Please explain the technical term "twiddling". :-)
On Wed, Mar 5, 2014 at 2:28 PM, Thomas Costigliola <[email protected]>wrote: > My lazy answer is: You can do some bit twiddling to remove the comparison. > > > On Wed, Mar 5, 2014 at 5:26 PM, Roger Hui <[email protected]> > wrote: > > > > min=-32768; > > > for(i=0;i<n;++i){if(min>*x)min=*x; ++x;} > > > > Typo: I mean min=32767;, of course. > > > > > > > > > > On Wed, Mar 5, 2014 at 2:19 PM, Roger Hui <[email protected]> > > 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
