Traversal of a subtree is slow, which affects prefix and range queries.

Transcription

1 Compac Tries Tries suffer from a large number nodes, Ω( R ) in he wors case. The space requiremen is large, since each node needs much more space han a single symbol. Traversal of a subree is slow, which affecs prefix and range queries. Compac ries reduce he number of nodes by replacing branchless pah segmens wih a single edge. Leaf pah compacion applies his o pah segmens leading o a leaf. The number of nodes is now O(dp(R)). Full pah compacion applies his o all pah segmens. Then every inernal node has a leas wo children. In such a ree, here is always more leaves han inernal nodes. Thus he number of nodes is O( R ). 41

2 Example 2.4: Compac ries for R = {po$, poao$, poery$, aoo$, empo$}. p o po $ aoo$ empo$ $ aoo$ empo$ ao$ ery$ ao$ ery$ The egde labels are facors of he inpu srings. Thus hey can be sored in consan space using poiners o he srings, which are sored separaely. In a fully compac rie, any subree can be raversed in linear ime in he number of leaves in he subree. Thus a prefix query for a sring S can be answered in O( S + r) ime, where r is he number of srings in he answer. 42

3 Ternary Tries The binary ree implemenaion of a rie suppors ordered alphabes bu awkwardly. Ternary rie is a simpler daa srucure based on symbol comparisons. Ternary rie is like a binary search ree excep: Each inernal node has hree children: smaller, equal and larger. The branching is based on a single symbol a a given posiion. The posiion is zero a he roo and increases along he middle branches. Ternary ries have varians similar o σ-ary ries: A basic ernary rie is a full represenaion of he srings. Compac ernary ries reduce space by compacing branchless pah segmens. 43

4 Example 2.5: Ternary ries for R = {po$, poao$, poery$, aoo$, empo$}. $ p o a o $ e r y a o o $ e m p o $ $ p o a o$ a oo$ ery$ empo$ $ p o a o$ a oo$ ery$ empo$ $ The sizes of ernary ries have he same asympoic bounds as he corresponding ries: O( R ), O(dp(R)) and O( R ). 44

5 A ernary rie is balanced if each lef and righ subree conains a mos half of he srings in is paren ree. The balance can be mainained by roaions similarly o binary search rees. b roaion d A B d b D E C D E A B C We can also ge reasonably close o balance by insering he srings in he ree in a random order. In a balanced ernary rie each sep down eiher moves he posiion forward (middle branch), or halves he number of srings remaining he he subree. Thus, in a balanced ernary rie soring n srings, any downward raversal following a sring S akes a mos O( S + log n) ime. 45

6 The ime complexiies of operaions in balanced ernary ries are he same as in he correspoding ries excep an addiional O(log n): Inserion, deleion, lookup and lcp query for a sring S akes O( S + log n) ime. Prefix query for a sring S akes O( S + log n + Q ) ime in an uncompac ernary rie and O( S + log n + Q ) ime in a compac ernary rie, where Q is he se of srings given as he resul of he query. In ries, where he child funcion is implemened using binary search ree, he ime complexiies could be O( S log σ), a muliplicaive facor insead of an addiive facor. 46

7 Sring Binary Search An ordered array is a simple saic daa srucure supporing queries in O(log n) ime using binary search. Algorihm 2.6: Binary search Inpu: Ordered se R = {k 1, k 2,..., k n }, query value x. Oupu: The number of elemens in R ha are smaller han x. (1) lef 0; righ n + 1 // final answer is in he range [lef..righ) (2) while righ lef > 1 do (3) mid lef + (righ lef)/2 (4) if k mid < x hen lef mid (5) else righ mid (6) reurn lef Wih srings as elemens, however, he query ime is O(m log n) in he wors case for a query sring of lengh m O(m + log n log σ n) on average for a random se of srings. 47

8 We can use he lcp comparison echnique o improve binary search for srings. The following is a key resul. Lemma 2.7: Le A B, B C be srings. Then lcp(b, B ) lcp(a, C). Proof. Le B min = min{b, B } and B max = max{b, B }. By Lemma 3.17, lcp(a, C) = min(lcp(a, B max ), lcp(b max, C)) lcp(a, B max ) = min(lcp(a, B min ), lcp(b min, B max )) lcp(b min, B max ) = lcp(b, B ) 48

9 During he binary search of P in {S 1, S 2,..., S n }, he basic siuaion is he following: We wan o compare P and S mid. We have already compared P agains S lef and S righ, and we know ha S lef P, S mid S righ. If we are using LcpCompare, we know lcp(s lef, P ) and lcp(p, S righ ). By Lemmas 1.18 and 2.7, lcp(p, S mid ) lcp(s lef, S righ ) = min{lcp(s lef, P ), lcp(p, S righ )} Thus we can skip min{lcp(s lef, P ), lcp(p, S righ )} firs characers when comparing P and S mid. 49

10 Algorihm 2.8: Sring binary search (wihou precompued lcps) Inpu: Ordered sring se R = {S 1, S 2,..., S n }, query sring P. Oupu: The number of srings in R ha are smaller han P. (1) lef 0; righ n + 1 (2) llcp 0; rlcp 0 (3) while righ lef > 1 do (4) mid lef + (righ lef)/2 (5) mlcp min{llcp, rlcp} (6) (x, mlcp) LcpCompare(S mid, P, mlcp) (7) if x = < hen lef mid; llcp mclp (8) else righ mid; rlcp mclp (9) reurn lef The average case query ime is now O(m + log n). The wors case query ime is sill O(m log n). 50

11 We can furher improve sring binary search using precompued informaion abou he lcp s beween he srings in R. Consider again he basic siuaion during sring binary search: We wan o compare P and S mid. We have already compared P agains S lef and S righ, and we know lcp(s lef, P ) and lcp(p, S righ ). The values lef and righ depend only on mid. In paricular, hey do no depend on P. Thus, we can precompue and sore he values LLCP [mid] = lcp(s lef, S mid ) RLCP [mid] = lcp(s mid, S righ ) 51

12 Now we know all lcp values beween P, S lef, S mid, S righ excep lcp(p, S mid ). The following lemma shows how o uilize his. Lemma 2.9: Le A B, B C be srings. (a) If lcp(a, B) > lcp(a, B ), hen B < B and lcp(b, B ) = lcp(a, B ). (b) If lcp(a, B) < lcp(a, B ), hen B > B and lcp(b, B ) = lcp(a, B). (c) If lcp(b, C) > lcp(b, C), hen B > B and lcp(b, B ) = lcp(b, C). (d) If lcp(b, C) < lcp(b, C), hen B < B and lcp(b, B ) = lcp(b, C). (e) If lcp(a, B) = lcp(a, B ) and lcp(b, C) = lcp(b, C), hen lcp(b, B ) max{lcp(a, B), lcp(b, C)}. Proof. Cases (a) (d) are symmerical, we show (a). B < B follows direcly from Lemma Then by Lemma 1.18, lcp(a, B ) = min{lcp(a, B), lcp(b, B )}. Since lcp(a, B ) < lcp(a, B), we mus have lcp(a, B ) = lcp(b, B ). In case (e), we use Lemma 1.18: lcp(b, B ) min{lcp(a, B), lcp(a, B )} = lcp(a, B) lcp(b, B ) min{lcp(b, C), lcp(b, C)} = lcp(b, C) Thus lcp(b, B ) max{lcp(a, B), lcp(b, C)}. 52

13 Algorihm 2.10: Sring binary search (wih precompued lcps) Inpu: Ordered sring se R = {S 1, S 2,..., S n }, arrays LLCP and RLCP, query sring P. Oupu: The number of srings in R ha are smaller han P. (1) lef 0; righ n + 1 (2) llcp 0; rlcp 0 (3) while righ lef > 1 do (4) mid lef + (righ lef)/2 (5) if LLCP [mid] > llcp hen lef mid (6) else if LLCP [mid] < llcp hen righ mid; rlcp LLCP [mid] (7) else if RLCP [mid] > rlcp hen righ mid (8) else if RLCP [mid] < rlcp hen lef mid; llcp RLCP [mid] (9) else (10) mlcp max{llcp, rlcp} (11) (x, mlcp) LcpCompare(S mid, P, mlcp) (12) if x = < hen lef mid; llcp mclp (13) else righ mid; rlcp mclp (14) reurn lef 53

14 Theorem 2.11: An ordered sring se R = {S 1, S 2,..., S n } can be preprocessed in O(dp(R)) ime and O(n) space so ha a binary search wih a query sring P can be execued in O( P + log n) ime. Proof. The values LLCP [mid] and RLCP [mid] can be compued in O(dp R (S mid )) ime. Thus he arrays LLCP and RLCP can be compued in O(dp(R)) ime and sored in O(n) space. The main while loop in Algorihm 2.10 is execued O(log n) imes and everyhing excep LcpCompare on line (11) needs consan ime. If a given LcpCompare call performs 1 + symbol comparisons, mclp increases by on line (11). Then on lines (12) (13), eiher llcp or rlcp increases by a leas, since mlcp was max{llcp, rlcp} before LcpCompare. Since llcp and rlcp never decrease and never grow larger han P, he oal number of exra symbol comparisons in LcpCompare during he binary search is O( P ). 54

15 Binary search can be seen as a search on an implici binary search ree, where he middle elemen is he roo, he middle elemens of he firs and second half are he children of he roo, ec.. The sring binary search echnique can be exended for arbirary binary search rees. Le S v be he sring sored a a node v in a binary search ree. Le S < and S > be he closes lexicographically smaller and larger srings sored a ancesors of v. The comparison of a query sring P and he sring S v is done he same way as he comparison of P and S mid in sring binary search. The roles of S lef and S righ are aken by S < and S >. If each node v sores he values lcp(s <, S v ) and lcp(s v, S > ), hen a search in a balanced search ree can be execued in O( P + log n) ime. Oher operaions including inserions and deleions ake O( P + log n) ime oo. 55

16 Auomaa Finie auomaa are a well known way of represening ses of srings. In his case, he se is ofen called a language. A rie is a special ype of an auomaon. Trie is generally no a minimal auomaon. Trie echniques including pah compacion and ernary branching can be applied o auomaa. Example 2.12: Compaced minimal auomaon for R = {po$, poao$, poery$, aoo$, empo$}. $ ery po ao a o $ emp 56

17 Auomaa are much more powerful han ries in represening languages: Infinie languages Nondeerminisic auomaa Even an acyclic, deerminisic auomaon can represen a language of exponenial size. Auomaa do no suppor all operaions of ries: Inserions and deleions Saellie daa, i.e., daa associaed o each sring. 57

18 3. Exac Sring Maching Le T = T [0..n) be he ex and P = P [0..m) he paern. We say ha P occurs in T a posiion j if T [j..j + m) = P. Example: P = aine occurs a posiion 6 in T = karjalainen. In his par, we will describe algorihms ha solve he following problem. Problem 3.1: Given ex T [0..n) and paern P [0..m), repor he firs posiion in T where P occurs, or n if P does no occur in T. The algorihms can be easily modified o solve he following problems oo. Is P a facor of T? Coun he number of occurrences of P in T. Repor all occurrences of P in T. 58

19 The naive, brue force algorihm compares P agains T [0..m), hen agains T [1..m + 1), hen agains T [2..m + 2) ec. unil an occurrence is found or he end of he ex is reached. Algorihm 3.2: Brue force Inpu: ex T = T [0... n), paern P = P [0... m) Oupu: posiion of he firs occurrence of P in T (1) i 0; j 0 (2) while i < m and j < n do (3) if P [i] = T [j] hen i i + 1; j j + 1 (4) else j j i + 1; i 0 (5) if i = m hen oupu j m else oupu n The wors case ime complexiy is O(mn). This happens, for example, when P = a m 1 b = aaa..ab and T = a n = aaaaaa..aa. 59

20 Knuh Morris Pra The Brue force algorihm forges everyhing when i moves o he nex ex posiion. The Morris Pra (MP) algorihm remembers maches. I never goes back o a ex characer ha already mached. The Knuh Morris Pra (KMP) algorihm remembers mismaches oo. Example 3.3: Brue force ainaisesi-ainainen ainai//nen (6 comp.) /ainainen (1) //ainainen (1) ai//nainen (3) /ainainen (1) //ainainen (1) Morris Pra ainaisesi-ainainen ainai//nen (6) ai//nainen (1) //ainainen (1) Knuh Morris Pra ainaisesi-ainainen ainai//nen (6) //ainainen (1) 60

21 MP and KMP algorihms never go backwards in he ex. When hey encouner a mismach, hey find anoher paern posiion o compare agains he same ex posiion. If he mismach occurs a paern posiion i, hen fail[i] is he nex paern posiion o compare. The only difference beween MP and KMP is how hey compue he failure funcion f ail. Algorihm 3.4: Knuh Morris Pra / Morris Pra Inpu: ex T = T [0... n), paern P = P [0... m) Oupu: posiion of he firs occurrence of P in T (1) compue f ail[0..m] (2) i 0; j 0 (3) while i < m and j < n do (4) if i = 1 or P [i] = T [j] hen i i + 1; j j + 1 (5) else i fail[i] (6) if i = m hen oupu j m else oupu n fail[i] = 1 means ha here is no more paern posiions o compare agains his ex posiions and we should move o he nex ex posiion. fail[m] is never needed here, bu if we waned o find all occurrences, i would ell how o coninue afer a full mach. 61

22 We will describe he MP failure funcion here. The KMP failure funcion is lef for he exercises. When he algorihm finds a mismach beween P [i] and T [j], we know ha P [0..i) = T [j i..j). Now we wan o find a new i < i such ha P [0..i ) = T [j i..j). Specifically, we wan he larges such i. This means ha P [0..i ) = T [j i..j) = P [i i..i). In oher words, P [0..i ) is he longes proper border of P [0..i). Example: ai is he longes proper border of ainai. Thus fail[i] is he lengh of he longes proper border of P [0..i). P [0..0) = ε has no proper border. We se fail[0] = 1. 62

23 Example 3.5: Le P = ainainen. i P [0..i) border f ail[i] 0 ε -1 1 a ε 0 2 ai ε 0 3 ain ε 0 4 aina a 1 5 ainai ai 2 6 ainain ain 3 7 ainaine ε 0 8 ainainen ε 0 The (K)MP algorihm operaes like an auomaon, since i never moves backwards in he ex. Indeed, i can be described by an auomaon ha has a special failure ransiion, which is an ε-ransiion ha can be aken only when here is no oher ransiion o ake. Σ a i n a i n e n

24 An efficien algorihm for compuing he failure funcion is very similar o he search algorihm iself! In he MP algorihm, when we find a mach P [i] = T [j], we know ha P [0..i] = T [j i..j]. More specifically, P [0..i] is he longes prefix of P ha maches a suffix of T [0..j]. Suppose T = #P [1..m), where # is a symbol ha does no occur in P. Finding a mach P [i] = T [j], we know ha P [0..i] is he longes prefix of P ha is a proper suffix of P [0..j]. Thus fail[j + 1] = i + 1. Algorihm 3.6: Morris Pra failure funcion compuaion Inpu: paern P = P [0... m) Oupu: array f ail[0..m] for P (1) i 1; j 0; fail[j] i (2) while j < m do (3) if i = 1 or P [i] = P [j] hen i i + 1; j j + 1; fail[j] i (4) else i fail[i] (5) oupu f ail When he algorihm reads fail[i] on line 4, fail[i] has already been compued. 64

25 Theorem 3.7: Algorihms MP and KMP preprocess a paern in ime O(m) and hen search he ex in ime O(n). Proof. I is sufficien o coun he number of comparisons ha he algorihms make. Afer each comparison P [i] = T [j] (or P [i] = P [j]), one of he wo condiional branches is execued: hen Here j is incremened. Since j never decreases, his branch can be aken a mos n + 1 (m + 1) imes. else Here i decreases since fail[i] < i. Since i only increases in he hen-branch, his branch canno be aken more ofen han he hen-branch. 65