CS213d Data Structures and Algorithms

Transcription

1 CS2d Data Structures and Algorithms Graphs-II Milind Sohoni IIT Bombay and IIT Dharwad March 7, 207 / 5

2 5,5M,5,5M 5,5M,5,5M 6 6 5,5M,5,5M 4,76 Belagavi 4,76 Hubballi Jn. 4,76 7,69 85 Shivamogga Town 5,,4 4,70 Amaravati Colony 5,,42 82 Chikkamagaluru 4, ,70,2,9 4,70 6,2,9 Vijayapura Hosapete Jn. 79 4, ,0,,2,,7 5,0,,,8 68 5,0,,,8 5,0,,,8 6 8 Mysuru Jn. 84 5,0,,, ,0, ,65,4 8 Tumakuru 70 5, ,6 0 24,92 8 5,74 6,0,9,6 Palani 62 7,,4,6 09 Raichur Ballari Jn ,7 4,7 4,7 4,7,6, ,72 7,6 7,6 26,08 2,9,4,9 58 Sri Pandurang Swami Road 4,7 4, ,0,6 9,46 7,9,50 04,4,9,6,7 6,72 6, , ,,05,05,4,9, ,46 25,07 7 6,0,6 Namakallu, ,46 7,50 8,57 8,57 Gadwal 8,57 8,60 29, ,86 20,87 9,90 Nandiyal Banaganapalli Nosam 96 Yerraguntla,94 8, ,8,50 9,8,5 Pattukottai 5 55a Mannargudi 67 9,90 0,05 8,97,0,5 56 7,8,50 9,8,5 2,2, , 7,8,48 8, ,5 88 2,2,86 2,22,99 54 Velankanni 88 20,29 8,97,,0,5 5 2,98 2,4 27, ,22,00 2,22,99 0,2,22 2,22, ,00 The Graph 5,, ,69 7,69 7, ,7 4,70 4,70 Nelamangala Bengaluru City,6,7,0, ,87 0,0 7,8, 7,6 7,6 March 7, / 5

3 Connected-ness For two vertices i, j V, we define the relation i j (and say that i is connected to j), if there is a path π = (v,..., v k ) such that v = i and v k = j. We also say that, there is a path from i to j. Lemma The relation on the vertices is an equivalence relation. MRJ(2) VSG(0) LD() UBL() JRU(5) BAY(4) RBB(6) March 7, 207 / 5

4 Proof Strategy Step I: If there is a trail π = (v,..., v n ), then there is a path µ = (w,..., w k ) such that v = w and v n = w k. Proof: Induction on n. Suppose that all vertices in π are distinct, then π is itself a path.. If not, suppose that v i = v j with i < j. Then construct the trail Now apply induction. π = (v,..., v i, v i, v j+,..., v n ) Step II. If v w and w x then there are paths π = (v = v,..., v k = w) and µ = (w = w,..., w r = x). Consider the trail: α = (v = v,..., v k = w,..., w r = x) This trail starts from v and ends at x. Now apply Step I above. This proves transitivity. Reflexivity and symmetry are easy. March 7, / 5

5 Distance Definition: If G(V, E) is a graph and v, w are vertices such that v w, then the distance d(v, w) is the length of the shortest path connecting v to w. If v w then we use d(v, w) =. Lemma For any vertices u, v, w V, we have d(u, v) + d(v, w) d(u, w). This is called the triangle inequality. Proof: Let π : u v implement d(u, v) and µ : v w implement d(v, w), then π µ is a trail from u to w. Thus, there is a path which is of the sum of the distances, or shorter. MRJ(2) VSG(0) LD() UBL() JRU(5) BAY(4) RBB(6) March 7, / 5

6 An application The minimum-change-over problem: For a railway network G(V, E), let π, π 2,..., π k be the paths of k trains in the network. Suppose that we want to go from station v to w with the fewest number of changes. Find the sequence of trains which will take me from v to w. MRJ(2) VSG(0) LD() UBL() JRU(5) BAY(4) RBB(6) Example: π = (2, ), π 2 = (0,,, 4), π = (, 5), π 4 = (4, 5, 6). Shortest route 2 6 is 2,,, 5, 6. Fewest changes π, π 2, π 6, i.e., 2,,, 4, 5, 6. March 7, / 5

7 The model Station-Train Graph: G (V, E ) defined as follows. (i) V is the stations of G. For v, v V, we say that (v, v ) is an edge if there is a train (say π i ) going from v to v. In which case, we put edata=i. MRJ(2) VSG(0) LD() UBL() JRU(5) BAY(4) RBB(6) Example: π = (2, ), π 2 = (0,,, 4), π = (, 5), π 4 = (4, 5, 6). Shortest route 2 6 is 2,,, 5, 6. Fewest changes π, π 2, π 6, i.e., 2,,, 4, 5, 6. Lemma: The shortest path µ from v to w in the train graph corresponds to the smallest sequence of changes. Prove. March 7, / 5

8 Shortening Loops Definition: For a graph G(V, E), let π = (v,..., v k ) be a loop. An index i is called a U-turn if v i = v i+. Lemma If π is a loop without U-turns then there is a subsequence µ of π which is a cycle. Problem. Given a sequence π check if π is a loop. Also, check if there is a subsequence µ of π which is a cycle. MRJ(2) VSG(0) LD() UBL() JRU(5) BAY(4) RBB(6) March 7, / 5

9 Connected components Each equivalence class is called a connected component of the graph. The graph is called connected if V is itself an (unique) equivalence class. Lemma Let G(V, E) be a graph, then V may be expresses as a disjoint union V = V... V k and E as a disjoint union E = E... E k, where E i ( V i ) 2, such that G(Vi, E i ) is connected, for all i. The quantity k is called the number of components of G. MRJ(2) 2 4 VSG(0) LD() UBL() RBB(6) JRU(5) BAY(4) 5 6 Connected Components: {,5}, {2,,6}, {4} March 7, / 5

10 Why connected components What happens if there is a transmission line failure? Which edges should be strengthened? How to put redundancy? March 7, / 5

11 Finding connected components Problem: Given a graph G(V, E), write a program to output its connected components. How do we output a connected component? As a bit vector [0 0 0 ] to denote {, 2, 5}. For a given vertex i, how do we find its connected component? Look at G.Vlist[i].eindex. Look at all tails of edges here. Then look at these vertices, and so on. Recursion?. vec[i]=; queue q; q.push(i); while (q.nos>0) { ii=q.pop(); get new neighbours jj of ii; vec[jj]=; q.push(jj);}; March 7, 207 / 5

12 Suppose we want to find all vertices connected with i for (j=0;j<g.n;j++) compo[j]=0; q.push(i); visited[i]=; while(!q.empty()) { jj=q.front(); q.pop(); // next vertex to be expande compo[jj]=; a=g.vlist[jj].eindex; if (a>=0){ k=a; while (G.Elist[k].head==jj&& k<2*g.m) { // push onto q when unvisited }; // of while };// of if a>0 };// of while q not empty March 7, / 5

13 Pushing onto the queue, if unvisited. kk=g.elist[k].tail; if (visited[kk]==0) { visited[kk]=; q.push(kk); cout << "pushing "<< kk << "\n"; }; k=k+; v queue VSG(0) MRJ(2) LD() UBL() RBB(6) JRU(5) BAY(4) March 7, 207 / 5

14 Some assertions (i) A phase begins with an empty stack and a vertex i which is unvisited.it begins with with an empty queue and i being pushed, and ends when the queue next gets empty. (ii) In this, there is a global visited which is updated and a compo which begins afresh. (iii) In each phase, there are several expansions, i.e., which begin with a pop and result in several vertices being added to the queue. (iv) For every vertex i which starts a phase, the vertices j which enter the queue in the phase are exactly the ones which (a) are labeled in compo, and (b) for which there is a path π i (j) from i to j. It is the last which needs proof. We prove by induction on the order in which the vertices enter the queue. March 7, / 5

15 Proof continued Clearly, when i enters, there is a path from i to i. A vertex kk enters the queue (i) during the expansion of jj which was already in the queue, and (ii) there was an edge (jj, kk). Thus, there already exists a path π i (jj) and π i (kk) = π i (jj) (jj, kk). But will every vertex in the connected component of i enter the queue? Let (i = v, v 2,..., v k, v k = kk) be a path i kk. Without loss of generality, we may assume that all v,..., v k are visited in the phase for i. Next, if v k has entered the queue, then clearly kk will be considered in the expansion of v k. Either it will enter the queue or kk was already visited!. March 7, / 5

16 Suppose we want to find the distance of all vertices connected to i for (j=0;j<g.n;j++) compo[j]=0; q.push(i); visited[i]=0; // distance is 0 while(!q.empty()) { jj=q.front(); q.pop(); // next vertex to be expande compo[jj]=; a=g.vlist[jj].eindex; if (a>=0){ k=a; while (G.Elist[k].head==jj&& k<2*g.m) { // update distance and push onto q }; // of while };// of if a>0 };// of while q not empty March 7, / 5

17 Pushing onto the queue, if unvisited. kk=g.elist[k].tail; if (visited[kk]==-) { visited[kk]=+visited[jj]; q.push(kk); cout << "pushing "<< kk << "\n"; }; k=k+; Output=[ ] MRJ(2) VSG(0) LD() UBL() JRU(5) BAY(4) RBB(6) Why does it work? March 7, / 5

18 If there is a path, there may be several shortest paths. If µ = (v = v, v 2,..., v r, v r = w) is a shortest path then µ a = (v,..., v a ) are all shortest paths to the respective vertices. Within a phase, at any moment in the queue (i) the distance vector of the elements are in increasing order, and (ii) no more than 2 consecutive numbers may appear. The path π i (jj) is of length visited[jj]. visited actually assigns the distance of a vertex from i, the leader of the phase. Again by induction, suppose that kk is at distance d and has been labeled r > d. Whence, there is a vertex jj at distance d which was labeled correctly. Whence kk would have been considered during the expansion of jj! See what happens when we change queue with a stack! March 7, / 5

19 An example bfs 6 dfs queue edges: (0,),(,2),(,8),(2,),(2,7),(8,9),(,4),(7,6),(4,5) stack edges: (0,),(,2),(,8),(8,7),(8,9),(7,6),(6,5),(5,4),(4,) March 7, / 5

20 Modeling The shortest-path algorithm (with a slight modification) also works for: Directed Graphs, i.e., where the edge may be only one-way. Edges with lengths, i.e., an edge may not be unit length but of some positive length. 5 2 Both of these give us lots of applications March 7, / 5

21 Application -room Guest-House Booking for period [,8]. arpit 5 usha srirang 4 7 om 2 4 baba 5 7 Maximize bookings. Service Graph: V = {b,..., b n }, bookings. E = {(b, b ) if b can follow b} u a 0 b s o cost(b,b ): number of days lost. Special start and end node. March 7, / 5

22 Application 2 The Professor Drop Problem : professors have to be dropped by a car which must start from IIT and come back to IIT. Find the shortest route for the car. This problem is called the Travelling Salesman Problem in literature. Let us see how to convert it into a shortest path problem. Note that this does not make the problem easier, it is just to show how to model! Let G(V, E) be the graph obtained by V = {0, a, b, c}, i.e., the locations of IIT, and the professors. Let d(v, w) be the length of the shortest route from v to w. 0 c 4 2 a b March 7, / 5

23 The layered model From this, we construct a gigantic graph GG(VV, EE), where VV is the collection of all partial sequences c 0ab 0abc a b 0a 0ac 0ba 0acb 0bac 0 0b 0bc 0bca 0c 0ca 0cab 0cb 0cba The Modeling Lesson Choice of vertices and edges important. What data to use and what problem to map to. March 7, / 5

24 Trees Definition. A graph G(V, E) is a tree if it is (i) connected, and (ii) acyclic, i.e., contains no cycles. (a) (b) (c) Recall: degree of a vertex is the number of edges coming out of it, i.e., the number of vertices adjacent to it. Lemma : If G(V, E) is a tree, then it has at least two vertices of degree. March 7, / 5

25 Proof: We choose v as the solitary vertex of degree, if there is only one such vertex, or an arbitrary vertex, if there is no vertex of degree. Next, we choose v 2 to be adjacent to v. We proceed thus, choosing v i+ to be adjacent to v i but distinct from v i. This can always be done since every v i with i 2 has degree at least 2. Since this sequence is potentially infinite, there must be a first r such that v r = v s, with s < r. Let us look at the sequence µ = (v s, v s+,..., v r, v r ). By the choice of r, there are no duplications in µ. Furthermore, every pair of adjacent vertices in the list actually make an edge. Thus µ is a cycle, which contradicts the hypothesis that G is acyclic. March 7, / 5

26 Proposition. (i) If G(V, E) is a tree then E = V. (ii) If G(V, E) is a connected graph with V edges, then it must be acyclic, and hence a tree. (iii) If G(V, E) is an acyclic graph with E = V edges, then it must be connected, and hence a tree. Proof: We prove this by induction on V. We know, by lemma, that there is a vertex v with only one edge e = (v, w) incident at v. Consider the graph G (V, E ) with V = V v, and E = E e. We claim that G is a tree. It is clearly acyclic, since E E and E allowed no cycles. What remains is to show that G is connected. For that, note that if x, y V, then x, y V as well. Since, G was connected, there was a path from π from x to y. We next argue that this path does not use the edge e, and thus lies completely in E. For if it did, then the path would terminate at v, since there is no exit from v. However, v V, and there v x, y. Thus π lies completely in E proving that x is connected to y in G. March 7, / 5

27 Thus G is a tree, and we have, inductively, E = V, which proves the same for G as well. Now we prove part (ii). First note that if α is a cycle in the graph G, and e α is an edge on the cycle, then G (V, E ) with E = E e is also connected. This is because any path using the edge e may now use the cycle α and go the other way. Whence, beginning with a connected graph G, we may sequentially delete edges from cycles, thus maintaining connectivity but reducing the number of edges. At some point this must stop and we will have a connected acyclic graph G (V, E ). Since G is then a tree, we have E = V = E. this means that G was acyclic to begin with. March 7, / 5

28 Finally, we prove (iii). Suppose G is not connected, we have, by lemma 4, a decomposition of G into its connected components. Whence, the identity E = V, tells us that G has a component G(V i, E i ) such that E i E i. Consequently, it is clear from part (i) that it cannot be a tree, and thus must contain a cycle. Thus G itself must contain a cycle, which contradicts the hypothesis. Definition: Given a connected graph G(V, E), a subset of edges E is called a spanning tree if G (V, E ) is a tree. March 7, / 5

29 An Example How do we know?. In this case, easy. ( 5 ) = 0. Two cycles of length. Thus 0 2 = 8. In General: The total is available through a deep theorem. March 7, / 5

30 bfs and dfs example Theorem: The edges printed by programs sptree or sptreestack give a spanning tree of G. These trees are called the breadth-first-search and depth-first-search respectively. jj=q.front(); q.pop()... kk=g.elist[k].tail; if (visited[kk]<0) { visited[kk]=+visited[jj]; q.push(kk); cout << "pushing "<< kk << "\n"; cout << "Edge ("<<jj << ","<< kk << ")\n"; }; k=k+; Proof : There are exactly n edges printed. Next, it is connected since π i (jj) is a path from i to every jj. March 7, / 5

31 Our Example bfs 6 dfs bfs edges: (0,),(,2),(,8),(2,),(2,7),(8,9),(,4),(7,6),(4,5) dfs edges: (0,),(,2),(,8),(8,7),(8,9),(7,6),(6,5),(5,4),(4,) Question: How do we detect if a given graph is a tree? Simple Answer: Check number of connected components, check number of edges. Deeper Answer: Does it have a non-tree edge? Check during expansion of a vertex. March 7, 207 / 5

32 sptree again jj=q.front(); q.pop()... kk=g.elist[k].tail; if (visited[kk]<0) { visited[kk]=+visited[jj]; q.push(kk); parent[kk]=j cout << "pushing "<< kk << "\n"; cout << "Edge ("<<jj << ","<< kk << ")\n"; }; if ((visited[kk]>=0)&&(parent[jj]!=kk) cout << "cycle detected\n"; k=k+; Can you print the cycle? March 7, / 5

33 Minimum Cost Spanning Trees Problem : Given a graph G(V, E) with edge costs, i.e., c : E R, find a spanning tree T E of minimum cost, i.e., c(t ) = e T c(e) is minimized March 7, 207 / 5

34 In Real Life March 7, / 5

35 March 7, / 5