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homework2_sol - Homework problems for CMSC423 1 Longest...

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Homework problems for CMSC423 1 Longest monotonically increasing subsequence Let’s ﬁnd the length of MIS ﬁrst. Suppose we know the length of a sequence ending at ( i - 1) th index — L[i-1]. If next number C i is greater than C i - 1 , then we can extend ( i - 1) th sequence with C i : Length( i ) = Length( C i - 1 ) + 1 . However, the sequence ending at i - 1 is not necessarily the longest sequence we have seen so far: there may have been a subsequence ending somewhere before i - 1 that was longer: 5 6 1 7 2 8 .. . i - 1 i In example above, sequence ending with 2 is of length 2 and we can extend it with 8 (2 < 8), but a sequence [5, 6, 7] is longer and it too can be extended with 8. To ﬁnd the longest increasing subsequence that ﬁts somewhere between 0 and i , we would need to consider all subsequences ending somewhere before i , and see if we can extend any of them: Algorithm 1 Finding locally long subsequence between 0 and i for j = 0 i - 1 do if C j < C i AND Length( j ) + 1 > Length( i ) then Length( i ) Length( j ) + 1 end if end for To compute the longest increasing subsequence within the whole string S , we need to start from the left and keep extending locally longest subsequences: Algorithm 2 Finding globally long subsequence in S for i = 0 N - 1 do for j = 0 i - 1 do if C j < C i AND Length( j ) + 1 > Length( i ) then Length( i ) Length( j ) + 1 end if end for end for 1

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We can also write down the problem as a recurrence: Length ( i ) = max j<i n Length ( j ) + 1 s.t. C j < C i Here, we go through all j less that i ; for j such that C j < C i , we consider a Length( j ) of the longest subsequence ending at or before that index and pick a maximum among those. 2 Guaranteed gap limit When k = 0, we are guaranteed that there will be no gaps in the alignment. The only two options left are mismatches and matches. Geometrically, that means we do not ever stray from the diagonal path from the lower left cell to the upper right in the global alignment matrix. Now, suppose we may have up to 1 gap. That implies that the optimal
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This note was uploaded on 01/13/2012 for the course CMSC 423 taught by Professor Staff during the Fall '07 term at Maryland.

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homework2_sol - Homework problems for CMSC423 1 Longest...

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