A couple of errors.

u&.v y   is not the same as    vi u v y 

I know some writers have written that it is, but it's not.

u&.:v y  is the same as  vi u v y  .

u&.v y   is the same as u&.:v"v y
 (which means you do the &.: thing on cells whose size is given
  by the monadic rank of v).


Use b. _1  to see the obverse:
   > b. _1
<


  (< i. >) 3 2 2 2

is not even close to right.  That is a fork, and produces
   (< 3 2 2 2) i. (> 3 2 2 2)
which is near nonsense.

  i.&.:>   is equivalent to  <@:i.@:>

so

  i.&.> 3 2 2 2  is equivalent to (<@:i.@:>)"> 3 2 2 2

   (<@:i.@:>)"> 3 2 2 2
+-----+---+---+---+
|0 1 2|0 1|0 1|0 1|
+-----+---+---+---+

Henry Rich


> -----Original Message-----
> From: [EMAIL PROTECTED] 
> [mailto:[EMAIL PROTECTED] On Behalf Of Terrence Brannon
> Sent: Tuesday, June 19, 2007 4:13 PM
> To: [email protected]
> Subject: [Jgeneral] 'age spec astig tear'=: i.&.>3 2 2 2 NB. 
> not clear onthis... help needed
> 
> Ok, 
> 'a b c' =: 1 2 3   NB. multiple assignments. this is easy
>                    NB. therefore 'age spec astig tear'=:  
>                    NB. is easy for me to understand
>                    NB. but the expression i.&.>3 2 2 2
>                    NB. has me confused.
> 
> I read up on &. and we are dealing with the monadic case...
> so:
> 
>    u&.v y IS THE SAME AS vi u v y
> 
> which basically means 3 verbs get applied in sequence to y
> 
> so:
>    i.&.>y  IS THE SAME AS vi i. > y
> where vi is the obverse of >
> 
> Is there a programmatic way to get the obverse?
> I'm guessing it is <  NB. box 
> And why is the obverse applied to 0-cells of y?
> 
> 
> So I think it actually is not to hard to figure out what this 
> is doing from a
> holistic standpoint, but 
> (< i. >) 3 2 2 2 
> 
> is not returning the same thing as 
> (i.&.>)3 2 2 2
> 
> which means I am making a mistake somewhere...
> 
> Help appreciated.
> 
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