That makes sense, I just recalled that u@v w was applied to the whole of w.
Thanks Henry > On 6 May 2019, at 10:22 pm, Henry Rich <henryhr...@gmail.com> wrote: > > My recollection of early J is that @ never had infinite rank, but the > arithmetic verbs did. Then the rank of arithmetic verbs was changed to 0, > and phrases like > > u@+ > > had to be rewritten as > > u@(+"_) > > That led to introduction of u@:v to mean (u@(v"_)) . > > Henry Rich > >> On 5/5/2019 10:23 PM, 'Rob Hodgkinson' via Programming wrote: >> Hi Thomas, apologies for the length here, but you have asked a good question. >> >> The operation to compose one verb with another is known as “atop” (as >> defined on Pages 12 and 17 in the Arithmetic paper you are reading)… (this >> is called function composition in linear algebra). >> >> In 2002 I believe this was implemented as “f@g” with infinite rank, but in >> 2019 is now implemented as “f@:g” and “f@g” produces a different result now >> (which is what you observed). I’ll try to explain briefly here. >> >> Using current J please note the 2 differences in your expression below: >> 3 (i.&1@<) i.7 >> 1 1 1 1 0 0 0 >> 3 (i.&1@:<) i.7 >> 4 >> >> In 2002 (I believe that) @ produced the 2nd result, but now it behaves >> differently. >> This change was required to deal with how “verb rank” impacted on “verb >> composition”. >> >> To the heart of your question: B2 was to encourage you to find the index of >> the first item where the left argument is < the right argument, for which >> the process is: >> 3 < i.7 >> 0 0 0 0 1 1 1 >> (i.&1)3 < i.7. NB. Find the first 1 in the result >> 4 >> >> All good here, so using “verb composition” to combine the 2 verbs as a new >> “derived verb”, and using current J, we now see: >> cl =: i.&1@< >> 3 cl i.7 >> 1 1 1 1 0 0 0 >> >> But what happened ? This does not match the result above of 4 ?? >> It is because the “Rank of <“ (a scalar, or rank 0, function) now causes the >> derived function cl to work on each item. >> So it actually produced the following result: >> 3 < i.7 >> 0 0 0 0 1 1 1 >> 0 0 0 0 1 1 1 (i."0)1. NB. Here I am applying i. to work on each item >> by forcing it to process “rank 0” (item wise)... >> 1 1 1 1 0 0 0 >> 0 i. 1 NB. We can see the individual results here >> 1 >> 1 i. 1 >> 0 >> >> In order for the derived function cl to use the whole of its argument (ie we >> call this infinite rank), then @: is now used: >> cli =: i.&1@:< >> 3 cli i.7 >> 4 >> >> If you view the Vocabulary in your current J implementation and look at the >> pages for: >> @ and @: you will see the distinction in the “verb rank” defined (@ inherits >> the ranks of g (mv lv rv) and @: is _ _ _ meaning infinite rank, so it >> applies the derived function to the whole argument). >> >> I hope I have clearly explained the distinction but you can experiment >> further to see how “verb rank” mixes with @ and @: here: >> >> box=:< NB. A “box” function to box the argument it sees ... >> 3 box@< i.7. NB. Note here that ‘box’ is given each item as it is >> processed >> ┌─┬─┬─┬─┬─┬─┬─┐ >> │0│0│0│0│1│1│1│ >> └─┴─┴─┴─┴─┴─┴─┘ >> 3 box@:< i.7 NB. Note here that ‘box’ sees all 7 items at once >> ┌─────────────┐ >> │0 0 0 0 1 1 1│ >> └─────────────┘ >> >> Perhaps Henry, Roger or others can confirm or correct me on the very early >> behaviour of @, my recollection is a bit rusty, but I believe it was >> initially implemented with infinite rank. >> Notwithstanding, the explanation above will help you to understand exactly >> how it is now implemented. >> >> HTH, Rob H >> >> >>> On 6 May 2019, at 5:41 am, Thomas Bulka <thomas.bu...@posteo.de> wrote: >>> >>> Am 05.05.2019 20:52 schrieb Roger Hui: >>>> When applied dyadically, the verb finds the index of the first place where >>>> x is less than y >>>> The verb is not very meaningful when applied monadically. >>> Hello Roger, >>> >>> thank you very much for your quick reply. I think I got it now. But, just a >>> follow-up question, to consolidate my understanding: cl is defined with @ >>> and is thus subsequently applied to all members of y: >>> >>> cl =: i.&1@< >>> >>> 3 cl i.7 --> 1 1 1 1 0 0 0 >>> >>> That means, cl is semantically equivalent to 3 (<:~) i.7 - or is there any >>> difference? >>> >>> Thanks again and best regards, >>> >>> Thomas >>> >>> ---------------------------------------------------------------------- >>> For information about J forums see http://www.jsoftware.com/forums.htm >> ---------------------------------------------------------------------- >> For information about J forums see http://www.jsoftware.com/forums.htm > > > --- > This email has been checked for viruses by AVG. > https://www.avg.com > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
