Look into
1 */\. y
Henry Rich
On 10/30/2013 10:49 PM, Don Kelly 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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