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Q Q and is n 2 n 2 v f ( ( by f f n d is v ) ) - - ity of checking whether a word is not Hamming- is the length of u and u[i] are its letters. The suffix of isometric. Our approach is based on the character- index i on u, denoted by ui, is the word u[i] u[n 1] ization of 2-error borders of such words. The naive of length n i. A suffix (or prefix) of it is···proper− if algorithm runs clearly in quadratic time. We show it is distinct− from u itself. that known algorithms for matching patterns with Let k be a non-negative integer. We say that a mismatches can be used to solve this problem effi- word u has a k-error border if u has a proper suf- ciently. Pattern matching with k mismatches can be fix s that matches its prefix of the same length with solved by algorithms running in time O(nk) (see [13] exactly k differences. In other words, the Hamming and [7]). These algorithms are mostly based on a distance between the suffix and the prefix is k. technique called the Kangaroo method. This method computes the Hamming distance for every alignment Example 1 The word 1010011 has a 2-error border. in time O(k) by “jumping” from one error to the next Indeed, it has the prefix 101 and the suffix 011 ans error. A faster algorithm for pattern matching with the Hamming distance between 101 and 011 is 2. k mismatches runs in O(n√k log k) [1]. A simpler version of this algorithm is given in [15]. Let f be a finite word and n be a positive integer. We show two methods to check whether a word Then a word u is called f-free if it does not contain is not Hamming-isometric. The first one uses the f as a factor, and f is called n-Hamming-isometric if Kangaroo method which allows to derive an algo- for every f-free words u and v of length n, the follow- O kn O n rithm running in time ( ) and using ( ) space ing holds: u can be transformed into v by changing n k to check whether a word of size has a -error bor- one by one all the letters on which u differs from v, in der. The method has a preprocessing of linear time such a way that all of the new words obtained during and space for computing the suffix tree of the word this process are also f-free. Such a transformation is and to enhance it in order to answer lowest common called an f-free transformation from u to v. Eventu- ancestor queries in constant time. This overall leads ally, a word f is said to be Hamming-isometric if it to a linear-time and linear-space algorithm to check is n-Hamming-isometric for all positive integer n. whether a word is not Hamming-isometric, and hence The d-ary n-cube, denoted by Qd , is the graph also to check whether a binary word is Lee-isometric. n whose vertices are the words of length n over the The second method uses the computation of a k- alphabet Z = 0, 1,...,d 1 , and for which any prefix table that gives, for some word u and every d two words u and{ v are adjacent− } if and only if u and position on u, the length of the longest proper suffix v differ by one unit at exactly one position, say i, of u at this position that matches its prefix of the that is, u[i] = v[i] 1 mod d [2]. The d-ary n-cube same length with at most k differences [5]. The com- avoiding f, where ±f is a word over the alphabet Z putation of this k-prefix table is done in time O(kn) d is the graph Qd (f) obtained from Qd by deleting the using O(n) space. n n vertices containing f as a factor. We also use the Kangaroo method to derive an al- A word f over Z is said to be Lee-isometric if for gorithm running in time O(kn) and using O(n) space d all n 1, Qd (f) is an isometric subgraph of Qd . on a constant size alphabet to check whether a word ≥ n n of size n has a k-Lee-error border and thus check in linear time whether a word over an alphabet of size Example 2 The word 0301 on the alphabet Z4 = 4 is Lee-isometric. 0, 1, 2, 3 is non-Lee-isometric. Indeed, the words {u = 030001} and v = 030201, which do not contain the factor 0301, are at distance 2 but there is no path 4 2 Definitions and background of length 2 from u to v in Q6(0301) since any path of length 1 changing the symbol of index 3 of u goes Let A be a finite alphabet. A word u in A⋆ is a finite from u to 030101 or to 030301 and these two words sequence u[0]u[1] u[n 1] of letters in A, where n both have the word 0301 as factor. ··· −
2 It is shown in [2] that non-Hamming-isometric and trie containing all the suffixes of u by their keys and non-Lee-isometric words coincide for words on an al- positions on u as their values [6], [4]. The tree has a phabet of size at most three. But this property is no linear number of nodes and edges, each edge contain- more true for greater alphabets. ing a pair of integers identifying a factor of u, e.g. Hamming-isometric words have the following char- (position, length), hence the linear space complexity. acterization obtained in [12, 17] for binary alphabets Suffix arrays can also be used for this problem. They and in [2] for general alphabets. contain essentially the starting positions of suffixes of u sorted in lexicographic order. Proposition 3 A word is not Hamming-isometric if To get the overall running time, we need to an- and only if it has a 2-error border. swer Lowest Common Ancestor (LCA) queries in con- stant time [8], [16]. LCA queries give us the longest Example 4 For instance the words 11, 1n for n 1 common prefix between two suffixes of u, essentially ≥ are Hamming-isometric. The word 1010011 is not telling us where the first mismatch appears between a Hamming-isometric. suffix of u and its prefix of the same length. This can be performed by first constructing a Longest Com- mon Prefix (LCP) array. The LCP array stores the 3 Algorithms for checking length of the longest common prefix between two con- whether a word is Hamming- secutive suffixes in the suffix array (lexicographic con- secutive suffixes). This array can also be constructed isometric in linear time. To compute the length of the longest prefix common to any two suffixes in the suffix tree In this section, we use the characterization of non- (instead of consecutive suffixes), we need to use some Hamming-isometric words in terms of 2-error border range minimum query data structure. (Proposition 3) and assume that the alphabet A has Thus, we assume that our suffix tree is enhanced a constant size. Observe that a quadratic-time naive to answer LCA queries in constant time. This can algorithm can be obtained to check wether a word is be done in linear time and space. We denote by non-Hamming-isometric by computing the Hamming LCA(i, j) the query that returns in O(1) time the distance between each suffix of index i and the prefix length of the commun prefix between the suffix u of the same length. We show that checking if a word i and the suffix u of u. of length n has a k-error border can be done in time j For every index i, we try to find if the suffix of index O(kn) and space O(n). i has k-mismatches with its prefix of the same length. Proposition 5 It can be checked in time O(kn) and We first compute LCA(0,i). Let this length be ℓ0. space O(n) whether a word of length n has a k-error We skip the mismatching character in u0 and ui and border. try to find LCA(ℓ0 +1,i + ℓ0 +1). We repeat this to obtain k mismatches between ui and u[0] u[n i Proof. We give two algorithms for solving this prob- 1] or fail to obtain this condition. ··· − − lem. The first one is based on a technique called the The pseudo code of the technique is given in Al- Kangaroo method used for pattern matching with k gorithm 1. We maintain a variable ℓ which gives, mismatches in O(nk) time (see [13], [7] and [14]). after line 4 of Algorithm 1, the index of ui of the These algorithms compute the Hamming distance for current mismatch between ui and u0. A variable every alignment in O(k) time by “jumping” from one d contains the current Hamming distance between error to the next. We use the Kangaroo method to u[i] u[i + ℓ 1] and u[0] u[ℓ 1]. It is increased check for each index i on a word u of length n whether by 1··· at line 8− since a mismatch··· has− been found. it has a k-error border of length n i in time O(k). Since there are at most O(k) LCA queries for each To do so, we first compute in time− and space O(n) index i, this can be done in O(k) time. The overall the suffix tree of u. The suffix tree is a compacted time complexity is thus O(kn) and the space com-
3 plexity is O(n). at the first step of the loop of line 3 we obtain line 4 We now show a second method to check whether ℓ = LCA(0, 1) = 0; we set ℓ to 1 (”the jump”) and d a word of length n has a k-error border. We use to 1 line 8. At the second step of the loop of line 3 the computation of a k-prefix table as done in [5]. we obtain line 4 ℓ = ℓ + LCA(1, 2)= 1; we set ℓ to 2 For each position i, we compute a table πk for and d to 2 line 8. At the third step of the loop of line which πk(i) is the length ℓ of the longest word 3, we obtain line 4 ℓ = ℓ + LCA(2, 3) = 2 and break u[i] ...u[i + ℓ 1] such that the Hamming distance line 10. For i = 2 the loop of line 3 fails to return between u[0] ...u− [ℓ 1] and u[i] ...u[i + ℓ 1] is at true. For i = 3, at the first step of the loop of line 3 most k and u[0] ...u−[ℓ 1] is proper prefix− of u. we obtain line 4, ℓ = LCA(0, 3) = 0; we set ℓ to 1 and This computation can− be done in time O(kn) and d to 1 line 8. At the second step of the loop of line 3 space O(n) (see [5, Theorem 5]). It needs the compu- we obtain line 4 ℓ = ℓ + LCA(1, 4)= 1; we set ℓ to 2 tations of the prefix array of u and the longest prefix and d to 2 line 8. At the third step of the loop of line array preprocessed for range minimum queries. 3, we obtain line 4 ℓ = ℓ + LCA(2, 5) = 3 and, since The existence of a k-error border is then obtained d = 2 and ℓ = n i = 3, the algorithm returns true as follows. For k 1, a word u has a k-error border if line 6. The algorithm− has thus detected the 2-error and only if there≥ is a position i, 1 i Example 6 Let u = 101011. Let us check with Al- A combinatorial characterization of Lee-isometric gorithm 1 whether u has a 2-error border. For i = 1, words over an alphabet of size 4 has been obtained in 4 [2]. It uses the notion of Lee distance which is defined time complexity is thus O(kn) and the space com- as follows. The Lee distance, denoted by dL, between plexity is O(n). two letters of the alphabet Zd = 0, 1,...,d 1 is { − } dL(a,b) = min( a b , d a b ). | − | − | − | Word with a -Lee-error The Lee distance between two words u and v of length Algorithm 3: k border(u) n over Zd is Input: A non empty word u of length n, a −1 n non-negative integer k dL(u, v)= X dL(u[i], v[i]). Output: true if u has a k-Lee-error border i=0 1 for i 1 to n 1 do ← − A word has a k-Lee-error border if it has a suffix u 2 (ℓ, d) (0, 0); ← and a prefix v of same length satisfying dL(u, v)= k. 3 while d k do Z ≤ For words over 4, the Lee-isometric words are 4 ℓ ℓ+ LCA(ℓ,i + ℓ); ← characterized as follows in [2]. 5 if d = k and ℓ = n i then − 6 return true; Proposition 8 A word over a 4-letter alphabet is 7 if (d = k and ℓ 5 [3] Marcella Anselmo, Dora Giammarresi, Maria [12] Sandi Klavzar and Sergey V. Shpectorov. Madonia, and Carla Selmi. 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