On Thu, Mar 4, 2010 at 3:16 AM, Zsbán Ambrus <[email protected]> wrote:
> Of course this two dimensional deconvolution verb in my previous mail
> only works for right arguments that are actually convolutions of the
> left argument, it gives spectacularly bad results for right arguments
> that aren't.
...
> For this case, I recommend the following solution instead.
>
> decv2 =: 4 :'z $ (,y) %. |:,"_1 (-(#:,@:i.) z=.y>:@-&$x) |. ($y){.x'
Ok, and that works well for the 2 dimensional case. Or, translated
to tacit (and reversing the argument order, so that argument
order is similar to that of division):
dcv2=: >:@-&$ $ ,@[ %. -@(#: ,@:i.)@(1 + -&$) |:@:(,@|.) $...@[ {. ]
Of course this fails for the 1 dimensional case.
Can you show me how to deal with the 1 dimensional case using
this style of approach, so I can focus on the differences between
the two implementations?
Thank you,
--
Raul
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