Succinct Data Structures

1. Succinct Data Structures Ian Munro University of Waterloo Joint work with David Benoit, Andrej Brodnik, D, Clark, F. Fich, M. He, J. Horton, A. López-Ortiz, S. Srinivasa Rao, Rajeev Raman, Venkatesh Raman, Adam Storm … How do we encode a large tree or other combinatorial object of specialized information … even a static one in a small amount of space and still perform queries in constant time ???

2. Example of a Succinct Data Structure: The (Static) Bounded Subset Given: Universe of n elements [0,...n-1] and m arbitrary elements from this universe Create: a static structure to support search in constant time (lg n bit word and usual operations) Using: Essentially minimum possible # bits ... Operation: Member query in O(1) time (Brodnik & M.)

3. Focus on Trees .. Because Computer Science is .. Arbophilic - Directories (Unix, all the rest) - Search trees (B-trees, binary search trees, digital trees or tries) - Graph structures (we do a tree based search) - Search indices for text (including DNA)

4. A Big Patricia Trie / Suffix Trie 0 1 • Given a large text file; treat it as bit vector • Construct a trie with leaves pointing to unique locations in text that “match” path in trie (paths must start at character boundaries) • Skip the nodes where there is no branching ( n-1 internal nodes) 0 1 • 1 0 0 0 1 1

5. Space for Trees Abstract data type: binary tree Size:n-1 internal nodes, n leaves Operations: child, parent, subtree size, leaf data Motivation: “Obvious” representation of an n node tree takes about 6 n lg n words (up, left, right, size, memory manager, leaf reference) i.e. full suffix tree takes about 5 or 6 times the space of suffix array (i.e. leaf references only)

6. Succinct Representations of Trees Start with Jacobson, then others: There are about 4n/(πn)3/2 ordered rooted trees, and same number of binary trees Lower bound on specifying is about 2n bits What are the natural representations?

7. Arbitrary Ordered Trees • Use parenthesis notation • Represent the tree • As the binary string (((())())((())()())):traverse tree as “(“ for node, then subtrees, then “)” • Each node takes 2 bits

8. Heap-like Notation for a Binary Tree Add external nodes Enumerate level by level Store vector 11110111001000000 length2n+1 (Here don’t know size of subtrees; can be overcome. Could use isomorphism to flip between notations) 1 1 1 1 0 1 1 1 0 0 0 0 1 0 0 0 0

9. How do we Navigate? Jacobson’s key suggestion:Operations on a bit vector rank(x) = # 1’s up to & including x select(x) = position of xth1 So in the binary tree leftchild(x) = 2 rank(x) rightchild(x) = 2 rank(x) + 1 parent(x) = select(x/2)

10. Rank & Select Rank -Auxiliary storage ~ 2nlglg n / lg n bits #1’s up to each (lg n)2rd bit #1’s within these too each lg nth bit Table lookup after that Select -more complicated but similar notions Key issue: Rank & Select take O(1) time with lg n bit word (M. et al) Aside: Interesting data type by itself

11. Other Combinatorial Objects Planar Graphs (Lu et al) Permutations [n]→ [n] Or more generally Functions [n] → [n] But what are the operations? Clearly π(i), but also π-1(i) And then πk(i) and π-k(i) Suffix Arrays (special permutations) in linear space

12. Permutations: a Shortcut Notation Let P be a simple array giving π; P[i] = π[i] Also have B[i] be a pointer t positions back in (the cycle of) the permutation; B[i]= π-t[i] .. But only define B for every tth position in cycle. (t is a constant; ignore cycle length “round-off”) So array representation P = [8 4 12 5 13 x x 3 x 2 x 10 1] 1 2 3 4 5 6 7 8 9 10 11 12 13 2 4 5 13 1 8 3 12 10

13. Representing Shortcuts In a cycle there is a B every t positions … But these positions can be in arbitrary order Which i’s have a B, and how do we store it? Keep a vector of all positions 0 indicates no B 1 indicates a B Rank gives the position of B[“i”] in B array So: π(i) and π-1(i) in O(1) time & (1+ε)n lg n bits Theorem: Under a pointer machine model with space (1+ ε) n references, we need time 1/ε to answer πandπ-1 queries; i.e. this is as good as it gets.

14. Getting n lg n Bits: an Aside This is the best we can do for O(1) operations But using Benes networks: 1-Benes network is a 2 input/2 output switch r+1-Benes network … join tops to tops 1 2 3 4 5 6 7 8 3 5 7 8 1 6 4 2 R-Benes Network R-Benes Network

15. A Benes Network Realizing the permutation (3 5 7 8 1 6 4 2) 1 2 3 4 5 6 7 8 3 5 7 8 1 6 4 2

16. What can we do with it? Divide into blocks of lg lg n gates … encode their actions in a word. Taking advantage of regularity of address mechanism and also Modify approach to avoid power of 2 issue Can trace a path in time O(lg n/(lg lg n) This is the best time we are able get for π and π-1 in minimum space. Observe: This method “violates” the pointer machine lower bound by using “micropointers”.

17. Back to the main track: Powers of π Consider the cycles of π ( 2 6 8)( 3 5 9 10)( 4 1 7) Keep a bit vector to indicate the start of each cycle ( 2 6 8 3 5 9 10 41 7) Ignoring parentheses, view as new permutation, ψ. Note: ψ-1(i) is position containing i … So we have ψ and ψ-1 as before Use ψ-1(i) to find i, then bit vector (rank, select) to find πkorπ-k

18. Functions Now consider arbitrary functions [n]→[n] “A function is just a hairy permutation” All tree edges lead to a cycle

19. Challenges here Essentially write down the components in a convenient order and use the n lg n bits to describe the mapping (as per permutations) To get fk(i): Find the level ancestor (k levels up) in a tree Or Go up to root and apply f the remaining number of steps around a cycle

20. Level Ancestors There are several level ancestor techniques using O(1) time and O(n) WORDS. Adapt Bender & Farach-Colton to work in O(n) bits But going the other way …

21. f-k is a set Moving Down the tree requires care f-3( ) = ( ) The trick: Report all nodes on a given level of a tree in time proportional to the number of nodes, and Don’t waste time on trees with no answers

22. Final Function Result Given an arbitrary function f: [n]→[n] With an n lg n + O(n) bit representation we can compute fk(i) in O(1) time and f-k(i) in time O(1 + size of answer).

23. Back to Text … And Suffix Arrays Text T[1..n] over (a,b)*#(a<#<b) There are 2n-1 such texts, which of the n! suffix arrays are valid? 1 2 3 4 5 6 7 8 SA=4 7 5 1 8 3 6 2 is a b b a a b a # SA-1=4 8 6 1 3 7 2 5 M=4 7 1 5 8 2 3 6isn’t ..why?

24. Ascending to Max M is a permutation so M-1 is its inverse i.e. M-1[i] says where i is in M Ascending-to-Max:  1  i  n-2 • M-1[i] < M-1[n] and M-1[i+1] < M-1[n] M-1[i] < M-1[i+1] • M-1[i] > M-1[n] and M-1[i+1] > M-1[n] M-1[i] > M-1[i+1] 4 7 5 1 8 3 6 2 OK 4 7 1 5 82 3 6 NO

25. Non-Nesting Non-Nesting: 1  i,j  n-1 andM-1[i]<M-1[j] • M-1[i] < M-1[i+1] and M-1[j] < M-1[j+1] M-1[i+1] < M-1[j+1] • M-1[i] > M-1[i+1] and M-1[j] > M-1[j+1] M-1[i+1] < M-1[j+1] 4 7 5 1 8 3 6 2 OK 4 7 1 582 36NO

26. Characterization Theorem for Suffix Arrays on Binary Texts Theorem:Ascending to Max & Non-nesting  Suffix Array Corollary: Clean method of breaking SA into segments Corollary: Linear time algorithm to check whether SA is valid

27. Cardinality Queries T= a b a a a b b a a a b a a b b # Remember lengths longest run of a’s and of b’s SA (broken by runs, but not stored explicitly) 8 3 | 9 4 12| 1 10 5 13|16 | 7 2 11 15|6 14 Ba, bit vector .. If SA-1[i-1] in an “a” section store 1 in Ba,[SA-1[i]], else 0 Ba 0 0 1 1 0 0 1 1 1 0 0 1 1 0 1 1 Create rank structure on Ba, and similarlyBb, (Note these are reversed except at #) Algorithm Count(T,P) s ← 1; e ←n; i ← m; while i>0 and se do if P[i[=a then s← rank1(Ba,s-1)+1; e←rank1(Ba,e) else s← na + 2 + rank1(Bb,s-1); e←na + 1 +rank1(Bb,e) i ← i-1 Return max(e-s+1,0) Time: O(length of query)

28. Listing Queries Complex methods Key idea: for queries of length at least d, index every dth position .. For T and forT(reversed) So we have matches for T[i..n] and T[1,i-1] View these as points in 2 space (Ferragina & Manzini and Grossi & Vitter) Do a range query (Alstrup et al) Variety of results follow

29. General Conclusion Interesting, and useful, combinatorial objects can be: Stored succinctly … O(lower bound) +o() So that Natural queries are performed in O(1) time (or at least very close) This can make the difference between using them and not …