SUBSTRING SEARCH BBM ALGORITHMS TODAY DEPT. OF COMPUTER ENGINEERING. Substring search applications. Substring search.

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1 M LGORITHMS TODY Substring search DPT. OF OMPUTR NGINRING rute force Knuth-Morris-Pratt oyer-moore Rabin-Karp SUSTRING SRH cknowledgement: The course slides are adapted from the slides prepared by R. Sedgewick and K. Wayne of Princeton University. 1 2 Substring search Substring search applications Goal. Find pattern of length M in a text of length N. Goal. Find pattern of length M in a text of length N. typically N >> M pattern N D L text I N H Y S T K N D L I typically N >> M N pattern N D L text I N H Y S T K N match D L I N match Substring search Substring search omputer forensics. Search memory or disk for signatures, e.g., all URLs or RS keys that the user has entered

2 Substring search applications Substring search applications Goal. Find pattern of length M in a text of length N. typically N >> M lectronic surveillance. Need to monitor all internet traffic. (security) pattern text N D L I N H Y S T K N D L I N No way! (privacy) match Well, we re mainly interested in TTK T DWN Identify patterns indicative of spam. PROFITS L0S W1GHT There is no catch. This is a one-time mailing. This message is sent in compliance with spam regulations. TTK T DWN substring search machine OK. uild a machine that ust looks for that. 5 found 5 Substring search applications Screen scraping. xtract relevant data from web page. x. Find string delimited by <b> and </b> after first occurrence of pattern Last Trade:. <tr> <td class= "yfnc_tablehead1" width= "48%"> Last Trade: </td> <td class= "yfnc_tabledata1"> <big><b>452.92</b></big> </td></tr> <td class= "yfnc_tablehead1" width= "48%"> Trade Time: </td> <td class= "yfnc_tabledata1">... Screen scraping: Java implementation Java library. The indexof() method in Java's string library returns the index of the first occurrence of a given string, starting at a given offset. public class StockQuote public static void main(string[] args) String name = " In in = new In(name + args[0]); String text = in.readll(); int start = text.indexof("last Trade:", 0); int from = text.indexof("<b>", start); int to = text.indexof("</b>", from); String price = text.substring(from + 3, to); StdOut.println(price); % ava StockQuote goog % ava StockQuote msft

3 rute force Knuth-Morris-Pratt oyer-moore Rabin-Karp SUSTRING SRH rute-force substring search heck for pattern starting at each text position. i i txt D R R pat R entries in red are mismatches R R entries in gray are for reference only R entries in black match the text R 4 10 R return i when is M match rute-force substring search: Java implementation heck for pattern starting at each text position. rute-force substring search: worst case rute-force algorithm can be slow if text and pattern are repetitive. i i D R D R D R public static int search(string pat, String txt) int M = pat.length(); int N = txt.length(); for (int i = 0; i <= N - M; i++) int ; for ( = 0; < M; ++) if (txt.chart(i+)!= pat.chart()) break; index in text where if ( == M) return i; pattern starts return N; not found i i txt pat rute-force substring search (worst case) match Worst case. ~ M N char compares

4 ackup In many applications, we want to avoid backup in text stream. Treat input as stream of data. bstract model: standard input. TTK T DWN substring search machine found rute-force algorithm needs backup for every mismatch. matched chars mismatch backup pproach 1. Maintain buffer of last M characters. pproach 2. Stay tuned. shift pattern right one position 13 rute-force substring search: alternate implementation Same sequence of char compares as previous implementation. i points to end of sequence of already-matched chars in text. stores number of already-matched chars (end of sequence in pattern). i D R 7 3 D R 5 0 D R public static int search(string pat, String txt) int i, N = txt.length(); int, M = pat.length(); for (i = 0, = 0; i < N && < M; i++) if (txt.chart(i) == pat.chart()) ++; else i -= ; = 0; if ( == M) return i - M; else return N; backup lgorithmic challenges in substring search SUSTRING SRH rute-force is not always good enough. Theoretical challenge. Linear-time guarantee. Practical challenge. void backup in text stream. fundamental algorithmic problem often no room or time to save text rute force Knuth-Morris-Pratt oyer-moore Rabin-Karp Now is the time for all people to come to the aid of their party. Now is the time for all good people to come to the aid of their party. Now is the time for many good people to come to the aid of their party. Now is the time for all good people to come to the aid of their party. Now is the time for a lot of good people to come to the aid of their party. Now is the time for all of the good people to come to the aid of their party. Now is the time for all good people to come to the aid of their party. Now is the time for each good person to come to the aid of their party. Now is the time for all good people to come to the aid of their party. Now is the time for all good Republicans to come to the aid of their party. Now is the time for all good people to come to the aid of their party. Now is the time for many or all good people to come to the aid of their party. Now is the time for all good people to come to the aid of their party. Now is the time for all good Democrats to come to the aid of their party. Now is the time for all people to come to the aid of their party. Now is the time for all good people to come to the aid of their party. Now is the time for many good people to come to the aid of their party. Now is the time for all good people to come to the aid of their party. Now is the time for a lot of good people to come to the aid of their party. Now is the time for all of the good people to come to the aid of their party. Now is the time for all good people to come to the aid of their attack at dawn party. Now is the time for each person to come to the aid of their party. Now is the time for all good people to come to the aid of their party. Now is the time for all good Republicans to come to the aid of their party. Now is the time for all good people to come to the aid of their party. Now is the time for many or all good people to come to the aid of their party. Now is the time for all good people to come to the aid of their party. Now is the time for all good Democrats to come to the aid of their party

5 Knuth-Morris-Pratt substring search Intuition. Suppose we are searching in text for pattern. Suppose we match 5 chars in pattern, with mismatch on th char. We know previous chars in text are. Don't need to back up text pointer! text after mismatch on sixth char brute-force backs pattern up to try this and this and this and this and this but no backup is needed i assuming, alphabet Knuth-Morris-Pratt algorithm. lever method to always avoid backup. (!) Deterministic finite state automaton (DF) DF is abstract string-searching machine. Finite number of states (including start and halt). xactly one transition for each char in alphabet. ccept if sequence of transitions leads to halt state. internal representation pat.chart() graphical representation, ,, If in state reading char c: if is halt and accept else move to state dfa[c][] DF simulation DF simulation pat.chart() pat.chart() ,, , ,, 19,

6 DF simulation DF simulation pat.chart() pat.chart() ,, , ,, 21, DF simulation DF simulation pat.chart() pat.chart() ,, , ,, 23,

7 DF simulation DF simulation pat.chart() pat.chart() ,, , ,, 25, DF simulation DF simulation pat.chart() pat.chart() ,, , ,, 27,

8 DF simulation DF simulation pat.chart() pat.chart() ,, , ,, 29, DF simulation DF simulation pat.chart() pat.chart() ,, , , substring found, 31,

9 Interpretation of Knuth-Morris-Pratt DF Q. What is interpretation of DF state after reading in txt[i]?. State = number of characters in pattern that have been matched. x. DF is in state 3 after reading in txt[0..]. txt, 7 8 suffix of text[0..] pat prefix of pat[] , i length of longest prefix of pat[] that is a suffix of txt[0..i] Knuth-Morris-Pratt substring search: Java implementation Key differences from brute-force implementation. Need to precompute dfa[][] from pattern. Text pointer i never decrements. public int search(string txt) int i,, N = txt.length(); for (i = 0, = 0; i < N && < M; i++) = dfa[txt.chart(i)][]; if ( == M) return i - M; else return N; Running time. Simulate DF on text: at most N character accesses. uild DF: how to do efficiently? [warning: tricky algorithm ahead] no backup, Knuth-Morris-Pratt substring search: Java implementation Key differences from brute-force implementation. Need to precompute dfa[][] from pattern. Text pointer i never decrements. ould use input stream. Knuth-Morris-Pratt construction Include one state for each character in pattern (plus accept state). public int search(in in) int i, ; for (i = 0, = 0;!in.ismpty() && < M; i++) = dfa[in.readhar()][]; if ( == M) return i - M; else return NOT_FOUND; no backup pat.chart() onstructing the DF for KMP substring search for , ,,

10 Knuth-Morris-Pratt construction Match transition. If in state and next char c == pat.chart(), go to +1. first characters of pattern next char matches now first +1 characters of have already been matched pattern have been matched Knuth-Morris-Pratt construction Mismatch transition: back up if c!= pat.chart(). pat.chart() pat.chart() onstructing the DF for KMP substring search for onstructing the DF for KMP substring search for, Knuth-Morris-Pratt construction Mismatch transition: back up if c!= pat.chart(). Knuth-Morris-Pratt construction Mismatch transition: back up if c!= pat.chart(). pat.chart() pat.chart() onstructing the DF for KMP substring search for onstructing the DF for KMP substring search for,, ,

11 Knuth-Morris-Pratt construction Mismatch transition: back up if c!= pat.chart(). Knuth-Morris-Pratt construction Mismatch transition: back up if c!= pat.chart(). pat.chart() pat.chart() onstructing the DF for KMP substring search for onstructing the DF for KMP substring search for,, , , 41, Knuth-Morris-Pratt construction Knuth-Morris-Pratt construction Mismatch transition: back up if c!= pat.chart(). pat.chart() pat.chart() onstructing the DF for KMP substring search for onstructing the DF for KMP substring search for,, , ,, 43,

12 How to build DF from pattern? Include one state for each character in pattern (plus accept state). How to build DF from pattern? Match transition. If in state and next char c == pat.chart(), go to +1. first characters of pattern have already been matched next char matches now first +1 characters of pattern have been matched pat.chart() pat.chart() How to build DF from pattern? Mismatch transition. If in state and next char c!= pat.chart(), then the last -1 characters of input are pat[1..-1], followed by c. To compute dfa[c][]: Simulate pat[1..-1] on DF and take transition c. Running time. Seems to require steps. still under construction (!) How to build DF from pattern? Mismatch transition. If in state and next char c!= pat.chart(), then the last -1 characters of input are pat[1..-1], followed by c. state X To compute dfa[c][]: Simulate pat[1..-1] on DF and take transition c. Running time. Takes only constant time if we maintain state X. x. dfa[''][5] = 1; dfa[''][5] = 4 simulate ; simulate ; take transition '' take transition '' = dfa[''][3] = dfa[''][3] pat.chart() x. dfa[''][5] = 1; dfa[''][5] = 4; from state X, from state X, take transition '' take transition '' = dfa[''][x] = dfa[''][x] X'= 0 from state X, take transition '' = dfa[''][x], simulation of, X , ,, 47,

13 Knuth-Morris-Pratt construction (in linear time) Include one state for each character in pattern (plus accept state). Knuth-Morris-Pratt construction (in linear time) Match transition. For each state, dfa[pat.chart()][] = +1. first characters of pattern now first +1 characters of have already been matched pattern have been matched pat.chart() pat.chart() onstructing the DF for KMP substring search for onstructing the DF for KMP substring search for Knuth-Morris-Pratt construction (in linear time) Mismatch transition. For state 0 and char c!= pat.chart(), set dfa[c][0] = 0. Knuth-Morris-Pratt construction (in linear time) Mismatch transition. For each state and char c!= pat.chart(), set dfa[c][] = dfa[c][x]; then update X = dfa[pat.chart()][x]. X = simulation of empty string pat.chart() pat.chart() onstructing the DF for KMP substring search for onstructing the DF for KMP substring search for,, X

14 Knuth-Morris-Pratt construction (in linear time) Mismatch transition. For each state and char c!= pat.chart(), set dfa[c][] = dfa[c][x]; then update X = dfa[pat.chart()][x]. Knuth-Morris-Pratt construction (in linear time) Mismatch transition. For each state and char c!= pat.chart(), set dfa[c][] = dfa[c][x]; then update X = dfa[pat.chart()][x]. X = simulation of X = simulation of pat.chart() pat.chart() onstructing the DF for KMP substring search for onstructing the DF for KMP substring search for,, , X X, Knuth-Morris-Pratt construction (in linear time) Mismatch transition. For each state and char c!= pat.chart(), set dfa[c][] = dfa[c][x]; then update X = dfa[pat.chart()][x]. Knuth-Morris-Pratt construction (in linear time) Mismatch transition. For each state and char c!= pat.chart(), set dfa[c][] = dfa[c][x]; then update X = dfa[pat.chart()][x]. X = simulation of X = simulation of pat.chart() pat.chart() onstructing the DF for KMP substring search for onstructing the DF for KMP substring search for,, X, X,, 55,

15 Knuth-Morris-Pratt construction (in linear time) Knuth-Morris-Pratt construction (in linear time) Mismatch transition. For each state and char c!= pat.chart(), set dfa[c][] = dfa[c][x]; then update X = dfa[pat.chart()][x]. X = simulation of pat.chart() pat.chart() onstructing the DF for KMP substring search for onstructing the DF for KMP substring search for,, X , ,, 57, onstructing the DF for KMP substring search: Java implementation For each state : opy dfa[][x] to for mismatch case. Set dfa[pat.chart()][] to +1 for match case. Update X. public KMP(String pat) this.pat = pat; M = pat.length(); dfa = new int[r][m]; dfa[pat.chart(0)][0] = 1; for (int X = 0, = 1; < M; ++) for (int c = 0; c < R; c++) dfa[c][] = dfa[c][x]; dfa[pat.chart()][] = +1; X = dfa[pat.chart()][x]; copy mismatch cases set match case update restart state KMP substring search analysis Proposition. KMP substring search accesses no more than M + N chars to search for a pattern of length M in a text of length N. Pf. ach pattern char accessed once when constructing the DF; each text char accessed once (in the worst case) when simulating the DF. Proposition. KMP constructs dfa[][] in time and space proportional to R M. Larger alphabets. Improved version of KMP constructs nfa[] in time and space proportional to M Running time. M character accesses (but space proportional to R M). KMP NF for

16 Knuth-Morris-Pratt: brief history Independently discovered by two theoreticians and a hacker. - Knuth: inspired by esoteric theorem, discovered linear-time algorithm - Pratt: made running time independent of alphabet size - Morris: built a text editor for the D 400 computer Theory meets practice. rute force Knuth-Morris-Pratt oyer-moore Rabin-Karp SUSTRING SRH SIM J. OMPUT. Vol., No. 2, June 1977 FST PTTRN MTHING IN STRINGS* DONLD. KNUTHf, JMS H. MORRIS, JR.:l: ND VUGHN R. PRTT bstract. n algorithm is presented which finds all occurrences of one. given string within another, in running time proportional to the sum of the lengths of the strings. The constant of proportionality is low enough to make this algorithm of practical use, and the procedure can also be extended to deal with some more general pattern-matching problems. theoretical application of the algorithm shows that the set of concatenations of even palindromes, i.e., the language can*, can be recognized in linear time. Other algorithms which run even faster on the average are also considered. Don Knuth Jim Morris Vaughan Pratt oyer Moore Intuition Scan the text with a window of M chars (length of pattern) Pattern in Text (M) Text oyer Moore Intuition (2) ase 3: Scan Window starts before the pattern starts Scan Window (M) ase 1: Scan Window is exactly on top of the searched pattern ase 4: Independent - Starting from one end check if all characters are equal. (We must check!) ase 2: Scan Window starts after the pattern starts. In case 4, simply shift window M characters void ase 2 onvert ase 3 to ase 1, by shifting appropriately

17 oyer Moore Intuition (3) If we can recognise the character in the scan window end-point, we can find how many characters to shift. D F G H d = 4 So, for example D is the 4th character, we must shift window 4 characters so that they overlap. D F G H oyer Moore Intuition (4) potential problem, the character in the text can repeat. For example, pattern = XXXX and the text is X X X X X X X X X X X Solution: be conservative, choose the instance with the least Shift (so we cannot miss the others). d = oyer Moore Intuition (5) X X X X X X X X X X X Search: XXXX So, for the example when it is at the endpoint we must shift for 2 characters. text: X we have a mismatch in last, now we must shift only once, so that we can check the configuration where the we found moves to middle. text: YXX we have a mismatch in Y, now we must shift 3 times as we know that the last 2 characters are in pattern and they can be repeating in the first 3 characters. oyer-moore: mismatched character heuristic Intuition. Scan characters in pattern from right to left. an skip as many as M text chars when finding one not in the pattern. - First we check the character in index pattern.length()-1 - It is N which is not, so we know that first 5 characters is not a match. Shift text 5 characters - S!= so shift 5, == so we can check for the pattern.length()-2, L!=N, skip 4. i text F I N D I N H Y S T K N D L I N 0 5 N D L pattern 5 5 N D L 11 4 N D L 15 0 N D L return i = 15 Mismatched character heuristic for right-to-left (oyer-moore) substring search

18 oyer-moore: mismatched character heuristic Q. How much to skip? ase 1. Mismatch character not in pattern. before i oyer-moore: mismatched character heuristic Q. How much to skip? ase 2a. Mismatch character in pattern. i before txt pat T L N D L txt pat N L N D L i i after after txt pat T L N D L txt pat N L N D L mismatch character 'T' not in pattern: increment i one character beyond 'T' mismatch character 'N' in pattern: align text 'N' with rightmost pattern 'N' oyer-moore: mismatched character heuristic Q. How much to skip? oyer-moore: mismatched character heuristic Q. How much to skip? ase 2b. Mismatch character in pattern (but heuristic no help). ase 2b. Mismatch character in pattern (but heuristic no help). i i before before txt pat L N D L txt pat L N D L i i aligned with rightmost? after txt pat L N D L txt pat L N D L mismatch character '' in pattern: align text '' with rightmost pattern ''? mismatch character '' in pattern: increment i by

19 oyer-moore: mismatched character heuristic oyer-moore: Java implementation Q. How much to skip?. Precompute index of rightmost occurrence of character c in pattern (-1 if character not in pattern). right = new int[r]; for (int c = 0; c < R; c++) right[c] = -1; for (int = 0; < M; ++) right[pat.chart()] = ; c N D L oyer-moore skip table computation right[c] D L M N public int search(string txt) int N = txt.length(); int M = pat.length(); int skip; for (int i = 0; i <= N-M; i += skip) skip = 0; for (int = M-1; >= 0; --) if (pat.chart()!= txt.chart(i+)) skip = Math.max(1, - right[txt.chart(i+)]); break; in case other term is nonpositive if (skip == 0) return i; return N; compute skip value match nother xample SRH FOR: XXXX X X X X X X X X X X X If the window scan points to an unrecognised character, we can skip past that character. For this example, for the initial step we first match X at the end, when check for previous character () which is not in the string we skip 3 steps. The X at the end, we matched can still be the first character of the pattern, so we do not skip that. oyer-moore: analysis Property. Substring search with the oyer-moore mismatched character heuristic takes about ~ N / M character compares to search for a pattern of length M in a text of length N. sublinear! Worst-case. an be as bad as ~ M N. i skip txt 0 0 pat oyer-moore-horspool substring search (worst case) oyer-moore variant. an improve worst case to ~ 3 N by adding a KMP-like rule to guard against repetitive patterns

20 rute force Knuth-Morris-Pratt oyer-moore Rabin-Karp SUSTRING SRH Rabin-Karp fingerprint search asic idea = modular hashing. ompute a hash of pattern characters 0 to M - 1. For each i, compute a hash of text characters i to M + i - 1. If pattern hash = text substring hash, check for a match. pat.chart(i) i % 997 = 13 txt.chart(i) i % 997 = % 997 = % 997 = % 997 = % 997 = % 997 = 929 match return i = % 997 = 13 asis for Rabin-Karp substring search fficiently computing the hash function Modular hash function. Using the notation ti for txt.chart(i), we wish to compute xi = ti R M-1 + ti+1 R M ti+m-1 R 0 (mod Q) Intuition. M-digit, base-r integer, modulo Q. fficiently computing the hash function hallenge. How to efficiently compute xi+1 given that we know xi. xi = ti R M 1 + ti+1 R M ti+m 1 R 0 xi+1 = ti+1 R M 1 + ti+2 R M ti+m R 0 Key property. an update hash function in constant time! xi+1 = ( xi t i R M 1 ) R + t i +M Horner's method. Linear-time method to evaluate degree-m polynomial. current value subtract leading digit multiply by radix add new trailing digit (can precompute R M 2 ) pat.chart() i % 997 = % 997 = (2*10 + ) % 997 = % 997 = (2*10 + 5) % 997 = % 997 = (25*10 + 3) % 997 = 59 R % 997 = (59*10 + 5) % 997 = 13 Q // ompute hash for M-digit key private long hash(string key, int M) long h = 0; for (int = 0; < M; ++) h = (R * h + key.chart()) % Q; return h; i current value text new value current value subtract leading digit * 1 0 multiply by radix add new trailing digit new value

21 Rabin-Karp substring search example Rabin-Karp: Java implementation i % 997 = 3 Q % 997 = (3*10 + 1) % 997 = % 997 = (31*10 + 4) % 997 = % 997 = (314*10 + 1) % 997 = % 997 = (150*10 + 5) % 997 = 508 RM R % 997 = (( *(997-30))*10 + 9) % 997 = % 997 = (( *(997-30))*10 + 2) % 997 = % 997 = (( *(997-30))*10 + ) % 997 = % 997 = (( *(997-30))*10 + 5) % 997 = 442 match % 997 = (( *(997-30))*10 + 3) % 997 = return i-m+1 = % 997 = (( *(997-30))*10 + 5) % 997 = 13 public class RabinKarp private long pathash; private int M; private long Q; private int R; private long RM; public RabinKarp(String pat) M = pat.length(); R = 25; Q = longrandomprime(); RM = 1; for (int i = 1; i <= M-1; i++) RM = (R * RM) % Q; pathash = hash(pat, M); // pattern hash value // pattern length // modulus // radix // R^(M-1) % Q private long hash(string key, int M) /* as before */ a large prime (but avoid overflow) precompute R M 1 (mod Q) Rabin-Karp substring search example public int search(string txt) /* see next slide */ Rabin-Karp: Java implementation (continued) Monte arlo version. Return match if hash match. public int search(string txt) check for hash collision int N = txt.length(); using rolling hash function int txthash = hash(txt, M); if (pathash == txthash) return 0; for (int i = M; i < N; i++) txthash = (txthash + Q - RM*txt.chart(i-M) % Q) % Q; txthash = (txthash*r + txt.chart(i)) % Q; if (pathash == txthash) return i - M + 1; return N; Las Vegas version. heck for substring match if hash match; continue search if false collision. Rabin-Karp analysis Theory. If Q is a sufficiently large random prime (about M N 2 ), then the probability of a false collision is about 1 / N. Practice. hoose Q to be a large prime (but not so large as to cause overflow). Under reasonable assumptions, probability of a collision is about 1 / Q. Monte arlo version. lways runs in linear time. xtremely likely to return correct answer (but not always!). Las Vegas version. lways returns correct answer. xtremely likely to run in linear time (but worst case is M N)

22 Rabin-Karp fingerprint search dvantages. xtends to 2d patterns. xtends to finding multiple patterns. Disadvantages. rithmetic ops slower than char compares. Las Vegas version requires backup. Poor worst-case guarantee. Substring search cost summary ost of searching for an M-character pattern in an N-character text. algorithm version operation count guarantee typical backup in input? correct? brute force M N 1.1 N yes yes 1 Knuth-Morris-Pratt full DF (lgorithm 5. ) mismatch transitions only extra space 2 N 1.1 N no yes MR 3 N 1.1 N no yes M full algorithm 3 N N / M yes yes R oyer-moore Rabin-Karp mismatched char heuristic only (lgorithm 5.7 ) Monte arlo (lgorithm 5.8 ) M N N / M yes yes R 7 N 7 N no yes 1 Las Vegas 7 N 7 N no yes yes 1 probabilisitic guarantee, with uniform hash function