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[Anders Logg <logg@xxxxxxxxx>]

On Tue, Apr 13, 2010 at 03:02:08PM +0200, Harish Narayanan wrote:
> On 4/13/10 2:47 PM, Anders Logg wrote:
> > On Tue, Apr 13, 2010 at 02:42:34PM +0200, Harish Narayanan wrote:
> >> On 4/13/10 1:32 PM, Anders Logg wrote:
> >>> What if T = 10 and dt = 0.9999999999. Then we get
> >>>
> >>>   ceil(T / dt) = 11.
> >>
> >> I am not sure I understand what is wrong with this.
> >>
> >> If dt = 1.0, then we get 10 time steps. If dt < 1.0, we get more time steps.
> >>
> >> Harish
> >
> > Isn't there a chance that one might specify the time step to be k and
> > the end time to be n*k and get T / k = n + eps?
>
> This, in my mind, is a common case. But the same problem would have
> existed in the earlier code as well. All I was doing was to replace the
> earlier code which always added one time step. Now:
>
> If T / k is an integer, dt = k.
> If T / k is not, then dt = T /(int(T / k) + 1)
>
> The earlier code was always using the second definition.

I'm not saying the old code was correct, but there should be some
clever solution that does the right thing. How about this:

n = ceil(T / dt - DOLFIN_EPS)

?

--
Anders

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