CS 3343: Analysis of Algorithms

CS 3343: Analysis of Algorithms
String Matching Algorithms

Definitions Text: a longer string T Pattern: a shorter string P
Exact matching: find all occurrence of P in T T length = m P Length = n

The naïve algorithm Length = n Length = m

Time complexity Worst case: O(mn) Best case: O(m) Average case?
aaaaaaaaaaaaaa vs. baaaaaaa Average case? Alphabet size = k Assume equal probability How many chars do you need to compare before find a mismatch? In average: k / (k-1) Therefore average-case complexity: mk / (k-1) For large alphabet, ~ m Not as bad as you thought, huh?

Real strings are not random
T: aaaaaaaaaaaaaaaaaaaaaaaaa P: aaaab Plus: O(m) average case is still bad for long strings! Smarter algorithms: O(m + n) in worst case sub-linear in practice how is this possible?

How to speedup? Pre-processing T or P
Why pre-processing can save us time? Uncovers the structure of T or P Determines when we can skip ahead without missing anything Determines when we can infer the result of character comparisons without actually doing them. ACGTAXACXTAXACGXAX ACGTACA

Cost for exact string matching
Total cost = cost (preprocessing) + cost(comparison) + cost(output) Overhead Minimize Constant Hope: gain > overhead

String matching scenarios
One T and one P Search a word in a document One T and many P all at once Search a set of words in a document Spell checking One fixed T, many P Search a completed genome for a short sequence Two (or many) T’s for common patterns Would you preprocess P or T? Always pre-process the shorter seq, or the one that is repeatedly used

Pattern pre-processing algs
Karp – Rabin algorithm Small alphabet and small pattern Boyer – Moore algorithm The choice of most cases Typically sub-linear time Knuth-Morris-Pratt algorithm (KMP) Aho-Corasick algorithm The algorithm for the unix utility fgrep Suffix tree One of the most useful preprocessing techniques Many applications

Algorithm KMP Not the fastest Best known Good for “real-time matching”
i.e. text comes one char at a time No memory of previous chars Idea Left-to-right comparison Shift P more than one char whenever possible

Intuitive example 1 T abcxabc mismatch P abcxabcde Naïve approach: T abcxabc ? abcxabcde abcxabcde abcxabcde abcxabcde Observation: by reasoning on the pattern alone, we can determine that if a mismatch happened when comparing P[8] with T[i], we can shift P by four chars, and compare P[4] with T[i], without missing any possible matches. Number of comparisons saved: 6

Intuitive example 2 Should not be a c T abcxabc mismatch P abcxabcde Naïve approach: T abcxabc ? abcxabcde ? abcxabcde abcxabcde abcxabcde abcxabcde abcxabcde Observation: by reasoning on the pattern alone, we can determine that if a mismatch happened between P[7] and T[j], we can shift P by six chars and compare T[j] with P[1] without missing any possible matches Number of comparisons saved: 7

KMP algorithm: pre-processing
Key: the reasoning is done without even knowing what string T is. Only the location of mismatch in P must be known. x T t z y P t’ t j i z y P t’ t j i Pre-processing: for any position i in P, find P[1..i]’s longest proper suffix, t = P[j..i], such that t matches to a prefix of P, t’, and the next char of t is different from the next char of t’ (i.e., y ≠ z) For each i, let sp(i) = length(t)

KMP algorithm: shift rule
x T t z y P t’ t j i z y P t’ t 1 sp(i) j i Shift rule: when a mismatch occurred between P[i+1] and T[k], shift P to the right by i – sp(i) chars and compare x with z. This shift rule can be implicitly represented by creating a failure link between y and z. Meaning: when a mismatch occurred between x on T and P[i+1], resume comparison between x and P[sp(i)+1].

Failure Link Example P: aataac
If a char in T fails to match at pos 6, re-compare it with the char at pos 3 (= 2 + 1) a a t a a c sp(i) aa at aat aac

Another example P: abababc
If a char in T fails to match at pos 7, re-compare it with the char at pos 5 (= 4 + 1) a b a b a b c Sp(i) ab ab abab abab ababa ababc

KMP Example using Failure Link
t a a c T: aacaataaaaataaccttacta aataac ^^* Time complexity analysis: Each char in T may be compared up to n times. A lousy analysis gives O(mn) time. More careful analysis: number of comparisons can be broken to two phases: Comparison phase: the first time a char in T is compared to P. Total is exactly m. Shift phase. First comparisons made after a shift. Total is at most m. Time complexity: O(2m) aataac .* aataac ^^^^^* Implicit comparison aataac ..* aataac .^^^^^

KMP algorithm using DFA (Deterministic Finite Automata)
P: aataac If a char in T fails to match at pos 6, re-compare it with the char at pos 3 Failure link a a t a a c If the next char in T is t after matching 5 chars, go to state 3 a t a a t a a c 1 2 3 4 5 DFA 6 a a All other inputs goes to state 0.

DFA Example T: aacaataataataaccttacta 1201234534534560001001
1 2 3 4 5 DFA 6 a a T: aacaataataataaccttacta Each char in T will be examined exactly once. Therefore, exactly m comparisons are made. But it takes longer to do pre-processing, and needs more space to store the FSA.

Difference between Failure Link and DFA
Preprocessing time and space are O(n), regardless of alphabet size Comparison time is at most 2m (at least m) DFA Preprocessing time and space are O(n ||) May be a problem for very large alphabet size For example, each “char” is a big integer Chinese characters Comparison time is always m.

The set matching problem
Find all occurrences of a set of patterns in T First idea: run KMP or BM for each P O(km + n) k: number of patterns m: length of text n: total length of patterns Better idea: combine all patterns together and search in one run

A simpler problem: spell-checking
A dictionary contains five words: potato poetry pottery science school Given a document, check if any word is (not) in the dictionary Words in document are separated by special chars. Relatively easy.

Keyword tree for spell checking
This version of the potato gun was inspired by the Weird Science team out of Illinois p s o c h o o l e 5 t i e t a t r n t e y c o r e y 3 1 4 2 O(n) time to construct. n: total length of patterns. Search time: O(m). m: length of text Common prefix only need to be compared once. What if there is no space between words?

Aho-Corasick algorithm
Basis of the fgrep algorithm Generalizing KMP Using failure links Example: given the following 4 patterns: potato tattoo theater other

Keyword tree p t t o h e h t a r e t a a 4 t t t o e o r 1 o 2 3

Keyword tree potherotathxythopotattooattoo p t t o h e h t a r e t a a
p t t o h e h t a r e a t a 4 t t t o e o r 1 o 2 3 potherotathxythopotattooattoo

Keyword tree potherotathxythopotattooattoo O(mn) p t t o h e h t a r e
p t t o h e h t a r e t a a 4 t t t o e o r 1 o 2 3 potherotathxythopotattooattoo O(mn) m: length of text. n: length of longest pattern

Keyword Tree with a failure link
p t t o h e h t a r e t a a 4 t t t o e o r 1 o 2 3 potherotathxythopotattooattoo

Keyword Tree with all failure links
p t t o h e h t a r e t a a 4 t t t o e o r 1 o 3 2

Example potherotathxythopotattooattoo p t t o h e h t a r e t a a 4 t
p t t o h e h t a r e t a a 4 t t t o e o r 1 o 3 2 potherotathxythopotattooattoo

Aho-Corasick algorithm
O(n) preprocessing, and O(m+k) searching. n: total length of patterns. m: length of text k is # of occurrence. Can create a DFA similar as in KMP. Requires more space, Preprocessing time depends on alphabet size Search time is constant

Suffix Tree All algorithms we talked about so far preprocess pattern(s) Karp-Rabin: small pattern, small alphabet Boyer-Moore: fastest in practice. O(m) worst case. KMP: O(m) Aho-Corasick: O(m) In some cases we may prefer to pre-process T Fixed T, varying P Suffix tree: basically a keyword tree of all suffixes

Suffix tree T: xabxac Suffixes: xabxac abxac bxac xac ac c
1 b b x x c 6 4 a a c c 5 2 3 Naïve construction: O(m2) using Aho-Corasick. Smarter: O(m). Very technical. big constant factor Difference from a keyword tree: create an internal node only when there is a branch

Suffix tree implementation
Explicitly labeling seq end T: xabxa T: xabxa$ x a x b a a x a b a x a $ 1 1 b b $ b b x x $ x x a 4 a a a $ 5 $ 2 3 2 3

Suffix tree implementation
Implicitly labeling edges T: xabxa$ 1:2 x a b 3:$ a x a 2:2 $ 1 1 b $ $ b $ $ x x 3:$ 3:$ 4 4 a a $ 5 5 $ 2 2 3 3

Suffix links Similar to failure link in a keyword tree
Only link internal nodes having branches x a b xabcf a b c f c d d e e f f g g h h i i j j

Suffix tree construction
acatgacatt 1:$ 1

acatgacatt 1:$ 2:$ 1 2

acatgacatt a 2:$ 2:$ 4:$ 3 1 2

acatgacatt a 4:$ 2:$ 2:$ 4:$ 4 3 1 2

5:$ acatgacatt 5 a 4:$ 2:$ 2:$ 4:$ 4 3 1 2

5:$ acatgacatt 5 a 4:$ c a 2:$ t 4:$ 4 t 5:$ $ 6 3 1 2

5:$ acatgacatt 5 a c 4:$ a c t a t 4:$ 4 t 5:$ 5:$ t $ 7 6 3 1 2

5:$ acatgacatt 5 a c 4:$ a c t t a t 4 t 5:$ 5:$ 5:$ t t $ 7 6 8 3 1 2

5:$ acatgacatt 5 t a c a t 5:$ c t a t 9 t 4 t 5:$ 5:$ 5:$ t t $ 7 6 8 3 1 2

5:$ acatgacatt 5 t a c $ 10 a c t 5:$ t a t 9 t 4 t 5:$ 5:$ 5:$ t t $ 7 6 8 3 1 2

ST Application 1: pattern matching
Find all occurrence of P=xa in T Find node v in the ST that matches to P Traverse the subtree rooted at v to get the locations x a b a x a c c c 1 b b x x c 6 4 a a c c 5 T: xabxac 2 3 O(m) to construct ST (large constant factor) O(n) to find v – linear to length of P instead of T! O(k) to get all leaves, k is the number of occurrence. Asymptotic time is the same as KMP. ST wins if T is fixed. KMP wins otherwise.

ST Application 2: set matching
Find all occurrences of a set of patterns in T Build a ST from T Match each P to ST x a b a x a c c c 1 b b x x c 6 4 a a c c 5 T: xabxac P: xab 2 3 O(m) to construct ST (large constant factor) O(n) to find v – linear to total length of P’s O(k) to get all leaves, k is the number of occurrence. Asymptotic time is the same as Aho-Corasick. ST wins if T fixed. AC wins if P’s are fixed. Otherwise depending on relative size.

ST application 3: repeat finding
Highly repeated substrings often have some meaning Computer virus replicates themselves after infection Poorly designed software may contain many duplicated code segments Genome contains repeated DNA sequences with interesting functions Length and number of repeats may vary Goal: find all repeats that are at least k-chars long and appear at least p times in a string

Repeats finding at least k-chars long and appear at least p times in a string Phase 1: top-down, count label lengths (L) from root to each node Phase 2: bottom-up: count # of leaves descended from each internal node For each node with L >= k, and N >= p, output all leave IDs O(m) to traverse tree (L, N)

Repeats finding Find repeats with at least 3 chars and 2 occurrences
acatgacatt 5 t a c $ 10 a 5:e c t t a t (3, 2) 9 t 4 (4, 2) t 5:e 5:e 5:e t t 7 6 8 3 1 2 Find repeats with at least 3 chars and 2 occurrences cat acat aca acatgacatt

Maximal repeats Right-maximal repeat Left-maximal repeat
S[i+1..i+k] = S[j+1..j+k], but S[i+k+1] != S[j+k+1] Left-maximal repeat S[i+1..i+k] = S[j+1..j+k] But S[i] != S[j] Maximal repeat But S[i] != S[j], and S[i+k+1] != S[j+k+1] acatgacatt 1. cat (right-maximal) 2. aca (left-maximal) 3. acat (maximal)

Maximal repeats finding
acatgacatt 5 t a c $ 10 a t 5:e c a t t 9 t 4 t 5:e 5:e 5:e t t 7 6 8 3 1 2 Left char = [] g c c a a How to find maximal repeat? A right-maximal repeats with different left chars

ST application 4: word enumeration
Find all length-k patterns that occur at least p times Compute (L, N) for each node L: total label length from root to node N: # leaves Find nodes v with L>=k, and L(parent)<k, and N>=y Traverse sub-tree rooted at v to get the locations L<k L=k L = K L>=k, N>=p

Joint Suffix Tree Build a ST for more than two strings
Two strings S1 and S2 S* = S1 & S2 Build a suffix tree for S* in time O(|S1| + |S2|) The separator will only appear in the edge ending in a leaf

S1 = abcd S2 = abca S* = abcd&abca$ & a b c d useless a d b c d & a c
2,4 a a 1,4 c a 2,3 b 2,1 c 2,2 1,1 d 1,3 1,2

To Simplify & a b c d useless a d & b b c d & a c a a d c b c $ c b b c d d d d c & a $ a & d a d 1,4 b 2,4 a a a 1,4 c a 2,3 2,4 a a b 2,1 1,1 2,3 c 2,1 2,2 1,1 d 1,3 2,2 1,2 1,3 1,2 We don’t really need to do anything, since all edge labels were implicit. The right hand side is more convenient to look at

Application of JST Longest common substring
For each internal node v, keep a bit vector B B[1] = 1 if a child of v is a suffix of S1 Find all internal nodes with B[1] = B[2] = 1 Report the one with the longest label Can be extended to k sequences. Just use a longer bit vector. Not subsequence a d b c d c b c $ d d 1,4 a a 2,4 a 1,1 2,3 2,1 1,3 2,2 1,2

Application of JST Given K strings, find all length-k patterns that appear in at least d strings L< k L >= k B = (1, 0, 1, 1) cardinal(B) >= d 4,x 1,x 3,x 3,x

Many other applications
Reproduce the behavior of Aho-Corasick Recognizing computer virus A database of known computer viruses Does a file contain virus? DNA finger printing A database of people’s DNA sequence Given a short DNA, which person is it from? … Catch Large constant factor for space requirement Large constant factor for construction Suffix array: trade off time for space

Summary One T, one P One T, many P One fixed T, many varying P
Boyer-Moore is the choice KMP works but not the best One T, many P Aho-Corasick Suffix Tree One fixed T, many varying P Suffix tree Two or more T’s Suffix tree, joint suffix tree, suffix array Alphabet independent Alphabet dependent

Pattern pre-processing algs
Karp – Rabin algorithm Small alphabet and small pattern Boyer – Moore algorithm The choice of most cases Typically sub-linear time Knuth-Morris-Pratt algorithm (KMP) Aho-Corasick algorithm The algorithm for the unix utility fgrep Suffix tree One of the most useful preprocessing techniques Many applications

Karp – Rabin Algorithm Let’s say we are dealing with binary numbers
Text: Pattern: Convert pattern to integer = 2^5 + 2^3 + 2^2 = 44

Karp – Rabin algorithm Text: 01010001011001010101001
Pattern: = 44 decimal = 2^ ^3 + 2^2 + 2^1 = 46 = 46 * 2 – 64*1 + 1 = 29 = 29 * 2 –64*0 + 1 = 59 = 59 * 2 – 64*1 + 0 = 54 = 54 * 2 – 64*1 + 0 = 44 Θ(m+n)

Karp – Rabin algorithm 10111011001010101001 = 46 (% 13 = 7)
What if the pattern is too long to fit into a single integer? Pattern: What if each word in our computer has only 4 bits? Basic idea: hashing. 44 % 13 = 5 = 46 (% 13 = 7) = 46 * 2 – = 29 (% 13 = 3) = 29 * = 59 (% 13 = 7) = 59 * = 54 (% 13 = 2) = 54 * = 44 (% 13 = 5) Θ(m+n) expected running time

Boyer – Moore algorithm
Three ideas: Right-to-left comparison Bad character rule Good suffix rule

Boyer – Moore algorithm
Right to left comparison x y Skip some chars without missing any occurrence. y But how?

Bad character rule 0 1 12345678901234567 T:xpbctbxabpqqaabpq
T:xpbctbxabpqqaabpq P: tpabxab *^^^^ What would you do now?

Bad character rule 0 1 12345678901234567 T:xpbctbxabpqqaabpq
T:xpbctbxabpqqaabpq P: tpabxab *^^^^ P: tpabxab

Bad character rule 0 1 123456789012345678 T:xpbctbxabpqqaabpqz
T:xpbctbxabpqqaabpqz P: tpabxab *^^^^ P: tpabxab * P: tpabxab

Basic bad character rule
tpabxab char Right-most-position in P a 6 b 7 p 2 t 1 x 5 Pre-processing: O(n)

k T: xpbctbxabpqqaabpqz P: tpabxab *^^^^ When rightmost T(k) in P is left to i, shift pattern P to align T(k) with the rightmost T(k) in P i = 3 Shift 3 – 1 = 2 P: tpabxab char Right-most-position in P a 6 b 7 p 2 t 1 x 5

k T: xpbctbxabpqqaabpqz P: tpabxab * When T(k) is not in P, shift left end of P to align with T(k+1) i = 7 Shift 7 – 0 = 7 P: tpabxab char Right-most-position in P a 6 b 7 p 2 t 1 x 5

k T: xpbctbxabpqqaabpqz P: tpabxab *^^ When rightmost T(k) in P is right to i, shift pattern P one pos i = 5 5 – 6 < 0. so shift 1 P: tpabxab char Right-most-position in P a 6 b 7 p 2 t 1 x 5

Extended bad character rule
k T: xpbctbxabpqqaabpqz P: tpabxab *^^ Find T(k) in P that is immediately left to i, shift P to align T(k) with that position i = 5 5 – 3 = 2. so shift 2 P: tpabxab char Position in P a 6, 3 b 7, 4 p 2 t 1 x 5 Preprocessing still O(n)

Extended bad character rule
Best possible: m / n comparisons Works better for large alphabet size In some cases the extended bad character rule is sufficiently good Worst-case: O(mn) What else can we do?

T:prstabstubabvqxrst P: qcabdabdab *^^
T:prstabstubabvqxrst P: qcabdabdab *^^ According to extended bad character rule

(weak) good suffix rule
T:prstabstubabvqxrst P: qcabdabdab *^^

(Weak) good suffix rule
T t Preprocessing: For any suffix t of P, find the rightmost copy of t, denoted by t’. How to find t’ efficiently? y P t’ t y P t’ t

(Strong) good suffix rule
T:prstabstubabvqxrst P: qcabdabdab *^^

T:prstabstubabvqxrst P: qcabdabdab *^^ P: qcabdabdab

In preprocessing: For any suffix t of P, find the rightmost copy of t, t’, such that the char left to t ≠ the char left to t’ z y P t’ t z ≠ y z y P t’ t Pre-processing can be done in linear time If P in T, searching may take O(mn) If P not in T, searching in worst-case is O(m+n)

Example preprocessing
qcabdabdab Bad char rule Good suffix rule char Positions in P a 9, 6, 3 b 10, 7, 4 c 2 d 8,5 q 1 q c a b d a b d a b dab cab Where to shift depends on T Does not depend on T

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