Approximate Online Palindrome Recognition and Applications Amihood Amir

Approximate On-line Palindrome Recognition, and Applications Amihood Amir Benny Porat

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Palindrome Recognition - Voz'mi-ka slovo ropot, - govoril Cincinnatu ego shurin, ostriak, -- I prochti obratno. A? Smeshno poluchaetsia? Vladimir Nabokov, Invitation to a Beheading (1) "Take the word ropot [murmur], " Cincinnatus' brother-in-law, the wit, was saying to him, "and read it backwards. Eh? Comes out funny, doesn't it? " [--› topor: the axe] A palindrome is a string that is the same whether read from right to left or from left to right: Examples: доход A man, a plan, a cat, a ham, a yak, a yam, a hat, a canal-Panama!

Palindrome Example Ibn Ezra: Medieval Jewish philosopher, poet, Biblical commentator, and mathematician. Was asked: " "אבי אל חי שמך למה מלך משיח לא יבא [ My Father, the Living God, why does the king messiah not arrive? ] His response: " שוב אליכם כי בא מועד , "דעו מאביכם כי לא בוש אבוש [ Know you from your Father that I will not be delayed. I will return to you when the time will come ]

Palindromes in Computer Science Great programming exercise in CS 101. Example of a problem that can be solved by a RAM in linear time, but not by a 1 -tape Turing machine. (Can be done in linear time by a 2 -tape TM)

Palindrome Concatenation We may be interested in finding out whether a string is a concatenation of palindromes of length > 1. Example: ABCCBABBCCBCAACB Why would we be interested in such a funny problem? – we’ll soon see Exercise: Do this in linear time…

Stream 2 - Approximations As in exact matching, there may be errors. Find the minimum number of errors that, if fixed, will give a string that is a concatenation of palindromes of length > 1 ABCCBABBCCBCAACB Example: ABCCBCBBCCBCABCB For Hamming distance: A-Porat [ISAAC 13]: Algorithm of time O(n 2)

Stream 3 - Reversals Why is this funny problem interesting? Sorting by reversals: In the evolutionary process a substring may “detach” and “reconnect” in reverse: ABCABCDAABCBAD CBAADCB

Sorting by Reversals What is the minimum number of reversals that, when applied to string A, result in string B? History: Introduced: Bafna & Pevzner [95] NP-hard: Carpara [97] Approximations: Christie [98] Berman, Hannenhalli, Karpinski [02] Hartman [03]

Sorting by Reversals – Polynomial time Relaxations Signed reversals: Hannenhalli & Pevzner [99] Kaplan, Shamir, Tarjan [00] Tannier & Sagot [04]. . . Disjointness: Swap Matching Muthu [96] Two constraints: 1. The length of the reversed substring is limited to 2. 2. All swaps are disjoint.

Pattern Matching with Disjoint Reversals • Reversal Distance (RD): – The RD between s 1 and s 2 is the minimum number k, such that there exist s 2’ , where HAM(s 1, s 2’) =k, and s 1 reversal match s 2. S 1: S 2: A D C B A E D B A A D A A B C E RD(S 1, S 2) = 2

Connection between Reversal Matching and Palindrome Matching Interleave Strings: S 1: S 2: A D C B A E D B A C D A A B D E A C D D C A B A A B E A D B B D A E

On-line Input Suppose that we get the input a byte at a time: For the palindrome problem: A C D D C ABAABEADBBDAE

On-line Input Suppose that we get the input a byte at a time: For the reversal problem: A C D D C A B A A B E A D B B D A E

Main Idea – Palindrome Fingerprint The Rabin Karp Fingerprint Φ(S)=r 1 s 0+ r 2 s 1+… rmsm-1 mod (p) s 0, s 1, s 2, …sm-1 ΦR(S)=r-1 s 0+ r-2 s 1+… r-msm-1 mod (p) The Reversal Fingerprint If rm+1ΦR(S) = Φ(S) => S is a palindrome. w. h. p.

Palindrome Fingerprint If rm+1ΦR(S) = Φ(S) => S is a palindrome. Φ(S)=r 1 s 0+ r 2 s 1+… rmsm-1 mod (p) Example: r 6ΦR(S)=r-1 s 0+ r-2 s 1+… r-msm-1 mod (p) S=ABCBA r 6 (1/r A + 1/r 2 B + 1/r 3 C + 1/r 4 B + 1/r 5 A) = r 5 A + r 4 B + r 3 C + r 2 B + r A = Φ(S)

Simple Online Algorithm for Finding a Palindrome in a Text t 1, t 2, t 3, … ti, ti+1, ti+2 , …ti+m, ti+m+1 , … tn Φ=r 1 ti+ r 2 ti+1+… rmti+m mod (p) If rm+1ΦR =Φ => ΦR=r-1 ti+ r-2 ti+1+… r-mti+m mod (p) there is a palindrome starting in the i-th position. If not, then for the next position: Φ= Φ + rm+1 t i+m+1 mod (p) ΦR=ΦR + r-(m+1)ti+m+1 mod (p) Note: This algorithm finds online whether the prefix of a text is a permutation. For finding online whether the text is a concatenation of permutations, assume even-length permutations, otherwise, every text is a concatenation of length-1 permutations.

Palindrome with mismatches Start with 1 mismatch case.

1 -Mismatch S= s 0, s 1, s 2, … sm-1 Choose l prime numbers q 1, …, ql < m such that

1 -Mismatch S= S 2, 0= S 2, 1= s 0, s 2, s 4 … S 3, 0= S 3, 1= S 3, 2= s 0, s 3, s 6 … s 1 s 3, s 5 … s 1, s 4, s 7 … s 2, s 5, s 8 … s 0, s 1, s 2, … mod 2 mod 3 sm-1 Examples: q 1=2, q 2=3 For each qi construct qi subsequences of S as follows: subsequence Sqi, j is all elements of S whose index is j mod qi.

Example S= S 2, 0= s 0, s 2, s 4 S 2, 1= s 1 s 3, s 5 S 3, 0= s 0, s 3 S 3, 1= s 1, s 4 S 3, 2= s 2, s 5 s 0, s 1, s 2, s 3, s 4, s 5 mod 2 mod 3

1 -Mismatch • We need to compare: s 0 , s 1 , s 2 , sm-1, sm-2, sm-3 … sm-2 , sm-1 … s 1 , s 0 • We prove that in the partitions strings: Sq, j= SRq, (m-1 -j)mod q

Example S 2, 0= s 0, s 2, s 4 S 2, 1= s 1 s 3, s 5 S 3, 0= s 0, s 3 S 3, 1= S 3, 2= S= s 0, s 1, s 2, s 3, s 4, s 5 SR = s 5, s 4, s 3, s 2, s 1, s 0 S 3, 0= s 0, s 3 SR 3, 2= s 5, s 2 s 1, s 4 S 3, 1= s 1, s 4 s 2, s 5 SR 3, 1= s 4, s 1 S 2, 0= s 0, s 2, s 4 SR 2, 1= s 5 s 3, s 1

Exact Matching Lemma: S=SR Sq, j = SRq, (m-1 -j) mod for all q and all 0 ≤ j ≤ q. q

1 -Mismatch Lemma: There is exactly one mismatch There is exactly one subpattern in each group that does not match. C. R. T

Chinese Remainder Theorem Let n and m two positive integers. In our case: if two different indices, i and j, have an error, and only one subsequence is erroneous, since the product of all q’s > m, it means that i=j.

Complexity There exists a constant c such that, for any x<m, there at least x/log m prime numbers between x and cx. Therefore, choose prime numbers between log m and c log m.

Complexity • For each qi we compute 2 qi different fingerprints: • Overall space: • Each character participates in exactly two fingerprints (the regular and the reverse). • Overall time:

Online All fingerprint calculations can be done online We know the m at every input character, to compute the comparisons. Conclude: Our algorithm is online.

k-Mismatches Use Group testing…

k-Mismatches Group Testing • Given n items with some positive ones, identify all positive ones by a small number of tests. • Each test is on a subset of items. • Test outcome is positive iff there is a positive item in the subset.

k-Mismatch • Group: partition of the text. • Test: distinguish between: algorithm) – match – 1 -mismatch – more then 1 -mismatch (using the 1 -mismatch

k-Mismatches S= S 2, 0= S 2, 1= s 0, s 2, s 4 … S 3, 0= S 3, 1= S 3, 2= s 0, s 3, s 6 … s 1 s 3, s 5 … s 1, s 4, s 7 … s 2, s 5, s 8 … s 0, s 1, s 2, … mod 2 mod 3 sm-1 Each Sq, j is a group in our group testing Similar to the 1 -mismatch algorithm just with more prime numbers…

Our tests • We define The reversal pair of Sq, j to be SRq, (m-1 -j)mod q • Each partition is “tested against” its reversal pair.

Correctness s 0, s 1, s 2, … sj …. sm-1 i 2 i 9 i 5 i 7 i For any group of k character i 1, i 2, . . ik There exists a partition where sj appears alone C. R. T

Correctness s 0, s 1, s 2, … sj …. sm-1 i 2 i 9 i 5 i 7 i If sj invokes a mismatch we will catch it.

Complexity • Overall space: • Overall time:

Approximate Reversal Distance Using the palindrome up to k-mismatches algorithm, can be solved in time, and space.