DAA- Unit II By Dr. A.S.Alvi Review of graphs A graph represents relationship among items. A graph consists of a set of vertices and a set of edges that connect the vertices. G = (V, E) Review of graphs (cont.) V: the set of vertices (or nodes) E: the set of pairs of edges that connect the vertices Vertex V0 V1 Edge V2 V3 Example: computing the fastest route through mass transportation computing the fastest way for routing electronic mail through a network of computers Problem: Laying Telephone Wire Central office Wiring: Naive Approach Central office Expensive! Wiring: Better Approach Central office Minimize the total length of wire connecting the customers Minimum-cost spanning trees Suppose you have a connected undirected graph with a weight (or cost) associated with each edge The cost of a spanning tree would be the sum of the costs of its edges A minimum-cost spanning tree is a spanning tree that has the lowest cost A 19 16 21 11 33 E F 18 B 5 14 D A 6 C 10 A connected, undirected graph 16 11 F E 18 B 5 6 C D A minimum-cost spanning tree Prim’s Algorithm In Kruskal algorithm the selection function chooses edges in increasing order of length without worrying too much about their connection to previously chosen edges, except that we are careful never to form a cycle. In Prim’s algorithm, on the other , the minimum spanning tree grows in a natural way, starting from the arbitrary root. At each stage we add a new branch to the tree already constructed. The algorithm stops when all the nodes have been reached. Prim’s Algorithm Let B be a set of nodes and T a set of edges. Initially B contain a single arbitrary node and T is empty. At each step Prim’s algorithm looks for the shortest possible edge {u,v} such that u B and v N\B. It then add v to B and {u,v} to T. In this way the edges in T form at any instant a minimum spanning tree for the nodes in B. We continue thus as long as B is not equal to N. Prim’s algorithm Start form any arbitrary vertex Find the edge that has minimum weight form all known vertices Stop when the tree covers all vertices The execution of Prim's algorithm(moderate part) 8 the root vertex 7 c b 4 d 9 2 11 a 7 8 h i 6 2 8 f 7 c b 4 e 10 g 1 14 4 d 9 2 11 a 7 8 h i 1 14 4 6 e 10 g 2 f 8 7 c b 4 d 9 2 11 a 7 8 h i 6 2 8 f 7 c b 4 e 10 g 1 14 4 d 9 2 11 a 7 8 h i 1 14 4 6 e 10 g 2 f 8 7 c b 4 d 9 2 11 a 7 8 h i 6 2 8 f 7 c b 4 e 10 g 1 14 4 d 9 2 11 a 7 8 h i 1 14 4 6 e 10 g 2 f 8 7 c b 4 d 9 2 11 a 7 8 h i 6 2 8 f 7 c b 4 e 10 g 1 14 4 d 9 2 11 a 7 8 h i 1 14 4 6 e 10 g 2 f 8 7 c b 4 d 9 2 11 a 7 8 h i 1 14 4 6 e 10 g 2 f Bottleneck spanning tree: A spanning tree of G whose largest edge weight is minimum over all spanning trees of G. The value of the bottleneck spanning tree is the weight of the maximum-weight edge in T. Theorem: A minimum spanning tree is also a bottleneck spanning tree. (Challenge problem) Example 1 1 1 4 6 4 3 2 2 4 5 3 5 6 8 7 4 3 7 6 Example 2 9 a 2 5 a 6 2 d 4 4 5 c 9 b 5 5 6 d 4 4 e 5 c b 5 e Example 3 Prim’s Algorithm function Prim (G= < N,A >:graph; length: A R+ ) ;set of edges {initialiazation} TØ B { an arbitrary member of N} while B ± N do find e ={u,v} of minimum length such that u B and v N\B T T {e} B B {v} return T Minimum Connector Algorithms Kruskal’s algorithm 1. Select the shortest edge in a network 2. Select the next shortest edge which does not create a cycle 3. Repeat step 2 until all vertices have been connected Prim’s algorithm 1. Select any vertex 2. Select the shortest edge connected to that vertex 3. Select the shortest edge connected to any vertex already connected 4. Repeat step 3 until all vertices have been connected Dijkstra Animated Example Dijkstra Animated Example Dijkstra Animated Example Dijkstra Animated Example Dijkstra Animated Example Dijkstra Animated Example Dijkstra Animated Example Dijkstra Animated Example Dijkstra Animated Example Dijkstra Animated Example Dijkstra's algorithm - Pseudocode dist[s] ←0 for all v ∈ V–{s} do dist[v] ←∞ S←∅ Q←V (distance to source vertex is zero) (set all other distances to infinity) (S, the set of visited vertices is initially empty) (Q, the queue initially contains all vertices) (while the queue is not empty) (select the element of Q with the min. distance) (add u to list of visited vertices) while Q ≠∅ do u ← mindistance(Q,dist) S←S∪{u} for all v ∈ neighbors[u] do if dist[v] > dist[u] + w(u, v) (if new shortest path found) then d[v] ←d[u] + w(u, v) (set new value of shortest path) (if desired, add traceback code) return dist An Example 2 0 4 4 2 2 1 1 2 6 3 4 2 3 Initialize Select the node with the minimum temporary distance label. 3 5 2 Update Step 2 0 4 4 2 2 1 1 2 6 3 4 2 3 4 3 5 Choose u such that N_(u) S 2 2 0 4 4 2 2 1 1 2 6 3 4 2 3 4 3 5 Update Step 6 2 2 0 4 4 2 2 1 1 2 4 2 3 The predecessor of node 3 is now node 2 6 3 3 5 4 3 4 Choose u Such That N_(u) 2 2 0 6 4 4 2 2 1 1 2 6 3 4 2 3 3 3 5 4 S Update 2 2 0 6 4 4 2 2 1 1 2 4 2 3 3 d(5) is not changed. 6 3 3 5 4 Choose u s.t . N_(u) 2 2 0 6 4 4 2 2 1 1 S 2 6 3 4 2 3 3 3 5 4 Update 2 2 0 6 4 4 2 2 1 1 2 4 2 3 3 d(4) is not changed 6 6 3 3 5 4 Choose u s.t. N_(u) 2 2 0 6 4 4 2 2 1 1 S 2 3 4 6 6 2 3 3 3 5 4 Update 2 2 0 6 4 4 2 2 1 1 2 3 4 2 3 3 d(6) is not updated 6 6 3 5 4 Choose u s.t. N_(u) 2 2 0 6 4 4 2 2 1 1 S 2 3 4 6 6 2 3 3 There is nothing to update 3 5 4 End of Algorithm 2 2 0 6 4 4 2 2 1 1 2 3 4 6 6 2 3 3 3 5 4 All nodes are now permanent The predecessors form a tree The shortest path from node 1 to node 6 can be found by tracing back predecessors 50 Kruskal’s Algorithm Starts with each vertex in its own component. Repeatedly merges two components into one by choosing a light edge that connects them (i.e., a light edge crossing the cut between them). Scans the set of edges in monotonically increasing order by weight. Uses a disjoint-set data structure to determine whether an edge connects vertices in different components. Example 1 1 1 4 6 4 3 2 2 4 5 3 5 6 8 7 4 3 7 6 Example 3 b 5 a 3 7 c 1 -3 11 d 0 e 2 f b 5 a 1 3 d 0 c -3 e f Example 4 A cable company want to connect five villages to their network which currently extends to the market town of Avonford. What is the minimum length of cable needed? 5 Brinleigh Cornwell 3 4 6 8 8 Avonford 7 Donster Fingley 5 4 2 Edan Ans: B 5 All vertices have been connected. C 3 6 8 The solution is 4 8 A D F 7 5 4 2 E ED 2 AB 3 CD 4 AE 4 EF 5 Total weight of tree: 18 Kruskal’s Algorithm : 2 8 14 21 19 25 9 17 5 13 1 Kruskal’s Algorithm F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E Sort the edges by increasing edge weight F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv edge dv (D,E) 1 (B,E) 4 (D,G) 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 Select first |V|–1 edges which do not generate a cycle F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv (B,E) 4 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) Select first |V|–1 edges which do not generate a cycle F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv (B,E) 4 (B,F) 4 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) 2 (E,G) Select first |V|–1 edges which do not generate a cycle F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv (B,E) 4 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) Accepting edge (E,G) would create a cycle Select first |V|–1 edges which do not generate a cycle F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv (B,E) 4 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) Select first |V|–1 edges which do not generate a cycle F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv (B,E) 4 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) Select first |V|–1 edges which do not generate a cycle F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv (B,E) 4 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) Select first |V|–1 edges which do not generate a cycle F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv (B,E) 4 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) Select first |V|–1 edges which do not generate a cycle F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv (B,E) 4 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) Select first |V|–1 edges which do not generate a cycle F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv (B,E) 4 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) Select first |V|–1 edges which do not generate a cycle F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv (B,E) 4 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) Select first |V|–1 edges which do not generate a cycle F 10 A 3 C 4 3 4 8 5 6 B 4 H 1 2 3 D 4 G 3 E edge dv (B,E) 4 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) Select first |V|–1 edges which do not generate a cycle 3 F C A 3 4 5 B H 2 3 G D 1 E edge dv (B,E) 4 2 (B,F) 4 (E,G) 3 (B,H) 4 (C,D) 3 (A,H) 5 (G,H) 3 (D,F) 6 (C,F) 3 (A,B) 8 (B,C) 4 (A,F) 10 edge dv (D,E) 1 (D,G) Done Total Cost = dv = 21 } not considere d 5 A 4 6 2 C B 2 1 3 E 3 D 2 4 F minimum- spanning tree A B 2 2 C D 1 3 E 2 F Example 1 1 1 4 6 4 3 2 2 4 5 3 5 6 8 7 4 3 7 6 function Kruskal(G = (N, A): graph; length: A - R+ ): set of edges {initialization} Sort A by increasing length n the number of nodes in N T Ø {will contain the edges of the minimum spanning tree} initialize n sets, each containing a different element of N {greedy loop} repeat e {u, v} shortest edge not yet considered ucomp find(u) vcomp find (v) if ucomp vcompthen merge(ucomp, vcomp) T T {e} until T contains n -1 edges return T The activity selection problem Problem: n activities, S = {1, 2, …, n}, each activity i has a start time si and a finish time fi, si fi. Activity i occupies time interval [si, fi]. i and j are compatible if sj fi. The problem is to select a maximum-size set of mutually compatible activities Example: i si 1 1 2 3 3 3 4 5 5 3 6 5 7 6 8 8 9 8 10 2 11 12 fi 4 5 6 7 8 9 10 11 12 13 14 The solution set = {1, 4, 8, 11} Solution of the example: i 1 si 1 fi 4 accept Yes 2 3 4 5 3 0 5 3 5 6 7 8 No No Yes No 7 8 9 6 8 8 10 11 12 No Yes No 10 2 13 No 11 12 14 Yes Solution = {1, 4, 8, 11} JOB SEQUENCING WITH DEADLINES The problem is stated as below. There are n jobs to be processed on a machine. Each job i has a deadline di≥ 0 and profit pi≥0 . pi is earned iff the job is completed by its deadline. The job is completed if it is processed on a machine for unit time. Only one machine is available for processing jobs. Only one job is processed at a time on the machine. JOB SEQUENCING WITH DEADLINES (Contd..) A feasible solution is a subset of jobs J such that each job is completed by its deadline. An optimal solution is a feasible solution with maximum profit value. Example : Let n = 4, (p1,p2,p3,p4) = (100,10,15,27), (d1,d2,d3,d4) = (2,1,2,1) JOB SEQUENCING WITH DEADLINES (Contd..) Sr.No. Feasible Solution (i) (1,2) (ii) (1,3) (iii) (1,4) (iv) (2,3) (v) (3,4) (vi) (1) (vii) (2) (viii) (3) (ix) (4) Processing Sequence (2,1) (1,3) or (3,1) (4,1) (2,3) (4,3) (1) (2) (3) (4) Profit value 110 115 127 25 42 100 10 15 27 is the optimal one JOB SEQUENCING WITH DEADLINES (Contd..) Example 2 : With n = 4 and the following values , (p1,p2,p3,p4) = (50,10,15,30) and (d1,d2,d3,d4) = (2,1,2,1) Find the Optimal solution. Example 2 : Let n = 4 , profit vector P = (30,35,10,25) , deadline vector D = (2,1,2,1) .Find the Optimal solution. GREEDY ALGORITHM TO OBTAIN AN OPTIMAL SOLUTION Consider the jobs in the non increasing order of profits subject to the constraint that the resulting job sequence J is a feasible solution. In the example considered before, the non-increasing profit vector is (100 27 p1 p4 15 10) p3 p2 (2 1 2 1) d1 d4 d3 d2 GREEDY ALGORITHM TO OBTAIN AN OPTIMAL SOLUTION (Contd..) J = { 1} is a feasible one J = { 1, 4} is a feasible one with processing sequence ( 4,1) J = { 1, 3, 4} is not feasible J = { 1, 2, 4} is not feasible J = { 1, 4} is optimal The Fractional Knapsack Problem Given: A set S of n items, with each item i having bi - a positive benefit wi - a positive weight Goal: Choose items with maximum total benefit but with weight at most W. If we are allowed to take fractional amounts, then this is the fractional knapsack problem. In this case, we let xi denote the amount we take of item i Objective: maximize b (x / w ) iS Constraint: x iS i i W i i Example Given: A set S of n items, with each item i having bi - a positive benefit wi - a positive weight Goal: Choose items with maximum total benefit but with weight at most W. “knapsack” Solution: • 1 ml of 5 • 2 ml of 3 • 6 ml of 4 • 1 ml of 2 Items: 1 2 3 4 5 Weight: 4 ml 8 ml 2 ml 6 ml 1 ml Benefit: $12 $32 $40 $30 $50 3 4 20 5 50 Value: ($ per ml) 10 ml Greedy solution for Fractional Knapsack Given a set of item I: Weight Cost w1 c1 w2 c2 w3 c3 w4 c4 Let P be the problem of selecting items from I, with weight K, such that the resulting cost(value) is maximum. 1. Calculate vi = ci / wi for i=1,2,3………n. Sort the items by decreasing vi. Let the sorted item sequence be 1,2,3…….n, and the corresponding v and weight be vi and wi respectely. 2. Greedy solution for Fractional Knapsack 3. Let K be the current weight limit (initially , k = K). In each iteration, we choose item I from the head of the unselected list. If k>=Wi , we take item I , and k=k-wi, then consider the next unselected item. If k<wi, we take a fraction f of item I, i.e. only take f=k/wi (<1) of item I, which weights exactely k. Then the algorithm if finished. function Kruskal(G = (N, A): graph; length: A - R+ ): set of edges {initialization} Sort A by increasing length n the number of nodes in N T Ø {will contain the edges of the minimum spanning tree} initialize n sets, each containing a different element of N {greedy loop} repeat e {u, v} shortest edge not yet considered ucomp find(u) vcomp find (v) if ucomp vcompthen merge(ucomp, vcomp) T T {e} until T contains n -1 edges return T Prim’s Algorithm function Prim (G= < N,A >:graph; length: A R+ ) ;set of edges {initialiazation} TØ B { an arbitrary member of N} while B ± N do find e ={u,v} of minimum length such that u B and v N\B T T {e} B B {v} return T Dijkstra's algorithm - Pseudocode dist[s] ←0 for all v ∈ V–{s} do dist[v] ←∞ S←∅ Q←V (distance to source vertex is zero) (set all other distances to infinity) (S, the set of visited vertices is initially empty) (Q, the queue initially contains all vertices) (while the queue is not empty) (select the element of Q with the min. distance) (add u to list of visited vertices) while Q ≠∅ do u ← mindistance(Q,dist) S←S∪{u} for all v ∈ neighbors[u] do if dist[v] > dist[u] + w(u, v) (if new shortest path found) then d[v] ←d[u] + w(u, v) (set new value of shortest path) (if desired, add traceback code) return dist Tower of Hanoi There are three towers The disks, with decreasing sizes, placed on the first tower You need to move all of the disks from the first tower to the second tower Larger disks can not be placed on top of smaller disks The third tower can be used to temporarily hold disks a b c Tower of Hanoi a b c Problem Characteristics Is the problem decomposable? Can solution steps be ignored or undone? Is the universe predictable? Is a good solution absolute or relative? Is the solution a state or a path? What is the role of knowledge? Does the task require human-interaction? Is the problem decomposable? Can the problem be broken down to smaller problems to be solved independently? Decomposable problem can be solved easily. 2 (x2 + 3x + sin2x.cos x) dx 2 x dx x3/3 3x dx 3x dx 2 2 sin x.cos x dx 2 (1− cos x)cos2xdx 3x2/2 cos2xdx − cos4xdx Can solution steps be ignored or undone? Ignorable (e.g. Theorem Proving), in which solution steps can be ignored. Recoverable (e.g. 8-puzzle), in which solution steps can be undone. Irrecoverable (e.g. Chess), in which solution steps can be undone. Is the universe predictable? The 8-Puzzle: Every time we make a move, we know exactly what will happen. Certain outcome! Playing Bridge: We cannot know exactly where all the cards are or what the other players will do on their turns. Uncertain outcome! Is a good solution absolute or relative? 1. Marcus was a man 2. Marcus was a Pompeian 3. Marcus was born in 40 A.D 4. All men are mortal. 5. All Pompeians died when the volcano erupted in 79 A.D 6. No mortal lives longer than 150 years. 7. It is now 2008 A.D Question: Is Marcus alive? What is the role of knowledge? Playing Chess Knowledge is important only to constrain the search for a solution. Reading Newspaper Knowledge is required even to be able to recognize a solution. Does the task require human-interaction? Is the solution a state or a path? The Water Jug Problem • The path that leads to the goal must be reported. A path-solution problem can be reformulated as a state- solution problem by describing a state as a partial path to a solution. The question is whether that is natural or not.

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