With multiprocessors, the calculus changes. Sort is still O(n) (I think
it can be O(lgn), but with bad constant factors), but hashing is O(1).
And there are interesting cases where you do have n processors (hpc,
gpgpu). In effect, once you have enough processors, they serve to hide
the latenc
I'm with Raul on this. A hashtable is a fine idea when memory is fast,
but if the hashtable greatly exceeds D3$ every hash lookup will be a
page miss which is hideous. A merge sort would look pretty good then.
Henry Rich
On 4/9/2022 3:23 PM, Elijah Stone wrote:
If ~.!.1 is used, then how wou
If ~.!.1 is used, then how would we denote a -. operating on sorted
arguments?
I build the hash table concurrently, using a multiprocessor; this means
the elements are processed out of order. Two ways of proceeding, not yet
sure which is better. The first is for each thread to accumulate the
On Sat, Apr 9, 2022 at 5:27 AM Elijah Stone wrote:
> Hashing is O(1) (or, if you prefer, O(#y) for ~.y, same as sorting). A
> sufficiently smart(tm) hash function will avoid inordinate collision
> rates, so I am not sure what worst case behaviour you are referring to.
Collision rates are indeed
I would favor ~.!.1. No mnemonic value, but less likely to be forgotten.
What is your fast algorithm?
Henry Rich
On 4/9/2022 4:03 AM, Elijah Stone wrote:
Suppose that ~. (and perhaps some relatives) can be implemented much
more efficiently if no guarantee is made about the order of the
resul
On Sat, 9 Apr 2022, Raul Miller wrote:
People will say that certain algorithms, such as hashing, are highly
efficient. But these assertions are quite often not accompanied by
adequate benchmarking on large datasets. And, these approaches often
have inefficient worst case behavior.
Hashing i
On Sat, Apr 9, 2022 at 4:04 AM Elijah Stone wrote:
> Suppose that ~. (and perhaps some relatives) can be implemented much more
> efficiently if no guarantee is made about the order of the result. Is it
> too much of an abuse of notation to use ({~?~@#)@~. as a special
> combination to invoke such
Hmm...
({.% */@}.) 1 2 3 4 5
would actually work if we replaced 1 2 3 4 5 with an arbitrary
different list of numbers.
It would not work, though, if we replaced % with a different verb.
Thanks,
--
Raul
On Thu, Oct 3, 2019 at 5:22 PM Nollaig MacKenzie
wrote:
>
> I'm guessing
>
> %*/2 3 4
I'm guessing
%*/2 3 4 5
wouldn't count
On 2019.10.02 23:43:40, you,
the extraordinary Skip Cave, spake thus:
>
> I can write:
> 1%(2%(3%(4%5)))
>
> 1.875
>
>
> Using insert, I can simplify:
>
> %/1 2 3 4 5
>
> 1.875
>
>
> Now I can write:
>
> (((1%2)%3)%4)%5
>
> 0.008
Thanks Roger. I have run into this issue a couple of times, and I knew J
must have a solution. I just need to remember the double reverse scheme
(reverse the noun order & reverse the dyad execution).
Skip
On Thu, Oct 3, 2019 at 12:18 AM Roger Hui wrote:
>%~/ |. 1 2 3 4 5
> 0.0083
>
>
>
%~/ |. 1 2 3 4 5
0.0083
On Wed, Oct 2, 2019 at 9:44 PM Skip Cave wrote:
> I can write:
> 1%(2%(3%(4%5)))
>
> 1.875
>
>
> Using insert, I can simplify:
>
> %/1 2 3 4 5
>
> 1.875
>
>
> Now I can write:
>
> (((1%2)%3)%4)%5
>
> 0.00833
>
>
> How can I use insert to simplify this?
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