# Space-Efficient Static Trees and Graphs Guy Jacobson IEEE Symposium on Foundations of Computer Science, 1989 Speaker: 吳展碩.

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Space-Efficient Static Trees and Graphs Guy Jacobson IEEE Symposium on Foundations of Computer Science, 1989 Speaker: 吳展碩

A pointer needs lg n bits to address n different locations.
1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 nil A pointer needs lg n bits to address n different locations. Using pointers to represent the linking relation of a graph will therefore occupy (n lg n) bits.

Outline To store a binary tree in asymptotically optimal space
Represent a tree in O(n) bits Efficient tree-traversal in space-efficient trees To store planar graphs in asymptotically optimal space

Binary Trees in 2n+1 bits Mark all the nodes of the tree with 1.
Add external nodes to the tree, and mark them all with 0. Read off the makes of nodes of the tree in level-order.

Binary Trees in 2n+1 bits Mark all the nodes of the tree with 1.
Add external nodes to the tree, and mark them all with 0. Read off the makes of nodes of the tree in level-order.

How to compute the linking relations in a space-efficient tree?

Rank and Select Define two operations rank(m) and select(m) as follows: rank(m): Counts the number of 1s from position 1 up to position m in a binary string select(m): Finds the m-th 1s in a binary string Example: rank(10) = 7 select(7) = 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

left-child(m)  2  rank(m) right-child(m)  2  rank(m)  1 parent(m)  select(m2) 1 1 2 3 2 3 4 5 6 7 4 5 6 8 9 10 11 12 13 7 8 14 15 16 17 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

How to compute rank(m) and select(m) efficiently? (e.g. constant time)

Compute Rank(m) and Select(m)
The rank directory: Conceptually break the bit-string into blocks of length lg2n. Keep a table containing the number of 1s up to the last position in each block. This takes n / lg n bits. Break each block into sub-blocks of length ½lg n. Keep a table containing the number of 1s within the block up to the last position in each sub-block. This takes n lglg2n / lg n bits. Keep a pre-computed table giving the number of 1s up to every possible position in every possible distinct sub-block.

n bits 1 2 3 n/lg2n 101100 110011 001... ..11 000010 101000 lg2n 3 7 55 56 lg n n / lg n bits ½lg n lglg2n 2n lglg2n / lg n bits

2½lg n  ½lg n  lg½lg n bits
1 2 3 n/lg2n 101100 110011 001... ..11 000010 101000 lg2n 3 7 55 56 lg n n / lg n bits ½lg n Precomputed Table 2½lg n  ½lg n  lg½lg n bits lglg2n 2n lglg2n / lg n bits

Planar Graphs in O(n) Space
Represent a special case of planar graphs called one-page graphs in O(n) bits k-page graphs can be represented in O(kn) bits Any planar graph can be embedded in a four-page graph. Yannakakis, M. "Four pages are necessary and sufficient for planar graphs." Proceedings of the 18th ACM Symposium on Theory of Computing, pages , 1986.

One-page graph One-page Graph:
All edges are lying to one side and can not cross.

One-page graph in O(n) Space
All edges are lying to one side and can not cross. | ((( | )( | ( | )) | )( | ))

How to compute the linking relations in the parenthesis string?

Finding the close parenthesis which match the open one
First, break the string of parentheses into blocks of length lgn.

Definitions Far parenthesis:
An open parenthesis p is called a far parenthesis if and only if p's matching parenthesis lies outside its own block. Pioneer: A far parenthesis is a pioneer if and only if its matching parenthesis lies in a different block that of the previous far parenthesis. The number of pioneers is at most 2lgn. Dotted lines denote the matches of far parentheses Red ones denote the matches of pioneers

Use a Directory Structure of Size O(n) bits for Matching Parentheses

Matching Parentheses For a parenthesis p, its matching parenthesis q can be found out as follows: Case1: p and q are in the same block Using precomputed table lgn

Matching Parentheses Case2: p and q are not in the same block
Find the pioneer of p and use it to locate the block containing q Compute the position of q via nesting depths

O(n) bits + number of pioneers  lg n bits

Conclusion References
A space-efficient data structure achieve almost optimal space while supporting the required operations almost as efficient as using pointers to represent it. References Dinesh P. Mehta and Sartaj Sahni. Handbook of Data Structures and Applications. Chapman & Hall/CRC, 2005.

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