Showing posts with label Data Structures. Show all posts
Showing posts with label Data Structures. Show all posts

Saturday, 17 June 2017

Minimal Spanning Tree

What is a Spanning Tree?

Given an undirected and connected graph G=(V,E), a spanning tree of the graph G is a tree that spans G (that is, it includes every vertex of G) and is a subgraph of G (every edge in the tree belongs to G)

What is a Minimum Spanning Tree?

The cost of the spanning tree is the sum of the weights of all the edges in the tree. There can be many spanning trees. Minimum spanning tree is the spanning tree where the cost is minimum among all the spanning trees. There also can be many minimum spanning trees.
Minimum spanning tree has direct application in the design of networks. It is used in algorithms approximating the travelling salesman problem, multi-terminal minimum cut problem and minimum-cost weighted perfect matching. Other practical applications are:
  1. Cluster Analysis
  2. Handwriting recognition
  3. Image segmentation
enter image description here


Reference: https://www.hackerearth.com/practice/algorithms/graphs/minimum-spanning-tree/tutorial/

DSA Lab 10 - Prim's Algorithm

Task:

In this task you have to implement the the prim's algorithm in you "MyGraph" class which you have created in lab 8.


Helping Materiel:

Prim’s Algorithm

Prim’s Algorithm also use Greedy approach to find the minimum spanning tree. In Prim’s Algorithm we grow the spanning tree from a starting position. Unlike an edge in Kruskal's, we add vertex to the growing spanning tree in Prim's.
Algorithm Steps:
  • Maintain two disjoint sets of vertices. One containing vertices that are in the growing spanning tree and other that are not in the growing spanning tree.
  • Select the cheapest vertex that is connected to the growing spanning tree and is not in the growing spanning tree and add it into the growing spanning tree. This can be done using Priority Queues. Insert the vertices, that are connected to growing spanning tree, into the Priority Queue.
  • Check for cycles. To do that, mark the nodes which have been already selected and insert only those nodes in the Priority Queue that are not marked.
Consider the example below:
enter image description here
In Prim’s Algorithm, we will start with an arbitrary node (it doesn’t matter which one) and mark it. In each iteration we will mark a new vertex that is adjacent to the one that we have already marked. As a greedy algorithm, Prim’s algorithm will select the cheapest edge and mark the vertex. So we will simply choose the edge with weight 1. In the next iteration we have three options, edges with weight 2, 3 and 4. So, we will select the edge with weight 2 and mark the vertex. Now again we have three options, edges with weight 3, 4 and 5. But we can’t choose edge with weight 3 as it is creating a cycle. So we will select the edge with weight 4 and we end up with the minimum spanning tree of total cost 7 ( = 1 + 2 +4).
Implementation:
#include <iostream>
#include <vector>
#include <queue>
#include <functional>
#include <utility>

using namespace std;
const int MAX = 1e4 + 5;
typedef pair<long long, int> PII;
bool marked[MAX];
vector <PII> adj[MAX];

long long prim(int x)
{
    priority_queue<PII, vector<PII>, greater<PII> > Q;
    int y;
    long long minimumCost = 0;
    PII p;
    Q.push(make_pair(0, x));
    while(!Q.empty())
    {
        // Select the edge with minimum weight
        p = Q.top();
        Q.pop();
        x = p.second;
        // Checking for cycle
        if(marked[x] == true)
            continue;
        minimumCost += p.first;
        marked[x] = true;
        for(int i = 0;i < adj[x].size();++i)
        {
            y = adj[x][i].second;
            if(marked[y] == false)
                Q.push(adj[x][i]);
        }
    }
    return minimumCost;
}

int main()
{
    int nodes, edges, x, y;
    long long weight, minimumCost;
    cin >> nodes >> edges;
    for(int i = 0;i < edges;++i)
    {
        cin >> x >> y >> weight;
        adj[x].push_back(make_pair(weight, y));
        adj[y].push_back(make_pair(weight, x));
    }
    // Selecting 1 as the starting node
    minimumCost = prim(1);
    cout << minimumCost << endl;
    return 0;
}
Time Complexity:
The time complexity of the Prim’s Algorithm is O((V+E)logV) because each vertex is inserted in the priority queue only once and insertion in priority queue take logarithmic time.


Reference: https://www.hackerearth.com/practice/algorithms/graphs/minimum-spanning-tree/tutorial/

DSA Lab 9 - Kruskal's Algorithm

Task:

In this task you have to implement the kruskal's algorithm for minimal spanning tree. You can use the class "MyGraph" which you have created in your previous lab.


Helping Materiel:



Kruskal’s Algorithm

Kruskal’s Algorithm builds the spanning tree by adding edges one by one into a growing spanning tree. Kruskal's algorithm follows greedy approach as in each iteration it finds an edge which has least weight and add it to the growing spanning tree.
Algorithm Steps:
  • Sort the graph edges with respect to their weights.
  • Start adding edges to the MST from the edge with the smallest weight until the edge of the largest weight.
  • Only add edges which doesn't form a cycle , edges which connect only disconnected components.
So now the question is how to check if 2 vertices are connected or not ?
This could be done using DFS which starts from the first vertex, then check if the second vertex is visited or not. But DFS will make time complexity large as it has an order of O(V+E) where V is the number of vertices, E is the number of edges. So the best solution is "Disjoint Sets":
Disjoint sets are sets whose intersection is the empty set so it means that they don't have any element in common.
Consider following example:
enter image description here
In Kruskal’s algorithm, at each iteration we will select the edge with the lowest weight. So, we will start with the lowest weighted edge first i.e., the edges with weight 1. After that we will select the second lowest weighted edge i.e., edge with weight 2. Notice these two edges are totally disjoint. Now, the next edge will be the third lowest weighted edge i.e., edge with weight 3, which connects the two disjoint pieces of the graph. Now, we are not allowed to pick the edge with weight 4, that will create a cycle and we can’t have any cycles. So we will select the fifth lowest weighted edge i.e., edge with weight 5. Now the other two edges will create cycles so we will ignore them. In the end, we end up with a minimum spanning tree with total cost 11 ( = 1 + 2 + 3 + 5).
Implementation:
#include <iostream>
#include <vector>
#include <utility>
#include <algorithm>

using namespace std;
const int MAX = 1e4 + 5;
int id[MAX], nodes, edges;
pair <long long, pair<int, int> > p[MAX];

void initialize()
{
    for(int i = 0;i < MAX;++i)
        id[i] = i;
}

int root(int x)
{
    while(id[x] != x)
    {
        id[x] = id[id[x]];
        x = id[x];
    }
    return x;
}

void union1(int x, int y)
{
    int p = root(x);
    int q = root(y);
    id[p] = id[q];
}

long long kruskal(pair<long long, pair<int, int> > p[])
{
    int x, y;
    long long cost, minimumCost = 0;
    for(int i = 0;i < edges;++i)
    {
        // Selecting edges one by one in increasing order from the beginning
        x = p[i].second.first;
        y = p[i].second.second;
        cost = p[i].first;
        // Check if the selected edge is creating a cycle or not
        if(root(x) != root(y))
        {
            minimumCost += cost;
            union1(x, y);
        }    
    }
    return minimumCost;
}

int main()
{
    int x, y;
    long long weight, cost, minimumCost;
    initialize();
    cin >> nodes >> edges;
    for(int i = 0;i < edges;++i)
    {
        cin >> x >> y >> weight;
        p[i] = make_pair(weight, make_pair(x, y));
    }
    // Sort the edges in the ascending order
    sort(p, p + edges);
    minimumCost = kruskal(p);
    cout << minimumCost << endl;
    return 0;
}
Time Complexity:
In Kruskal’s algorithm, most time consuming operation is sorting because the total complexity of the Disjoint-Set operations will be O(ElogV), which is the overall Time Complexity of the algorithm.



Reference: This content is taken from https://www.hackerearth.com/practice/algorithms/graphs/minimum-spanning-tree/tutorial/ 

Sunday, 21 May 2017

Graph Data Structure

A graph is a pictorial representation of a set of objects where some pairs of objects are connected by links. The interconnected objects are represented by points termed as vertices, and the links that connect the vertices are called edges.
Formally, a graph is a pair of sets (V, E), where V is the set of vertices and E is the set of edges, connecting the pairs of vertices. Take a look at the following graph −
Graph Basics
In the above graph,
V = {a, b, c, d, e}
E = {ab, ac, bd, cd, de}

Graph Data Structure

Mathematical graphs can be represented in data structure. We can represent a graph using an array of vertices and a two-dimensional array of edges. Before we proceed further, let's familiarize ourselves with some important terms −
  • Vertex − Each node of the graph is represented as a vertex. In the following example, the labeled circle represents vertices. Thus, A to G are vertices. We can represent them using an array as shown in the following image. Here A can be identified by index 0. B can be identified using index 1 and so on.
  • Edge − Edge represents a path between two vertices or a line between two vertices. In the following example, the lines from A to B, B to C, and so on represents edges. We can use a two-dimensional array to represent an array as shown in the following image. Here AB can be represented as 1 at row 0, column 1, BC as 1 at row 1, column 2 and so on, keeping other combinations as 0.
  • Adjacency − Two node or vertices are adjacent if they are connected to each other through an edge. In the following example, B is adjacent to A, C is adjacent to B, and so on.
  • Path − Path represents a sequence of edges between the two vertices. In the following example, ABCD represents a path from A to D.
graph

Basic Operations

Following are basic primary operations of a Graph −
  • Add Vertex − Adds a vertex to the graph.
  • Add Edge − Adds an edge between the two vertices of the graph.
  • Display Vertex − Displays a vertex of the graph.

Depth First Traversal

Depth First Search (DFS) algorithm traverses a graph in a depthward motion and uses a stack to remember to get the next vertex to start a search, when a dead end occurs in any iteration.
Depth First Travesal
As in the example given above, DFS algorithm traverses from A to B to C to D first then to E, then to F and lastly to G. It employs the following rules.
  • Rule 1 − Visit the adjacent unvisited vertex. Mark it as visited. Display it. Push it in a stack.
  • Rule 2 − If no adjacent vertex is found, pop up a vertex from the stack. (It will pop up all the vertices from the stack, which do not have adjacent vertices.)
  • Rule 3 − Repeat Rule 1 and Rule 2 until the stack is empty.
StepTraversalDescription
1.Depth First Search Step OneInitialize the stack.
2.Depth First Search Step TwoMark S as visited and put it onto the stack. Explore any unvisited adjacent node from S. We have three nodes and we can pick any of them. For this example, we shall take the node in an alphabetical order.
3.Depth First Search Step ThreeMark A as visited and put it onto the stack. Explore any unvisited adjacent node from A. Both S and D are adjacent to A but we are concerned for unvisited nodes only.
4.Depth First Search Step FourVisit D and mark it as visited and put onto the stack. Here, we have B and C nodes, which are adjacent to D and both are unvisited. However, we shall again choose in an alphabetical order.
5.Depth First Search Step FiveWe choose B, mark it as visited and put onto the stack. Here B does not have any unvisited adjacent node. So, we pop B from the stack.
6.Depth First Search Step SixWe check the stack top for return to the previous node and check if it has any unvisited nodes. Here, we find D to be on the top of the stack.
7.Depth First Search Step SevenOnly unvisited adjacent node is from D is C now. So we visit C, mark it as visited and put it onto the stack.
As C does not have any unvisited adjacent node so we keep popping the stack until we find a node that has an unvisited adjacent node. In this case, there's none and we keep popping until the stack is empty.

Breadth First Traversal

Breadth First Search (BFS) algorithm traverses a graph in a breadthward motion and uses a queue to remember to get the next vertex to start a search, when a dead end occurs in any iteration.
Breadth First Traversal
As in the example given above, BFS algorithm traverses from A to B to E to F first then to C and G lastly to D. It employs the following rules.
  • Rule 1 − Visit the adjacent unvisited vertex. Mark it as visited. Display it. Insert it in a queue.
  • Rule 2 − If no adjacent vertex is found, remove the first vertex from the queue.
  • Rule 3 − Repeat Rule 1 and Rule 2 until the queue is empty.
StepTraversalDescription
1.Breadth First Search Step OneInitialize the queue.
2.Breadth First Search Step TwoWe start from visiting S (starting node), and mark it as visited.
3.Breadth First Search Step ThreeWe then see an unvisited adjacent node from S. In this example, we have three nodes but alphabetically we choose A, mark it as visited and enqueue it.
4.Breadth First Search Step FourNext, the unvisited adjacent node from S is B. We mark it as visited and enqueue it.
5.Breadth First Search Step FiveNext, the unvisited adjacent node from S is C. We mark it as visited and enqueue it.
6.Breadth First Search Step SixNow, S is left with no unvisited adjacent nodes. So, we dequeue and find A.
7.Breadth First Search Step SevenFrom A we have D as unvisited adjacent node. We mark it as visited and enqueue it.
At this stage, we are left with no unmarked (unvisited) nodes. But as per the algorithm we keep on dequeuing in order to get all unvisited nodes. When the queue gets emptied, the program is over.