Storing Graphs
Because I am a computer scientist I don't just care about using graphs but also about how to store them. There are multiple ways of storing graphs. Depending on the type of graph and the requirements a certain storage method might be preferred as it might be more efficient in terms of memory or time complexity.
Adjacency Matrix
An adjacency matrix is the most common and the most simple way of storing graphs. If we have a graph with vertices, we create an adjacency matrix with dimensions of . As the name suggests this matrix stores the adjacency of vertices i.e. the relationship between the vertices i.e. the edges.
Below you can see an example of a very simple adjacency matrix of an undirected unweighted graph.
Weighted and Unweighted Graphs
If we have a weighted graph where the weights are integer values we can create the following matrix:
int[][] G = new int[n][n]
The weight of the edge from vertex to vertex would then be stored at G[y][x]
.
If there is no edge between 2 vertices then there are multiple ways to indicate this. The simplest way would be to set
the value to 0 (which it already is at initialization) another common approach is to set the weight to
Integer.MAX_VALUE
.
The last possibility but the worst in space complexity is to use a null
value. This is only possible if we use an
object array instead of a primitive array. This would mean that instead of using int[][]
we use Integer[][]
.
This would however mean that we would end up using a lot more memory as you can read more about
here (opens in a new tab).
If the graph is unweighted we can use the same int array and just store all edge weights as 1 or 0. We could however also use a 2D boolean array which interestingly does use less space in Java than an int array. In Java, a normal boolean variable uses 32 bits like an int. However, in an array each boolean value only takes up 8 bits because that is what the CPU likes to use internally. This means that a boolean array uses 4 times less space than an int array.
Directed and Undirected Graphs
If the graph is undirected then we commonly set the value at G[y][x]
and G[x][y]
to the same value. This means that
mathematically i.e. the matrix is symmetric along the diagonal, and that we could get away with only storing
the upper or lower triangle of the matrix, which would half the memory usage. However, this would make the code more
complicated.
If the graph is directed then we only set the value at G[y][x]
and not at G[x][y]
. So is no longer
guaranteed.
The biggest problem with storing graphs with an adjacency matrix is that its space complexity is . What makes this worse is that a lot of the space is wasted as in most cases there are only a few edges between vertices making it a sparse matrix. Very rarely do we need to store a complete graph.
Implementation
As mentioned although the adjacency matrix isn't the most efficient way of storing graphs it is the most common and very simple to implement. Below you can see a simple implementation for an undirected unweighted graph.
public class UndirectedUnweightedGraph {
final boolean[][] adjMatrix;
final int n;
public GraphI(int numNodes) {
if (numNodes < 1) throw new IllegalArgumentException();
this.adjMatrix = new boolean[numNodes][numNodes];
this.n = numNodes;
}
public boolean addEdge(int x, int y) {
if (0 <= x && x < n && 0 <= y && y < n) {
if (adjMatrix[y][x]) return false; // already set
adjMatrix[y][x] = adjMatrix[x][y] = true;
return true;
}
throw new IndexOutOfBoundsException();
}
}
Edge List
Another simple but less common way of storing graphs is to just store a list of the edges. The edges could have the following structure:
class Edge {
int from, to, weight;
}
If it is an unweighted graph the weight attribute could just be omitted and if it is an undirected graph you can either have two entries for each edge or handle from and to the same way when processing.
The advantage of this solution is that it only uses memory with being the number of edges. The disadvantage of this storage solution is that you can not quickly find out how many vertices are in the graph and what they are. This could however be solved by just adding another list containing all the vertices. This solution would then be very similar to the formal definition of a graph . We would then have a memory usage of with being the number of vertices and the number of edges.
Implement this
Adjacency Lists
An adjacency list stores for each vertex has a list of its edges. An adjacency list can just be a simple array but the list storing the edges is most commonly a linked list due to the storage and length being dynamic. If the graph is undirected you can again either handle it by just storing an edge in one of the list, for example always in the source vertex, or you can also store it additionally in the destination vertexes list. This structure uses just like the edge table memory with being the number of vertices and the number of edges.
Implementation
Is this really correct and they way I want it?
public class UnweightedGraph<K> {
private static class Vertex<K> {
K data;
int indegree, deg = 0;
boolean visited;
List<Vertex<K>> adjList = new LinkedList<Vertex<K>>();
Vertex(K value) {
data = value;
}
boolean addEdgeTo(Vertex <K> to) {
return (adjList.contains(to)) ? false : adjList.add(to);
}
}
private Map<K, Vertex<K>> vertices;
private int nOfEdges = 0;
public UnweightedGraph() {
this(false);
}
public UnweightedGraph(boolean directed) {
super(directed);
vertices = new HashMap<K, Vertex<K>>();
}
public UnweightedGraph(UnweightedGraph<K> orig) { // copy constructor
this(orig.isDirected());
for (K k: orig.vertices.keySet()) {
addVertex(k);
}
for (Vertex<K> v: orig.vertices.values()) {
for (Vertex<K> w: v.adjList) {
addEdge(v.data, w.data);
}
}
}
public boolean addVertex(K vertex) {
if (vertex != null && !vertices.containsKey(vertex)) {
vertices.put(vertex, new Vertex<K>(vertex));
return true;
} else {
return false;
}
}
public boolean addEdge(K from, K to) {
Vertex<K> vf = vertices.get(from);
Vertex<K> vt = vertices.get(to);
if (vf != null && vt != null && vf.addEdgeTo(vt)) {
vt.indegree++;
if (!isDirected()) {
vt.addEdgeTo(vf);
vf.indegree++;
}
nOfEdges++;
return true;
} else {
return false;
}
}
}