9:46:47 AM UTC-8, Ralf Stephan wrote:
What is here?
AMD Phenom 3GHz, 8GB RAM, no other big jobs
On Intel(R) Core(TM) i7-2600 CPU @ 3.40GHz I'm getting the same times as
Vincent. That's on 6.5beta4 or 5.
The difference you're reporting is very large. You
On Mon, Jan 19, 2015 at 10:32 AM, Nils Bruin nbr...@sfu.ca wrote:
On Monday, January 19, 2015 at 9:46:47 AM UTC-8, Ralf Stephan wrote:
What is here?
AMD Phenom 3GHz, 8GB RAM, no other big jobs
On Intel(R) Core(TM) i7-2600 CPU @ 3.40GHz I'm getting the same times as
Vincent. That's on
On Monday, January 19, 2015 at 10:35:42 AM UTC-8, William wrote:
Nils, did you specifically try this **exact input**??
Full session:
sage: sage: K.sqrt3 = QuadraticField(3)
sage: sage: R.a1,a2,a3,a4,a5 = K[]
sage: sage: timeit((a1+a2+a3+a4+sqrt3*a5)^25)
5 loops, best of 3: 79.9 ms per
The computation in pari (directed from Sage):
sage: x=pari(x)
sage: y=pari(y)
sage: sqrt3=pari(Mod)(x, x^2-3)
sage: sqrt5=pari(Mod)(y, y^2-5)
sage: a1=pari(a1)
sage: a2=pari(a2)
sage: a3=pari(a3)
sage: a4=pari(a4)
sage: a5=pari(a5)
sage: time f = (a1+a2+a3+sqrt5*a4+sqrt3*a5)**18
CPU times: user
Your example can be reduced to polynomials
sage: K.sqrt3 = QuadraticField(3)
sage: R.a1,a2,a3,a4,a5 = K[]
sage: timeit((a1+a2+a3+a4+sqrt3*a5)^25)
5 loops, best of 3: 81 ms per loop
(And just for completeness, the symbolic expansion on my laptop took 4.54s)
Vincent
2015-01-18 7:26 UTC+01:00,
Hi,
2015-01-18 18:03 UTC+01:00, Ondřej Čertík ondrej.cer...@gmail.com:
Can you invent an example, that can't be converted to polynomials?
Perhaps (a1+a2+a3+sqrt(5)*a4+sqrt(3)*a5)^25?
Still doable. You need to involve log, exp, cos or similar
transcendental functions.
Vincent
--
You received
Hi Vincent,
On Sun, Jan 18, 2015 at 1:18 AM, Vincent Delecroix
20100.delecr...@gmail.com wrote:
Your example can be reduced to polynomials
sage: K.sqrt3 = QuadraticField(3)
sage: R.a1,a2,a3,a4,a5 = K[]
sage: timeit((a1+a2+a3+a4+sqrt3*a5)^25)
5 loops, best of 3: 81 ms per loop
That's cool,
Hi,
I was wondering what the fastest way is to do this benchmark in Sage:
┌┐
│ Sage Version 6.4, Release Date: 2014-11-14 │
│ Enhanced for SageMathCloud.│
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