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ECS124_Lecture6

ECS124_Lecture6 - ECS 124 Theory and practice of...

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ecture 6: CSub B Search ECS 124: Theory and practice of bioinformatics Lecture 6: LCSub, DB Search Instructor: IliasTagkopoulos Office: Kemper 3063 and GBSF 5313 4/21/2010 UC Davis 1

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LAST TIME: Free end-gaps and LCS c DP LCS Algorithm: Let T(i,j) = LCS(S1 i ,S2 j ) c For the first row: c T(1,j) = 1 if S1(1) matches something in S2 c 0 otherwise milarly for T(i,1) 4/21/2010 2 c Similarly for T(i,1) c For the rest: c T(i,j) = 1 + T(i-1,j-1) if S1(i) = S2(j) c T(i,j) = max{T(i-1,j), T(I,j-1)} S1: TTAAAGAA S2: TTGCAACC P = length of matching prefix X i = 1 if S2(i) is used, else 0 So: P = X 1 + X 2 +…+ X n And E(P) = n/#characters
LAST TIME: Expected LCS value c Assume you have a string S c Create a random string S2, length n c Probability of any given character in any string position = 1/#characters ets find a lower bound of the expected LCS by finding the 4/21/2010 UC Davis 3 c Lets find a lower bound of the expected LCS by finding the longest prefix of S: c In reality, for #char=4, E(LCS) > 0.54n S1: TTAAAGAA S2: TTGCAACC P = length of matching prefix X i = 1 if S2(i) is used, else 0 So: P = X 1 + X 2 +…+ X n And E(P) = n/#characters

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Substring c Assume strings/sequences S1 and S2, with length m c Again, if each char in S1 is picked randomly from an alphabet of size c, then probability of any char is p=1/c c probability of observing S1 is p m =(1/c) m , and similarly probability of observing any substring of S1 with length n 4/21/2010 UC Davis 4 (n<=m) is p n =(1/c)
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ECS124_Lecture6 - ECS 124 Theory and practice of...

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