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MST

Course: ECE 665, Fall 2009
School: UMass (Amherst)
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Spanning Minimum Trees 2704 867 849 ORD 1846 SFO 337 LAX 1464 DFW 1235 1121 MIA 2342 802 1391 740 621 187 JFK 184 BWI 1090 946 144 1258 BOS PVD Minimum Spanning Trees 1 Outline and Reading Minimum Spanning Trees (7.3) The PrimJarnik Algorithm (7.3.2) Kruskal's Algorithm (7.3.1) Baruvka's Algorithm (7.3.3) Minimum Spanning Trees 2 Definitions A crucial fact Minimum Spanning Tree Spanning subgraph Spanning...

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Spanning Minimum Trees 2704 867 849 ORD 1846 SFO 337 LAX 1464 DFW 1235 1121 MIA 2342 802 1391 740 621 187 JFK 184 BWI 1090 946 144 1258 BOS PVD Minimum Spanning Trees 1 Outline and Reading Minimum Spanning Trees (7.3) The PrimJarnik Algorithm (7.3.2) Kruskal's Algorithm (7.3.1) Baruvka's Algorithm (7.3.3) Minimum Spanning Trees 2 Definitions A crucial fact Minimum Spanning Tree Spanning subgraph Spanning tree Subgraph of a graph G containing all the vertices of G Spanning subgraph that is itself a (free) tree Spanning tree of a weighted graph with minimum total edge weight Communications networks Transportation networks ORD 1 10 PIT 9 6 7 3 5 DEN 4 Minimum spanning tree (MST) STL 8 DCA 2 Applications DFW ATL 3 Minimum Spanning Trees Cycle Property Cycle Property: f 2 6 8 8 9 3 7 e 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 C 4 7 Replacing f with e yields a better spanning tree f 2 6 8 9 3 8 7 e 4 C 7 4 Minimum Spanning Trees Partition Property Partition Property: U f 7 V 4 Consider a partition of the vertices of 9 5 G into subsets U and V 2 8 Let e be an edge of minimum weight 3 8 across the partition e There is a minimum spanning tree of 7 G containing edge e Replacing f with e yields Proof: another MST Let T be an MST of G If T does not contain e, consider the U V 7 cycle C formed by e with T and let f f 4 be an edge of C across the partition By the cycle property, 9 5 2 weight(f) weight(e) 8 Thus, weight(f) = weight(e) 3 8 e We obtain another MST by replacing f with e 7 Minimum Spanning Trees 5 PrimJarnik'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 Minimum Spanning Trees 6 PrimJarnik's Algorithm (cont.) A priority queue stores the vertices outside the cloud Locatorbased methods Key: distance Element: vertex We store three labels with each vertex: insert(k,e) returns a locator replaceKey(l,k) changes the key of an item 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 G.vertices() if v = s setDistance(v, 0) else setDistance(v, ) setParent(v, ) l Q.insert(getDistance(v), v) setLocator(v,l) while Q.isEmpty() u Q.removeMin() for all e G.incidentEdges(u) z G.opposite(u,e) r weight(e) if r < getDistance(z) setDistance(z,r) setParent(z,e) Q.replaceKey(getLocator(z),r) 7 Minimum Spanning Trees Example 2 2 A B 5 8 7 2 2 A 7 5 8 7 5 C 9 8 E 7 7 8 C 9 8 E 7 7 4 F 3 0 2 2 A B 5 8 7 D 4 F 3 0 A 2 2 B 5 8 7 7 5 C 9 8 E 7 8 7 5 C 9 8 D 7 4 F 3 7 0 E B D D 7 4 F 3 4 0 Minimum Spanning Trees Example (contd.) 2 2 A B 5 8 7 7 5 C 9 8 E 3 D 7 4 F 3 2 2 A 7 5 8 7 5 C 9 8 E 3 9 4 0 B D 7 4 F 3 4 0 Minimum Spanning Trees Analysis Graph operations Label operations Method incidentEdges is called once for each vertex We set/get the distance, parent and locator labels of vertex z O(deg(z)) times Setting/getting a label takes O(1) time 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 Priority queue operations PrimJarnik's algorithm runs in O((n + m) log n) time provided the graph is represented by the adjacency list structure The running time is O(m log n) since the graph is connected Minimum Spanning Trees 10 Recall that v deg(v) = 2m Kruskal's Algorithm A priority queue stores the edges outside the cloud At the end of the algorithm Key: weight Element: edge 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) of {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 = T.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 Minimum Spanning Trees 11 Data Structure for Kruskal Algortihm The algorithm a maintains 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 Minimum Spanning Trees 12 Representation of a Partition 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(nu,nv), where nu and nv 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 Minimum Spanning Trees 13 PartitionBased Implementation A partitionbased 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 Running time: Add (u,v) to T O((n+m)log n) P.union(u,v) return T Minimum Spanning Trees 14 Kruskal Example 1846 SFO 337 LAX 1464 2704 867 849 ORD 740 621 187 JFK 184 1391 BWI BOS PVD 144 1258 802 1090 DFW 1235 1121 MIA 2342 Minimum Spanning Trees 15 946 Example 1846 SFO 337 LAX 1464 2704 867 849 ORD 740 621 187 JFK 184 1391 BWI BOS PVD 144 1258 802 1090 DFW 1235 1121 MIA 2342 Minimum Spanning Trees 16 946 Example 1846 SFO 337 LAX 1464 2704 867 849 ORD 740 621 187 JFK 184 1391 BWI BOS PVD 144 1258 802 1090 DFW 1235 1121 MIA 2342 Minimum Spanning Trees 17 946 Example 1846 SFO 337 LAX 1464 2704 867 849 ORD 740 621 187 JFK 184 1391 BWI BOS PVD 144 1258 802 1090 DFW 1235 1121 MIA 2342 Minimum Spanning Trees 18 946 Example 1846 SFO 337 LAX 1464 2704 867 849 ORD 740 621 187 JFK 184 1391 BWI BOS PVD 144 1258 802 1090 DFW 1235 1121 MIA 2342 Minimum Spanning Trees 19 946 Example 1846 SFO 337 LAX 1464 2704 867 849 ORD 740 621 187 JFK 184 1391 BWI BOS PVD 144 1258 802 1090 DFW 1235 1121 MIA 2342 Minimum Spanning Trees 20 946 Example 1846 SFO 337 LAX 1464 2704 867 849 ORD 740 621 187 JFK 184 1391 BWI BOS PVD 144 1258 802 1090 DFW 1235 1121 MIA 2342 Minimum Spanning Trees 21 946 Example 1846 SFO 337 LAX 1464 2704 867 849 ORD 740 621 187 JFK 184 1391 BWI BOS PVD 144 1258 802 1090 DFW 1235 1121 MIA 2342 Minimum Spanning Trees 22 946 Example 1846 SFO 337 LAX 1464 2704 867 849 ORD 740 621 187 JFK 184...

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ECE 122Exception Handling Exception handling provides a graceful way of handling exceptional cases and errors. The system will not crash when such exception happens. The control will be transferred to exception handling code, which will correct th
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UMass (Amherst) - ECE - 122
1. Most students did very well this time. Most required functions are implemented, and many completed the extra points section.2. Still there are several groups forgot to submit the supporting classes. 10 points were deducted according t
UMass (Amherst) - ECE - 122
Account Project4= =A36563 100A37395 70A73448A93030 0B15582B30534 100B62000 100B63878 100B87334 60C05516C07107 90C16660C31579 80C76512
UMass (Amherst) - ECE - 122
Account Project6= =A36563A37395 90A73448A93030 100B15582B30534 110B62000 110B63878 100B87334 107C05516C07107 110C16660C31579C76512 110C92057
UMass (Amherst) - ECE - 122
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Grading Crietria: Basic requirement: total 100 points. 1. project1a 50 points 2. project1b 50 points Extra requirment: total 10 points 1. readDouble 5 points 2. readString
UMass (Amherst) - ECE - 122
Account Project2=A36563 75A37395A73448A93030 60B15582B30534 105B62000 105B63878B87334 110C05516C07107 103C16660C31579 100C76512 110C92057 100D08322 100D14101D78830 110D847
UMass (Amherst) - ECE - 122
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Account Project3= =A36563 100A37395 90A73448A93030 110B15582B30534 110B62000 110B63878B87334 100C05516C07107 100C16660C31579 50C76512 85C9
UMass (Amherst) - ECE - 122
General comments:1. Many students did not submit Constant.java, NegatePipe.javaand Pipe.java with their work. Your Programs will not compileunless you submit them. We had to supply the necessary classfiles to compile them. Though, they have not
UMass (Amherst) - ECE - 122
About grading:1. Some students' programs do not compile. According to the grading policy, you should have got 0s. 2. Many students did not submit the SID in their codes. Please do so from next time.3. If there was no proper documentation, you wi
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General comments:1. Many students did not have graduateDate variable and its set/get methods in Student.java. Some of them did not have set/get methods to access private variables in Student.java/Teacher.java/Course.java.2. Some students did not
UMass (Amherst) - ECE - 122
About grading:1. Some students' programs do not compile. According to the grading policy, you should have got 0s. However, dependingon how well your code was written, we gave 50 - 70 points for your effort. Next time we may strictly follow the
UMass (Amherst) - ECE - 122
Grading Crietria: Basic requirement: 1. Part(a): 50 points. 2. Part(b): 50 points. Extra requirment: total 10 points 1. getUniqueStudentId method to return an unique integer as student ID. If your program doe
UMass (Amherst) - ECE - 122
Account Project5= =A36563A37395 100A73448A93030 100B15582B30534 100B62000 100B63878 65B87334 85C05516C07107 100C16660C31579 95C76512 100C92057 100D08322
UMass (Amherst) - ECE - 122
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UMass (Amherst) - ECE - 122
Download Eclipse 1. Download Eclipse IDE from www.eclipse.org by clicking on one of the download links on page, http:/www.eclipse.org/downloads/index.php , Choose version 3.01. Choose a site that nears us. Save the file in a folder on your local comp
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UMass (Amherst) - ECE - 122
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UMass (Amherst) - ECE - 122
ECE 122We know thisCat cat = new Cat(); Animal animal = new Animal();and this? Animal animal = new Cat(); An object reference of super class type points to an object of sub class type. It is legal and it is called polymorphism.Motivation L
UMass (Amherst) - ECE - 122
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UMass (Amherst) - ECE - 122
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UMass (Amherst) - ECE - 122
ECE122Feb. 17. 2005Introduction to Class and Object Java is an object-oriented language. Javas view of the world is a collection of objects and their interactions. Class is a template, or blueprint to build objects. Objects are instantiated (c
UMass (Amherst) - ECE - 122
ECE 122Exception Handling Exception handling provides a graceful way of handling exceptional cases and errors. The system will not crash when such exception happens. The control will be transferred to exception handling code, which will correct th