Re: filtering logical array

[Robert T. Short]

On 06/26/2012 01:51 PM, Mike Miller wrote:
> On Tue, Jun 26, 2012 at 3:53 PM, Robert T. Short
> <address@hidden> wrote:
> > On 06/26/2012 11:48 AM, Jordi Gutiérrez Hermoso wrote:
> > > a=[0 0 1 1 1 1 1 0 1 0 1 0 0 1 0 0 1 1 0 0 0 0 0 1 1 1 0]
> > Kind of a fun problem.  Here is another solution.
> More fun than what I was doing...
>
> > x = conv(a,[1 1 1])
> > z = x>2
> > z = z | shift(z,-1) | shift(z,-2)
> > z = z(1:length(a))
> >
> > There is probably a much more elegant way to do the last two lines but it
> > isn't coming to me.
> I noticed your shift-and-or of z is really sort of a convolution.  How about:
>
> x = conv(a, [1 1 1])
> y = conv(x, [1 1 1])
> z = y >= 6
> z = z(3:end-2)
Ahh!  But convolution is commutative so we can do this

y = (conv(a,conv([1 1 1],[1 1 1]))>5)(3:end-2)

8-)

Actually, if we have convolution over GF(2) the answer is cleaner but I 
am not aware of such a function.

Bob