It always pleases me when debug removes the @ and @:'s.

The two examples you point to could also be defined:

f=: 13 :'x*((([:%: +)/)@:*:) y'
   f2=: 13 :'x * ([:(([:%:  +)/)*:) y'
   (2 f 1 2 3)-:2 f2 1 2 3
1
   
   g=: 13 :'x * ((%: @ (+/))@:*:) y'
   g2=: 13 :'x * ([:([:%: +/)*:) y'
   (2 g 1 2 3)-:2 g2 1 2 3
1

Linda


-----Original Message-----
From: Programming [mailto:[email protected]] On Behalf 
Of Henry Rich
Sent: Saturday, December 2, 2017 1:29 PM
To: [email protected]
Subject: Re: [Jprogramming] APL Exercises

There are no exceptions, only rules: see

http://code.jsoftware.com/wiki/Vocabulary/Modifiers#Grouping_And_Order_Of_Execution

Henry Rich

On 12/2/2017 1:02 PM, Andrew Dabrowski wrote:
> On 12/02/2017 02:17 AM, Roger Hui wrote:
>
>>>> SC =: 3 : '(3 3$4>i.5) ,./^:2@(*/)^:y ,.1'
>>> SC confuses me.  I would have thought that
>>>
>>> (3 3$4>i.5) (*/)^:y ,.1
>> The left operand of the power operator ^: is ,./^:2@(*/) .
>>
> Oh, I parsed it as
>
> (3 3$4>i.5) (,./^:2) @ ((*/)^:y) ,.1
>
> but it should be
>
> (3 3$4>i.5) (,./^:2 @ */)^:y ,.1
>
> is that right?  I'm certainly not au courant in all the exceptions to 
> right-association.
>
> Btw, does the cut function ;._3 have an inverse that could be used to 
> solve this problem?
> ----------------------------------------------------------------------
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