On Thu, Nov 28, 2013 at 10:45:39AM -0800, William Stein wrote:
> On Thu, Nov 28, 2013 at 10:27 AM, Georgi Guninski
> wrote:
> > On Thu, Nov 28, 2013 at 07:48:24AM -0800, William Stein wrote:
> >> On Nov 28, 2013 4:36 AM, "Volker Braun" wrote:
> >> >
> >> > It would be just as easy for a compromi
On 2013-11-28, Volker Braun wrote:
> It would be just as easy for a compromised cloud ssh to download your
> personal private key than to log your password. The only solution is to
> make a different account on your own machine to log into, assuming that
> your university allows remote logins t
Hey David,
thank you very much for your answer. Finally, i decided to solve it
numerically just implementing an implicit Euler rule.
Thanks a lot!
Jose
On 28/11/13 14:14, David Joyner wrote:
On Thu, Nov 28, 2013 at 4:37 AM, Jose Guzman wrote:
Hi everybody,
I want to solve the following O
On 28 November 2013 18:27, Georgi Guninski wrote:
>
> please don't spread disinformation and don't
Please learn some etiquette.
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On Thu, Nov 28, 2013 at 10:45 AM, William Stein wrote:
> On Thu, Nov 28, 2013 at 10:27 AM, Georgi Guninski
> wrote:
>> On Thu, Nov 28, 2013 at 07:48:24AM -0800, William Stein wrote:
>>> On Nov 28, 2013 4:36 AM, "Volker Braun" wrote:
>>> >
>>> > It would be just as easy for a compromised cloud s
On Thu, Nov 28, 2013 at 10:27 AM, Georgi Guninski wrote:
> On Thu, Nov 28, 2013 at 07:48:24AM -0800, William Stein wrote:
>> On Nov 28, 2013 4:36 AM, "Volker Braun" wrote:
>> >
>> > It would be just as easy for a compromised cloud ssh to download your
>> personal private key than to log your pass
On Thu, Nov 28, 2013 at 07:48:24AM -0800, William Stein wrote:
> On Nov 28, 2013 4:36 AM, "Volker Braun" wrote:
> >
> > It would be just as easy for a compromised cloud ssh to download your
> personal private key than to log your password.
>
> I always protect my ssh keys by passphrase protecting
Of course once the attacker has access to all of your files he doesn't
really need your private key (passphrase or not) to log into your machine.
He can install his own keys, put a modified ssh/ssh-agent binary in the
PATH, etc..
On Thursday, November 28, 2013 3:48:24 PM UTC, William wrote:
>
On Nov 28, 2013 4:36 AM, "Volker Braun" wrote:
>
> It would be just as easy for a compromised cloud ssh to download your
personal private key than to log your password.
I always protect my ssh keys by passphrase protecting them, so just
downloading the private key does not trivially give access.
On Thu, Nov 28, 2013 at 04:36:47AM -0800, Volker Braun wrote:
> It would be just as easy for a compromised cloud ssh to download your
> personal private key than to log your password. The only solution is to
> make a different account on your own machine to log into, assuming that
> your univers
On Thu, Nov 28, 2013 at 4:37 AM, Jose Guzman wrote:
> Hi everybody,
>
> I want to solve the following ODE:
>
> dp/dt = (1-p)/tau - u*p*delta(t-tOn)
>
> In principle, desolve_odeint should be able to do it. However, in sage
>
> sage: import numpy as np
> sage: from sage.calculus.desolvers import de
On Thu, Nov 28, 2013 at 06:01:05AM -0600, Jason Grout wrote:
> On 11/27/13 9:31 AM, Georgi Guninski wrote:
> >A minor disadvantage of this solution is cloud.sagemath
> >sees the password/ssh key of...@foo.edu, so I wouldn't
> >use this if I care about the account.
>
> Not if you use passwordless l
It would be just as easy for a compromised cloud ssh to download your
personal private key than to log your password. The only solution is to
make a different account on your own machine to log into, assuming that
your university allows remote logins to personal machines at all (e.g.
Oxford doe
On 11/27/13 9:31 AM, Georgi Guninski wrote:
A minor disadvantage of this solution is cloud.sagemath
sees the password/ssh key of...@foo.edu, so I wouldn't
use this if I care about the account.
Not if you use passwordless logins (i.e., copy your sagemath public key
to your personal server's aut
Hi everybody,
I want to solve the following ODE:
dp/dt = (1-p)/tau - u*p*delta(t-tOn)
In principle, desolve_odeint should be able to do it. However, in sage
sage: import numpy as np
sage: from sage.calculus.desolvers import desolve_odeint
sage: p = var('p')
sage: t = var('t')
sage: tau = 2
sa
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