I just noticed that

/:~ : /:

can replaced by

/:~~

bivalently.

Henry Rich

On 10/27/2011 4:03 PM, Kip Murray wrote:
> I like especially your second form I =: ([:<   [: /: ":)"0
>
> For the third form I =:<  @ /: @ ": " 0 note that [: f g and f @: g are
> always equivalent, but f @: g and f @ g are not when g has rank 0 --
> conjunction @: always uses sequential processing, but conjunction @ uses
> parallel processing when g has rank 0, as shown below.
>
>
>      (+/ @: *:) 3 4
> 25
>
>      (+/ @ *:) 3 4
> 9 16
>
>
> In the first case above the right to left flow chart is
>
> 25<-- +/<-- 9 16<-- *:<-- 3 4  (sequential processing)
>
> while in the case involving @ the flow chart is
>
>               <--  9<-- *:<-- 3
> 9 16<-- +/                   (parallel processing because *: is rank 0)
>               <-- 16<-- *:<-- 4
>
>
> As I nearly always want sequential processing I use
>
>      [: f [: g h   (read "the f the g h")
>
> or
>
>      f @: g @: h   (read math's "f o g o h")
>
>
> Check:
>
>      ([: +/ *:) 3 4
> 25
>
>
> On 10/27/2011 5:21 AM, Raul Miller wrote:
>> Also, for the domain in question, we are not using>   for anything but its 
>> rank.
>>
>> Thus we could simplify:
>>
>>      I =: ([:<   [: /: [: ": ])"0
>>
>> Also, since we are always using this as a monad, we could further simplify:
>>
>>
>>      I =: ([:<   [: /: ":)"0
>>
>> Though, personally, I find myself comfortable using @
>>
>>      I =:<@/:@":"0
>>
>> Or, going back to the original message, and applying @ to achieve what
>> the dictionary was talking about:
>>
>>      <@([: /: ":)@>a,b
>>
>> Or, using "0 to replace @>
>>
>>      <@([: /: ":)"0 a,b
>>
>> But if you are using trains in boxes, maybe it's better to state that
>> explicitly, and that could also get rid of any of the @ conjunctions:
>>
>>      ([: /: ":) L:0<"0 a,b
>>
>> That said, when you replace a shorter expression with a longer one, I
>> think you should expect the longer one to lose some of the grace of
>> the original.
>>
>> I hope this helps.
>>
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