ANNOUNCEMENTS

HW4 out on web, due 04/20/1999 in class
Reading - Chapter 9
 
 

REVIEW

Disjoint set - solving equivalence problem
What is equivalence problem?  Given equivalence relation ~ defined on set S, a, bES, a~b?
What is equivalence relation?

• Reflexive aRa
• Symmetric aRb iff bRa
• Transitive aRb iff bRc

Approach

• Elements all suparate disjoint set
• State equivalencee ==> union
• Do find - find(a) == find(b)? a~b?

• Use trees to represent sets
• Root is used to name set
• Only need parent info -> can use an array
• S[i] = parent of element i = -1 if i == root (special case)

TODAY

Implement unions (find)
Smarter algorithm
Running time
Example

Union by size - smaller tree subtree of larger tree, S[root] = -size

[ We provided the code to implement this algorithm, as went through an illustrative example ]

Alternative - union by height

• Make shallower tree subtree of larger tree
• Height only changes when 2 equally deep trees joined
• Store 1-height in root

• Path compression
• When do find(x), tough call nodes from x to root
• Change every nodes parent to be root
• We need to modify find

OPTIMIZATIONS

Union by size
Union by height
Path compression

Can we combine these methods?

• Union by size and path compression compatable
• Union by height and path compressino not compatable because we change height

ANALYSIS

Analysis of union/find algorithm

• Hard
• Algorithm almost linear (O(M)) in worst case
• Specifically O(M*p*(M,N))  where p*(M,N) is the functional inverse of Akerman's function

• A(i,j) = 2^j for j>=1
• A(i,1) = A(i-1,2) for i>=2
• A(i,j) = A(i-1, A(i,j-1)) for i,j >= 2

Grows very fast, thus inverse grows very slowly
p*(M,N) grows slower than log^*N, where log^*N is the number of times you log N until N<=1

• log^*(2^6536) = 5, even though it is a 20,000 digit number, but p*(M,N) grows slower

BUILDING MAZES

• A by B maze
• Each cell is an element
• Initially walls between all elements
• Randomly break down walls - union
• Stop when find(first) = find(last)