From Suffix Trie to Suffix Tree

Transcription

1 Outline Exact String Matching Suffix tree an extremely powerful data structure for string algorithms Input: P and S. Output: All occurrences of P in S. Time: O( P + S ) Technique: Z values of PS. Z(i + P ) P iff P = S[i i + P ]. i+ P i+ P +d- P S 200/5/4 200/5/4 2 To P or not to P... Preprocessing P Gusfield Boyer-Moore Knuth-Morris-Pratt Preprocessing S Suffix tree From Suffix Trie to Suffix Tree 200/5/ /5/4 4

2 Suffixes of S S = a S[ 8]= a S[2 8]= a S[3 8]= a S[4 8]= a S[5 8]= a S[6 8]= a S[7 8]= S[8 8]= st suffix 2nd suffix 3rd suffix 4th suffix 5th suffix 6th suffix 7th suffix 8th suffix KEY: P occurs in S iff P is a prefix of a suffix of S. S = a S[ 8]= a S[2 8]= a S[3 8]= a S[4 8]= a S[5 8]= a S[6 8]= a S[7 8]= S[8 8]= st suffix 2nd suffix 3rd suffix 4th suffix 5th suffix 6th suffix 7th suffix 8th suffix 200/5/ /5/4 6 T = Suffix Trie of S Why suffix trie? a a a a a a The following statements are equivalent. P occurrs in S. P is a prefix of a suffix of S. P corresponds to a path of T starting from the root of T. 200/5/ /5/4 8 2

3 P = a P = a a a a a a a a a a a a a a P occurs in S! P doesn t occur in S! 200/5/ /5/4 0 P = a a a a a a a a Q: Where does P occur in S? P doesn t occur in S! 200/5/4 200/5/4 2 3

4 Output: P = a a a a a a a a Solving the Sustring Prolem using suffix trie? Question: If we are given a suffix trie of S, what is the time complexity for finding an occurrence of P in S? Answer: O( P ) time. 200/5/ /5/4 4 Constructing ti Suffix Trie Q: Time complexity for constructing the suffix trie T of S? ( S ) ( S log S ) ( S 2 ) ( S 3 ) 6 Time = O( S 2 ) a a a a a a /5/ /5/4 6 4

5 Worst-case time = ( S 2 ) S = a a a How to estalish a lower ound? Answer: Showing an example which takes ( S 2 ) time for any algorithm. 200/5/ /5/4 8 Summary Suffix trie is good in solving Sustring Prolem, ut may require ( S 2 ) time and space. Question: is there a compact representation of suffix trie that needs only O( S ) time and space? Suffix Tree A compact representation of suffix trie 200/5/ /5/4 20 5

6 Oservations on Trie T of S S = a a a T has at most S leaves. Why? T has at most S ranching nodes. Why? Keeping leaves and ranching nodes only. compact representation of edge laels [4,8] [2,2] [3,3] [,] ] [5,8] [5,8] [5,8] [5,8] 200/5/ /5/4 22 S = a a a S = a [,] [5,8] [2,2] [5,8] [3,3] [5,8] [4,8] [5,8] 200/5/ /5/4 24 6

7 S = a S = a [3,3] [,] [3,3] [,] [7,8] [3,3] [2,3] [3,3] [,] [7,8] [7,8] [4,8] [3,3] [2,3] [4,8] [7,8] [7,8] [4,8] [3,3] [2,3] [7,8] [4,8] [7,8] [4,8] [4,8] [7,8] [4,8] [7,8] [4,8] [4,8] 200/5/ /5/4 26 Question The space complexity of suffix tree O( S ) O( S log S ) O( S 2 ) O( S 3 ) Why? Numer of nodes = O( S ). Numer of edges = O( S ). Space required y each edge = O(). The challenge Constructing Suffix Tree in Linear Time 200/5/ /5/4 28 7

8 History of Suffix Tree Algorithms [Weiner, IEEE FOCS 973] Linear time ut expensive in space. D. E. Knuth: the algorithm of 973. [McCreight, J. ACM 976] Linear time and quadratic space. [Ukkonen, Algorithmica 995] Linear time and linear space. Much etter readaility. Esko Ukkonen, University of Helsinki, Finland 200/5/ /5/4 30 Ukkonen s Algorithm Let T(k) denote the suffix tree of S[ k]. Framework compute T(); for k = 2 to S do compute T(k) from T(k-); S iterations k-th iteration has k steps S = a S[ 8]= a S[2 8]= a S[3 8]= a S[4 8]= a S[5 8]= a S[6 8]= a S[7 8]= S[8 8]= 200/5/ /5/4 32 8

9 Ukkonen s approach on Suffix Trie a a a a a a The corresponding process of growing suffix tree S = a [3,3] [3,3] [3,4] [3,5] [3,6] [,] [7,7] [7,8] [4,7] [4,8] [3,3] [,] [2,3] [,2] [3,3] [3,4] [3,5] [3,6] [2,3] [2,4] [2,5] [2,6] [7,7] [7,8] [4,7] [4,8] Oservation: The tree structure does not change very often, i.e., only O( S ) times. [7,7] [7,8] [4,7] [4,8] 200/5/ /5/4 34 Oservation At the eginning of the k-th iteration, there are exactly k growing points ( 生長點 ), all with different heights. The k-th iteration iteratively grows k edges with lael S[k] at those k growing points. Ukkonen s approach on Suffix Trie a a a a a a 200/5/ /5/4 36 9

10 Growing Suffix Trie Three cases while growing trie Case : growing an edge at a leaf. Case 2: growing a new ranch of leaf. Case 3: does not change the tree structure. Ukkonen s approach on Suffix Trie a a a a a a Case : 長此以往 200/5/ /5/4 38 三階段定理 Those k steps in the k-th iteration have the following pattern: some (at least one) Case- steps, followed y some (could e zero) Case-2 steps, followed y some (could e zero) Case-3 steps. Ukkonen s approach on Suffix Trie a a a a a a Case : 長此以往 200/5/ /5/4 40 0

11 Proving 三階段定理 () 在同一個 iteration 之中, 長此以往 之前 ( 下 ) 的 step 一定也是 長此以往 At the end of each iteration, if a growing point is a leaf, then all previous (lower) growing points are also leaves. Why? An illustration a a a a a a 200/5/ /5/4 42 Key: what does a leaf growing point mean? The i-th growing point = the end of the i-th suffix S[i k] of the current prefix S[ k]. Argument: the i-th (i > ) growing point is a leaf. Neither S[i k]a nor S[i...k] is a sustring of S[ k]. Neither S[i k]a nor S[i...k] is a sustring of S[ k]. The (i )-st growing point is a leaf. An illustration a a a a a a 200/5/ /5/4 44

12 What does a Case- Step mean? At the eginning of the current iteration, its corresponding growing point is a leaf. An illustration a a a a a a 200/5/ /5/4 46 三階段定理 (): 長此以往 之前一定是 長此以往 At the end of each iteration, if a growing point is a leaf, then all previous (lower) growing points are also leaves. Therefore, in the next iteration, if a step is a 長此以往, then all previous steps are also 長此以往. Proving 三階段定理 (2) 在同一個 iteration 之中, 若無其事 之後 ( 上 ) 的 step 一定也是 若無其事 At the end of each iteration, if a growing point is an internal node, then all latter (higher) growing points are also internal. Why? 200/5/ /5/4 48 2

13 An illustration a a a a a a Key: what does an internal growing point mean? The i-th growing point = the end of the i-th suffix S[i k] of the current prefix S[ k]. Argument: the i-th (i < k) growing point is internal. S[i k]a or S[i...k] is a sustring of S[ k]. S[i+ k]a or S[i+...k] is a sustring of S[ k]. The (i+)-st growing point is internal. 200/5/ /5/4 50 An illustration a a a a a a What does a Case-3 Step mean? At the end of the current iteration, its corresponding growing point is still an internal node. 200/5/ /5/4 52 3

14 An illustration a a a a a a 三階段定理 (2): 若無其事 之後一定是 若無其事 At the end of each iteration, if a growing point is an internal node, then all latter (higher) growing points are also internal. Therefore, in this iteration, if a step is 若無其事, then all its following steps are also 若無其事. 200/5/ /5/4 54 三階段定理 得證 Those k steps in the k-th iteration have the following pattern: some (at least one) Case- steps, followed y some (could e zero) Case-2 steps, followed y some (could e zero) Case-3 steps. This is aout the status within the same iteration The fate of growing points in different iterations 葉子的宿命 一日為葉子, 終身為葉子 200/5/ /5/4 56 4

15 Ukkonen s approach on Suffix Trie a a a a a a Case : 長此以往 Oservation Only O( S ) Case-2 steps throughout the execution. Why? 200/5/ /5/4 58 What s remaining directly working on suffix tree 葉子生長點之所對應的 edge lael 斷頭指標 Taking a closer look Case- Step 200/5/ /5/4 60 5

16 Closer look at Case- step An idea for Case- steps Always occurs at a leaf (growing point). At the eginning of iteration k, the path aove a leaf has lael [?, k ]. Each Case- stepin iteration k simply changes the lael [?, k ] to [?, k]. [?, k ] [?, k] Use [?, -] to lael the path aove each leaf. Thus, no need to do anything for each case- step. [?, -] 200/5/ /5/4 62 Using [?,-] for each leaf Saving a lot of efforts S = a [3,3] [3,-] [,] [7,-] [4,-] [3,3] [,-] [2,3] [3,-] [2,-] [7,-] [4,-] 2 [7,-] [4,-] We can simply ignore all Case- steps. Recall that the numer of Case-2 steps is at most S. Q: Is this good enough? a a a a a a Case : 長此以往 200/5/ /5/4 64 6

17 Closer look at Case-2 step Taking a closer look Case-2 Steps: 節外生枝 Always occurs at a growing point that is not a leaf, which is not necessarily an internal node. When the growing point is an internal node Lael = [k, -], where k is the current iteration index. [k, -] 200/5/ /5/4 66 Closer look at Case-2 step When the growing point is not an internal node [i, i+t-] [i, j] [k, -] [i+t, j] t Taking a closer look Case-3 Steps: 若無其事 200/5/ /5/4 68 7

18 Closer look at Case-3 step Closer look at Case-3 step Always occurs at a growing point that is not a leaf, which is not necessarily an internal node. When the new position of the growing point is an internal node [i, j] t 0 When the new position of the growing point is not an internal node [i, j] t t + 200/5/ /5/4 70 Exercise Give a sequence S such that the numer of Case-3 steps throughout the process of growing its suffix trie (or suffix tree) is ( S 2 ). How does Ukkonen overcome the prolem of too many Case-3 ( 若無其事 ) steps? Completely ignore them 以 若無其事 的態度處理 若無其事 的狀況 200/5/ /5/4 72 8

19 Why Case-3 steps? Why Case-3 steps? [i, j] t t + [i, j] t 0. For correctly [i, i+t-] maintaining the position of each [i, j] growing point. [k, -] (Why?) [i+t, j] t 2. For correctly running Case-2 steps. (By what?) 200/5/ /5/4 74 Saving even more efforts What if Saving the ook keeping efforts in all Case-3 steps a a a a a a Case : 長此以往 200/5/ /5/4 76 9

20 Rough idea Only one growing point Just keep one current growing point throughout the execution. Deriving the new position of the current growing point from its previous position (with the helpusing suffix links ( 斷頭指標 ) The challenges: How do we derive the position of the current growing point? Vertically (case 2) Horizontally y( (case 3) Q: Which one is easier? a a a a a a 200/5/ /5/4 78 Horizontally, Horizontal movement Moving from iteration k to iteration k. The growing point does not move! This is the easier case a a a a a a a a a a a a Case : 長此以往 200/5/ /5/

21 Vertically, Moving from Step i to Step i+ in the same iteration. The growing point moves dramatically. This is the tougher case a a a a a a Vertical movement a a a a a a Case : 長此以往 200/5/ /5/4 82 前人種樹後人涼 斷頭指標 (suffix link) 前人種樹後人涼 的哲學 每次千辛萬苦找到 vertical movement 的目的時, 把這個 movement 的起點與終點用一個 link 記錄下來. 下回遇到這個起點時, 就可以直接走到終點去, 不用再重新找一次. 這些 link 就叫做 suffix link ( 斷頭指標 ). 200/5/ /5/4 84 2

22 為何稱為 斷頭指標? 斷頭指標 的性質 () 終點所對應的字串, 是起點所對應之字串的 斷頭字串 (second suffix) a a a a a a 每個 斷頭指標 的起點一定是個 internal node, 不會葉子 不會是某個 suffix tree edge 的中間 Why? a a a a a a 200/5/ /5/4 86 斷頭指標 的性質 (2) 運用 斷頭指標 每個 internal node 一定是某個 斷頭指標 的起點 Why? a a a a a a S = a [3,3] [3,-] [,] [4,-] [7,-] [3,3] [,-] [2,3] [3,-] [2,-] [7,-] [4,-] 2 [7,-] [4,-] 200/5/ /5/

23 Traversal with the help of suffix links: phase () Goinguptoaclosest internal node (whose suffix link must e availale). Suppose this upward traversal passes through t characters. Following the suffix link that starts from this internal node. [i, j] t Traversal with the help of suffix links: phase (2) Going down y matching the t-character sustring S[i, i + t ] of S. [i, j] t 200/5/ /5/4 90 Running Time? Overall Time = O( S ) Naïvely: O(t). Cleverly: O(+ d), where d is the numer of internal nodes eing went through during phase (2). [i, j] t Suppose d i is the d in the i-th Case-2-step traversal. It suffices to show d +d 2 + +d S =O( S ) a a a a a a 200/5/ /5/

24 = the slack of the growing point The slack means the distance etween a position P and the closest internal node aove P. [i, j] t case-3 traversal Each case-3 traversal (i.e., horizontal movement) can only increase the value of y at most one. (It can even decrease the value of.) a a a a a a 200/5/ /5/4 94 case-2 traversal d +d 2 + +d S = O( S ) The i-th case-2 traversal (i.e., vertical movement) decreases the value of y at least d i a a a a a a Initial = O(). can e increased y one for at most S times (ecause there are at most S horizontal movements (i.e., case-3 traversals). Since is always non-negative, the aove ound is proved. 200/5/ /5/

25 運用 斷頭指標 S = a [3,3] [3,-] [,] [4,-] [7,-] [3,3] [,-] [2,3] [3,-] [2,-] [7,-] [4,-] 2 [7,-] [4,-] 200/5/