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 ???

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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
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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)

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Succinct Representations of Trees

• Start with Jacobson, then others
• There are about 4n/(pn)3/2 ordered rooted trees,
  and same number of binary trees
• Lower bound on specifying is about 2n bits
• What are the natural representations?

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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

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Heap-like Notation for a Binary Tree
Add external nodes Enumerate level by
level Store vector 11110111001000000
length2n1 (Here dont 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
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How do we Navigate?

• Jacobsons key suggestionOperations on a bit
  vector
• rank(x) 1s up to including x
• select(x) position of xth 1
• So in the binary tree
• leftchild(x) 2 rank(x)
• rightchild(x) 2 rank(x) 1
• parent(x) select(?x/2?)

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Rank Select

• Rank -Auxiliary storage 2nlglg n / lg n bits
• 1s up to each (lg n)2 rd bit
• 1s 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

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Other Combinatorial Objects

• Planar Graphs (Lu et al)
• Permutations n? n
• Or more generally
• Functions n ? n
• But what are the operations?
• Clearly p(i), but also p -1(i)
• And then p k(i) and p -k(i)
• Suffix Arrays (special permutations) in linear
  space

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Permutations a Shortcut Notation

• Let P be a simple array giving p Pi pi
• Also have Bi be a pointer t positions back in
  (the cycle of) the permutation
• Bi p-ti .. 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
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Representing Shortcuts

• In a cycle there is a B every t positions
• But these positions can be in arbitrary order
• Which is 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 Bi in B array
• So p(i) and p -1(i) in O(1) time (1e)n lg n
  bits
• Theorem Under a pointer machine model with space
  (1 e) n references, we need time 1/e to answer p
  and p -1 queries i.e. this is as good as it gets.

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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
• r1-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
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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
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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 p and
  p-1 in minimum space.
• Observe This method violates the pointer
  machine lower bound by using micropointers.

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Back to the main track Powers of p

• Consider the cycles of p
• ( 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 4 1 7)
• Ignoring parentheses, view as new permutation, ?.
• Note ?-1(i) is position containing i
• So we have ? and ?-1 as before
• Use ?-1(i) to find i, then bit vector (rank,
  select) to find pk or p-k

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Functions

• Now consider arbitrary functions n?n
• A function is just a hairy permutation
• All tree edges lead to a cycle

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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

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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

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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
• Dont waste time on trees with no answers

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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).

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Back to Text And Suffix Arrays

• Text T1..n over (a,b) (altltb)
• 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 6 isnt ..why?

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Ascending to Max

• M is a permutation so M-1 is its inverse
• i.e. M-1i says where i is in M
• Ascending-to-Max ? 1 ? i ? n-2
• M-1i lt M-1n and M-1i1 lt M-1n ? M-1i
  lt M-1i1
• M-1i gt M-1n and M-1i1 gt M-1n ? M-1i
  gt M-1i1
• 4 7 5 1 8 3 6 2 OK
• 4 7 1 5 8 2 3 6 NO

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Non-Nesting

• Non-Nesting? 1 ? i,j ? n-1 and M-1iltM-1j
• M-1i lt M-1i1 and M-1j lt M-1j1 ?
  M-1i1 lt M-1j1
• M-1i gt M-1i1 and M-1j gt M-1j1 ?
  M-1i1 lt M-1j1
• 4 7 5 1 8 3 6 2 OK
• 4 7 1 5 8 2 3 6 NO

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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

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Cardinality Queries

• T a b a a a b b a a a b a a b b
  
• Remember lengths longest run of as and of bs
• SA (broken by runs, but not stored explicitly)
• 8 3 9 4 12 1 10 5 1316 7 2 11
  156 14
• Ba, ? bit vector .. If SA-1i-1 in an a
  section store 1 in Ba,SA-1i, 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 similarly Bb,
  (Note these are reversed except at )
• Algorithm Count(T,P)
• s ? 1 e ?n i ? m
• while igt0 and s?e do
• if Pia 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-s1,0)
• Time O(length of query)

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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 Ti..n and T1,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

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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