Need help with function..

**nadroj** · 09-03-2008

ive never studied or heard of the proof for that. not that i dont believe there is one, ive just never heard mention of it before. do you have a link to one? preferably not in form of a thesis.

**~~master5001~~** · 09-03-2008

Just think of it with pure logic. Take a pile of puzzle pieces (preferably a small one since this is a simple concept). Don't try to put it together, but instead just consider what the maximum speed you could possibly put it together.

You will need to do the following:
1) Have something to put together.
2) Identify the existance of each individual piece.
3) Arrange the individual piece into the larger collection.
4) Go to the next piece until the operation is complete.

Now, if you tell a computer to do the same, it will follow similar logic, right? Unless there is some sort of predefined pattern to the puzzle pieces (or a bunch of them fell out of the box already put together) then you have no way of short cutting this system. What makes you assume your program can do otherwise?

**dwks** · 09-03-2008

I always liked this explanation (it was for sorting algorithms, but it works here too):

Say you have compared all of the numbers in a set except for one with each other and found their minimum. How can you know whether this number is in fact the minimum without checking if it is less than the last number?

So in other words, you can't get any better than n-1 comparisons, which is O(n). (You have to look at every number in order to find the lowest in the bunch.)

. . . or something like that. Knuth said it better.

**nadroj** · 09-03-2008

i never said anything about a program, let alone a program i made.

i dont understand how your analogy is to explain the proof. a funny thing about logic (in the sense your talking about) is that it is ambiguous and discussions can be made about the smallest topic that go on forever. i have read pseudocode and proofs that prove an algorithms complexity using only mathematics (and basic english terms, not 'logic' though). i am interested in seeing the (mathematical) proof that you mentioned. also if a sorting algorithm (which i think the best today are O(n*log(n)) are possible to be improved significantly, ie better than O(n), then you could use that algorithm and take the first element of that list (which would be O(1)) and have that the same complexity for a min algorithm as the sorting algorithm.

**dwks** · 09-03-2008

Why not read my post? It's not exactly a proof, but I'm sure you can find something like that if you google around. Or, if you have Knuth's book on sorting algorithms, you can look it up in there. It's near the beginning. Right where he describes the classes of sorting algorithms, and asks that you inform him immediately if you discover a new super-sort.

**~~master5001~~** · 09-03-2008

In what way was what me and nadroj too convoluted for you to understand? Since you must identify ALL components of whatever you are sorting, you can at best only be as fast as looking at each item once. If you do not look at all elements of whatever it is that you are sorting, you potentially leave something unsorted. So your best case sort in sorting is to more or less proof read something that is already sorted and just look at each piece and acknowledge that it is sorted (i.e. a function that doesn't even really do anything other than read pre-sorted data).

What is debateable is what can most rapidly confirm that the data is already sorted. You have the web at your access to my friend. I am not going to search stuff you are capable of looking up on your own. You were given satisfactory explanation to the logic. Even if you have multiple CPU's each reading part of a pile of data, there is always going to be a point where things are only processed on an individual basis. No matter what. I applaud that you are so boldy trying to make something "better."

**nadroj** · 09-03-2008

dwks, my reply was to master5001, i never said anything about you. also, after i posted your post didnt show up until after i refreshed.

i realize that this sort of thing is logical and only makes sense that you would have to look at each number (otherwise you could be missing the actual min, etc).

the super-sort, is that the paragraph before or after the "typo $1 cheque" thing? (or was that Dijkstra). anyways, Knuth's collection scares me!

**~~master5001~~** · 09-03-2008

Given your reply, and your understanding of the process and dwks' alternative explanation then I think we are good.

**nadroj** · 09-03-2008

ya, in what way was what me and master said too convoluted for you?

well you are correct in that i do have the web to access. i only figured that since you mentioned the mathematical proof in your post that i was interested in seeing it. im not asking you to go find it for me, and im not going to look anyways, but it just made it seem that you had it available.

**dwks** · 09-03-2008

dwks, my reply was to master5001, i never said anything about you. also, after i posted your post didnt show up until after i refreshed.

I know, I could tell by the time you posted and the length of your post. That's why I suggested you read it. Sorry if I came across as implying you had ignored my post and skipped over it.

[edit] That's what just happened to this post, too. [/edit]

the super-sort, is that the paragraph before or after the "typo $1 cheque" thing? (or was that Dijkstra). anyways, Knuth's collection scares me!

I don't remember. I think Knuth has something like that: you know, offering $1 in whatever year for the first error discovered in his book, $2 the second year, $4 the third, etc. You could probably make a small fortune by now if you managed to actually find an error in one. :P

**nadroj** · 09-03-2008

hes going in powers of 2? wow.. i thought it was constant $1 USD.

but thats ok.. it did seem like you were being smart and saying "well why not read MYpost then, smrtas?"

**dwks** · 09-03-2008

Yeah, I'm pretty sure it's powers of two.
[edit] I guess it's powers of two, but actually cents, not dollars.

You are entitled to a reward of at least $2.56 if you are the first person to report a bona-fide error not on those lists. Each page tells you how to report an error for the book in question.

http://www-cs-faculty.stanford.edu/~knuth/books.html [/edit]

I guess I didn't put enough s

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**nadroj** · 09-03-2008

lol, cheers

**dwks** · 09-03-2008

End note: actually, it's $2.56 for one error, $5.12 for the next, and so on. Not by year, by error number.
http://www.upto11.net/generic_wiki.p...h_reward_check
http://www.nationmaster.com/encyclop...h-reward-check

**matsp** · 09-03-2008

Originally Posted by nadroj

you mean there isnt one yet!

For a list that contains unordered numbers, there is no other way than to compare every single item. Obviously, if you have a sorted list, you can find things in it much quicker. Unfortunately, it take more than O(n) to sort the list, so there's no gain in that for a single find. (And you can do "max and min" in O(n) as well - because you just check for both largest and smallest on every item, and that is also O(n)).

--
Mats