Finding All Approximate Palindromes

The 25th Workshop on Combinatorial Mathematics and Computation Theoryfocus on algorithms to find the edit distancebetween two strings, and many of them [1, 3, 4, 6,7, 10, 11, 12, 14, 15, and 16] use the followingrule: Consider the two strings A A 1A2andB B 1B 2, if A2 B2and ED ( A1 , B1) k , thenED ( A,B) k .There are two approaches to solve theapproximate string matching problem. The firstapproach [12, 2, 10, 11 and 15] is a local approachwhich computes ED ( P,S)where S is asubstring of T starting from each location i .That is, for every possible position i , we testwhether there exists a substring S of Tstarting from this location such that ED( P,S) k .Another approach [16], named Approach 2, is aglobal approach which can solve the problem inmO( nk ) time-complexity where w is the word wsize. We shall use the technique in [16] to solvethe problem.In the above, we only talked about one case.In the following, we give the precise definition ofS1and S 2for all cases.Case 1: We find even approximate palindromes. If1 i n2, let S1be T ( 1, i)and S2beT ( i 1, min{2i k,n}); otherwise, let S1beT (max{ 2i n k 1,1},i)and S2beT ( i 2, n).Case 2: We find odd approximate palindromes.If 1 i n2 , let S 1be T ( 1, i)and S 2beT ( i 2,min{2i k 1,n}); otherwise, let S 1beT (max{ 2i n k 2,1}, i)and S2beT ( i 2, n).Essentially, for each location i in T , wecompute the edit distance between S 1and S 2.Given two strings S 1 p 1p 2 p m',S t t tand k , we define2 1 2 n'M k1,if ED(S1(1, j),S2(1, i)) k( i,j) ,0,otherwise.where 0 i n'and 0 j m'.We now use dynamic approach to find M k ( i,j).Let M k ( i,j)I, M k ( i,j)Dand M k ( i,j)Sdenotethe M k( i,j)related to insertion, deletion andsubstitution respectively, wherek 11, if t p , M ( i 1,j) 1i jkkM ( i,j) or ti pjand M ( i 1,j 1) 1 0, otherwise.MI,kDandMkSk 11, if t p , M ( i,j 1) 1i jk( i,j) or ti pjand M ( i 1,j 1) 1 0, otherwise.k1, if t p and Mi j( i,j) or ti pjand M 0, otherwise.1k( i 1,j 1) 1( i 1,j 1) 1After every M k ( i,j)I, M k ( i,j)DandM k S( i,j)are found, we can determine M k( i,j)by M k ( i,j)= M k I( i,j)or M k D( i,j)or ( i,j)M k Swhere “or” is the “logical or” operation .Our algorithm is presented in the following:Input: A string T t1t2... tn.and k.Output: All occurrences of approximate palindromesof T with error k ' .for i=1 to n dofor 0 k' k dofor 0 i' n'and 0 j' m'dok 'M(0, j') 1, if 0 j' k'k ' k 'M ( i', j') M( i',0) 1, if 0 i' k'0 otherwise'M k '( i', j') M k '( i', j') or ( i', j')IM k D'or M k S( i', j') ;end forend forend for2-243-