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How to analyze an efficient algorithm?

Posted on 2008-06-16
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Last Modified: 2008-08-28
Bob has a set A of n nuts and a set B of n bolts, such that each nut in A has a unique matching bolt in B.
Unfortunately, the nuts in A all look the same, and the bolts in B all look the same as well. The only kind of a comparison that Bob can make is to take a nut-bolt pair (a,b) such that a is in B, and test it to see if the threads of a are larger, smaller, or a perfect match with the threads of b. Describe and analyze an efficient algorithm for Bob to match up all of his nuts and bolts.
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Question by:secondcup
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Frosty555 earned 168 total points
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Well, the rigorous solution is to try every nut with every bolt, and use the ones that work, ignoring the extra information you have of knowing whether a nut is too small or too large for the bolt.

That would run in O(N^2) time. So when the question says "describe an efficient algorithm", it wants something better than the obvious.
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by:sdstuber
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Compare bolt x with nut y.  If y is larger then put in "large" pile.  if y is smaller put in "small" pile.
continue until matching y is found.

Get bolt x2,  compare with y1.  if y1 is larger go to "small" pile and search.  dividing into small 1 and small 2.  If y1 is smaller then go to "large" pile and search, dividing into large 1 and large 2.

If not found in already searched piles, go to remainder of original pile of nuts.  again dividing into large and small.

get bolt x3,  compare to y2 and y1 to determine which of the prior semi-sorted nut piles to look through.  

continue through each bolt,  compare to previous bolt's nuts to try find smaller piles to search.  As you go through the bolts the piles should get smaller but more numerous with each being closer and closer to a full sort by size.

If you can get any kind of inter-bolt comparison or inter-nut comparison (which it sounds like you've already eliminated) then you can make it more efficient.  but if you can only compare bolts to nuts then I think semi-sorted piles is about as good as you'll get as it will let you derive a sorting that will help over a simple n^2 comparison
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by:Infinity08
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Is this an assignment ? How far did you get ? What are you unsure about ?
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by:NovaDenizen
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Use partitioning.  The expected performance is exactly the same as quicksort.
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by:philipjonathan
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The performance can be better than O(n^2), it should be O(n log n)
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