I had the hardest time learning how dyad u/ works, but I was very proud 
to figure out u"l r"l _ .  Now that I get u/, I use u"l r/ instead.

Henry Rich

On 11/4/2011 10:28 AM, Roger Hui wrote:
> When "table" x f/y was introduced (at the very beginning) we did not
> fully comprehend the power of the rank operator.  If I had realized
> that table was just x f"(lr,_) y, where lr is the left rank of f, I
> might have tried to argue strongly for using the dyad f/ for something
> else.
>
>
>
> On Fri, Nov 4, 2011 at 2:15 AM, Henry Rich<[email protected]>  wrote:
>> Just a quibble with terminology: 'item' means _1-cell.  That concept
>> does not apply here.  If you define 'l-cell' to be the left rank of the
>> verb, and 'r-cell' the right rank, you could say
>>
>> ...it loops through all the l-cells of x, then for each of those, all
>> the r-cells of y...
>>
>> Henry Rich
>>
>> On 11/4/2011 12:13 AM, Marshall Lochbaum wrote:
>>> * has rank 0 already, so it isn't necessary. The definition of / is that it
>>> applies the verb with rank (left rank),_ . Essentially, this means that it
>>> loops through all the items of x, then for each of those, all the items of
>>> y, where an "item" is an item with the left rank of the verb. If you try
>>> magic/~ a with magic having rank _ (because it's a fork), you just get (a
>>> magic"_ _"_ _ a), which simply applies magic regularly.
>>>
>>> Marshall
>>>
>>> On Thu, Nov 3, 2011 at 11:57 PM, Ricardo 
>>> Forno<[email protected]>wrote:
>>>
>>>> I have this verb:
>>>> magic =: * %>:@(+:@*) - +
>>>> that I use only as a dyad, and, say,
>>>> a =: 0.1 * i. 10
>>>> If I want to get a table of the * verb, I  write:
>>>> a * / a
>>>> If I write
>>>> a magic / a
>>>> I dont get a table. To get a table, I have to write:
>>>> a magic"0 / a
>>>> Why is it so, since both * and magic may be used as dyads?
>>>> Thanks.
>>>>
>>>>
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