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![Adjacency List: Implementation
No. of nodes : N
No of edges : M
Vector<int> array[N+1];
While(M--)
{
cin>>a>>b;
Array[a].push_back(b);
Array[b].push_back(a)
}](https://image.slidesharecdn.com/graphtheory-250920073056-66fe90d9/85/Analysis-of-Algorithms-Graph-theory-pptx-10-320.jpg)
![Void dfs(int v)
{
visited[v] = 1;
cout<< v << “ -> “;
for(int i=0; i< array[v].size(); i++)
{
int child = array[v][i];
if(visited[child] == 0)
dfs(child);
}
}
Depth first Search Implementation](https://image.slidesharecdn.com/graphtheory-250920073056-66fe90d9/85/Analysis-of-Algorithms-Graph-theory-pptx-21-320.jpg)
Analysis of Algorithms Graph theory.pptx
- 1. Graph theory LOREM IPSUMDOLOR SIT AMET, CONSECTETUER ADIPISCING ELIT
- 2. Course Overview 1.Graph representation Dept first search Finding connected components Bipartite Graph Test(Two Coloring) Cycle Detection In/Out time of Nodes Finding Diameter of graph/Tree Finding edges (Cut Edge) Finding Articulation points (Cut Vertex) Finding Euler Circuits Breadth First Search Cycle detection using BFS Finding Shortest path from a given node (in unweighted graph) Finding strongly CC( Kosaraju’s Algorithm) Finding Strongly CC( Tojan’s Algorithm)
- 3. Each of thegraphs above is an independent connected component
- 4. These are nodes,which if removed from single connected components Creates multiple connected components (Red nodes in above diagram)
- 5. The edge whichif removed will create multiple connected components
- 6.
- 7. Graph Representation Wewill look at 2 types of graph representation techniques 1. Adjacency Matrix 2. Adjacency List
- 8. Problems with AdjacencyMatrix Method 1. Space Requirements This is easy technique but has O(n2 )space complexity If n=105 then space required will be = (105 )2 = 1010 4 x 1010 Byte = 4x107 KB = 4x104 =40GB 2. BFS or DFS traversal is of order O(n2 ) which is too slow 3. Adding a new node is not easy
- 9. 2. Adjacency List Canbe implemented using vectors in C++ or ArrayList in Java
- 10. Adjacency List: Implementation No. of nodes : N No of edges : M Vector<int> array[N+1]; While(M--) { cin>>a>>b; Array[a].push_back(b); Array[b].push_back(a) }
- 11. Depth First Search Each node can be assigned a level, for e.g Node 1 can be assigned level 0 Node 2 is only node at level 1 In DFS each level is traversed then it goes to another level How to do it? You visit a node and mark it 1 from 0 in visited array, to show its visited using adjacency list. We use stacks for operation
- 13. We mark 1as 1 in visited array and check nodes connected to node 1, we see too and it shows That it is not visited in visited arrya
- 14. So now weadd 2 to the stack and 1 will be in pause state at the moment
- 15. We reach node3 and mark it visited add it to stack and see if its connected nodes i.e. 2 is Visited or not, so we find it is visited, and has no more nodes so we pop it out from stack, and Process the top node now, which will be node 2 and checks another node, which is 4
- 18. Now node 4was put in stack, and checked its connected node, which is 2, and is already visited, so It goes to next node 5, which is not visited, so it adds 5 to stack and check its connected nodes, which We see is node 4, and shows its visited, so node 5 will be poped out and node 4’s next connection 6 Will be taken into consideration.
- 20. Node 6 hasnode 4 which is already visited, so 6 will pop out, then 4 will pop out as the pointer at 6 Shows no other element exists at 4, so it will pop out 4, then 2 then 1 and all nodes would have Been traversed by now
- 21. Void dfs(int v) { visited[v]= 1; cout<< v << “ -> “; for(int i=0; i< array[v].size(); i++) { int child = array[v][i]; if(visited[child] == 0) dfs(child); } } Depth first Search Implementation
- 22. Connected Components Connected componentis a set of vertices such that if you choose any two nodes from the set of Vertices then there exists path between the two. By definition a single node itself is a connected Component. Hence the graph contains three Connected components.
- 23. Strongly connected components Herewe have directed graph, where We have 3 strongly connected components. Strongly connected components are components Where when you chose two, there exists a path Between them (in directed graphs). Choosing any two blue nodes will have a path Back to it, but red to blue will not have blue to red Path or blue to yellow will not have reverse path. If we assume that graph is not directed, so it becomes One weakly connected component of a directed graph.
- 24. Counting number ofConnected Components in undirected graphs We have two CC, using dfs call to any of the node, the whole cc will Be traversed. We will can iterate from one node and iterate count, so in the end we will have two cc
- 25.
- 26.