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
