On Thu, Oct 30, 2014 at 9:25 AM, Jon Hough <[email protected]> wrote:
>
>
> But the result of doing that is clearly not what is wanted.


Can you elaborate on what you wanted?


This is what's happening:

Let's start with addition:

m =: 2 2 $ 3 5 7 1
v1=: (+/ . +)~ m
v2=: 2 2 $ ((+/ (3+3),(5+7)),(+/ (3+5),(5+1)),(+/ (7+3),(1+7)),(+/ (7+5),(1+1)))

]v1-:v2
1

If we move on to the the floor, I think it becomes clear:

v1=:(<./ . +)~ m
v2=:2 2 $ ((<./ (3+3),(5+7)),(<./ (3+5),(5+1)),(<./ (7+3),(1+7)),(<./
(7+5),(1+1)))

]v1-:v2
1

]v2
6 6
8 2

Or more explicitly:
a=:<./ (3+3),(5+7)
b=:<./ (3+5),(5+1)
c=:<./ (7+3),(1+7)
d=:<./ (7+5),(1+1)

] (2 2 $ a,b,c,d) -: v2
1

The minimum of a is 6, b is 6, c is 8, and d is 2

Does that help?
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