Speaker: 吳展碩

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.

• Using pointers to represent the linking relation of a graph will therefore occupy (n lg n) bits.

1

2 3

4 5 6

7 8

1 2 3 4 5 6 7 8

2 3 4 5 nil 6 nil nil 7 nil 8 nil nil nil nil nil

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

1. Mark all the nodes of the tree with 1.

2. Add external nodes to the tree, and mark them all with 0.

3. Read off the makes of nodes of the tree in level-order.

Binary Trees in 2n+1 bits

1. Mark all the nodes of the tree with 1.

2. Add external nodes to the tree, and mark them all with 0.

3. Read off the makes of nodes of the tree in level-order.

Binary Trees in 2n+1 bits

• 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 stringselect(m): Finds the m-th 1s in a binary string

• Example:

rank(10) = 7select(7) = 10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

1 1 1 1 1 0 1 0 0 1 0 1 0 0 0 0 0

1 2 3 4 5 6 7 8

Algorithm to Compute Linking Relations

1

2 3

4 5 6

7 8

1

2 3

4 5

9 12

6 7

8 10 11 13

14 15 16 17

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

1 1 1 1 1 0 1 0 0 1 0 1 0 0 0 0 0

1 2 3 4 5 6 7 8

left-child(m) 2 rank(m)right-child(m) 2 rank(m) 1parent(m) select(m2)

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

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.

Compute Rank(m) and Select(m)

1 2 3 n/lg2n

101100 110011 001... ... ... ..11 000010 101000

0 3 7 ... ... ... ... 55 56

lg2n

lg n

0

½lg n

lglg2n

n / lg n bits

2n lglg2n / lg n bits

n bits

1 2 3 n/lg2n

101100 110011 001... ... ... ..11 000010 101000

0 3 7 ... ... ... ... 55 56

lg2n

lg n

0

½lg n

lglg2n

n / lg n bits

2n lglg2n / lg n bits

n bits

Precomputed Table

2½lg n ½lg n lg½lg n bits

Planar Graphs in O(n) Space• Represent a special case of planar graphs called one-p

age 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 gra

ph.Yannakakis, M. "Four pages are necessary and sufficient for planar graphs." Proceedings of the 18th ACM Symposium on Theory of Computing, pages 104-108, 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

• One-page Graph: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 bloc

ks 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 parenthesesRed ones denote the matches of pioneers

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

arentheses

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

O(n) bits

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

Conclusion

• A space-efficient data structureachieve 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 Structur

es and Applications. Chapman & Hall/CRC, 2005.