Re: "Just use a hash": An overworked mantra?

[vkon]

in reply to "Just use a hash": An overworked mantra?

But just because hash key lookups/increments are O(1) operations, that doesn't make them computationally cheap.
......
An array lookup is also O(1), but it turns out it's a much computationally cheaper task to perform these array operations.

not correct

Hash is not O(1).

Also, this example of 0..999 integers looks too artifical to me, and it is too obvious to use an array here, so - for me - it does not deserve to even start the discussion, in the first place :)

I do appreciate good English of the original post, however, which I, personally, do not possess
:) :)

[davido]

Hash is not O(1)

Are you sure?

• "Many hash table designs also allow arbitrary insertions and deletions of key-value pairs, at constant average (indeed, amortized) cost per operation." -- http://en.wikipedia.org/wiki/Hash_table
• http://en.wikipedia.org/wiki/Big_O_notation
• "Thus, everyone knows that hash table queries run in amortized constant time. That is, as the number of keys increases, the average time necessary to recover a key-value pair does not increase." -- http://lemire.me/blog/archives/2009/08/18/do-hash-tables-work-in-constant-time/
• "Hash tables, sometimes also known as associative arrays, or scatter tables, are a data structure that offers fast lookup, insertion, and deletion of (key, value) pairs. In a well-designed implementation of hash tables, all of these operations have a time complexity of O(1), rather than the O(n) that linked lists, association lists, and vectors would require for some of them." -- http://www.math.utah.edu/~beebe/software/hash/hash.html
• "Perl hashes provide constant-time lookups. They are implemented as true hash tables (with automatic retuning as necessary), not associative arrays." -- http://stackoverflow.com/questions/3774904/what-is-the-time-complexity-for-looking-up-an-item-in-a-perl-hash
• http://www.sysarch.com/Perl/sort_paper.html

I'll grant that an individual lookup might consume greater than O(1) time, but in the aggregate it's fairly well established and accepted that hash operations are O(1) operations, where 'n' is the number of elements in the hash.

[vkon]

yes. I am sure.

first, the third link you've pointed reads:
Thus—formally—hash tables often run in time O(log n).

next, if the question is "is it possible to create an algorithm that, given arbitrary strings of finite length, make associative store/lookup for O(1)" - then the answer is "yes" but these are non-realistic algorithms.
but - back to reality - real-life alhorithms are O(log N) in best case, but I afraid Perl's one is worse than that
and - yet - have you ever heard of so-called "hash complexity attack" that was security-fixed by seed randomization in 5.8.0?
To inform you, pre-5.8.0 were looking like O(1) but actually were O(n) (or maybe even worse in worst case scenario)

next, your "stackoverflow.com" link contains simply wrong information.
IOW, I do not see any proof there that convinces me.

And - your last link does not even mention hashes.
Bother explain what is doing here? TIA!

and finally, do you see something strange in your phrase
hash operations are O(1) operations, where 'n' is the number of elements in the hash
?
I could be a bit joking, but I do not trust phrases like that.
you're explaining parameter 'n' that is not presents in your formulae - what other mistakes/typos have you made?
:)

[BrowserUk]

real-life alhorithms are O(log N) in best case, but I afraid Perl's one is worse than that

I'd like to see you demonstrate that Perl's hashes are O(log N), let alone "worse than that"?

[Anonymous Monk]

And - your last link does not even mention hashes. Bother explain what is doing here? TIA!

Look at the table

Here are some well-known algorithms and their order functions (adapted from [2]):

Notation       Name       Example
O(1)       constant       array or hash index

...

2. Kernighan, B. W. & Pike, R., (1999). The Practice of Programming, p. 41. Addison-Wesley.

yes. I am sure.

Great, prove it :)