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Added Kann's algorithm #1676

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74e8ec2
feat:added kanns algorithm
Jul 8, 2024
9f3aa64
Updated kahn's algorithm
Jul 11, 2024
2f2c099
Updated kahn's algorithm
Jul 11, 2024
3e9c9e4
feat: added topological sorting algorithm
Jul 11, 2024
f0325a5
Merge remote-tracking branch 'origin/master'
Jul 11, 2024
99535c5
Delete Graphs/KahnsAlgorithm.js
Jul 11, 2024
35fe1c3
Delete Graphs/KannsAlgorithm.js
Jul 11, 2024
ef63f66
Delete Graphs/test/KahnsAlgorithm.test.js
Jul 11, 2024
154a330
Delete Graphs/test/KannsAlgorithm.test.js
Jul 11, 2024
191993f
Merge remote-tracking branch 'origin/master'
Jul 11, 2024
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44 changes: 44 additions & 0 deletions Graphs/KannsAlgorithm.js
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Original file line number Diff line number Diff line change
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/**
* @author {RaviSadam}
* @name kannsAlgorithm
* @description -
* Kann's Algorithm implementation in JavaScript
* @summary
* Kann's Algorithm is used for topological sorting in directed acyclic graphs
*
* @param graph - Graph [[v1,v2],[v3,v4,v5]..]
* @param n - number of vertices
* @returns {Array} - Empty array if cycle is detected or else result array;
*
*/

export function kannsAlgorithm(graph, n) {
if (n === null || n === undefined) throw Error('Invalid n was given')
const inorder = Array(n).fill(0)
const result = []
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(By the way, you could preallocate this array.)

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Preallocate could increase the time complexity. At the end we need to check array filled or not.

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@appgurueu appgurueu Jul 11, 2024 •
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Preallocate could increase the time complexity.

No, not really. It's still linear time in the size of the graph - O(n + m) where n is the number of nodes and m is the number of edges. (Allocation can also optimistically be assumed to be constant time.)

At the end we need to check array filled or not.

Yes, and that can be done in constant time, if you keep an index to the last element. It'd look something like this:

const topoOrder = Array(n) // preallocate
const idx = 0
...
// to push:
topoOrder[idx++] = node
...
if (idx < n) return null; // cyclic

Or another variant to write this would be:

const topoOrder = Array(n) // preallocate
for (let i = 0; i < n; ++i) {
    if (stack.length === 0) return null; // cyclic
    const node = stack.pop();
    topoOrder[i] = node;
    ...
}

The benefit is that when a topological order exists (which should usually be the case; this is the "happy path"), you can save some reallocations of the array.

But this is just a suggestion, I'm fine with pushing to keep the code simple as well. Though perhaps the preallocation variant is a bit more readable, since it makes explicit that we want to have all n nodes be in this array.

for (let entry of graph) {
for (let edge of entry) {
inorder[edge] += 1
}
}
const queue = []
console.log(inorder)
for (let i = 0; i < n; i++) {
if (inorder[i] === 0) {
queue.push(i)
}
}
while (queue.length != 0) {
const node = queue[0]
result.push(node)
queue.splice(0, 1)
for (let nei of graph[node]) {
inorder[nei] -= 1
if (inorder[nei] == 0) {
queue.push(nei)
}
}
}
if (result.length != n) return []
return result
}
13 changes: 13 additions & 0 deletions Graphs/test/KannsAlgorithm.test.js
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import { kannsAlgorithm } from '../KannsAlgorithm'
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You forgot to delete this file.


test('Test Case 1: Graph without cycle', () => {
const graph = [[], [], [3], [1], [0, 1], [0, 2]]

expect(kannsAlgorithm(graph, 6)).toEqual([4, 5, 0, 2, 3, 1])
})

test('Test Case 2: Graph with cycle', () => {
const graph = [[2], [], [3, 5], [0, 1], [0, 2]]

expect(kannsAlgorithm(graph, 6)).toEqual([])
})