We will now implement a graph in Java using adjacency matrices. Look back to the previous lesson to see our abstract base class Graph
.
One way to represent graphs is through adjacency matrices. Given a graph with V vertices, we create an V x V 2d matrix.
If there exists an edge from one vertex (column) to another (row), we place a 1 there. If not, then we place a 0. If we had a weighted graph, we can place any nonzero element in lieu of 1.
0 1 2 3 4 5 6

0  0 1 1 0 0 0 0
1  0 1 0 1 1 1 0
2  0 0 0 0 0 0 0
3  0 0 0 0 0 0 0
4  0 0 0 0 0 0 0
5  0 1 0 0 0 0 1
6  0 0 0 0 0 0 0
Let's consider some basic efficiencies. Storing this graph would take V^{2} space. Finding whether or not an edge exists, or adding an edge between two vertices would take O(1) time since all we have to do is a quick lookup.
Below are two functions that check whether two vertices are connected. The inEdgeExists(int v, int w)
method checks if there exists an edge that goes from w to v. The outEdgeExists(int v, int w)
function checks if an edge goes from w to v.
 public boolean inEdgeExists(int v, int w) throws VertexOutOfBoundsException {
 return outEdgeExists(w,v);
 }

 public boolean outEdgeExists(int v, int w) throws VertexOutOfBoundsException {
 int numV = getNumVertices();
 if (v >= numV  w >= numV) {
 throw new VertexOutOfBoundsException();
 }
 return adjMatrix[w][v] != 0;
 }
The runtime efficiency is just O(1) since we just need to access a single cell in the array.
Adding a vertex is more tedious. A naive approach to this problem would be creating a new 2d array of V+1 size each time a new vertex is added. We can then copy all the old elements to this new array.
A better way of doing this would be to anticipate the addition of more vertices and create a new array of 2*V size everytime we fill up more than 1/2 the current array. This will limit the number of instances where we have to resize our matrix.
public void addVertex() {
// If the number of vertices is more than half the size of our matrix,
// double the size of our matrix
int numV = getNumVertices();
if (numV > 0.5 * size) {
size = 2*size;
int[][] newAdjMatrix = new int[size][size];
for (int i = 0; i < adjMatrix.length; i++) {
for (int j = 0; j < adjMatrix[0].length; j++) {
newAdjMatrix[i][j] = adjMatrix[i][j];
}
}
adjMatrix = newAdjMatrix;
}
setNumVertices(numV+1);
}
Conversely, when deleting a vertex, we can shrink our matrix's dimensions to half its original size to save space.
public void removeVertex() throws VertexOutOfBoundsException {
int numV = getNumVertices();
if (numV == 0) {
throw new VertexOutOfBoundsException();
}
if (size < 0.5 * numV) {
size = (int) 0.5*size;
int[][] newAdjMatrix = new int[size][size];
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
newAdjMatrix[i][j] = adjMatrix[i][j];
}
}
adjMatrix = newAdjMatrix;
}
setNumVertices(numV1);
}
Both adding and removing a vertex can take up to V^{2} time.
Adding an edge is as simple as inserting a value into our array matrix.
public void addEdge(int v, int w) throws VertexOutOfBoundsException {
int numV = getNumVertices();
if (v >= numV  w >= numV) {
throw new VertexOutOfBoundsException();
}
adjMatrix[v][w] = 1;
setNumEdges(getNumEdges()+1);
}
To remove an edge, simply set that edge's cell to zero.
public void removeEdge(int v, int w) throws VertexOutOfBoundsException {
int numV = getNumVertices();
if (v >= numV  w >= numV) {
throw new VertexOutOfBoundsException();
}
adjMatrix[v][w] = 0;
setNumEdges(getNumEdges()1);
}
Finding the in or outdegree of a node takes just V time. Simply add 1 for every nonzero element in the corresponding column or row.
public int getInDegree(int v) throws VertexOutOfBoundsException {
int numV = getNumVertices();
if (v >= numV) {
throw new VertexOutOfBoundsException();
}
// Count the number of indegrees
int count = 0;
for (int i = 0; i < getNumVertices(); i++) {
if (adjMatrix[i][v] != 0) {
count++;
}
}
return count;
}
public int getOutDegree(int v) throws VertexOutOfBoundsException {
int numV = getNumVertices();
if (v >= numV) {
throw new VertexOutOfBoundsException();
}
// Count the number of indegrees
int count = 0;
for (int i = 0; i < getNumVertices(); i++) {
if (adjMatrix[v][i] != 0) {
count++;
}
}
return count;
}
The degree sequence of a graph is an ordered descending list containing the degrees of each vertex in the graph. The degree sequence is an invariant of the graph can may be used to analyze graph properties.
To find the degree sequence, we have to sum up the in and outdegrees, add them to a list, then sort them in descending order. We can place this in our abstract Graph
class.
public List<Integer> getDegreeSeq() throws VertexOutOfBoundsException {
List<Integer> degreeSeq = new ArrayList<Integer>();
int degrees = 0;
for (int i = 0; i < getNumVertices(); i++) {
degrees = getInDegree(i) + getOutDegree(i);
degreeSeq.add(degrees);
}
Collections.sort(degreeSeq);
Collections.reverse(degreeSeq);
return degreeSeq;
}
To find all neighbors of a vertex, find all nonzero elements in that vertex's row and column.
public List<Integer> getNeighbors(int v) throws VertexOutOfBoundsException {
int numV = getNumVertices();
if (v >= numV) {
throw new VertexOutOfBoundsException();
}
List<Integer> neighbors = new ArrayList<Integer>();
for (int i = 0; i < getNumVertices(); i++) {
if (adjMatrix[v][i] != 0) {
neighbors.add(i);
}
}
return neighbors;
}
To find all neighbors two nodes apart, we could use the neighbors()
method twice...or...we can use a special property of matrices and square the adjacent matrix. All nonzero elements in this squared matrix contains the twodistanced neighbors! Neat! We can even cube our matrix to find all threedistanced neighbors and so on.
public List<Integer> getNeighborsTwoApart(int v) throws VertexOutOfBoundsException {
int numV = getNumVertices();
if (v >= numV) {
throw new VertexOutOfBoundsException();
}
List<Integer> neighbors = new ArrayList<Integer>();
int[][] sqMatrix = new int[size][size];
for (int i = 0; i < numV; i++)
for (int j = 0; j < numV; j++)
for (int k = 0; k < numV; k++)
sqMatrix[i][j] += adjMatrix[i][k] * adjMatrix[k][j];
for (int i = 0; i < numV; i++) {
if (sqMatrix[i][v] != 0  sqMatrix[v][i] != 0) {
neighbors.add(i);
}
}
return neighbors;
}
Notice any ways you can improve this implementation? Comment below!
Jump start your Java education with Head First Java! This book provides clean diagram examples, with text that is easytounderstand in an almost toocasual language. Great for anyone new to Java or programming in general.
$ Check priceLearn all the different scenarios of . This book is jam packed with code snippets written in C++ and Java to give you the edge you need to ace your technical interviews. Problems come with detailed solutions as well as a hint in case you get stuck, and organized by difficulty level with several variants to help you apply your knowledge more widely.
$ Check priceAd