Won't it give the maximum though, instead of the minimum?

Thanks,

-- 
Raul



On Wed, Mar 5, 2014 at 8:43 PM, Roger Hui <rogerhui.can...@gmail.com> wrote:

> Yes, a break or return would be nice.  It's still OK without that, because
> it's looping a maximum of M times.
>
>
>
> On Wed, Mar 5, 2014 at 5:00 PM, Raul Miller <rauldmil...@gmail.com> wrote:
>
> > You probably want a return or break statement in that second loop?
> >
> > Thanks,
> >
> > --
> > Raul
> >
> >
> >
> > On Wed, Mar 5, 2014 at 7:13 PM, Roger Hui <rogerhui.can...@gmail.com>
> > wrote:
> >
> > > > it avoids any branching in the loop.  And no "comparisons".
> > >
> > >     I1 b[M];
> > >     memset(b,0x00,sizeof(b));
> > >     for(i=0;i<n;++i)b[*x++]=1;
> > >     for(i=0;i<M;++i)if(b[i])min=i;
> > >
> > > Well, there is branching and there is comparison, but it's in the M
> loop
> > > rather than the n loop.  M is small and fixed (65536 for 2-bytes)
> > whereas n
> > > can be millions, billions.  The J models are:
> > >
> > >     1 i.~ 1 x}M$0  NB. minimum
> > >     1 i:~ 1 x}M$0  NB. maximum
> > >
> > >
> > >
> > >
> > >
> > > On Wed, Mar 5, 2014 at 4:05 PM, Roger Hui <rogerhui.can...@gmail.com>
> > > wrote:
> > >
> > > > Zehr gut.  It's not just that it's O(n) (the conventional
> > > if(min>*x)min=*x
> > > > is also O(n)) but it avoids any branching in the loop.  And no
> > > > "comparisons".
> > > >
> > > > Bonus question:  Suppose M is really large and the cost of setting
> > count
> > > > to 0 is prohibitive.  How can you avoid that cost?  (Not saying it's
> > > > related to finding the min or the max).
> > > >
> > > >
> > > >
> > > >
> > > > On Wed, Mar 5, 2014 at 3:27 PM, Peter B. Kessler <
> > > > peter.b.kess...@oracle.com> wrote:
> > > >
> > > >> For the minimum in a small universe, use your "broken" sorting code
> > (:-)
> > > >>
> > > >>
> > > >>     I4 count[M];
> > > >>     memset(count,0x00,sizeof(count));
> > > >>     for(i=0;i<n;++i)count[*x++]=1;
> > > >>
> > > >> (O(n)) and then look for the first (lowest) 1 in count (also O(n)).
> > > >>
> > > >>                         ... peter
> > > >>
> > > >>
> > > >> On 03/05/14 14:19, Roger Hui wrote:
> > > >>
> > > >>> Good answers.  For /:~x vs. g{x, the explanations are:
> > > >>>
> > > >>>     - Indexing must check for index error.  Sorting does not.
> > > >>>     - Indexing uses random read access over a large chunk of memory
> > > (i.e.
> > > >>>
> > > >>>     x).  Sort does not.
> > > >>>
> > > >>> A more detailed explanation:  To sort over a small known universe
> > (and
> > > >>> characters definitely qualify), you basically compute #/.~x (the
> > > ordering
> > > >>> is wrong, but you get the idea).  In C:
> > > >>>
> > > >>> I4 count[M];
> > > >>> memset(count,0x00,sizeof(count));
> > > >>> for(i=0;i<n;++i)count[*x++]=1;
> > > >>>
> > > >>>
> > > >>> This is lightning fast on modern CPUs: sequential read access and
> no
> > > >>> branch
> > > >>> prediction fails.  (And the ordering is correct.)  Once having the
> > > >>> counts,
> > > >>> as Henry said, you can do count#a. or in C:
> > > >>>
> > > >>> for(i=0;i<M;++i){m=count[j]; for(j=0;j<m;++j)*z++=i;}
> > > >>>
> > > >>>
> > > >>> Also lightning fast with very localized reads.
> > > >>>
> > > >>> It's ironic that in school sorting is an application with heavy
> > > emphasis
> > > >>> on
> > > >>> comparisons, counting # of comparisons, etc.  In the method above,
> > > there
> > > >>> is
> > > >>> not a single comparison involving x.  I once told someone that I
> can
> > > sort
> > > >>> 4-byte integers and 8-byte IEEE floats in linear time.  He looked
> at
> > me
> > > >>> like I was crazy, presumably remembering from school that sorting
> was
> > > >>> PROVEN to take n log n comparisons.
> > > >>>
> > > >>> As for why sorting is faster than grading, see
> > > >>> http://www.jsoftware.com/jwiki/Essays/Sorting_versus_Grading
> > > >>>
> > > >>> Now, for those of you who know C (or other scalar language), is
> > there a
> > > >>> faster way to find the minimum of a vector of small integers x
> > (2-byte
> > > >>> integers, say) than the following:
> > > >>>
> > > >>> min=-32768;
> > > >>> for(i=0;i<n;++i){if(min>*x)min=*x; ++x;}
> > > >>>
> > > >>>
> > > >>> I know an alternative which is 70% faster.  No fancy SSE
> > instructions.
> > > >>>  No
> > > >>> multicore.  No loop unrolling.
> > > >>>
> > ----------------------------------------------------------------------
> > > >>> For information about J forums see
> > http://www.jsoftware.com/forums.htm
> > > >>>
> > > >>>
> >  ----------------------------------------------------------------------
> > > >> For information about J forums see
> > http://www.jsoftware.com/forums.htm
> > > >>
> > > >
> > > >
> > > ----------------------------------------------------------------------
> > > For information about J forums see http://www.jsoftware.com/forums.htm
> > >
> > ----------------------------------------------------------------------
> > For information about J forums see http://www.jsoftware.com/forums.htm
> >
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> For information about J forums see http://www.jsoftware.com/forums.htm
>
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