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Lecture overview ingle operation time complexity revious results ptimality proof F(k)-trees perations and complexity UF(k)-trees roperties and complexity
Lecture overview•Single operation time complexity
•Previous results
•Optimality proof
•UF(k)-trees
•Operations and complexity
•IUF(k)-trees
•Properties and complexity
Single operation complexityA data structure in which each UNION operation can be performed in O(k), and each FIND operation can be performed in time O(logkn).
k is a parameter, 2 k n .
Previous algorithms had operation complexity (log n).
If we choose k = log n / log log n,
UNION: O(k) = O(log n / log log n)
FIND: O(logkn) = O(log n / log log n)
Single operation complexity
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Union: O(k)
Find: O(logkn)
Optimal data structure
Theorem (Blum): Suppose that every UNION operation can be performed in O(k). Then there is a FIND operation that needs time
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Corollary: There is always either a UNION or a FIND operation that takes
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Optimal data structure
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UF(k)-treesLet k be an integer, 2 k n. A tree T is called a UF(k)-tree , if
The root of T has at most k sons.Each node in T has either 0 or more than k grandsons.
All leaves of T have the same depth.
UF(k)-tree structurename
sons
number of sons
height
leaf
UK(k)-tree structureElements are stored at the leaves
a b c d e f g h
To FIND an element, climb to the root
UK(k)-tree structure
a b c d e f
Initially all trees are of size 2.
UF(k)-trees - balanced
If level i contains any nodes, it contains at least ki/2 nodes.
The tree has at most n leaves
Height 2logkn
FIND : O(logkn)
UF(k)-trees - linear
UF(k)-trees - linear
UF(k) - case 1
Total number of sons k
UF(k) - case 1
Total number of sons k
UF(k) - case 2
Total number of sons > k
UF(k) - case 2
Total number of sons > k
UF(k) - case 3
UF(k) - case 3
UF(2)-trees - example
a b c d e f
UF(2)-trees - example
a b c d e f
UF(2)-trees - example
a b c d e f
UF(2)-trees - example
a b c d e f
UF(k) - case 1
Total number of sons k
UF(k) - case 2
Total number of sons > k
UF(k) - case 3
Complexity = O(k) = O(k+logkn)
IUF(k)-treesLet k be an integer, 2 k n. A tree T is called a IUF(k)-tree , if:
•The root of T has at most k sons.•Each node in T has either 0 or more than k grandsons.
IUF(k) trees - unbalanced
IUF(k) - linear size
IUF(k) - linear size
L - depth 0 (leaves)
M - depth 1
R - depth > 1
M L
#nodes (k+1)R
L+M+R (k+1)R
R (L+M)/K (L+M)/2 L
IUF(k) - case 1
Total number of sons k
IUF(k) - case 1
Total number of sons k
IUF(k) - case 2
Total number of sons > k
IUF(k) - case 2
Total number of sons > k
IUF(k) - case 3
IUF(k) - case 3
IUF(k) trees - balanced
It suffices to show than an IUF(k) tree has the same height as the corresponding UK(k) tree that stores the same set.
We perform the sequence of UNION operations on the two data structures in parallel.
•At each moment, the trees have the same height, and the roots have the same number of sons.
•If we are in case j for the UF structure, we are in case j for the IUF structure.
Case 3
Case 3
Summary
•Union: O(k) , Find: O(logkn)
•Optimality
•UF(k)-trees: balanced, linear.
•Three cases. Complexity: O(k+logkn)
•IUF(k)-trees. Linear, potentially unbalanced
•Similarity between data structures
References
•A Data Structure for the Union-Find Problem Having Good Single-Operation Complexity / Michiel H.M. Smid , 1989.
•Presentation: http://www.math.tau.ac.il/~ysh/seminar/uf.ppt
