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Re^3: Patience Sorting To Find Longest Increasing Subsequence

by demerphq (Chancellor)
on May 06, 2006 at 09:02 UTC ( #547804=note: print w/ replies, xml ) Need Help??


in reply to Re^2: Patience Sorting To Find Longest Increasing Subsequence
in thread Patience Sorting To Find Longest Increasing Subsequence

A merge sort is only O(N^2) worst case so no,

No, a merge sort is worst case O(n log n).

You are welcome to use a Van_Emde_Boas_tree which claims to be able to do the whole thing in O(N Log N)

Actually the paper by Bespamyatnikh & Segal contains a proof that you can do it in O(N log log N) time. I havent verified it tho.

But I doubt that the vEB based algorithm would in practice beat a simpler algorithm to do patience sorting. Unless I guess if you were dealing with a deck with tens of thousands or even millions or billions of cards. The overhead of maintaining a vEB tree is prohibitive for small datasets. The cost of doing binary operations on the keys, maintaining the vEB tree and etc, would most likely outweigh that of a simpler less efficient algorithm.

As bart said, sometimes a binary search algorithm is not as fast a scan, even though one is O(log N) and the other is O(N). The reason of course is that big-oh notation glosses over issues like cost per operation, and only focuses on the "overwhelming factor" involved. So in a binary search if it takes 4 units of work to perform an operation and in linear search it takes 1 then binary search only wins when 4 * log N < N, so for lists shorter than 13 elements there would be no win to a binary search. And I'd bet that in fact the ratio is probably something like 20:1 and not 4:1. Apply this kind of logic to a deck of 52 cards, and IMO its pretty clear that vEB trees are not the way to go for this, regardless of the proof.

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$world=~s/war/peace/g


Comment on Re^3: Patience Sorting To Find Longest Increasing Subsequence
Re^4: Patience Sorting To Find Longest Increasing Subsequence
by Limbic~Region (Chancellor) on May 06, 2006 at 14:11 UTC
    demerphq,
    I had taken bart's word for the merge sort in the CB. I later told him the math was wrong (in the CB) but didn't change it because I knew the math was also wrong for his desired method of finishing the sorting. The thing neither of us considered is that the problem space decreases with each pass. In any case, regardless of the accuracy of the math - the merge sort is still the most efficient given the data structure.

    Actually the paper by Bespamyatnikh & Segal contains a proof that you can do it in O(N log log N) time.
    What is the "it" that is O(N log log N) time though? The partial sort needed to obtain the LIS, obtaining the LIS itself, or completing the patience sort? What I understood from the paper, which I admittedly only read far enough to know that it was over my head, was that the O(N log log N) was not for a complete sort which Wikipedia agrees with.

    As far as the binary search is concerned - I have provided implementations to get to the partial sort using both methods so Benchmarking shouldn't be hard. Additionally, implementing a binary search & splice approach to bench against the merge sort is also straight forward.

    Cheers - L~R

      What is the "it" that is O(N log log N) time though?

      From what I understand its both finding the LIS and doing the patience sort. Actually, if i understand things correctly the B&S algorithm actually finds all increasing sequences in O(N log log N).

      Also I have an implementation of Patience sorting with backrefs for the LIS that I will post when I get a moment.

      ---
      $world=~s/war/peace/g

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