Minimum spanning trees

Transcription

1 CE 367R

2 MINIMUM SPANNING TREES

3 Minimum Spanning Tree Identify a subset of links which form a spanning tree (connecting all nodes), where the total cost of the links in the tree is minimized. Minimum Spanning Trees

4 In the minimum spanning tree problem, the direction of all links is ignored. (This is an undirected network.) Connections can go in either direction. Every link has a cost; all nodes must be connected by the chosen links. Minimum Spanning Trees

5 Applications Building roads in rural areas Providing utilities and other infrastructure Clustering Espionage networks, passing messages between spies Minimum Spanning Trees

6 Infrastructure network design Nodes indicate towns or other terminals that must be connected. Links indicate all possible infrastructure connections that can be built; the cost indicates the construction cost. Building the links in the minimum spanning tree is the cheapest way to connect all of the nodes. Minimum Spanning Trees

7 Clustering data By forming a minimum spanning tree, then deleting long edges, we can classify data into clusters based on their similarity. Minimum Spanning Trees

8 Espionage networks There are n spies who must communicate, passing messages through each other. Whenever a message is passed between spies i and j, the probability the message will be intercepted is p ij. How should a spy distribute a message to all other spies with minimum probability of detection? Create a node for each spy, a link between all pairs of spies, and set the cost of each link to log(1 p ij ). The minimum spanning tree is the best way to transmit the message. (Why?) Minimum Spanning Trees

9 How many links must be in a spanning tree? In any spanning tree, adding a link will create exactly one cycle. In a minimum spanning tree, any new link which is added must have a cost at least equal to the maximum cost of the other links in that cycle. Minimum Spanning Trees

10 Students often confuse shortest path and minimum spanning tree. What are the differences? Minimum Spanning Trees

11 PRIM S ALGORITHM

12 Notation Let G = (N, A) be the original network; let T = (N T, A T ) be the links in the spanning tree. At first T is empty, but over time N T and A T will grow. T is called a subnetwork of G, because N T N and A T A. A link is admissible if exactly one of its end nodes is in N T. Prim s Algorithm

13 Algorithm (Assumes that the network is connected.) 1 Arbitrarily choose some root node s. 2 Initialize N T = {s}, A T = 3 Repeat the following steps until A T has n 1 links: Identify all of the admissible links; Choose an admissible link (u, v) with minimum cost. (Assume u is in N T, but not v.) Add this link to the tree: NT = N T {v}, A T = A T (u, v) Prim s Algorithm

14 Correctness Does it terminate? Each iteration adds an admissible link to the tree. By doing so, one more node is added to the tree. After n 1 admissible links have been added, N T = N. There are no more admissible links at this point, so the algorithm must terminate. Prim s Algorithm

15 Correctness When it terminates, do we have a minimum spanning tree? There are three ways it can go wrong: at termination, T might not be a tree; or it might be a tree, but not a spanning tree; or it might be a spanning tree, but not a minimum cost one. Is T a tree? Only admissible links are added; admissible links have one end node not in N T, so no cycles are created. Is T a spanning tree? Since the network is connected, if N T N then there must be an admissible link. Prim s Algorithm

16 Must T be a minimum spanning tree? Here is a proof sketch (with some handwaving). Assume not, and let T be a minimum spanning tree. Let (u, v) be the first link chosen by the algorithm which is not part of T. Let (u, v ) be a link connected to u in T (with v v ) Since Prim s algorithm chose (u, v), c uv c uv. If c uv < c uv, it could be swapped into T to reduce its cost (eliminating another link on the cycle created.) However this is impossible since T is a minimum spanning tree. Therefore, c uv = c uv whenever the algorithm chooses a link not part of T, so its total cost is the same. Prim s Algorithm

17 Complexity There are O(n) iterations (technically n 1). At each iteration, we must identify all admissible links (of which there are at most m), and identify one with minimum cost (which again takes m steps). So, Prim s algorithm is O(nm). There are more clever ways of identifying admissible links and finding one with minimum cost, which can reduce the running time to O(m log n) or O(m+n log n). These do so by avoiding duplication of effort in subsequent iterations. Prim s Algorithm

18 KRUSKAL S ALGORITHM

19 Prim s algorithm builds the tree by growing it from a seed node. Kruskal s algorithm grows multiple trees simultaneously, eventually joining them all into one spanning tree. Idea: sort all link costs from low to high. Select the cheapest links, but avoid creating a cycle. Kruskal s Algorithm

20 Kruskal s algorithm 1 Sort all the arcs in increasing order of cost, let a = 0. 2 Create an empty set of arcs T. 3 Repeat the following steps until T = n 1: Would adding arc a to T create a cycle? If so, increase a by 1 and repeat this step. Add arc a to the tree. If there are n 1 arcs in the tree, stop: we are done. Otherwise increase a by 1 and return to step 3. Kruskal s Algorithm

21 Example Kruskal s Algorithm

22 CODING TIPS

23 Dictionaries So far we have used lists to store collections of items, which we can access with a numerical index: For a list population=[25,2000,1500] we can access the elements with population[0], population[1], and population[2] A dictionary allows us to refer to items with a more descriptive key which need not be a number. For instance: population[ Laramie ], population[ Austin ], and so forth. Coding tips

24 A dictionary stores pairs of keys and values, you can access each value using the appropriate key. If population={ Laramie : 25, Austin :1500} then we can access the values with population[ Laramie ] (which equals 25), and so forth. You can create new key-value pairs by assignment: population[ Seattle ] = 2000 Coding tips

25 You can iterate over all the keys in a dictionary using a for loop for city in population: print(population[city]) would print the three values 25, 2000, and 1500, not necessarily in that order since the keys in a dictionary are not ordered. Coding tips

26 One way to find the key corresponding to the smallest value in a dictionary is to manually loop over all the values: minpopulation = mincity = None for city in population: if population[city] < minpopulation: minpopulation = population[city] mincity = city Coding tips

27 To iterate over a dictionary in increasing order of the values, use this kind of loop: for city in sorted(population, key=population.get): print(population[city]) This will print the values, 25, 1500, and 2000 in that order. Coding tips

28 You can also use tuples for dictionary keys. For the min-spanning tree algorithms, you may find it helpful to create dictionaries where the keys are tuples representing links, and the values are costs of the links. cost[(5,4)] = self.cost[5][4] and so forth. Coding tips

29 The minspanningtree method asks you to find and return a minimum spanning tree, using a list treelinks. So, if the minimum spanning tree contains the links (1,0), (1,2), and (2,3), you should return the list [(1,0), (1,2), (2,3)]. Don t worry about ties if there are multiple minimum spanning trees, or the order of the nodes within each link (the network is undirected). Coding tips

30 Tips on Prim s 1 Arbitrarily choose some root node s. 2 Initialize N T = {s}, A T = 3 Repeat the following steps until A T has n 1 links: Identify all of the admissible links; Choose an admissible link (u, v) with minimum cost. (Assume u is in N T, but not v.) Add this link to the tree: NT = N T {v}, A T = A T (u, v) Coding tips

31 You can use the list treelinks to represent A T ; create another list to represent N T. You need to find the admissible link with the lowest cost; at each iteration create a dictionary of admissible links and costs. You need to add links and nodes to your lists representing A T and N T. Coding tips

32 Tips on Kruskal s 1 Sort all the arcs in increasing order of cost, let a = 0. 2 Create an empty set of arcs T. 3 Repeat the following steps until T = n 1: Would adding arc a to T create a cycle? If so, increase a by 1 and repeat this step. Add arc a to the tree. If there are n 1 arcs in the tree, stop: we are done. Otherwise increase a by 1 and return to step 3. Coding tips

33 You need to keep track of/do the following things: Loop over links in increasing order of cost. Know what links are currently in the forest (treelinks) Determine whether adding a link would create a cycle. One solution is to keep lists with the nodes in each forest. Adding a link only creates a cycle if its head and tail nodes are in the same tree. What do we do with these lists of trees when adding a link? Coding tips