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Topic: The Sums of a Set (Read 3853 times) |
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Dudidu
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The Sums of a Set
« on: Dec 7th, 2003, 9:36am » |
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You are given a set S [subseteq] {1, ..., n}.
Devise an efficient algorithm that outputs all the numbers (x + y) such that x, y [in] S.
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Dudidu
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Re: The Sums of a Set
« Reply #1 on: Dec 9th, 2003, 7:12am » |
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Can someone give an answer (even the trivial one) ?
Don't worry about how efficent it is, we'll improve it later... 
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towr
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Re: The Sums of a Set
« Reply #2 on: Dec 9th, 2003, 8:08am » |
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on Dec 9th, 2003, 7:12am, Dudidu wrote:| Can someone give an answer (even the trivial one) ? |
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like
f(S) = {x+y [in] S | (x,y) [in] S[times]S} ?
hmm.. on second thought I don't think I actually know a language that'll pass it (PVS comes close, though it's not so much a programming language as a theorem prover)
Prolog should be simple enough as well though (and reads much the same):
f(S, R):-setof(T, X^Y^(member(X,S), member(Y,S), T is X+Y, member(T,S)), R).
%% in case member isn't predefined
member(E, [E|_]).
member(E, [_|T]):-member(E, T).
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| « Last Edit: Dec 9th, 2003, 8:20am by towr » |
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Dudidu
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Re: The Sums of a Set
« Reply #3 on: Dec 10th, 2003, 12:17am » |
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If you didn't understand, you can think of the set S as an array of size m (so getting the elements is as simple as addressing some element in the array).
Now, its seems (to me  ) that its quite easy to give a pseudo-code for the trivial answer.
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Tom Wang
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Re: The Sums of a Set
« Reply #5 on: Dec 10th, 2003, 1:33am » |
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for (int i = 2; i <= 2n; i++)
cout << i << " ";
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Dudidu
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Re: The Sums of a Set
« Reply #6 on: Dec 10th, 2003, 3:09am » |
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on Dec 10th, 2003, 12:54am, towr wrote:| my answer isn't trivial enough? |
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No, firstly since I don't know Prolog and secondly (the more problematic issue) since I don't know how f(S, R) is practically calculated and thus, I don't know what is its time complexity.
Sorry... 
P.S. - towr, just to be sure does 'member(T,S)' mean that T is a member of S (since if it is, I think that the definition of 'f(S, R)' is incorrect (? ?).
Quote:for (int i = 2; i <= 2n; i++)
cout << i << " "; |
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Tom Wang, welcome to the forum  .
What did you mean by this loop ?
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| « Last Edit: Dec 10th, 2003, 3:16am by Dudidu » |
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Tom Wang
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Re: The Sums of a Set
« Reply #7 on: Dec 10th, 2003, 3:54am » |
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The program is supposed to output all the numbers z = x + y such that x and y are members of S, given the number n.
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Tom Wang
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Re: The Sums of a Set
« Reply #8 on: Dec 10th, 2003, 3:58am » |
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I was confused by the question; I thought S was the set from 1 to n.
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towr
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Re: The Sums of a Set
« Reply #9 on: Dec 10th, 2003, 4:21am » |
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on Dec 10th, 2003, 3:09am, Dudidu wrote:| No, firstly since I don't know Prolog and secondly (the more problematic issue) since I don't know how f(S, R) is practically calculated and thus, I don't know what is its time complexity. |
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it should be quadratic
Quote:I guess I just know too many languages.. And it's so tempting to solve problems in the language that solves them the easiest.
Quote:| P.S. - towr, just to be sure does 'member(T,S)' mean that T is a member of S (since if it is, I think that the definition of 'f(S, R)' is incorrect (?:-/?). |
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Yes that is what it means.
I thought you wanted the sum to be in S.. Apparantly I also misread the problem.. But you can easily remove it..
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| « Last Edit: Dec 10th, 2003, 6:16am by towr » |
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Dudidu
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Re: The Sums of a Set
« Reply #10 on: Dec 10th, 2003, 9:44am » |
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on Dec 10th, 2003, 3:58am, Tom Wang wrote:| I was confused by the question; I thought S was the set from 1 to n. |
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Don't be discouraged, everybody makes mistakes (sometimes)... keep on trying !
on Dec 10th, 2003, 4:21am, towr wrote:| it should be quadratic... |
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I tought so.
Quote:| I thought you wanted the sum to be in S.. Apparantly I also misread the problem.. But you can easily remove it... |
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You're right, it is not necessary that the sum will be in S.
Now, the question is how can you improve it (a hint will follow soon (if no one gets it before)...)
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| « Last Edit: Dec 10th, 2003, 9:44am by Dudidu » |
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Barukh
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Re: The Sums of a Set
« Reply #11 on: Dec 10th, 2003, 11:56am » |
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Here’s a more conventional code that will be easier to discuss the improvements on (at least for me ):
Code:for (sum = 2*S[1]; sum <= 2*S[m]; sum++)
{
for (i = 1; (i <= m) && (S[i] <= sum/2); i++)
{
if ( inS(sum – S[i]) ) { print( sum ); break; }
}
}
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The S[] is the array Dudidu talked about in his other mail; inS() is a membership check. The complexity of this code is O(N*m), which is quadratic in N in the worst case.
One way to improve this code would be to start the inner loop not at 1, but at index k such that sum - S[k] [le] N. However, this doesn’t imrpove the complexity.
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James Fingas
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Re: The Sums of a Set
« Reply #12 on: Dec 10th, 2003, 1:32pm » |
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You didn't happen to get the idea for this question from another thread, did you?
I was going to post a link, but I didn't want to give away the answer...
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| « Last Edit: Dec 10th, 2003, 1:33pm by James Fingas » |
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Barukh
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Re: The Sums of a Set
« Reply #13 on: Dec 10th, 2003, 2:49pm » |
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on Dec 10th, 2003, 1:32pm, James Fingas wrote:| You didn't happen to get the idea for this question from another thread, did you? |
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Personally, I haven't made any connection to any existing threads... Thus, the code is as poor as it is.
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Dudidu
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Re: The Sums of a Set
« Reply #14 on: Dec 11th, 2003, 4:07am » |
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on Dec 10th, 2003, 11:56am, Barukh wrote:| Here’s a more conventional code that will be easier to discuss the improvements on (at least for me ):... |
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Barukh hi,
I have some comments/questions about your code:
Quote:| inS() is a membership check... |
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1) How does it implemented ? - since if it is implemented as a naive search (i.e. looking for that element in S) then it will take O(m2N) and not O(mN)...
Quote:| quadratic in N in the worst case... |
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2) How can you improve it to be O(m2) (assuming that you fixed the inS() problem/feature )?
3) One remark: you assume that the array S is sorted (which isn't a must) - but this doesn't create a problem (since sorting it would take O(m) ?!).
Waiting for your reply...
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James Fingas
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Re: The Sums of a Set
« Reply #15 on: Dec 11th, 2003, 4:38am » |
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on Dec 10th, 2003, 2:49pm, Barukh wrote:
Personally, I haven't made any connection to any existing threads... Thus, the code is as poor as it is. |
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I was actually asking Dudidu about the original question. Your code seems fairly straightforward, although there are probably some easier ways to do it.
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Dudidu
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Re: The Sums of a Set
« Reply #16 on: Dec 11th, 2003, 6:02am » |
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on Dec 11th, 2003, 4:38am, James Fingas wrote:| I was actually asking Dudidu about the original question... |
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James hi,
I didn't get the idea for this question from another thread.
But if there is a similar thread and I'm recycling problems, I'm sorry...
P.S. - can you send me the link to the other thread (or post it (after this riddle is solved)) ?
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| « Last Edit: Dec 11th, 2003, 6:04am by Dudidu » |
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James Fingas
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Re: The Sums of a Set
« Reply #17 on: Dec 11th, 2003, 6:09am » |
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I guess I'm the only one who sees the similarity. The solution I'm thinking of works in O(n log n). Of course, that's not great if m << n, but pretty cool nonetheless.
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Barukh
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Re: The Sums of a Set
« Reply #18 on: Dec 11th, 2003, 1:11pm » |
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on Dec 11th, 2003, 4:07am, Dudidu wrote:| 1) How does it implemented ? - since if it is implemented as a naive search (i.e. looking for that element in S) then it will take O(m2N) and not O(mN)... |
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Since there were no restrictions on the space, the simplest way is to have a boolean array inS[] - it takes O(N) to build it. Or, even better, use a hash.
Quote:| 2) How can you improve it to be O(m2) (assuming that you fixed the inS() problem/feature )? |
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I wish I knew...
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TenaliRaman
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Re: The Sums of a Set
« Reply #19 on: Dec 11th, 2003, 10:43pm » |
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One thing i don't get in the question,
given any set S with #S=m one can easily find a O(m2) solution.(its just a matter of two loops)
I thought that the knowlege that S is a subset of {1,..,n} would help the case. I came up with O(n*m2log(m)) which hardly seems any better.And my friend seems to suggest that S is a subset of {1,..,n} can hardly ever make the algo efficient because he feels it makes the search space that much bigger.
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James Fingas
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Re: The Sums of a Set
« Reply #20 on: Dec 12th, 2003, 6:02am » |
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The trivial answer that is O(m[sup2]) is this one (pseudocode):
input array s of length m
result list Z = empty list
for i = 1 to m {
for j = i+1 to m {
Z.add( s[i] + s[j] )
}
}
This of course has duplicate entries, but if we don't care, it's a trivial O(m[sup2]) answer. I haven't yet thought of a method faster than O(m[sup2]logm) (sort them) to remove the duplicates, so that appears to be the limiting performance factor. So here is a listing of the best times so far:
my interesting solution: O(n log n)
Barukh's solution: O(mn)
my trivial solution: O(m[sup2] log m)
Which of these is best will depend on the relative sizes of m and n. My guess is that an answer running in O(m[sup2]) is possible.
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TenaliRaman
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Re: The Sums of a Set
« Reply #21 on: Dec 12th, 2003, 8:35am » |
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hmm on "re-analysis" it seems the method i came up with is O(nmlogm) so it maybe relevant after all.
the procedure is as follows,
Given,
N = {1,...,n}
S subset of N and #S=m
Generate a set A = {2,....,2n}
(Note:this set has all the possible (x+y)'s for x,y belong to N.)
Sort S which is m*logm
Let S' be the set with all the (x+y)'s for x,y belong to S.
int k=0;
for(i=0;i<2n-1;i++)
{
....for(j=0;j<m;j++)
....{
.........if(A[i]>S[j] && A[i]-S[j] is in S)
..............S'[k++] = A[i];
....}
}
the A[i]-S[j] is in S can be done using a binary search algorithm which is O(logm).
So the overall is O(nmlogm)
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James Fingas
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Re: The Sums of a Set
« Reply #22 on: Dec 12th, 2003, 2:34pm » |
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Another way to implement the inS function is just to make an array L (for lookup) that is N long, full of zeros, then do this:
for i = 1 to m {
L[ s[i] ] = 1
}
The inS(j) is L[j], with constant time.
And if you want indentation to work properly, post with indentation screwed up, then modify and resave your post (which fixes indentation).
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TenaliRaman
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Re: The Sums of a Set
« Reply #23 on: Dec 13th, 2003, 6:57am » |
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Thanks for that tip off James, that would improve my algo's efficiency to O(mn).
I just realised that my implementation is same as that of Barukh's, the only thing being Barukh has better efficiency since my average efficiency will be always O(mn) and Barukh's average efficiency will be < O(mn).
I think after looking at different directions for better algo i think i would rather vote for the simple O(m2) solution.
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Dudidu
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Re: The Sums of a Set
« Reply #24 on: Dec 14th, 2003, 1:04am » |
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Everybody hi,
Quote:| The trivial answer that is O(m2) is this one... |
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James hi,
This is the algorithm that I asked Barukh about. As you indicated it doesn't remove duplicates so... a trivial algorithm would do it (output the desired numbers and remove the duplicates) in an O(m2 + N) (or some may say O(max{ m2,N})) time complexity (which is better then O(mN)) - what is it ?
Nevertheless, I see that you all are fixed on the ~m2 time complexity area so try to improve it... (Big Hint: How polynomials can be used ?).
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