To get what you want, replace the */ in the denominator of the verbs I
defined with ]

But Henry has suggested a better approach for the numerator.

Thanks,

-- 
Raul


On Wed, Oct 30, 2013 at 10:49 PM, Don Kelly <[email protected]> wrote:

> I tried it but it doesn't work.
> What I want is result  D =(a*b +b*c +c*a) and get D/c, D/a, D/b
>
>  zij
> 3j1.5 5j1.5 7j_1.5
>
> F zij
> 0.586738j_0.159111   wrong answer
>
> d2y2=:+/ %~ ] * _1&|.    (Jasmin's version)
> 1.57426j0.242574 0.920792j0.707921 2.47855j_0.0478548   Correct answer
>
> Jasmin's answer took about 1/3 the time and space of the explicit original
> version. so did Day's version and Boss'  version using deltoya=: 13 :  vs 3
> :
> Not that time and space are important where this would be applied.
>
> Don Kelly
>
>
>
>
> On 30/10/2013 5:31 AM, Raul Miller wrote:
>
>> On Wed, Oct 30, 2013 at 2:51 AM, Don Kelly <[email protected]> wrote:
>>
>>  You are right but it is a bit of serendipity  What I wanted is (a*b+
>>> b*c+c*a ) % c a b .
>>> it doesn't matter how the pairs are formed in the numerator ac+ba+cb is
>>> the same.
>>>
>>>  Like this?
>>
>>     F=: +/ .* (</~i.3) +/ .* ] % */
>>
>> That's probably awful for speed, but it does express the idea.
>>
>> Thanks,
>>
>>
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