Skip to content

Instantly share code, notes, and snippets.

Embed
What would you like to do?

Prompt

Write a function that determines if a path exists between two vertices of a graph.

The graph will be represented as an object, each of whose keys represents a vertex(node) of the graph and whose value represents all vertices that can be reached from the aforementioned key. In the example below, there is a connection from vertex a to vertex b and a connection from vertex b to vertices c and d but not a connection from vertex b to vertex a.

{a: ['b'],
 b: ['c', 'd'],
 c: ['d']
 }

Refresher:

  • linear data structures: arrays, stacks, queues, linked lists
  • non-linear: (hierachial data structures) trees, graphs

Graph Terminology

A graph consists of a set of vertices connected together by a set of edges. A tree is a special type of graph. In this example, a vertex is a node of that tree, and an edge is a link between a parent and one of it's children.

Graph vs. Tree

tree: connections are bound to be in a certain way, in a tree there are rules dictating the direction among the nodes. In a tree with N nodes, we must have exactly N-1 edges (one edge for each parent-child relationship)

  • In a tree, all nodes much be reachable from the root
  • Must be exactly one possible path from root to node

In graph: No rules for connections

  • collection of nodes where each node might point to another node
  • each node must have some identification.. obvi; naming is NOT indicative of any order.
  • edges are uniquely identified by its two endpoints

2 types of edges: 1) directed: connection is ONE way street 2) undirected: connection is TWO way street

Usually, a graph either directed or undirected, but it's possible for there to be both.

All trees are graphs, but not all graphs are trees

Approach

This problem is essentially a DFS/BFS problem. Either algorithm is sufficient. The only catch is that graphs can be cyclic. In other words, it's possible for a loop to exist in the graph. The graph below showcases this problem:

{ a: ['a', 'c'], c: ['s', 'r'], r: ['a'], s: [] }

What's the big deal? Imagine we started travsering at vertex a. Eventually we would reach vertex r, but notice that it points right back to vertex a. If you were to unleash an unmodified BFS/DFS traversal algorithm on this input, the algorithm would proceed in an infinite loop. This means that the algorithm must be changed to keep track of all vertices that it has seen. If a vertex has been seen, we know to not consider it's edges a second time. The algorithm completes when either it has found the target vertex or when it has exhausted all possible vertices without finding it's target.

Discussion

The data structure seen above used to represent the graph is called an ajacency list. An alternative data structure exists for representing graphs called adjacency matrices. The cyclic graph above could have been modeled as follows using an adjacency matrix:

    a  c  s  r
  a 1  1  0  0
  c 0  0  1  1
  s 0  0  0  0
  r 1  0  0  0

In javascript, this table would be represented using an array of arrays or object of objects. A 1 indicates that a given vertex has an edge pointing to another vertex, and a 0 indicates that it does not.

This table reads as follows:

a -> a
a -> c
c -> s
c -> r
r -> a

Consider the tradeoffs between using one of these data structres of the other. How do they compare for the following ways:

Attribute Answer
Testing if a given edge is in the graph adjacency matrix
Finding the degree (number of edges of) a vertex adjacency list
Insertion/deletion of edges adjacency matrix O(1) vs O(d) where d is the degree of the vertex
Memory usage for sparse graphs adjacency list (m + n) vs (n^2)
Memory usage for dense graphs adjacency matrix
Graph traversal adjacency list O(m + n) vs O(n^2)
Better overall adjacency list

Comparison from The Algoritm Design Manual, Skiena - second Edition - page 152

Solution

See a heavily-commented repl

DFS

var graph = {a: ['c'],
 b: ['c'],
 c: ['s', 'r'],
 d: ['a'],
 s: ['a', 'c'],
 r: ['d'],
 z: ['z']
 };

var doesPathExist = function(graph, visited, start, target) {
  visited[start] = true;
  return graph[start].some(function(vertex){
    if (start === target) {
      return true;
    } else if (!visited[vertex]) {
      return doesPathExist(graph, visited, vertex, target);
    } else {
      return false;
    }
  });
};

console.log(doesPathExist(graph, {}, 'a', 'c')) // true
console.log(doesPathExist(graph, {}, 'a', 's')) // true
console.log(doesPathExist(graph, {}, 'a', 'b')) // false
console.log(doesPathExist(graph, {}, 'b', 'a')) // true
console.log(doesPathExist(graph, {}, 'a', 'd')) // true
console.log(doesPathExist(graph, {}, 's', 'r')) // true
console.log(doesPathExist(graph, {}, 'z', 'z')) // true
console.log(doesPathExist(graph, {}, 'c', 'c')) // true

BFS

var graph = {a: ['c'],
 b: ['c'],
 c: ['s', 'r'],
 d: ['a'],
 s: ['a', 'c'],
 r: ['d'],
 z: ['z']
 };


function doesPathExist(graph, start, target) {
    var queue = [], node = start, visited = {};
    
    while (node) {
        visited[node] = true;
        
        if(node == target) return true;
        
        if(graph[node]){
          graph[node].forEach(vertex => {
            if(!visited[vertex]){
              queue.push(vertex)
            }
          })
        }
        node = queue.shift();
    }
    return false;
}
console.log(doesPathExist(graph, 'a', 'c')) // true
console.log(doesPathExist(graph, 'a', 's')) // true
console.log(doesPathExist(graph, 'a', 'b')) // false
console.log(doesPathExist(graph, 'b', 'a')) // true
console.log(doesPathExist(graph, 'a', 'd')) // true
console.log(doesPathExist(graph, 's', 'r')) // true
console.log(doesPathExist(graph, 'z', 'z')) // true
console.log(doesPathExist(graph, 'c', 'c')) // true

real world implementations:

  • social network (undirected graph because friendship is mutual relationship; connection is two-way!)
  • internet (directed graph! if pageA has link to pageB doesn't mean pageB automatically has link to pageA)
    • web crawler: graph traversal

Extra resources: [youtube](https://www.youtube.com/watch?v=gXgEDyodOJU, https://www.youtube.com/watch?v=zaBhtODEL0w)

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment
You can’t perform that action at this time.