I'm aware of branch mis predict costs.
It begs the question why the bounds check is inside the loop. Can't you
check against shape?
On Mar 5, 2014 4:57 PM, "Roger Hui" <rogerhui.can...@gmail.com> wrote:
> Compare the two:
>
> Sorting (fixed typos!):
>
> I4 count[M];
> memset(count,0x00,sizeof(count));
> for(i=0;i<n;++i)count[*x++]++;
> for(i=0;i<M;++i){m=count[i]; for(j=0;j<m;++j)*z++=i;}
>
>
> Indexing:
>
> #define ASSERT(p,stmt) {if(!(p)){stmt;}
> for(i=0;i<n;++i){j=*g++; ASSERT(0<=j&&j<n,INDEX_ERROR); *z++=x[j];}
>
>
> Branching in modern CPUs is relatively
> expensive<https://en.wikipedia.org/wiki/Branch_predictor>.
> Apparently the rest of the world is catching up to APL/J. See also
> Knuth's Two Notes on
> Notation<http://arxiv.org/PS_cache/math/pdf/9205/9205211v1.pdf> (he
> calls it Iverson's convention).
>
> In the sort program there isn't a branch in sight, other than possibly in
> the looping, and the C compiler is pretty good at that. In the indexing,
> you have an if statement, in which (in this case) 100% of the time the
> branch is not taken. A branch which is 100% (or 0%? I forget) not taken is
> particularly slow.
>
>
>
>
>
> On Wed, Mar 5, 2014 at 2:41 PM, Yike Lu <yikelu.h...@gmail.com> wrote:
>
> > Roger, why is index lookup so slow in the g2 case (see my message
> earlier)?
> > Are you really saying that bounds checking makes it so that the lookup is
> > only as fast as the sort?
> >
> >
> > On Wed, Mar 5, 2014 at 4:35 PM, Roger Hui <rogerhui.can...@gmail.com>
> > wrote:
> >
> > > > I4 count[M];
> > > > memset(count,0x00,sizeof(count));
> > > > for(i=0;i<n;++i)count[*x++]=1;
> > >
> > > More typos:
> > >
> > > for(i=0;i<n;++i)count[*x++]++;
> > >
> > >
> > >
> > >
> > > On Wed, Mar 5, 2014 at 2:19 PM, Roger Hui <rogerhui.can...@gmail.com>
> > > 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.
> > > >
> > > >
> > > >
> > > ----------------------------------------------------------------------
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> > >
> > ----------------------------------------------------------------------
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> >
> ----------------------------------------------------------------------
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