I might say:
+/ , M * M = +/&> {i.&.> $M
6
If I were under pressure (e.g. with Dijkstra in the audience
asking the question) and wanted to avoid any fumbling,
I might say:
+/ , M * M = (i.#M) +/ i.{:$M
6
And then from that to successively "better" answers
+/ , M * M = (#M) +/&i. {:$M
+/ , M * M = +/&i./ $M
+/ , M * M = +/&> {i.&.> $M
not forgetting to point out that the last expression
works for M of any rank.
Dijkstra's question in 1963 was: How would you represent a
more complex operation, for example, the sum of all elements of
a matrix which are equal to the sum of the corresponding row
and column indices?
----- Original Message -----
From: Oleg Kobchenko <[EMAIL PROTECTED]>
Date: Wednesday, November 7, 2007 17:00
Subject: Re: [Jchat] why did Dijkstra dislike APL so much?
To: Chat forum <[email protected]>
> What is the standard J answer?
>
> ((= +/@#&, [) +/&i./@$) i.3 4
> 6
> (+/@#~&, (= +/&i./@$)) i.3 4
> 6
> (+/@,@:* (= +/&i./@$)) i.3 4
> 6
> +/,M*M=+/&i./$ M =. i.3 4
> 6
>
>
> --- Roger Hui <[EMAIL PROTECTED]> wrote:
>
> > I don't know of any and I suspect Ken was too smart to respond.
> >
> > There was an encounter between KEI and Dijkstra recorded in
> > http://keiapl.info/anec/#Dijkstra
> >
> >
> >
> > ----- Original Message -----
> > From: Richard Hill <[EMAIL PROTECTED]>
> > Date: Wednesday, November 7, 2007 15:21
> > Subject: Re: [Jchat] why did Dijkstra dislike APL so much?
> > To: Chat forum <[email protected]>
> >
> > > Is there any record, or even an anecdote, of KEI's
> > > own response to Dijkstra's comment?
> > > Richard Hill
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