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Lecture 11. Graph Algorithms. Definitions. Graph is a set of vertices V, with edges connecting some of the vertices (edge set E). . An edge can connect two vertices. An edge between vertex u and v is denoted as (u, v).

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## Lecture 11

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**Lecture 11**Graph Algorithms**Definitions**Graph is a set of vertices V, with edges connecting some of the vertices (edge set E). An edge can connect two vertices. An edge between vertex u and v is denoted as (u, v). • In a directed graph (digraph) all edges have directions. In an undirected graph, an edge does not have a direction. • Unless otherwise mentioned a graph is undirected**A**B E F C D Real life examples where graphs are useful Cities and Roads • Networks: • Routers are vertices • Links are edges • Shortest path between vertices. Constraint representation: If A is there, B can not be there, etc.**A**A B B E E F F C C D D A vertex v is adjacent to vertex u, if there is an edge (u, v), e.g., A, B In an undirected graph, existence of edge (u, v) means both u and v are adjacent to each other. In a digraph, existence of edge (u, v) does not mean u is adjacent to v. E is adjacent to B, but B is not adjacent to E**A**B E F C D An edge may have a weight, e.g., distance of a highway 0.4 0.5 0.1 0.1 0.3 0.2 A path in a graph is a sequence of vertices w1 w2 …..wP such that consecutive vertices wi wi+1 have an edge between them, i.e., wj+1 is adjacent to wj ABE is a path**A path in a graph is simple if all vertices are distinct,**except possibly the first and the last one, e.g., ABE, but ABEB is not a simple path, unless otherwise stated paths are simple. Length of a path is the number of edges in the path. What is the length of a simple path of path ABE? 2 A cycle is a path of length at least 1 such that the first and the last vertices are equal. Example? BCDEB A cycle is simple if the path is simple, e.g., BCDEB For undirected graph, we require a cycle to have distinct edges. Length of a cycle is the number of edges in the cycle. What is the length of a simple cycle of P vertices? P**A**A B B E E F F C C D D Connected Graphs A graph is connected if there is a path from every vertex to every other vertex. A directed graph with this property is strongly connected. If a directed graph is not strongly connected, but underlying undirected graph is connected then the directed graph is weakly connected. Connected Weakly connected, cant go from E to A**Subgraphs and Components**A subgraph of a graph is a graph which has a subset of vertices and a subset of edges of the original graph. • A component is a subgraph which satisfies two properties • is connected, and • Maximal with respect to this property. ``Connected ‘’ will be replaced by ``strongly connected’’ for digraphs.**A**A B B E E F F C C D D H G Entire graph is a component Two components: ABCDEF GH**A**B E F C D 3 components BCDE A F**B**E C D Complete Graph A graph which has edge between any vertex pair is complete, e.g., BCDE A complete digraph has directed edges between any two vertices. How many edges can a complete graph of N vertices have? N(N-1)/2 How about a digraph? N(N-1)**Representation of Graphs**Adjacency matrix: Let there be N vertices,0, 1,….N-1 Declare a N x N array A(adjacency matrix) A[j][k] = 1 if there is an edge (j, k) = 0 otherwise Storage? O(N2)**A**B E F C D Adjacency matrix: A B C D E F A 0 1 0 0 0 0 B 1 0 1 0 1 0 C 0 1 0 1 0 0 D 0 0 1 0 1 0 E 0 1 0 1 0 1 F 0 0 0 0 1 0**A**B E F C D Adjacency matrix: A B C D E F A 0 1 0 0 0 0 B 0 0 1 0 1 0 C 0 0 0 1 0 0 D 0 0 0 0 1 0 E 0 0 0 0 0 1 F 0 0 0 0 0 0**What will be a the structure of A for an undirected graph?**A[j][k] = A[k][j] symmetric Weighted graph A[j][k] = weight of edge (j, k) if edges have weights = a very large or very small value if edge (j, k) does not exist**Adjacency List**If the graph is complete or almost complete, then adjacency matrix representation is fine, otherwise O(N2) storage is used even though there are fewer edges Adjacency list is a more efficient storage. Every vertex has a linked list of the vertices which are adjacent to it. Read Section 9.1**A**B E F C D A B B A,C,E C B,D D C,E E B,D,F F E**A**B E F C D A B B C,E C D D E E F F**Storage: O(V+E)**However, problem with adjacency list is that one may have to traverse the entire link list corresponding to a vertex in order to locate an edge.**Sparse and Dense Graphs**Sparse graphs have (V) edges. Dense graphs have (V2) edges. (V) What is the storage in adjacency list for sparse graphs? dense graphs? adjacency matrix for sparse graphs? dense graphs? (V2) (V2) (V2) Adjacency list is better for sparse graphs, adjacency matrix for dense graphs**A**B E F C D Degree Relations Number of edges incident from a vertex is the degree of the vertex in a graph. 1 Deg(A) ? 3 Deg(B) ? Number of edges ending at a vertex is the indegree of the vertex in a digraph. Number of edges originating from a vertex is the outdegree of the vertex in a digraph.**A**B E F C D Indeg(A) ? 0 1 Indeg(B) ? Outdeg(A) ? 1 2 Outdeg(B) ?**A**A B B E E F F C C D D For a graph, Sum of degrees of all vertices = 2. Number of edges Sum of degrees ? 12 Number of edges ? 6 For a digraph, sum of indegrees of all vertices = sum of outdegrees of all vertices = sum of edges Sum of indegrees ? 6 Sum of outdegrees ? 6**Graph Traversal**Breadth First Search • Starts from a vertex s (source), and discovers all vertices which are reachable from s • A vertex v is reachable from s if there is a path from s to v. Start from a vertex v Go to all vertices adjacent to v For each of these adjacent vertices, go to all vertices adjacent to these**A**B E F C D Start from C Go to B, D From B go to A, E From E go to F Vertices are discovered in order of increasing shortest path lengths from start vertex Order of discovery starting from C: B,D,A,E,F Shortest path lengths from C: B,D(1), A,E(2), F(3)**We use a FIFO queue in the search which is initially empty.**Initially all vertices are colored white. When a vertex is ``discovered’’, it is colored gray and put in the FIFO queue First source is discovered, then the vertices adjacent to source, etc. Remove vertices from the queue in FIFO order, discover its undiscovered neighbors and put them in the FIFO queue When all vertices adjacent to a vertex v have been discovered, the algorithm finishes processing v, and colors it black.**A**A A B B B E E E F F F C C C D D D Discover source C, enqueue it Queue = C Dequeue C Discover B, D, make them gray Queue = B,D Blacken C Dequeue B, Discover A E, make them gray Queue = D,A, E, Blacken B Dequeue D, nothing new to discover Queue = A,E, Blacken D**A**A B B E E F F C C D D Dequeue A, Nothing new to discover Queue = E, Blacken A Dequeue E, Discover F, make it gray Queue = F, Blacken E Dequeue F, Nothing new to discover Queue = empty, Blacken F**Notation:**d[u] is the length of the shortest path from s to u color[u] is the color of vertex u pred[u] is the predecessor of u in the search BFS is used to measure the shortest paths by maintaining d[u] BFS generates a tree by maintaining the predecessors pred(C ) = Null, pred(B ) = pred(D ) = C , pred(A ) = B, pred(E ) = B, pred(F ) = E**Pseudocode**BFS(G,s) { For each v in V, {color[v]=white; d[u]= INFINITY; pred[u]=NULL} color[s] = gray; d[s]=0; Queue = {s}; While Queue is nonempty { u = Dequeue[Q];**For each v in Adj[u], {**if (color[v] = white) /*if v is discovered*/ { color[v] = gray; /*Discover v*/ d[v] = d[u] + 1; /*Set distance of v*/ pred[v] = u; /*Set pred of v*/ Enqueue(v); /*put v in Queue*/ } } Color[u] = black; /*done with u*/ } } Work out predecessors and distance labels in he previous example.**Complexity Analysis**• A vertex is visited once. • Thus the while loop is executed at most V times. Complexity of operations inside the for loop is constant. We want to compute the number of times the for loop is executed. For each vertex v the for loop is executed at most (deg v + 1) times. The factor 1 comes as for a 0 degree vertex we need a constant complexity**Thus the for loop is executed v (deg v + 1) times**This equals V + 2E Initialization complexity is V Thus overall we have complexity V + V + 2E, i.e. O(V+E)**Depth First Search**Another graph traversal process. We want to visit all rooms in a castle. Start from a room Move from room to room till you reach an undiscovered room Draw a graffiti in each undiscovered room Once you reach a discovered room take a door which you have not taken before.**A**B E F C D Start from C Go to B From B go to A Return to B From B go to E From E go to F Return to E, B, C Go to D**Will have 3 possible colors for a vertex:**white for an undiscovered vertex gray for a discovered vertex black for a finished vertex Will store predecessor Will store 2 numbers for each vertex (timestamps) When we first discover a vertex store a counter d[u] When you finish off store another f[u] d[u] is not the distance**Pseudocode**DFS(G) { For each v in V, {color[v]=white; pred[u]=NULL} time=0; For each u in V If (color[u]=white) DFSVISIT(u) } time is a global variable**DFSVISIT(u)**{ color[u]=gray; d[u] = ++time; For each v in Adj(u) do If (color[v] = white) { pred[v] = u; DFSVISIT(v); } color[u] = black; f[u]=++time; }**A**A A A B B B B E E E E F F F F C C C C D D D D Discover source C, d[C] = 1 Discover B, make it gray d[B] = 2 Discover A, make it gray d[A] = 3 Finish A, f[A] = 4, Blacken A Discover E, make it gray, d[E]=5**A**A A B B B E E E F F F C C C D D D Discover F, d[F] = 6 Finish F, f[F]=7 Discover D, d[D]=8 Finish D, f[D]=9, Finish E f[E] = 10, Finish B, f[B] = 11, Finish C, f[C] = 12 pred[B]=C, pred[A]=B, pred[E]=B, pred[F]=E, pred[D] = E**Complexity Analysis**Note down example from board There is only one DFSVISIT(u) for each vertex u. Let us analyze the complexity of a DFSVISIT(u) Ignoring the recursion calls the complexity is O(deg(u)+1) We consider the recursive calls in separate DFSVISIT(v) Initialization complexity is O(V) Overall complexity is O(V + E)**A**B E F C D DFS Tree Structure Consider a directed graph. Observe that if u is predecessor of v in DFS, there is an edge (u, v) in the graph.All such edges are predecessor edges or tree edges. The predecessor edges constitute an acyclic graph (DFS tree)**A**B E F C D If there is a path from v to u in the DFS tree, then u is an ancestor of v in DFS tree and v is a descendant of u. B is ancestor of F F is descendant of B**A**B E F A C C D B An edge (u, v) in the graph such that v is an ancestor of u is a ``back edge’’, e.g., edge CD An edge (u, v) where v is a direct descendant of u is a forward edge, e.g., BE An edge (u, v) where u is not an ancestor nor descendant of v is a cross-edge. Do cross edges exist in undirected graphs? No Are all edges either forward, cross and back? No in digraph (e.g., AC), yes in graph Other edges are called tree edges**2,5**1,6 3,4 7,10 8,9 A C B D E DFS starts from A, order A,B,C,E,D Forward edge: AB, BC, ED Tree edge: AC Cross edge: DB Backward edge: CA d[A] = 1, d[B] = 2, d[C]=3, f[C]=4, f[B]=5, f[A]=6, d[E]=7,d[D]=8,f[D]=9,f[E]=10**Relation between timestamps and ancestry**u is an ancestor of v if and only if [d[u], f[u]] [d[v], f[v]] u is a descendant of v if and only if [d[u], f[u]][d[v], f[v]] u and v are not related if and only if [d[u], f[u]] and [d[v], f[v]] are disjoint These relations are called parenthesis lemma**Applications of DFS**DFS can be used to find out whether a graph or a digraph contains a cycle. Consider a digraph. It has a cycle if and only if the graph has a back edge. The same holds for graphs. Run DFS Check the nature of every edge (How do you know whether an edge is a back edge or not?) If there is a back edge, then the graph has a cycle. Complexity? O(V + E)**Now we show that a digraph has a cycle if and only if there**is a back edge. If there is a back edge there is a cycle. We show that if there is a cycle, there is a back edge. Consider an edge (u, v) in a digraph. If it is a back edge, then f[u] f[v]. Otherwise (for tree, forward, cross edges) f[u] > f[v] We show it as follows. For tree, back and forward edges, the result follows intuitively as well as from the parenthesis lemma.**For cross edge (u, v) note that the intervals [d[u], f[u]]**and [d[v], f[v]] are disjoint. When we were processing u, v was not white otherwise (u, v) will be a predecessor edge Thus processing v started before processing u. Thus d[v] < d[u]. Since the intervals are disjoint, this means f[v] < f[u]. Now we show that if there is a cycle, there is a back edge.**Suppose there is no back edge.**Move along any path. All edges are tree forward or cross edges. Thus the finish times decrease monotonically. Hence we don’t come back to the same vertex. Thus there is no cycle.

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