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Prof. Swarat Chaudhuri COMP 482: Design and Analysis of Algorithms Spring 2013 Lecture 4

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Q3: Analyzing an algorithm You have an array A with integer entries A[1],…, A[n] Output a 2-D array B such that B[i,j] contains the sum A[i] + A[i+1] + … + A[j] Here is an algorithm: For i=1, 2, …, n For j = i+1, 2, …, n { Add up entries A[i] through A[j] Store result in B[i,j] } Obtain an upper bound and a lower bound for the algorithm. 2

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The lower bound is the more interesting one Consider the times during the execution of the algorithm when i ≤ n/4 and j ≥ 3n/4. In these cases, j − i + 1 ≥ 3n/4 − n/4 + 1 > n/2. Therefore, adding up the array entries A[i] through A[j] would require at least n/2 operations, since there are more then n/2 terms to add up. How many times during the execution of the given algorithm do we encounter such cases? There are (n/4) 2 pairs (i, j) with i ≤ n/4 and j ≥ 3n/4. the given algorithm enumerates over all of them, and as shown above, it must perform at least n/2 operations for each such pair. Therefore, the algorithm must perform at least n/2.(n/4) 2 = n 3 /32 operations. This is Ω(n 3 ), as desired. 3

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Graphs 3.1 Basic Definitions and Applications

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5 Undirected Graphs Undirected graph. G = (V, E) n V = nodes. n E = edges between pairs of nodes. n Captures pairwise relationship between objects. n Graph size parameters: n = |V|, m = |E|. V = { 1, 2, 3, 4, 5, 6, 7, 8 } E = { 1-2, 1-3, 2-3, 2-4, 2-5, 3-5, 3-7, 3-8, 4-5, 5-6 } n = 8 m = 11

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6 Some Graph Applications transportation Graph street intersections NodesEdges highways communicationcomputersfiber optic cables World Wide Webweb pageshyperlinks socialpeoplerelationships food webspeciespredator-prey software systemsfunctionsfunction calls schedulingtasksprecedence constraints circuitsgateswires

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7 World Wide Web Web graph. n Node: web page. n Edge: hyperlink from one page to another. cnn.com cnnsi.com novell.comnetscape.com timewarner.com hbo.com sorpranos.com

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Terrorist Network Social network graph. n Node: people. n Edge: relationship between two people. Reference: Valdis Krebs,

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9 Ecological Food Web Food web graph. n Node = species. n Edge = from prey to predator. Reference:

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10 Graph Representation: Adjacency Matrix Adjacency matrix. n-by-n matrix with A uv = 1 if (u, v) is an edge. n Two representations of each edge. n Space proportional to n 2. n Checking if (u, v) is an edge takes (1) time. n Identifying all edges takes (n 2 ) time

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11 Graph Representation: Adjacency List Adjacency list. Node indexed array of lists. n Two representations of each edge. n Space proportional to m + n. n Checking if (u, v) is an edge takes O(deg(u)) time. n Identifying all edges takes (m + n) time degree = number of neighbors of u 37

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12 Paths and Connectivity Def. A path in an undirected graph G = (V, E) is a sequence P of nodes v 1, v 2, …, v k-1, v k with the property that each consecutive pair v i, v i+1 is joined by an edge in E. Def. A path is simple if all nodes are distinct. Def. An undirected graph is connected if for every pair of nodes u and v, there is a path between u and v.

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13 Cycles Def. A cycle is a path v 1, v 2, …, v k-1, v k in which v 1 = v k, k > 2, and the first k-1 nodes are all distinct. cycle C =

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14 Trees Def. An undirected graph is a tree if it is connected and does not contain a cycle. Theorem. Let G be an undirected graph on n nodes. Any two of the following statements imply the third. n G is connected. n G does not contain a cycle. n G has n-1 edges.

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15 Rooted Trees Rooted tree. Given a tree T, choose a root node r and orient each edge away from r. Importance. Models hierarchical structure. a tree the same tree, rooted at 1 v parent of v child of v root r

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16 Phylogeny Trees Phylogeny trees. Describe evolutionary history of species.

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17 GUI Containment Hierarchy Reference: GUI containment hierarchy. Describe organization of GUI widgets.

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3.2 Graph Traversal

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19 Connectivity s-t connectivity problem. Given two nodes s and t, is there a path between s and t? s-t shortest path problem. Given two nodes s and t, what is the length of the shortest path between s and t? Applications. n Facebook. n Maze traversal. n Erdos number. n Kevin Bacon number. n Fewest number of hops in a communication network.

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20 Breadth First Search BFS intuition. Explore outward from s in all possible directions, adding nodes one "layer" at a time. Effect: find “shallow” paths to nodes. BFS algorithm. n L 0 = { s }. n L 1 = all neighbors of L 0. n L 2 = all nodes that do not belong to L 0 or L 1, and that have an edge to a node in L 1. n L i+1 = all nodes that do not belong to an earlier layer, and that have an edge to a node in L i. Theorem. For each i, L i consists of all nodes at distance exactly i from s. There is a path from s to t iff t appears in some layer. s L1L1 L2L2 L n-1

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Implementing BFS Q: What’s a good way to implement the above algorithm? A: Use a queue for the “frontier” 21

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22 Breadth First Search Property. Let T be a BFS tree of G = (V, E), and let (x, y) be an edge of G. Then the level of x and y differ by at most 1. L0L0 L1L1 L2L2 L3L3

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23 Breadth First Search: Analysis Theorem. The above implementation of BFS runs in O(m + n) time if the graph is given by its adjacency list representation. Pf. n Easy to prove O(n 2 ) running time: – at most n lists L i – each node occurs on at most one list; for loop runs n times – when we consider node u, there are n incident edges (u, v), and we spend O(1) processing each edge n Actually runs in O(m + n) time: – when we consider node u, there are deg(u) incident edges (u, v) – total time processing edges is u V deg(u) = 2m ▪ each edge (u, v) is counted exactly twice in sum: once in deg(u) and once in deg(v)

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24 Connected Component Connected component. Find all nodes reachable from s. Connected component containing node 1 = { 1, 2, 3, 4, 5, 6, 7, 8 }.

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Q1: Finding connected components Give an algorithm to find the set of all connected components of an undirected graph. 25

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26 Connected Component Connected component. Find all nodes reachable from s. Theorem. Upon termination, R is the connected component containing s. n BFS = explore in order of distance from s. s uv R it's safe to add v

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27 Q2: Flood Fill Flood fill. Given lime green pixel in an image, change color of entire blob of neighboring lime pixels to blue. recolor lime green blob to blue

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28 Flood Fill Flood fill. Given lime green pixel in an image, change color of entire blob of neighboring lime pixels to blue. recolor lime green blob to blue

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29 Flood Fill Flood fill. Given lime green pixel in an image, change color of entire blob of neighboring lime pixels to blue. n Node: pixel. n Edge: two neighboring lime pixels. n Blob: connected component of lime pixels. recolor lime green blob to blue

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Depth-first search Use recursion DFS intuition. Explore outward from s along one path as far as possible, and backtrack when you cannot progress. Effect: find faraway nodes. DFS(u): Mark u as “Explored” and add u to R For each edge (u,v) incident to u If v is not marked “Explored” then Recursively call DFS(v) 30

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Depth-first search Property. For a given recursive call DFS(u), all nodes marked “Explored” between the beginning and end of this recursive call are descendants of u in T. Theorem. Let T be a depth-first search tree, let x and y be nodes in T, and let (x,y) be an edge of G that is not an edge of T. Then one of x or y is an ancestor of the other. 31

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Q3: BFS and DFS trees We have a connected graph G = (V, E) and a specific vertex u. Suppose we compute a DFS tree rooted at u, and obtain a tree T that includes all nodes of G. Suppose we then compute a BFS tree rooted at u, and obtain the same tree T. Prove that G = T. 32

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Answer Suppose G has an edge e = {a, b} that does not belong to T. As T is a DFS tree, one of the two ends must be an ancestor of the other—say a is an ancestor of b. (*) Since T is a BFS tree, the distance of the two nodes from u in T can differ at most by one. But if a is an ancestor of b, and (*) holds, then a must be the direct parent of b. This means that {a, b} is an edge in T. Contradiction. 33

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Q4: Finding a cycle Given a graph G, determine if it has a cycle. If so, the algorithm should output this cycle. Answer: Assume that G is connected; otherwise work on the connected components. Run BFS from an arbitrary node s, and obtain a BFS tree T. If every edge of G appears in the tree, then G = T and there is no cycle. Otherwise, there is an edge e = (v, w) that is in G but not in T. Consider the least common ancestor u of v and w in T. We get a cycle from edge e and paths u-v and u-w in T. 34

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