No, it's not fair.
- You need to sort 1-byte characters to approximate operations on 1-byte
integers.
- You need to ditch the conversions to and from booleans. In J such
conversions are relatively costly (compared to /:~, f'instance); in C, the
conversion is free.
On Thu, Mar 6, 2014 at 11:47 PM, Linda Alvord <lindaalv...@verizon.net>wrote:
> Isn't this a fair timing which shows that <./ is fastest? I still think
> /:~#:A is awesome!
>
> A=:?1000000$10000000
> 10 timer '{.#./:~#:A'
> 0.861233
> 10 timer '{./:~A'
> 0.152205
> 10 timer '<./A'
> 0.00360548
>
> {.#./:~#:A
> 7
> {./:~A
> 7
> <./A
> 7
>
> Linda
>
>
> -----Original Message-----
> From: programming-boun...@forums.jsoftware.com
> [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Linda
> Alvord
> Sent: Thursday, March 06, 2014 10:55 PM
> To: programm...@jsoftware.com
> Subject: Re: [Jprogramming] a good bar bet!
>
> You warned me that dyadic /: was spectacular!
>
> Linda
>
> -----Original Message-----
> From: programming-boun...@forums.jsoftware.com
> [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Roger Hui
> Sent: Thursday, March 06, 2014 10:39 PM
> To: Programming forum
> Subject: Re: [Jprogramming] a good bar bet!
>
> Kind of, except I don't think you need to convert to bits and then convert
> back.
>
> It boggles my mind that taking the first element of sort to find the
> minimum is competitive with the conventional coding for finding the
> minimum:
>
> x=: a.{~1e7 ?@$ 256
> y=: 1e7 ?@$ 2e9
> timer=: 6!:2
> 10 timer '{./:~x'
> 0.0134736
> 10 timer '<./y'
> 0.0126129
>
> I can not do a direct comparison in J because J doesn't have 1-byte
> integers. In Dyalog APL, which does have 1-byte integers, the times
> are 0.0124262
> for the conventional coding and 0.0104232 for the first element of the
> sort. Unbelievable.
>
>
>
> On Thu, Mar 6, 2014 at 6:45 PM, Linda Alvord <lindaalv...@verizon.net
> >wrote:
>
> > Is this considered fair?
> >
> > ]A=:?20$1000
> > 951 524 574 923 841 986 242 790 695 936 140 901 900 366 33 309 244 327
> 817
> > 680
> >
> > f=: 13 :'{.#.(#:y)/:#:y'
> > f
> > [: {. [: #. #: /: #:
> >
> > f A
> > 33
> >
> > Linda
> >
> > -----Original Message-----
> > From: programming-boun...@forums.jsoftware.com
> > [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Roger Hui
> > Sent: Thursday, March 06, 2014 8:50 PM
> > To: Programming forum
> > Subject: Re: [Jprogramming] a good bar bet!
> >
> > You are not guaranteed sortedness, but nobody promised that it'd be
> sorted.
> >
> >
> > On Thu, Mar 6, 2014 at 5:42 PM, Yike Lu <yikelu.h...@gmail.com> wrote:
> >
> > > Problem with the hash table is how do you guarantee sortedness when you
> > > spit the result back out?
> > > On Mar 6, 2014 6:49 PM, "Roger Hui" <rogerhui.can...@gmail.com> wrote:
> > >
> > > > Well, if you use a hash table it'd be O(1) expected time to read and
> > > write.
> > > > The method I know is O(1) worst case time to read and write.
> > > >
> > > >
> > > > On Thu, Mar 6, 2014 at 2:54 PM, Yike Lu <yikelu.h...@gmail.com>
> wrote:
> > > >
> > > > > Hmm, maybe not. It has to basically be able to insert ordered in
> > > constant
> > > > > time. Everything I remember seems to be log time for one or the
> > other.
> > > > >
> > > > > I'm starting to lean towards a multi-pass / multi-array solution.
> > > > >
> > > > >
> > > > > On Thu, Mar 6, 2014 at 4:42 PM, Roger Hui <
> rogerhui.can...@gmail.com
> > >
> > > > > wrote:
> > > > >
> > > > > > The method I know offers O(1) time to read and write. Can your
> > > > > dictionary
> > > > > > do the same? (And I think we are talking about _how_ you
> implement
> > a
> > > > > > dictionary.)
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > > On Thu, Mar 6, 2014 at 2:34 PM, Yike Lu <yikelu.h...@gmail.com>
> > > wrote:
> > > > > >
> > > > > > > Using a dictionary is another way.
> > > > > > >
> > > > > > >
> > > > > > > On Thu, Mar 6, 2014 at 3:30 PM, Roger Hui <
> > > rogerhui.can...@gmail.com
> > > > >
> > > > > > > wrote:
> > > > > > >
> > > > > > > > Yes, that's the cost. The trick is to avoid initializing
> count
> > > > > > > altogether,
> > > > > > > > because for this subproblem M is much greater than n. The
> > answer
> > > > is
> > > > > > > > algorithmic and not a matter of using this or that C
> operation.
> > > > > > > >
> > > > > > > > Recently I had occasion to use the trick for M=65536 and n
> > about
> > > > > 5000.
> > > > > > > As
> > > > > > > > originally posed, I think the intention was that M would be
> > > > zillions.
> > > > > > > >
> > > > > > >
> > > >
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