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Please help me in improving the time complexity of the following algorithm.

Hasse Diagram(Skip this section if you already know what is Hasse Diagram, Please directly go to next section):

Consider a partially ordered set (poset, for short) (A,⊆), where A is a set and ⊆ a partial order. Each node of the diagram is an element of the poset, and if two elements x and y are connected by a line then x ⊆ y or y ⊆ x . The position of the elements and the connections are drawn respecting the following rules:

  1. If x ⊆ y in the poset, then the point corresponding to x appears below the point corresponding to y.

  2. The transitivity of the poset is graphically omitted, that is, if x ⊆ y and y ⊆ z, then, by transitivity of the partial order ⊆, x ⊆ z. In this case the connection x-z is omitted.

  3. Similarly reflexivity is graphically omitted.

Hasse Diagram representation of the Poset(S={{1,2,3,5}, {2,3}, {5}, {3}, {1,3}, {1,5}}, ⊆) is as follows(only the edges are reported)

{1,2,3,5}->{2,3}

{1,2,3,5}->{1,3}

{1,2,3,5}->{1,5}

{2,3}->{3}

{1,3}->{3}

{1,5}->{5}

My initial thought

The only algorithm i could think of is O(N^2) as follows:

Read the first element is S and insert it as a first element in Hasse Diagram. As we read the next elements insert them in the already constructed diagram either in the right position (Suppose the diagram constructed till now has K elements, then it takes O(K) time to insert a new element in the right place). This way O(N ^ 2) is evident.

But i'm thinking whether sorting the elements of poset S could help but sorted order for complete elements in S cannot be built as ⊆ may not hold for all the pair of elements(Example, consider {2,3} & {1,3}).

Any ideas for improving the worst case complexity is welcomed!!

Thanks.

P.S: This is not a homework problem!!

hivert
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algo_freek
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  • Partial order usually correspond to a [topological sorting](http://en.wikipedia.org/wiki/Topological_sorting) which can be done in O(M+N) for graphs with M vertices and N edges. But as I understand your question, your mian problem is that of finding the correct edges, so I'm not sure whether this will be of any use. – MvG Jul 10 '12 at 09:57
  • You do not have a graph to start with!! But i see ur thought! – algo_freek Jul 10 '12 at 10:05
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    You probably could improve average case behaviour by inserting elements in order of ascending cardinality, and start by comparing new elements with maximal elements in the current set, descending to children only if there was a partial but no complete match. But this will not improve worst case behaviour. – MvG Jul 10 '12 at 10:19

2 Answers2

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This problem is usually called transitive reduction. I believe that your algorithm is incorrect; while I do not have enough details to tell you exactly how, there is an efficient reduction from transitive closure to transitive reduction, and the best known algorithms for the former run in time O(nω) for some exponent ω > 2.

redoubtable
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if the set is {1,2,3,5}, the subset will be ({Æ},{1},{2},{3},{5},{1,2},{1,3],{1,5},{2,3}{2,5},{3,5},{1,2,3}{1,3,5},{1,2,5},{2,3,5},{1,2,3,5}). after taking off transitive relations we will have ({{Æ}},{1},{2},[3},{4},{5},{1,2},{2,3}{1,5},{2,5},{3,5},{1,2,3},{1,3,5},{1,2,5},{1,2,3,5}). now start with empty set and draw lines to 1,2,3,5.. then draw line to 1,2 from the points 1 and 2.draw line from 2 and 3 to join 2,3.then draw lines to 5 from 1,2,3... then from there draw lines to {1,2,3} from [1,2},{2,3}. then draw lines to {1,3,5} from {3,5}and {1,5} . then again draw a line to next set , and atlast draw lines from all the previous sbset to the last subset...

Priya
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