Minimum Spanning Trees (MST) Graph Algorithms Minimum Spanning Trees (MST) Union - Find Dana Shapira

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

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 1 3 3 6 6 9 1 7 5 8 3 2 7 4

How To Build A Minimum Spanning Tree General strategy: 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 AF. 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)) A= ; while (A is not a spanning tree of G) do choose a safe edge e=(u,v)E A=A{e} return A

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.

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. 1 4 9 3 2

A Cut Theorem 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. 1 9 2 4 3 4

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. 12 9 4

A Cut Theorem u e e v f T w(e) ≤ w(f) w(e) ≤ w(f) w(T') ≤ w(T)

Proof: Let T be a MST such that AT. 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).

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

Kruskal’s Algorithm Kruskal(G) 1 A ← ∅ 2 for every edge e = (v, w) of G, sorted by weight 3 do if v and w belong to different connected components of (V, A) 4 then add edge e to A a a 9 9 1 1 b b d d 3 3 4 4 c c 5 5 (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 e e 1 1 3 3 12 12 8 8 6 6 f f 5 5 i i 2 2 g g 2 2 1 1 h h

Correctness Proof ei ei Sorted edge sequence: e1, e2, e3, e4, e5, e6, …, ei, ei + 1, ei + 2, ei + 3, …, en Every edge ej that cross the cut have a weight w(ej) ≥ w(ei). Hence, edge ei is safe.

Union-Find Data Structures Given a set S of n elements, maintain a partition of S into subsets S1, S2, …, Sk Support the following operations: Union(x, y): Replace sets Si and Sj such that x  Si and y  Sj with Si  Sj in the current partition Find(x): Returns a member r(Si) of the set Si 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.

Kruskal’s Algorithm Using Union-Find Data Structure Kruskal(G,w) A   for each vertex vV do Make-Set(v) sort the edges in E in non-decreasing weight order w for each edge (u,v)E do if Find-Set(u) ≠ Find-Set(v) then A  A  {(u,v)} Union(u,v) return A

Kruskal’s Algorithm Using Union-Find Data Structure Analysis: O(|E| log |E|) time for everything except the operations on S Cost of operations on S: O(α(|E|,|V|)) amortized time per operation on S |V| – 1 Union operations |E| Find operations Total: O((|V| + |E|)α(|E|,|V|)) running time Total running time: O(|E| lg |E|).

Prim’s Algorithm Prim(G) 1 for every vertex v of G 2 do label v as unexplored 3 for every edge e of G 4 do label e as unexplored and non tree edge 5 s ← some vertex of G 6 Mark s as visited 7 Q ← Adj(s) 8 while Q is not empty 9 do (u, w) ← DeleteMin(Q) 10 if (u, w) is unexplored 11 then if w is unexplored 12 then mark edge (u, w) as tree edge 13 mark vertex w as visited 14 Insert(Q, Adj(w)) a 9 1 b d 3 4 c 5 e 1 3 12 8 6 f 5 i 2 g 2 1 h

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.

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.

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 nm

First Attempt Initially, arr[i]=i (Make-Set(i)) 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

Analysis Find(i) takes O(1) time Union(Si,Sj) takes (n) time 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

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)?

Analysis Find(i) takes O(1) time Make-Set(i) takes O(1) time 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?)

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 H X F A W B R H, A and W belong to the same set. H is the representative X, B, R and F are in the same set. X is the representative

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); } Example: V={a,b,c,d}, E ={} union(a,b) union(c,a) union(d,c)

Union Union is more complicated. Make one representative element point to the other, but which way? Does it matter?

Union(H, X) H X F X points to H B, R and F are now deeper A W B R H X H points to X A and W are now deeper A W B R

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)

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. Example: V = {0,1,2,3,4,5,6,7} array: -1, -1, -1, -1, -1, -1, -1, -1, -1 union(4,5) union(6,7) union(4,6) now, E={(5,4),(7,6),(6,4)} array: -1, -1, -1, -1, -1, 4, 4, 6

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 be find(x) { if (A[x] < 0) return(x); else return(find(A[x])); Performance: ???

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…

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

Union by Weight Heuristic Make-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; Example: before: -1, 0, 0, -1, 3 union(0,3) after: -1, 0, 0, 0, 3

Union by Weight Heuristic Let’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]; Example: before: -1, 0, 0, -1, 3 union(0,3) after: -1, 0, 0, 0, 3

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

Implementation 1 2 3 4 5 6 7 8 9 -2 -1 8 1 8 2 3 4 2 7 4 4 6 5 9 8 -7 7

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. Mathematical proof available in Tom A’s notes

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.

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.

Proof (continued) h  log2(m-a)  log2m The height of Tk 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

Example 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) 1 2 3 4 5 6 7

Example 2 4 6 1 3 5 7

Example 4 1 2 5 6 3 7

Example 1 2 4 3 5 6 7

Analysis Worst case: Amortized 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

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. Example: Starting array: -1, -1, -1, -1, -1, 4, 4, 6 find(7) find(6) find(4) resulting array: -1, -1, -1, -1, -1, 4, 4, 4

Find(7) 1 2 4 1 2 4 6 7 3 5 6 3 5 7

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

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)). Example: N lg*N 2 1 lg(2) = 1 4 2 lg(lg(4)) = 1 8 3 lg(lg(lg(8)=3)=1.58) = .664 16 3 lg(lg(lg(16) = 4)=2)=1 65536 4 lg(lg(lg(lg(65536)=16)=4)=2)=1 2^65536 5 lg*(2^65536) = 1 + lg*(65536)

Ackerman’s Function A(1,j) = 2j for j  1 A(i,1) = A(i-1,2) for i  2 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

(p,q) (m,n) = min{z  1 | A(z, m/n) > log2n} 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

Analysis T(f,u) = (n+(f+n,n)) 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