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graphs | bipartite graph question added
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‎README.md

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@@ -187,6 +187,10 @@ it gets converted to i-1,j-1 or we take min or max of i,j-1 and i-1,j
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- Most of the graph questions revolves around finding the adjacent nodes and doing operations on it. Finding adjacent nodes is easy as you can get it from the matrix easily by seeing if the value corresponding to that vertex and some i is 1 and that it has to be unvisited. So basically we either do DFS or BFS.
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- When path between two vertices is to be found, they need not be connected directly. There may be some vertices/ nodes in between.
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- Sometimes a given matrix can be assumed to be a graph with a set of different rules and DFS or BFS can be applied on it.
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- Check question6 grapsh to check trick on how to access all surrounding elements of a cell in matrix using
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two arrays
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- In a bipartite graph a cycle will always be of even length
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## Advanced data
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@@ -443,6 +447,9 @@ In case of linked list find takes O(n) time and union also takes O(n) time and c
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- [Write a program to do topological sort in a graph](/graphs/question3.c)
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- [Find if there is a path between two vertices Vi and Vj in a directed graph](/graphs/question4.c)
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- [Given an undirected graph, find if it has a cycle or not](/graphs/question5.c)
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- [Given a 2D boolean matrix, find the number of islands](/graphs/question6.c)
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- [Check whether a given graph is biparite or not](/graphs/question7.c)
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- [Detect a cycle in a directed graph](/graphs/question8.c)
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### Pattern Matching

‎graphs/question5.c

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@@ -98,7 +98,7 @@ int dfs(int index, int parent){
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if(visited[temp->data] != 1){
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parent = index;
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dfs(temp->data, parent);
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}else if(visited[temp->data] == 1 && temp->data != parent){
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}else if(temp->data != parent){
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return 1;
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}
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temp = temp->next;

‎graphs/question6.c

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/*
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Given a 2D boolean matrix, find the number of islands
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One island is a single element or a group of ones. Group should be traversable from 1 to another in all directions
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but should be contigous
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Eg;
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101
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010
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101
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forms a single island only
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1010001
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1100000
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one with 101
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11
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and other with 1
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forms two islands and so on
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METHOD:
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We assume this matrix to be a graph where each cell is a node and adjacent nodes are all surrounding ones
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at 1 distance from it. At max a cell can have 8 adjacent nodes. There will only exist an edge between a node
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and its adjacent nodes if the value is 1 at the node surrounding it assumed to be the adjacent node.
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Time complexity: O(n^2 + 8n^2) //for each cell in worst case 8 other nodes will be traversed
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*/
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#include <stdio.h>
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#include <stdlib.h>
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#include <stdbool.h>
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int rows, columns;
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int isSafe(int arr[rows][columns], int i, int j, bool visited[rows][columns]){
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return (i >= 0) && (i < rows) && (j >= 0) && (j < columns) && (arr[i][j] && !visited[i][j]);
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}
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void dfs(int arr[rows][columns], int i, int j, bool visited[rows][columns]){
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static int rowNbr[] = {-1,-1,-1,0,0,1,1,1};
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static int colNbr[] = {-1,0,1,-1,1,-1,0,1};
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visited[i][j] = true;
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int k;
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for(k=0;k<8;k++){
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if(isSafe(arr, i+rowNbr[k],j+colNbr[k],visited)){
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dfs(arr, i+rowNbr[k], j+ colNbr[k], visited);
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}
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}
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}
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int findIslands(int arr[rows][columns]){
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bool visited[rows][columns];
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int i,j;
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int count = 0;
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for(i=0;i<rows;i++){
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for(j=0;j<columns;j++){
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if(arr[i][j] && !visited[i][j]){
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dfs(arr,i,j,visited);
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count++;
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}
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}
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}
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return count;
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}
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int main(){
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int arr[][5] = {
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{1, 1, 0, 0, 0},
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{0, 1, 0, 0, 1},
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{1, 0, 0, 1, 1},
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{0, 0, 0, 0, 0},
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{1, 0, 1, 0, 1}
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};
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rows = sizeof(arr)/sizeof(arr[0]);
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columns = sizeof(arr[0])/sizeof(arr[0][0]);
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printf("There are in total %d islands in the given matrix\n", findIslands(arr));
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return 0;
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}
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‎graphs/question7.c

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/*
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Check whether a given graph is bipartite or not
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A bipartitie graph is one whose vertices can be divided into two independent sets U and V such that every (U,V)
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either connects a vertex from U to V or V to U
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Or a graph which if represented in the form of a tree, alternating levels belong to two different sets
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METHOD:
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We will do a BFS
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1) A graph containing odd length of cycles is not bipartite.
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2) We can start coloring the vertex starting from any vertex. The start vertex lets say is blue, all its
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neighbours must be given yellow. Then all the ones with yellow its neighbours should be given blue.
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At any given point if an already colored vertex is colored again in an opposite color, then the graph is
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NOT bipartite else it is bipartite
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Time complexity: O(E+V)
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*/
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#include <stdio.h>
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#include <stdlib.h>
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#include <stdbool.h>
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#include <math.h>
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#define MAX 100
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int queue[MAX];
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int rear = -1, front = 0;
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int queueElementCount = 0;
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int visited[MAX];
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int color[MAX];
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struct AdjListNode{
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int data;
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struct AdjListNode *next;
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};
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struct AdjList{
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struct AdjListNode *head;
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};
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struct Graph{
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int vertices, edges;
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struct AdjList *map;
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} *newGraph;
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int dequeue(){
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queueElementCount--;
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return queue[front++];
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}
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void enqueue(int data){
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queueElementCount++;
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queue[++rear] = data;
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}
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bool isQueueEmpty(){
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return queueElementCount == 0;
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}
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struct AdjListNode *createNode(int data){
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struct AdjListNode *temp = (struct AdjListNode *)malloc(sizeof(struct AdjListNode));
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temp->data = data;
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temp->next = NULL;
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return temp;
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}
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void addEdge(int start, int end){
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struct AdjListNode *newNode = createNode(end);
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if(newGraph->map[start].head){
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newNode->next = newGraph->map[start].head;
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newGraph->map[start].head = newNode;
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}else{
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newGraph->map[start].head = newNode;
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}
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newNode = createNode(start);
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if(newGraph->map[end].head){
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newNode->next = newGraph->map[end].head;
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newGraph->map[end].head = newNode;
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}else{
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newGraph->map[end].head = newNode;
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}
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}
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void displayAdjList(){
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int i;
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for(i=0;i<newGraph->vertices;i++){
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struct AdjListNode *temp = newGraph->map[i].head;
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printf("%d --> ", i);
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while(temp){
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printf("%d --> ", temp->data);
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temp = temp->next;
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}
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printf("\n");
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}
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}
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void initColors(){
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int i;
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for(i=0;i<newGraph->vertices;i++){
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color[i] = -1;
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}
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}
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bool isBipartite(){
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initColors();
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enqueue(0);
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color[0] = 1;
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while(!isQueueEmpty()){
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int temp = dequeue();
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struct AdjListNode *tempVertex = newGraph->map[temp].head;
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while(tempVertex){
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if(color[tempVertex->data] == -1){
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enqueue(tempVertex->data);
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}
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if(color[temp] == 1){
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if(color[tempVertex->data] != -1 && color[tempVertex->data] != 2){
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return false;
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}
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color[tempVertex->data] = 2;
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}else{
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if(color[tempVertex->data] != -1 && color[tempVertex->data] != 1){
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return false;
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}
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color[tempVertex->data] = 1;
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}
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tempVertex = tempVertex->next;
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}
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}
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return true;
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}
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int main(){
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int vertices = 5;
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newGraph = (struct Graph *)malloc(sizeof(struct Graph));
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newGraph->vertices = vertices;
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newGraph->map = (struct AdjList *)calloc(sizeof(struct AdjList),vertices);
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addEdge(0,2);
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addEdge(0,1);
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addEdge(2,3);
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addEdge(1,3);
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displayAdjList();
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if(isBipartite()){
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printf("Graph is bipartite\n");
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}else{
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printf("Graph is not bipartite\n");
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}
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int i;
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for(i=0;i<vertices;i++){
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printf("%d ", color[i]);
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}
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return 0;
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}
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‎graphs/question8.c

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/*
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Detect a cycle in a directed graph
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Will be done once back edges are done
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*/

‎nextquestions.md

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- Graph topological sort method2 and matrix implementation of topo
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- Write a program to detect a cycle in a graph (question5) method2
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- breadth and depth first traversal in a tree
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- Graphs question6 union and find method to be done
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- Graphs question8 to be done along with backedges
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