Hello everybody
I have been willing for some time to implement a recognition algorithm
for Interval Graphs [1], and I ended up forgetting to sleep one
evening to have it done in the morning. :-)
What I now have is an algorithm which uses PQ-Trees [2] and is able to
tell, given a graph, wheth
That did it, thanks!
On Jun 7, 4:48 pm, William Stein wrote:
> On Mon, Jun 7, 2010 at 1:37 PM, Brion Keagle wrote:
> > Hello - I'm the IT guy trying to get Sage up and running for our Math
> > department. I downloaded the VMWare virtual appliance and got it set
> > up on the network properly.
Hello,
I would like to do the following computation concerning a subring of a
quotient of a polynomial ring:
The first task is easily solvable: compute the dimension of the
coinvariant ring of the symmetric group:
sage: R. = QQ['x1, x2, x3']
sage: I = R.ideal( x1+x2+x3, x1^2+x2^2+x3^2, x1^3+x2^3
Hi Marco,
On Fri, Jun 11, 2010 at 5:03 AM, Marco Boretto wrote:
> I'm tryng to to simple thing like this:
>
> m=[0.6158, 0.5893, 0.5682, 0.51510, 0.4980, 0.4750, 0.5791,
> 0.5570,0.5461, 0.4970, 0.4920, 0.4358, 0.422, 0.420]
> m.count
>
> i want to know the number of the object but the answer is
On Thu, Jun 10, 2010 at 12:03 PM, Marco Boretto wrote:
> I'm tryng to to simple thing like this:
>
> m=[0.6158, 0.5893, 0.5682, 0.51510, 0.4980, 0.4750, 0.5791,
> 0.5570,0.5461, 0.4970, 0.4920, 0.4358, 0.422, 0.420]
> m.count
>
> i want to know the number of the object but the answer is
You use l
> m=[0.6158, 0.5893, 0.5682, 0.51510, 0.4980, 0.4750, 0.5791,
> 0.5570,0.5461, 0.4970, 0.4920, 0.4358, 0.422, 0.420]
> m.count
len(m) does the job, you should probably look into the tutorial at
http://www.sagemath.org/doc/tutorial/ for this kind of questions...
m.count is a function returning the
I'm tryng to to simple thing like this:
m=[0.6158, 0.5893, 0.5682, 0.51510, 0.4980, 0.4750, 0.5791,
0.5570,0.5461, 0.4970, 0.4920, 0.4358, 0.422, 0.420]
m.count
i want to know the number of the object but the answer is
Want does it mean?
Another similar thing, i want to multiply the all the e
Hi, I don't think Sage has the right flavor of tools for the kinds of
thing that XPP is intended for. XPP is a really great learning tool
for numerically solving systems of nonlinear ODEs and to study the
dynamics of small systems, and very appropriate for computational
neuroscience. A pythonic alt
Dear Sage users,
I am going to take a course on computational neuroscience soon. Tutors
suggested us to use xppaut (or similar) to solve ODEs. Since I am a
little bit more familiar with Sage, I was wondering if I could simply
use Sage in stead of xppaut. I would really appreciate if somebody w
Hi
On Thu, Jun 10, 2010 at 11:01:38PM +0200, Jose Guzman wrote:
> I am going to take a course on computational neuroscience soon. Tutors
> suggested us to use xppaut (or similar) to solve ODEs. Since I am a
> little bit more familiar with Sage, I was wondering if I could simply
> use Sage in
On Jun 10, 2010, at 12:22 PM, Christian Stump wrote:
m=[0.6158, 0.5893, 0.5682, 0.51510, 0.4980, 0.4750, 0.5791,
0.5570,0.5461, 0.4970, 0.4920, 0.4358, 0.422, 0.420]
m.count
len(m) does the job, you should probably look into the tutorial at
http://www.sagemath.org/doc/tutorial/ for this kind o
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