1. www.monash.edu.au 1 prepared from lecture material © 2004 Goodrich & Tamassia COMMONWEALTH OF AUSTRALIA Copyright Regulations 1969 WARNING This material has been reproduced and communicated to you by or on behalf of Monash University pursuant to Part VB of the Copyright Act 1968 (the Act). The material in this communication may be subject to copyright under the Act. Any further reproduction or communication of this material by you may be the subject of copyright protection under the Act. Do not remove this notice.

2. www.monash.edu.au FIT2004 Algorithms & Data Structures L15: Minimum Spanning Trees Prepared by: Bernd Meyer from lecture materials © 2004 Goodrich & Tamassia March 2007

3. www.monash.edu.au 3 prepared from lecture material © 2004 Goodrich & Tamassia Minimum Spanning Trees (Goodrich & Tamassia § 12.7) Spanning subgraph –Subgraph of a graph G containing all the vertices of G Spanning tree –Spanning subgraph that is itself a (free) tree Minimum spanning tree (MST) –Spanning tree of a weighted graph with minimum total edge weight Applications –Communications networks –Transportation networks ORD PIT ATL STL DEN DFW DCA 10 1 9 8 6 3 2 5 7 4

4. www.monash.edu.au 4 prepared from lecture material © 2004 Goodrich & Tamassia Cycle Property Cycle Property: –Let T be a minimum spanning tree of a weighted graph G –Let e be an edge of G that is not in T and let C be the cycle formed by e with T –For every edge f of C, weight(f)  weight(e) Proof: –By contradiction –If weight(f)  weight(e) we can get a spanning tree of smaller weight by replacing e with f 8 4 2 3 6 7 7 9 8 e C f 8 4 2 3 6 7 7 9 8 C e f Replacing f with e yields a better spanning tree

5. www.monash.edu.au 5 prepared from lecture material © 2004 Goodrich & Tamassia Partition Property Partition Property: –Consider a partition of the vertices of G into subsets U and V –Let e be an edge of minimum weight across the partition –There is a minimum spanning tree of G containing edge e Proof: –Let T be an MST of G –If T does not contain e, consider the cycle C formed by e with T and let f be an edge of C across the partition –By the cycle property, weight(f)  weight(e) –Thus, weight(f)  weight(e) –We obtain another MST by replacing f with e UV 7 4 2 8 5 7 3 9 8 e f 7 4 2 8 5 7 3 9 8 e f Replacing f with e yields another MST UV

6. www.monash.edu.au 6 prepared from lecture material © 2004 Goodrich & Tamassia How to compute a MCST Try to apply an inductive approach -what does the induction run over? -what is the base case? -what is the induction hypothesis? -can you prove it?

7. www.monash.edu.au 7 prepared from lecture material © 2004 Goodrich & Tamassia How to compute a MCST Three basic ideas: 1.Kruskal’s Algorithm: start with no edges and successively add edges in order of increasing cost (making sure that we don’t insert edges that create cycles) 2.Prim’s algorithm: start with any node and iteratively grow a tree from it. At each step add the node (and associated edge) that is the cheapest extension to the tree 3.Reverse Deletion Algorithm: start with the full graph and delete edges in order of decreasing cost (making sure that we don’t disconnect the graph) Note that all of these are greedy approaches ! Why do they work?

8. www.monash.edu.au 8 prepared from lecture material © 2004 Goodrich & Tamassia Kruskal’s Algorithm A priority queue stores the edges outside the cloud Key: weight Element: edge At the end of the algorithm We are left with one cloud that encompasses the MST A tree T which is our MST Algorithm KruskalMST(G) for each vertex V in G do define a Cloud(v)  {v} let Q be a priority queue. Insert all edges into Q using their weights as the key T   while T has fewer than n-1 edges do edge e = Q.removeMin() Let u, v be the endpoints of e if Cloud(v)  Cloud(u) then Add edge e to T Merge Cloud(v) and Cloud(u) return T

9. www.monash.edu.au 9 prepared from lecture material © 2004 Goodrich & Tamassia Kruskal Example

10. www.monash.edu.au 10 prepared from lecture material © 2004 Goodrich & Tamassia Data Structure for Kruskal Algortihm The algorithm maintains a forest of trees An edge is accepted it if connects distinct trees We need a data structure that maintains a partition, i.e., a collection of disjoint sets, with the operations: -find(u): return the set storing u -union(u,v): replace the sets storing u and v with their union

11. www.monash.edu.au 11 prepared from lecture material © 2004 Goodrich & Tamassia List-based Partition Implementation Each set is stored in a sequence represented with a linked-list Each node should store an object containing the element and a reference to the set name

12. www.monash.edu.au 12 prepared from lecture material © 2004 Goodrich & Tamassia Runtime of union/find Each set is stored in a sequence Each element has a reference back to the set –operation find(u) takes O(1) time, and returns the set of which u is a member. –in operation union(u,v), we move the elements of the smaller set to the sequence of the larger set and update their references –the time for operation union(u,v) is min(n u,n v ), where n u and n v are the sizes of the sets storing u and v Whenever an element is processed, it goes into a set of size at least double, hence each element is processed at most log n times

13. www.monash.edu.au 13 prepared from lecture material © 2004 Goodrich & Tamassia Partition-Based Implementation A partition-based version of Kruskal’s Algorithm performs cloud merges as unions and tests as finds. Algorithm Kruskal(G): Input: A weighted graph G. Output: An MST T for G. Let P be a partition of the vertices of G, where each vertex forms a separate set. Let Q be a priority queue storing the edges of G, sorted by their weights Let T be an initially-empty tree while Q is not empty do (u,v)  Q.removeMinElement() if P.find(u) != P.find(v) then Add (u,v) to T P.union(u,v) return T Running time: O((n+m)log n)

14. www.monash.edu.au 14 prepared from lecture material © 2004 Goodrich & Tamassia Prim-Jarnik’s Algorithm Similar to Dijkstra’s algorithm (for a connected graph) We pick an arbitrary vertex s and we grow the MST as a cloud of vertices, starting from s We store with each vertex v a label d(v) = the smallest weight of an edge connecting v to a vertex in the cloud At each step: We add to the cloud the vertex u outside the cloud with the smallest distance label We update the labels of the vertices adjacent to u

15. www.monash.edu.au 15 prepared from lecture material © 2004 Goodrich & Tamassia Prim-Jarnik’s Algorithm (cont.) A priority queue stores the vertices outside the cloud –Key: distance –Element: vertex Priority queue should be implemented with a little trick: Locator-based. Each element keeps a pointer (index, locator) to its position in the queue. This allows to use replacekey without having to search the queue. We store three labels with each vertex: –Distance –Parent edge in MST –Locator in priority queue Algorithm PrimJarnikMST(G) Q  new heap-based priority queue s  a vertex of G for all v  vertices(G) if v  s setDistance(v, 0) else setDistance(v,  ) setParent(v,  ) l  insert(Q, getDistance(v), v) while  isEmpty(Q) u  min(Q); Q  removeMin(Q) for all e  incidentEdges(G, u) z  opposite(G,u,e) r  weight(e) if r  getDistance(z) setDistance(z,r) setParent(z,e) replaceKey(Q,z,r)

16. www.monash.edu.au 16 prepared from lecture material © 2004 Goodrich & Tamassia Example B D C A F E 7 4 2 8 5 7 3 9 8 0 7 2 8   B D C A F E 7 4 2 8 5 7 3 9 8 0 7 2 5  7 B D C A F E 7 4 2 8 5 7 3 9 8 0 7 2 5  7 B D C A F E 7 4 2 8 5 7 3 9 8 0 7 2 5 4 7

17. www.monash.edu.au 17 prepared from lecture material © 2004 Goodrich & Tamassia Example (contd.) B D C A F E 7 4 2 8 5 7 3 9 8 0 3 2 5 4 7 B D C A F E 7 4 2 8 5 7 3 9 8 0 3 2 5 4 7

18. www.monash.edu.au 18 prepared from lecture material © 2004 Goodrich & Tamassia Analysis Graph operations –Method incidentEdges is called once for each vertex Label operations –We set/get the distance, parent and locator labels of vertex z O(deg(z)) times –Setting/getting a label takes O(1) time Priority queue operations –Each vertex is inserted once into and removed once from the priority queue, where each insertion or removal takes O(log n) time –The key of a vertex w in the priority queue is modified at most deg(w) times, where each key change takes O(log n) time (this is for reheap by percolating) Prim-Jarnik’s algorithm runs in O((n  m) log n) time provided the graph is represented by the adjacency list structure –Recall that  v deg(v)  2m The running time is O(m log n) since the graph is connected

19. www.monash.edu.au 19 prepared from lecture material © 2004 Goodrich & Tamassia Baruvka’s Algorithm Like Kruskal’s Algorithm, Baruvka’s algorithm grows many “clouds” at once. Each iteration of the while-loop halves the number of connected compontents in T. –The running time is O(m log n). Algorithm BaruvkaMST(G) T  V {just the vertices of G} while T has fewer than n-1 edges do for each connected component C in T do Let edge e be the smallest-weight edge from C to another component in T. if e is not already in T then Add edge e to T return T

20. www.monash.edu.au 20 prepared from lecture material © 2004 Goodrich & Tamassia Appendix: a better union/find structure: Tree-based Implementation Each element is stored in a node, which contains a pointer to a set name A node v whose set pointer points back to v is also a set name Each set is a tree, rooted at a node with a self- referencing set pointer For example: The sets “1”, “2”, and “5”: 1 74 2 63 5 108 12 119

21. www.monash.edu.au 21 prepared from lecture material © 2004 Goodrich & Tamassia Union-Find Operations To do a union, simply make the root of one tree point to the root of the other To do a find, follow set- name pointers from the starting node until reaching a node whose set- name pointer refers back to itself 2 63 5 108 12 11 9 2 63 5 108 12 11 9

22. www.monash.edu.au 22 prepared from lecture material © 2004 Goodrich & Tamassia Union-Find Heuristic 1 Union by size: –When performing a union, make the root of smaller tree point to the root of the larger Implies O(n log n) time for performing n union-find operations: –Each time we follow a pointer, we are going to a subtree of size at least double the size of the previous subtree –Thus, we will follow at most O(log n) pointers for any find. 2 63 5 108 12 11 9

23. www.monash.edu.au 23 prepared from lecture material © 2004 Goodrich & Tamassia Path compression: –After performing a find, compress all the pointers on the path just traversed so that they all point to the root Implies O(n log * n) time for performing n union-find operations: –Proof is complex… (in Weiss 8.6.1, Theorem 8.1) Union-Find Heuristic 2 2 63 5 108 12 11 9 2 63 5 108 12 11 9

24. www.monash.edu.au 24 prepared from lecture material © 2004 Goodrich & Tamassia log* is an amazingly slow growing function. It is the inverse of the tower-of-twos function (ie. the number of time you can draw the logarithm of n before the result is less than 2) so far practically relevant numbers, O(n log* n) is not much worse than O(n) and the constants (which Big-O neglects) is probably more important. Log* n 22 =4 2 2 2 =16 2 2 =65536 2222222222 log* n 12345

25. www.monash.edu.au 25 prepared from lecture material © 2004 Goodrich & Tamassia Proof of log* n Amortized Time For each node v that is a root –define n(v) to be the size of the subtree rooted at v (including v) –identified a set with the root of its associated tree. We update the size field of v each time a set is union’ed into v. Thus, if v is not a root, then n(v) is the largest the subtree rooted at v can be, which occurs just before we union v into some other node whose size is at least as large as v ’s. For any node v, then, define the rank of v, which we denote as r (v), as r (v) = [log n(v)]: Thus, n(v) ≥ 2 r(v). Also, since there are at most n nodes in the tree of v, r (v) = [logn], for each node v.

26. www.monash.edu.au 26 prepared from lecture material © 2004 Goodrich & Tamassia Proof of log* n Amortized Time (2) For each node v with parent w: –r (v ) < r (w ) Claim: There are at most n/ 2 s nodes of rank s. Proof: –Since r (v) < r (w), for any node v with parent w, ranks are monotonically increasing as we follow parent pointers up any tree. –Thus, if r (v) = r (w) for two nodes v and w, then the nodes counted in n(v) must be separate and distinct from the nodes counted in n(w). –If a node v is of rank s, then n(v) ≥ 2 s. –Therefore, since there are at most n nodes total, there can be at most n/ 2 s that are of rank s.

27. www.monash.edu.au 27 prepared from lecture material © 2004 Goodrich & Tamassia Proof of log* n Amortized Time (3) Definition: Tower of two’s function: –t(i) = 2 t(i-1) Nodes v and u are in the same rank group g if –g = log*(r(v)) = log*(r(u)): Since the largest rank is log n, the largest rank group is –log*(log n) = (log* n)-1

28. www.monash.edu.au 28 prepared from lecture material © 2004 Goodrich & Tamassia Proof of log* n Amortized Time (4) Charge 1 cyber-dollar per pointer hop during a find: –If w is the root or if w is in a different rank group than v, then charge the find operation one cyber-dollar. –Otherwise (w is not a root and v and w are in the same rank group), charge the node v one cyber-dollar. Since there are most (log* n)-1 rank groups, this rule guarantees that any find operation is charged at most log* n cyber-dollars.

29. www.monash.edu.au 29 prepared from lecture material © 2004 Goodrich & Tamassia Proof of log* n Amortized Time (5) After we charge a node v then v will get a new parent, which is a node higher up in v ’s tree. The rank of v ’s new parent will be greater than the rank of v ’s old parent w. Thus, any node v can be charged at most the number of different ranks that are in v ’s rank group. If v is in rank group g > 0, then v can be charged at most t(g)-t(g-1) times before v has a parent in a higher rank group (and from that point on, v will never be charged again). In other words, the total number, C, of cyber- dollars that can ever be charged to nodes can be bound as

30. www.monash.edu.au 30 prepared from lecture material © 2004 Goodrich & Tamassia Proof of log* n Amortized Time (end) Bounding n(g):Returning to C: