1 graph algorithms minimum spanning trees (mst) union - find dana shapira

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Graph Algorithms

Minimum Spanning Trees (MST)Union - Find

Dana Shapira

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A spanning tree of G is a subset T E of edges, such that the sub-graph G'=(V,T) is connected and acyclic.

Spanning tree

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Minimum Spanning Tree

Given a graph G = (V, E) and an assignment of weights w(e) to the edges of G, a minimum spanning tree T of G is a spanning tree with minimum total edge weight

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How To Build A Minimum Spanning Tree

General strategy:1. Maintain a set of edges A such that (V, A) is a spanning

forest of G and such that there exists a MST (V, F) of G such that AF.

2. As long as (V, A) is not a tree, find an edge that can be added to A while maintaining the above property.

Generic-MST(G=(V,E))1. A= ;2. while (A is not a spanning tree of G) do3. choose a safe edge e=(u,v)E4. A=A{e}5. return A

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Cuts

A cut (X, Y) of a graph G = (V, E) is a partition of the vertex set V into two sets X and Y = V \ X.

An edge (v, w) is said to cross the cut (X, Y) if v X and w Y.

A cut (X, Y) respects a set A of edges if no edge in A crosses the cut.

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A Cut Theorem

Theorem: Let A be a subset of the edges of some minimum spanning tree of G; let (X, Y) be a cut that respects A; and let e be a minimum weight edge that crosses(X, Y). Then A {e} is also a subset of the edges of a minimum spanning tree of G; edge e is safe.

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Theorem: Let A be a subset of the edges of some minimum spanning tree of G; let (X, Y) be a cut that respects A; and let e be a minimum weight edge that crosses(X, Y). Then A {e} is also a subset of the edges of a minimum spanning tree of G; edge e is safe.

A Cut Theorem

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A Cut Theorem

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Theorem: Let A be a subset of the edges of some minimum spanning tree of G; let (X, Y) be a cut that respects A; and let e be a minimum weight edge that crosses(X, Y). Then A {e} is also a subset of the edges of a minimum spanning tree of G; edge e is safe.

A Cut Theorem

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A Cut Theorem

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T w(e) ≤ w(f)w(e) ≤ w(f)

w(T') ≤ w(T)

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Let T be a MST such that AT. If e = (u,v) T, add e to T. The edge e = (u,v) forms a cycle with edges on the path p from u

to v in T. Since u and v are on opposite sides of the cut, there is at least one edge f = (x,y) in T on the path p that also crosses the cut.

f A since the cut respects A. Since f is on the unique path from u to v in T, removing it breaks T into two components.

w(e) ≤ w(f) (why?) Let T ' = T – {f} {e} w(T ') ≤ w(T).

Proof:

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Proof: The cut (VC, V–VC) respects A, and e is a light edge for this cut. Therefore, e is safe.

Corollary: Let G=(V,E) be a connected undirected graph and A a subset of E included in a minimum spanning tree T for G, and let C=(VC,EC) be a tree in the forest GA=(V,A). If e is a light edge connecting C to some other component in GA, then e is safe for A.

A Cut Theorem

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Kruskal’s Algorithm

Kruskal(G)1 A ← ∅2 for every edge e = (v, w) of G, sorted by weight3 do if v and w belong to different connected components of (V, A)4 then add edge e to A

(a, d):1 (h, i):1 (c, e):1 (f, h):2 (g, h):2(b, c):3 (b, f):3 (b, e):4 (c, d):5 (f, g):5(e, i):6 (d, g):8 (a, b):9 (c, f):12

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Correctness Proof

ei

ei

Sorted edge sequence: e1, e

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3, e

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i, e

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i + 2, e

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Every edge ej that cross the cut have a weight w(e

j) ≥ w(e

i).

Hence, edge ei is safe.

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Union-Find Data Structures

Given a set S of n elements, maintain a partition of S into subsets S

1, S

2, …, S

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Support the following operations:Union(x, y): Replace sets S

i and S

j such that x S

i and y S

j

with Si S

j in the current partition

Find(x): Returns a member r(Si) of the set S

i that contains x

In particular, Find(x) and Find(y) return the same element if and only if x and y belong to the same set.It is possible to create a data structure that supports the above operations in O(α(n)) amortized time, where α is the inverse Ackermann function.

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Kruskal’s Algorithm Using Union-Find Data Structure

Kruskal(G,w)A for each vertex vV do

Make-Set(v)sort the edges in E in non-decreasing weight order wfor each edge (u,v)E do

if Find-Set(u) ≠ Find-Set(v) then A A {(u,v)}

Union(u,v) return A

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Kruskal’s Algorithm Using Union-Find Data Structure

Analysis:O(|E| log |E|) time for everything except the operations on SCost of operations on S:

O(α(|E|,|V|)) amortized time per operation on S|V| – 1 Union operations|E| Find operationsTotal: O((|V| + |E|)α(|E|,|V|)) running time

Total running time: O(|E| lg |E|).

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Prim’s AlgorithmPrim(G)1 for every vertex v of G2 do label v as unexplored3 for every edge e of G4 do label e as unexplored and non tree edge5 s ← some vertex of G6 Mark s as visited7 Q ← Adj(s)8 while Q is not empty9 do (u, w) ← DeleteMin(Q)10 if (u, w) is unexplored11 then if w is unexplored12 then mark edge (u, w) as tree edge13 mark vertex w as visited14 Insert(Q, Adj(w))

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Correctness Proof

Observation: At all times during the algorithm, the set of tree edges defines a tree that contains all visited vertices; priority queue Q contains all unexplored edges incident to these vertices.

Corollary: Prim’s algorithm constructs a minimum spanning tree of G.

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Union/Find

Assumptions: The Sets are disjoint. Each set is identified by a representative of the set.

Initial state: A union/find structure begins with n elements, each considered

to be a one element set. Functions:

Make-Set(x): Creates a new set with element x in it. Union(x,y): Make one set out of the sets containing x and y. Find-Set(x): Returns a pointer to the representative of the set

containing x.

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Basic Notation

The elements in the structure will be numbered 0 to n-1 Each set will be referred to by the number of one of the

element it contains Initially we have sets S0,S1,…,Sn-1

If we were to call Union(S2,S4), these sets would be removed from the list, and the new set would now be called either S2 or S4

Notations: n Make-Set operations m total operations nm

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First Attempt

Represent the Union/Find structure as an array arr of n elements

arr[i] contains the set number of element i Initially, arr[i]=i (Make-Set(i))

Find-Set(i) just returns the value of arr[i] To perform Union(Si,Sj):

For every k such that arr[k]=j, set arr[k]=i

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Analysis

The worst-case analysis: Find(i) takes O(1) time Union(Si,Sj) takes (n) time

A sequence of n Unions will take (n2) time

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Second Attempt

Represent the Union/Find structure using linked lists. Each element points to another element of the set. The representative is the first element of the set. Each element points to the representative. How do we perform Union(Si,Sj)?

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Analysis

The worst-case analysis: Find(i) takes O(1) time Make-Set(i) takes O(1) time Union(Si,Sj) takes (n) time (Why?)

A sequence of n Unions-Find will take (n2) time (Example?)

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Up-Trees A simple data structure for implementing disjoint sets is the up-

tree. We visualize each element as a node A set will be visualized as a directed tree

Arrows will point from child to parent The set will be referred to by its root

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H, A and W belong to the same set. H is the representative

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X, B, R and F are in the same set. X is the representative

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Operations in Up-Trees

Follow pointer to representative element.

find(x) {

if (x≠p(x)) // not the representative

then p(x)find(p(x));return p(x);

}

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Union

Union is more complicated.

Make one representative element point to the other, but which way?Does it matter?

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Union(H, X)

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W B

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X points to H

B, R and F are now deeper

H points to X

A and W are now deeper

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A worst case for Union

Union can be done in O(1), but may cause find to become O(n)

A B C D E

Consider the result of the following sequence of operations:

Union (A, B)Union (C, A)Union (D, C)Union (E, D)

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Array Representation of Up-tree

Assume each element is associated with an integer i=0…n-1. From now on, we deal only with i.

Create an integer array, A[n] An array entry is the element’s parent A -1 entry signifies that element i is the representative element.

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Array Representation of Up-tree

Now the union algorithm might be:Union(x,y) {

A[y] = x; // attaches y to x}

The find algorithm would befind(x) {

if (A[x] < 0)return(x);

elsereturn(find(A[x]));

}Performance: ???

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Analysis

Worst case: Union(Si,Sj) take O(1) time Find(i) takes O(n) time

Can we do better in an amortized analysis? What is the maximum amount of time n operations could

take us? Suppose we perform n/2 unions followed by n/2 finds

The n/2 unions could give us one tree of height n/2-1 Thus the total time would be n/2 + (n/2)(n/2) = O(n2)

This strategy doesn’t really help…

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Array Representation of Up-tree

There are two heuristics that improve the performance of union-find. Union by weight Path compression on find

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Union by Weight HeuristicMake-Set(x) {

p(x)x;rank(x)=0;

}Always attach smaller tree to larger.

union(x,y) {LINK(FIND-Set(x),Find-Set(y))

}LINK(x,y){

if (rank(x) > rank(y)) { p(y)x;else p(x)y;if(rank(x)=rank(y)){

rank(x) = rank(y)+1;}

}

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Union by Weight HeuristicLet’s change the weight from rank to number of nodes: union(x,y) {

rep_x = find(x);rep_y = find(y);if (weight[rep_x] < weight[rep_y]) {

A[rep_x] = rep_y;weight[rep_y] += weight[rep_x];

}else {

A[rep_y] = rep_x;weight[rep_x] += weight[rep_y];

}}

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Implementation

Still represent this with an array of n nodes If element i is a “root node”, then arr[i]= -s, where s is

the size of that set. Otherwise, arr[i] is the index of i’s “parent”

If arr[i] < arr[j], then set arr[i] to arr[i]+arr[j] and set arr[j]to i

Else, set arr[j] to arr[i]+arr[j] and set arr[j] to i

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Implementation

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Performance w/ Union by Weight If unions are done by weight, the depth of any element is never

greater than lg N. Initially, every element is at depth zero. When its depth increases as a result of a union

operation (it’s in the smaller tree), it is placed in a tree that becomes at least twice as large as before (union of two equal size trees).

How often can each union be done? -- lg n times, because after at most lg n unions, the tree will contain all n elements.

Therefore, find becomes O(lg n) when union by weight is used.

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New Bound on h

Theorem:Assume we start with a Union/Find structure where each set has 1 node, and perform a sequence of Weighted Unions. Then any tree T of m nodes has a height no greater than log2 m.

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Proof

Base case: If m=1, then this is clearly true Assumption: Assume it is true for all trees of size m-1 or less Proof: Let T be a tree of m nodes created by a sequence of

Weighted Unions. Consider the last union: Union(Sj,Sk). Assume Sj is the smaller tree. If Sj has a nodes, then Sk has m-a nodes, and 1 a m/2.

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Proof (continued)

The height of T is either:

The height of Tk

One more than the height of Tj

Since a m-a m-1, the assumptions applies to both Tk and Tj

If T has the height of Tk, then

h log2(m-a) log2m If T is one greater than the height of Tj:

h log2a+1 log2m/2)+1 log2m

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Example

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Is the bound tight?Yes: “pair them off”

Union(S0,S1), Union(S2,S3), Union(S4,S5), Union(S6,S7), Union(S0,S2), Union(S4,S6), Union(S0,S4)

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Example

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Analysis

Worst case: Union is still O(1) Find is now O(log n)

Amortized case: A “worst amortized case” can be achieved if we perform n/2

unions and n/2 finds Take O(n log n) time

Conclusion: This is better, but we can improve it further

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Path Compression Each time we do a find on an element x, we make all elements on path

from root to x be immediate children of root by making each element’s parent be the representative.find(x) {

if (A[x]<0)return(x);

A[x] = find(A[x]);return (A[x]);

} When path compression is done, a sequence of m operations takes

O(m lg n) time. Amortized time is O(lg n) per operation.

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Find(7)

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Analysis

The worst case analysis does not change In fact, we are going to have to increase the worst-case

time of Find by a constant factor The amortized analysis does get better

we need to define Ackerman’s function

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Performance with Both Optimizations When both optimizations are performed, for a sequence

of m operations (m n) (unions and finds), it takes no more than O(m lg* n) time. lg*n is the iterated (base 2) logarithm of n. The

number of times you take lg n before n becomes 1. Example:

lg*16=3 lg*65536=4 lg*265536=5

Union-find is essentially O(m) for a sequence of m operations (Amortized O(1)).

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*log 2 1n

n

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Ackerman’s Function

Ackerman’s A(i,j) is defined as follows:

A(1,j) = 2j 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 Some values:

A(2,1) = 4 A(2,2) = 16 A(2,3) = 65536 A(2,4) = > 10^100 (probably much greater than this) A(4,4) = 2^(2^(2^65536)) - 3

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(p,q)

The function (m,n) is related to the inverse of Ackerman function

For m n 1:

(m,n) = min{z 1 | A(z, m/n) > log2n} This is a very slow growing function:

(m,n) 3 if n < 216

n has to be huge to make (m,n) = 4

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Analysis

Lemma (Tarjan and Van Leeuwen): If we start with a Union/Find structure with one element per set (n elements), and let T(f,u) be the time to process any sequence of f finds and u unions. If u n/2 then

T(f,u) = (n+(f+n,n)) Translation: This is basically linear

In practice, you will never have n large enough to give (f+n,n) a value of even 4