A graph is a data structure that consists of a set of nodes connected by edges. Graphs are used to simulate many real-world problems, such as paths in cities, circuit networks, and social networks. This is graph implementation part 2 – weighted graph as adjacency list.
Table of Content
- Terminology
- Define classes
- Add node and edge
- Remove node and edge
- Search by node and edge
- Find path with DFS
- Find path with BFS
- Print weighted graph as adjacency list
- Traverse with DFS
- Traverse with BFS
- Free download
Terminology
A node in the graph is also called vertex. A line between two nodes is edge. Two nodes are adjacent (or neighbors) if they are connected to each other through an edge. Path represents a sequence of edges between the two nodes.
In a directed graph, all of the edges represent a one-way relationship. In an undirected graph, all edges are bi-directional.
If the edges in the graph have weights, the graph is said to be a weighted graph. If the edges do not have weights, the graph is said to be unweighted.
The weight of the edges might represent the distances between two cities, or the cost of flights etc. When we include weight as a feature of graph’s edges, some interesting questions arise. The problems such as finding shortest path or longest path are applied to weighted graphs.
Weighted graph can be directed or undirected. For example, the minimum spanning tree is undirected graph. The famous Dijkstra’s algorithm to find shortest path is for directed graphs.
Graph can be presented as adjacency list or adjacency matrix. An adjacency list is an array of edges or nodes. Adjacency list is used for representation of the sparse graphs. An adjacency matrix is a square matrix with dimensions equivalent to the number of nodes in the graph. Adjacency matrix is preferred when the graph is dense.
Define classes
First we define an Edge class. It has two fields: connectedVertex and weight. The connectedVertex is the node at the other end of the edge. weight is the value associated with the edge.
The GraphWeighted class has two fields: adj and directed. adj is a HashMap in which the key is the node at the start of the edge, the value is all its neighbors. The value is represented as linked list of the edges. directed is a boolean variable to specify whether the graph is directed or undirected. By default, it is undirected.
The nodes can be any data type, for example primitive data type, such as integer or string. Or it can be an object, such as graphNode.
Java
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class Edge<T> { T connectedVetex; //connected vertex int weight; //weight //Constructor, Time O(1) Space O(1) public Edge(T v, int w) { this.connectedVetex = v; this.weight = w; } //Time O(1) Space O(1) @Override public String toString() { return "(" + connectedVetex + "," + weight + ")"; } } public class GraphWeighted<T> { Map<T, LinkedList<Edge<T>>> adj = new HashMap<>() ; boolean directed; //Constructor, Time O(1) Space O(1) public GraphWeighted () { directed = false; //default, Undirected graph } //Constructor, Time O(1) Space O(1) public GraphWeighted(boolean d) { directed = d; } } |
Javascript
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class Edge { //Constructor, Time O(1) Space O(1) constructor(v, w) { this.connectedVetex = v; this.weight = w; } //Time O(1) Space O(1) toString(){ return "(" + this.connectedVetex + "," + this.weight + ")"; } } class GraphWeighted { //Constructor, Time O(1) Space O(1) constructor(directed) { this.adj = new Map(); this.directed = directed; //true or false } } |
Python
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class Edge : #Constructor, Time O(1) Space O(1) def __init__(self, v, w) : self.connectedVetex = v self.weight = w #Time O(1) Space O(1) def __str__(self): return "(" + str(self.connectedVetex) + "," + str(self.weight) + ")" class GraphWeighted : #Constructor, Time O(1) Space O(1) def __init__(self, directed) : self.adj = {} self.directed = directed #true or false |
Add node and edge
To add a node to the graph is to add a key in the hashmap. To add an edge is to add an item in this key’s value. The following method addEdge includes both adding a node and adding an edge. For a directed graph, we add edge from a to b. For undirected graph, we also add edge from b to a. Note the weight is one of the input and used to create edge object.
Java
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//Add edges including adding nodes, Time O(1) Space O(1) public void addEdge(T a, T b, int w) { adj.putIfAbsent(a, new LinkedList<>()); //add node adj.putIfAbsent(b, new LinkedList<>()); //add node Edge<T> edge1 = new Edge<>(b, w); adj.get(a).add(edge1); //add edge if (!directed) { //undirected Edge<T> edge2 = new Edge<>(a, w); adj.get(b).add(edge2); } } |
Javascript
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//Add edges including adding nodes, Time O(1) Space O(1) addEdge(a, b, w) { if (this.adj.get(a) == null) this.adj.set(a, new Array()); //add node if (this.adj.get(b) == null) this.adj.set(b, new Array()); //add node var edge1 = new Edge(b, w) this.adj.get(a).push(edge1); //add edge if (!this.directed) { let edge2 = new Edge(a, w); this.adj.get(b).push(edge2); //add edge } } |
Python
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#Add edges including adding nodes, Time O(1) Space O(1) def addEdge(self, a, b, w) : if a not in self.adj: self.adj[a] = [] if b not in self.adj: self.adj[b] = [] edge1 = Edge(b, w) self.adj[a].append(edge1) if (self.directed == False) : edge2 = Edge(a, w) self.adj[b].append(edge2) |
Remove node and edge
Remove operation includes remove edge and remove node. Before we continue, let’s create a utility method to find the edge between two nodes. Use one node as key to find its neighbors. Then loop through the neighbors to find the other node. Return the edge object with the weight. This method will be used in following operations.
To remove edge, we use the node as key to find its neighbors in the hashmap. Then remove the other node from its neighbors. For a directed graph, we just need to remove edge from a to b. For an undirected graph, we also need to remove the edge from b to a.
Remove node has more work to do than remove edge. We have to remove all connected edge before remove the node itself. For an undirected graph, first we get all neighbors of the node. Then for each of its neighbors, remove itself from the value list. For a directed graph, we search all keys in the hashmap for their values, and check whether this node exists in their neighbors. If it does, remove it. The last step is to remove the node as the key in the hashmap. Then this node is no longer in the hashmap’s key set.
Java
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//Find the edge between two nodes, Time O(n) Space O(1), n is number of neighbors private Edge<T> findEdgeByVetex(T a, T b) { LinkedList<Edge<T>> ne = adj.get(a); for (Edge<T> edge: ne) { if (edge.connectedVetex.equals(b)) { return edge; } } return null; } //Remove direct connection between a and b, Time O(n) Space O(1) public void removeEdge(T a, T b) { LinkedList<Edge<T>> ne1 = adj.get(a); LinkedList<Edge<T>> ne2 = adj.get(b); if (ne1 == null || ne2 == null) return; Edge<T> edge1 = findEdgeByVetex(a, b); ne1.remove(edge1); if (!directed) {//undirected Edge<T> edge2 = findEdgeByVetex(b, a); ne2.remove(edge2); } } //Remove a node including all its edges, Time O(V) Space O(1), V is number of vertics in graph public void removeNode(T v) { if (!directed) { //undirected LinkedList<Edge<T>> ne1 = adj.get(v); for (Edge<T> edge: ne1) { Edge<T> edge1 = findEdgeByVetex(edge.connectedVetex, v); adj.get(edge.connectedVetex).remove(edge1); } } else { //directed for (T key: adj.keySet()) { Edge<T> edge2 = findEdgeByVetex(key, v); if (edge2 != null) adj.get(key).remove(edge2); } } adj.remove(v); } |
Javascript
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//Find the edge between two nodes, Time O(n) Space O(1), n is number of neighbors findEdgeByVetex(a, b) { var ne = this.adj.get(a); for (let i = 0; i < ne.length; i++) { if (ne[i].connectedVetex == b) { return i; } } return -1; } //Remove direct connection between a and b, Time O(n) Space O(1) removeEdge(a, b) { var ne1 = this.adj.get(a); var ne2 = this.adj.get(b); if (ne1 == null || ne2 == null) return; var index1 = this.findEdgeByVetex(a, b); if (index1 >= 0) ne1.splice(index1, 1); if (!this.directed) { //undirected var index2 = this.findEdgeByVetex(b, a); if (index1 >= 0) ne2.splice(index2, 1) } } //Remove a node including all its edges, //Time O(V) Space O(1), V is number of vertics in graph removeNode(v) { if (!this.directed) { //undirected var ne1 = this.adj.get(v); for (let edge of ne1) { let list = this.adj.get(edge.connectedVetex); let index1 = this.findEdgeByVetex(edge.connectedVetex, v); if (index1 >= 0) list.splice(index1, 1); } } else { //directed for (let entry of this.adj.entries()) { let list = entry[1]; let index2 = this.findEdgeByVetex(entry[0], v); if (index2 >= 0) list.splice(index2, 1); } } this.adj.delete(v); } |
Python
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#Find the edge between two nodes, Time O(n) Space O(1), n is number of neighbors def findEdgeByVetex(self, a, b) : ne = self.adj.get(a) for edge in ne: if edge.connectedVetex == b : return edge return None #Remove direct connection between a and b, Time O(1) Space O(1) def removeEdge(self, a, b) : ne1 = self.adj[a] ne2 = self.adj[b] if ne1 == None or ne2 == None : return edge1 = self.findEdgeByVetex(a, b) ne1.remove(edge1) if (self.directed == False) : edge2 = self.findEdgeByVetex(b, a) ne2.remove(edge2) #Remove a node including all its edges, #Time O(v) Space O(1), V is number of vertics in graph def removeNode(self, a) : if self.directed == False: #undirected ne1 = self.adj[a] for edge in ne1 : edge1 = self.findEdgeByVetex(edge.connectedVetex, a) self.adj[edge.connectedVetex].remove(edge1) else : #directed for k, v in self.adj.items(): edge2 = self.findEdgeByVetex(k, a) if edge2 is not None: self.adj[k].remove(edge2); self.adj.pop(a) |
Search by node and edge
Search can be search node, edge or path. We can check whether there is a node existing in the graph. This can be done by simply checking the hashmap contains the key. We can also check whether there is a direct connection between two nodes (aka whether there is an edge). This can be done by checking whether the other node is in one node’s neighbors.
Java
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//Check whether there is node by its key, Time O(1) Space O(1) public boolean hasNode(T key) { return adj.containsKey(key); } //Check whether there is direct connection between two nodes, Time O(n), Space O(1) public boolean hasEdge(T a, T b) { Edge<T> edge1 = findEdgeByVetex(a, b); if (directed) {//directed return edge1 != null; } else { //undirected or bi-directed Edge<T> edge2 = findEdgeByVetex(b, a); return edge1 != null && edge2!= null; } } |
Javascript
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//Check whether there is node with the key, Time O(1) Space O(1) hasNode(key) { return this.adj.has(key); } //Check whether there is direct connection between two nodes, Time O(n), Space O(1) hasEdge(a, b) { var index1 = this.findEdgeByVetex(a, b) if (this.directed) //directed return index1 >=0; else { //undirected or bi-directed var index2 = this.findEdgeByVetex(b, a) return index1 >= 0 && index2 >= 0; } } |
Python
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#Check whether there is node by its key, Time O(1) Space O(1) def hasNode(self, key) : return key in self.adj.keys() #Check whether there is direct connection between two nodes, Time O(n), Space O(1) def hasEdge(self, a, b): edge1 = self.findEdgeByVetex(a, b) if self.directed : #directed return edge1 is not None else : #undirected or bi-directed edge2 = self.findEdgeByVetex(b, a) return edge1 is not None and edge2 is not None |
Find path with depth first search in weighted graph
More useful operation is to search path. A path is a sequence of edges. There can be more than one path between two nodes. There are two common approaches: depth first search (DFS) and breadth first search (BFS). In this section, we use DFS and BFS to find out whether there is path from one node to another. In “Print and traversal” section, we use them to find all reachable nodes from the source node in graph.
Depth First Search starts from the source node, and explores the adjacent nodes as far as possible before call back. DFS is usually implemented with recursion or stack. It is used to solve find path or detect cycle problems.
Java
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//Check there is path from src and dest //DFS, Time O(V+E), Space O(V) public boolean hasPathDFS(T src, T dest) { HashMap<T, Boolean> visited = new HashMap<>(); return dfsHelper(src, dest, visited); } //DFS helper, Time O(V+E), Space O(V) private boolean dfsHelper(T v, T dest, HashMap<T, Boolean> visited) { if (v == dest) return true; visited.put(v, true); for (Edge<T> edge : adj.get(v)) { T u = edge.connectedVetex; if (visited.get(u) == null) return dfsHelper(u, dest, visited); } return false; } |
Javascript
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//Check there is path from src and dest //DFS, Time O(V+E), Space O(V) hasPathDfs(src, dest) { var visited = new Map(); return this.dfsHelper(src, dest, visited); } //DFS helper, Time O(V+E), Space O(V) dfsHelper(v, dest, visited) { if (v == dest) return true; visited.set(v, true); for (let edge of this.adj.get(v)) { let u = edge.connectedVetex; if (visited.get(u) == null) return this.dfsHelper(u, dest, visited); } return false; } |
Python
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# Check there is path from src and dest # DFS, Time O(V+E), Space O(V) def hasPathDfs(self, src, dest) : visited = {} return self.dfsHelper(src, dest, visited) #DFS helper, Time O(V+E), Space O(V) def dfsHelper(self, v, dest, visited) : if v == dest: return True visited[v] = True for edge in self.adj[v] : u = edge.connectedVetex if u not in visited: return self.dfsHelper(u, dest, visited) return False |
Find path with breadth first search in weighted graph
Breath First Search starts from the source node, and explores all its adjacent nodes before going to the next level adjacent nodes. BFS is usually implemented with Queue. It is often used to solve shortest path problems.
Java
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//Check there is path from src and dest //BFS, Time O(V+E), Space O(V), V is number of vertices, E is number of edges public boolean hasPathBFS(T src, T dest) { if (!hasNode(src) || !hasNode(dest)) return false; HashMap<T, Boolean> visited = new HashMap<>(); Queue<T> q = new LinkedList<>(); visited.put(src, true); q.offer(src); while (!q.isEmpty()) { T v = q.poll(); if (v == dest) { return true; } for (Edge<T> edge: adj.get(v)) { T u = edge.connectedVetex; if (visited.get(u) ==null) { visited.put(u, true); q.offer(u); } } } return false; } |
Javascript
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//Check there is path from src and dest //BFS, Time O(V+E), Space O(V), V is number of vertices, E is number of edges hasPathBfs(src, dest) { if (!this.hasNode(src) || !this.hasNode(dest)) return false; var visited = new Map(); var q = new Array(); visited.set(src, true); q.push(src); while (q.length > 0) { let v = q.shift(); if (v == dest) return true; for (let edge of this.adj.get(v)) { let u = edge.connectedVetex; if (visited.get(u) == null) { q.push(u); visited.set(u, true); } } } return false; } |
Python
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#Check there is path from src and dest #BFS, Time O(V+E), Space O(V) def hasPathBfs(self, src, dest) : visited = {} q = [] visited[src] = True q.append(src) while q : v = q.pop(0); if v == dest: return True for edge in self.adj[v] : u = edge.connectedVetex if u not in visited: q.append(u) visited[u] = True return False |
Print weighted graph as adjacency list
Print is to visit all nodes in the graph and print the information stored. Three ways are introduced here.
Print all nodes and their neighbors in the hashmap. This can be done by looping through the key set of the hashmap. This method is used for debugging purpose.
Java
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//Print graph as hashmap, Time O(V+E), Space O(1) public void printGraph() { for (T key: adj.keySet()) { System.out.println(key + "," + adj.get(key)); } } |
JavaScript
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//Print graph as hashmap, Time O(V+E), Space O(1) printGraph() { for (let entry of this.adj.entries()) { console.log(entry[0] + "-" + entry[1]); } } |
Python
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# Print graph as hashmap, Time O(V+E), Space O(1) def printGraph(self) : for k, v in self.adj.items(): print(str(k) + "-", end ="") for edge in v: print(edge, end="") print() |
Traverse with DFS
DFS traversal: Use depth first search to visit nodes in the graph and print the node’s information. This is similar to DFS traversal in binary tree. Starting from the source node, we call recursive method to visit its neighbor’s neighbor until call back. Please node the source might be any node in the graph. Some nodes might not be reached in a directed graph. (In binary tree, we always start from the root and all nodes should be visited.)
Java
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//Traversal starting from src, DFS, Time O(V+E), Space O(V) public void dfsTraversal(T src) { HashMap<T, Boolean> visited = new HashMap<>(); helper(src, visited); System.out.println(); } //DFS helper, Time O(V+E), Space O(V) private void helper(T v, HashMap<T, Boolean> visited) { visited.put(v, true); System.out.print(v.toString() + " "); for (Edge<T> edge : adj.get(v)) { T u = edge.connectedVetex; if (visited.get(u) == null) helper(u, visited); } } |
JavaScript
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//Traversal starting from src, DFS, Time O(V+E), Space O(V) dfsTraversal(src) { var visited = new Map(); this.helper(src, visited); } //DFS helper, Time O(V+E), Space O(V) helper(v, visited) { visited.set(v, true); console.log(v.toString()); for (let edge of this.adj.get(v)) { let u = edge.connectedVetex; if (visited.get(u) == null) this.helper(u, visited); } } |
Python
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#Traversal starting from src, DFS, Time O(V+E), Space O(V) def dfsTraversal(self, src) : visited = {} self.helper(src, visited) print() #DFS helper, Time O(V+E), Space O(V) def helper(self, v, visited) : visited[v] = True print(str(v) +" ",end="") for edge in self.adj[v] : u = edge.connectedVetex if u not in visited: self.helper(u, visited) |
Traverse with BFS
BFS traversal: Use breadth first search to visit all nodes in the graph and print the node’s information. This is similar to BFS traversal in binary tree. Starting from the source, visit all its neighbors first before visiting neighbor’s neighbor. Please node the source might be any node in the graph. Some nodes might not be reached in a directed graph. (In binary tree, we always start from the root and all nodes should be visited.)
Java
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//Traversal starting from src, BFS, Time O(V+E), Space O(V) public void bfsTraversal(T src) { Queue<T> q = new LinkedList<>(); HashMap<T, Boolean> visited = new HashMap<>(); q.add(src); visited.put(src, true); while (!q.isEmpty()) { T v = q.poll(); System.out.print(v.toString() + " "); for (Edge<T> edge : adj.get(v)) { T u = edge.connectedVetex; if (visited.get(u) == null) { q.add(u); visited.put(u, true); } } } System.out.println(); } |
JavaScript
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//Traversal starting from src, BFS, Time O(V+E), Space O(V) bfsTraversal(src) { var q = new Array(); var visited = new Map(); q.push(src); visited.set(src, true); while (q.length > 0) { let v = q.shift(); console.log(v.toString() + " "); for (let edge of this.adj.get(v)) { let u = edge.connectedVetex; if (visited.get(u) == null) { q.push(u); visited.set(u, true); } } } } |
Python
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# Traversal starting from src, BFS, Time O(V+E), Space O(V) def bfsTraversal(self, src) : q = [] visited = {} q.append(src) visited[src] = True while (len(q) > 0) : v = q.pop(0) print(str(v) + " ",end=""); for edge in self.adj[v] : u = edge.connectedVetex if u not in visited: q.append(u); visited[u] = True print() |
Free download
Download WeightedGraph.java
Download WeightedGraph.js
Download WeightedGraph.py
Implement graph as adjacency list
