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Finding all Palindrome Subsequences in a String K.R. Chuang1, R.C.T. L Finding all Palindrome Subsequences in a String K.R. Chuang1, R.C.T. L

Finding all Palindrome Subsequences in a String K.R. Chuang1, R.C.T. L - PDF document

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Finding all Palindrome Subsequences in a String K.R. Chuang1, R.C.T. L - PPT Presentation

Let matched pair to denot where nji and we define palindrome subsequence to be a palindrome subsequence which has matched pairs of We use the notation finding all palindrome subsequence ID: 133571

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Finding all Palindrome Subsequences in a String K.R. Chuang1, R.C.T. Lee2 and C.H. Huang3* 1,2 Department of Computer Science, National Chi-Nan University, Puli, Nantou Hsieh, Taiwan 545 3 Department of Computer Science and Information Engineering, National Formosa University, 64, Wen-Hwa Rod, Hu-wei, Yun-Lin, Taiwan 632 A palindrome is a string of symbols that is read the same forward and backward. Palindrome also occurs in DNA. DNA palindromes appear frequently and are widespread in human cancers. Identifying them could help advance the understanding of genomic instability [2, 6]. The Palindrome subsequences detection problem is therefore an mputational biology. In this paper, we present an algorithm to find all palindrome subsequences. 1.Introduction Let matched pair, (), to denot where nji and we define -palindrome subsequence to be a palindrome subsequence which has matched pairs of . We use the notation ( finding all palindrome subsequences problem. In this algorithm, we find all palindrome subsequences form one palindrome subsequence to the longest palindrome subsequence. Given of length , let -palindrome where Step 1: We use incidence matrix to find all matched pairs () where njiadd them into , because each matched pair is 1-palindrome subsequence. Step 2: Wegenerate from where . For all -1-palindrome subsequences in -1-palindrome subsequence () ‘ () form and we check all 1-palindromes from whether there is a 1-palindrome () which satisfies the rule . If it is satisfied, we combine the -1-palindrome () ‘ (with the 1-palindrome () to be -palindrome ) ‘ () and add it into the set is generated, we can get the set which contains all palindrome subsequences of In the following, we present the algorithm ing all palindrome subsequences. findAllPalindromeSubsequences Input: A string Output: All palindrome subsequences of Step 1 /* Finding out matched pair for nji */ U1 := Š‹ = 1 +1 then w := (i, ) Š endfor endfor Step 2 /* Finding all palindrome subsequences of = 2 /2 do Uk := Š‹ -palindrome () from all 1-palindrome () fro if i † ik-1 and j  jk-1 then ik := i jk := j ) ‘ ( Š endif endfor endfor endfor U := U1U2‘Un/2 is the set of all palindrome subsequences of S */ 4.An Example Given a = ACGATGTAC, We now illustrate the whole procedure in detail. ACGATG T A C Step 1: We use incidence matrix to find all hed pairs () where njiTable 1 The incidence matrix for this string ACGATGTAC 12345 6 7 8 9 ACGAT G T A C 1A 0010 0 0 1 0 2C 000 0 0 0 1 3G 00 1 0 0 0 4A 0 0 0 1 0 5T 0 1 0 0 6G 0 0 0 7T 0 0 8A 0 9C After the incidence matrix is generated, we can get the = Š(1, 4), (1, 8), (2, 9), (3, 6), (4, 8), (5, 7)‹ Step 2: = 2, = Š(1, 4), (1, 8), (2, 9), (3, 6), (4, 8), There is no 1-palindrome which can be = Š(2, 9) (4, 8) (5, 7)‹ We take the 2-palindrome (4, 8) (5, 7) from Check all 1-palindrome from There is no 1-palindrome which can be = 4, = Š(1, 4), (1, 8), (2, 9), (3, 6), (4, 8), (5, 7)‹, = Š(1, 8) (3, 6), (1, 8) (5, 7), (2, 9) (3, 6), (2, 9) (4, 8), (2, 9) (5, 7), (4, 8) (5, 7)‹, Š(2, 9) (4, 8) (5, 7)‹, We take the 3-palindrome (2, 9) (4, 8) (5, 7) from Check all 1-palindrome from There is no 1-palindrome which can be Finally, we get the set which contains all palindrome subsequences of = Š(1, 4), (1, 8), (2, 9), (3, 6), (4, 8), (5, 7), (1, 3, 6), (1, 8) (5, 7), (2, 9) (3, 6), (2, 9) (4, 8), (2, 9) (5, 7), (4, 8) (5, 7), (2, 9) (4, 8) (5, 7)‹ The all palindrome subsequences of , 4) AA (1, 8) AA (2, 9) CC (3, 6) GG (4, 8) AA (5, 7) TT (1, 8) (3, 6) AGGA (1, 8) (5, 7) ACCA (2, 9) (3, 6) CGGC (2, 9) (4, 8) CAAC (2, 9) (5, 7) CTTC (4, 8) (5, 7) ATTA (2, 9) (4, 8) (5, 7) CATTAC 5.ConclusionsIn this paper, we proposed an algorithm to solve the finding all palindrome subsequences in a string. Palindrome subsequences occur frequently in DNA sequences and have been proved to be critical for some biological characteristics. Our algorithm provides an effective tool for the related research. References [1] Allison, L. (2004) Finding Approximate Palindromes in Strings Quickly and Simply [2] Choi, Charles Q (2005) DNA palindromes er. The Scientist [3] Gusfied, D. (1997) Algorithms on Strings, Trees, and Sequences: Computer Science and Computational Biology, Cambridge University Press, New York. [4] Manacher, D. (1975) A new Linear-Time Algorithm for Finding the Smallest Initial Palindrome of a String. J. Assoc. Comput. [5] Proto, A. H. L. and Barbosa V. C. (2002) Finding Approximate Palindromes in Strings. [6] Tanaka, Hisashi; BERGSTROM, Donald A; YAO, Meng-Chao and TAPSCOTT, Stephen J (2006) Large DNA form of structural chromosome aberrations in human cancers. Human Cell [7] Wen, W. H. (2006) Longest Palindrome and Tandem Repeat Subsequences