Some Modified Algorithms for Dijkstras Longest Upsequence Problem
Some Modified Algorithms for Dijkstras Longest Upsequence Problem
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None of this changes our basic approach. But now we propose to mix the two algorithms as follows: 1) Partition the set of all runs into two equal parts, a left one containing the leftmost half of the runs, and a right one, containing the righmost half. 2) Apply the first algorithm to the left part, and the reverse algorithm to the right part, to obtain two final runs. 3) Merge these two runs by using either the first algorithm or its reverse counterpart. 4) The length of the resulting run is th...en the length of the Page 6 longest upsequence. Considering the same example as before, we again have the following partition into runs : 13573/29/4 10/6 We merge rl with r2 'to the left', and merge r3 with r4 'to the right' yielding: 125789/4 10 Now we merge these runs, say, to the right: 1 2 5 7 8 9 10 ^^. ^^. , ^, and obtain 7 as the length of the longest upsequence. The correctness of this hybrid algorithm is more difficult to establish than that of t;he p, receding algorithms. It is proved as follows: .
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You can find similar books in the "Read Also" column, or choose other free books by Robert B K Dewar to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
None of this changes our basic approach. But now we propose to mix the two algorithms as follows: 1) Partition the set of all runs into two equal parts, a left one containing the leftmost half of the runs, and a right one, containing the righmost half. 2) Apply the first algorithm to the left part, and the reverse algorithm to the right part, to obtain two final runs. 3) Merge these two runs by using either the first algorithm or its reverse counterpart. 4) The length of the resulting run is th...en the length of the Page 6 longest upsequence. Considering the same example as before, we again have the following partition into runs : 13573/29/4 10/6 We merge rl with r2 'to the left', and merge r3 with r4 'to the right' yielding: 125789/4 10 Now we merge these runs, say, to the right: 1 2 5 7 8 9 10 ^^. ^^. , ^, and obtain 7 as the length of the longest upsequence. The correctness of this hybrid algorithm is more difficult to establish than that of t;he p, receding algorithms. It is proved as follows: .
What to read after Some Modified Algorithms for Dijkstras Longest Upsequence Problem?
You can find similar books in the "Read Also" column, or choose other free books by Robert B K Dewar to read onlineMoreLess
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