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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