Can we detect cycles in directed graph using Union-Find data structure?
Asked Answered
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I know that one can detect cycles in direct graphs using DFS and BFS. I want to know whether we can detect cycles in directed graphs using Union-Find or not?

  • If yes, then how? and
  • If we can't, then why?
Tourney answered 12/4, 2020 at 6:39 Comment(5)
Hey, that is about DFS vs Union-Find. Both can be used in detecting cycles in the undirected graph but what about the Directed graph? I have never seen detecting cycles using Union-Find for directed graphs. So, I am asking why? Is it cannot be implemented or what?Tourney
Can you suggest the best resources to learn graph data structures.. and also how/where did you learned graph data structures?Tourney
how to represent edge in an undirected graph. Is it like 0->1 or both (0->1 and 1->0)Tourney
Yes you push both the edges in case of undirected graph. There's a really good course called Algorithms 1 on Coursera by Princeton University. It's a great resource. Apart from that I mostly learnt by solving problems on various sites like hackerrank, codeforcesGrumble
Study Graphs: there is enough material here - geeksforgeeks.org/graph-data-structure-and-algorithms/?ref=ghmMilka
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No, we cannot use union-find to detect cycles in a directed graph. This is because a directed graph cannot be represented using the disjoint-set(the data structure on which union-find is performed).

When we say 'a union b' we cannot make out the direction of edge

  1. is a going to b? (or)
  2. is b going to a?

But, incase of undirected graphs, each connected component is equivalent to a set. So union-find can be used to detect a cycle. Whenever you try to perform union on two vertices belonging to the same connected component, we can say that cycle exists.

Grumble answered 25/4, 2020 at 6:32 Comment(0)
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No.

Let me give you an example:

  • Q1. Take an undirected graph:

Pic1

Is there a cycle in above undirected graph? Yes. And we can find the cycle using Union-Find algo.

  • Q2. Now look at the similar directed graph:

Pic2

Is there a cycle in above directed graph? No! BUT if you use Union-Find algo to detect cycle in above directed graph, it will say YES! Since union-find algo looks at the above diagram as below:

Pic3 OR Pic4

Is there a cycle in above diagram? Yes! But the original(Q2) question was tampered and this is not what was asked. So Union-find algo will give wrong results for directed graphs.

Henleigh answered 16/12, 2020 at 2:52 Comment(0)
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No, we cannot use union-find to detect cycles in a directed graph. This is because a directed graph cannot be represented using the disjoint-set(the data structure on which union-find is performed).

When we say 'a union b' we cannot make out the direction of edge

is a going to b? (or) is b going to a? But, incase of unordered graphs, each connected component is equivalent to a set. So union-find can be used to detect a cycle. Whenever you try to perform union on two vertices belonging to the same connected component, we can say that cycle exists.

Just wanted to add to @Cherubim's answer, I just thought of this question that why can't we make out the direction of 'a union b' like we can consider it as a is connected to b right ('a union b' only nodes which are a->b)?

Well that will also fail, consider two nodes x and y, they are connected x -> y and they already have parents in the disjoint set, x_root and y_root. So what happens when we say 'x union y' (x is connected to y), we make y_root the parent of x_root. well this tells us that:

  1. x_root -> y_root (could be opposite too)
  2. x_root and y_root are connected. (they could be disconnected)

So the main problem beside ambiguous direction, is also that not every node in a directed graph is connected to each other.

Lenny answered 15/4, 2022 at 15:27 Comment(0)
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No, the Union-Find algorithm is not directly applicable to directed graphs. Here's why:

Union-Find's Core Assumption: Union-Find fundamentally relies on the concept of equivalence relations. It assumes that if two elements are connected, they are equivalent in a bidirectional sense. In undirected graphs, edges represent mutual relationships: if A is connected to B, then B is also connected to A. This aligns with the equivalence concept. Directed Graphs Violate Equivalence:

In directed graphs, edges represent one-way relationships: a connection from A to B doesn't guarantee a connection from B to A. This breaks the equivalence assumption. Union-Find's operations (union and find) assume bidirectional connectivity, which doesn't hold true in directed graphs.

Example:

Consider a directed graph with edges A -> B and C -> B. If you unioned A and B, Union-Find would incorrectly assume that C is also connected to A, leading to incorrect results.

Sinusoidal answered 3/2, 2024 at 22:22 Comment(0)

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