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An undirected graph is complete if there is an edge between every pair of vertices. Given a complete undirected graph on $n$ vertices, in how many ways can you choose a direction for the edges so that there are no directed cycles?

1. $n$
2. $\frac{n(n-1)}{2}$
3. $n!$
4. $2^n$
5. $2^m, \: \text{ where } m=\frac{n(n-1)}{2}$

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n! ?
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@arjun Sir , @experienced folks what should i do if i cant solve these type of questions ie having a diificulty in counting/ permutations and combinarions.
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take any arbitrary value of n.

I've taken n = 3,and form cycles according to qsn. you will get a value & then match the value wth the given option

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Ans should be n! ....best approach is to draw graph with specifications given in question

They are asking to convert Complete Undirected graph into Directed graph without cycle by choosing direction for the edges.

See this $K_{3}$ graph- By this time you must have got the problem statement meaning- our resultant graph should be acyclic.

Lets say we have a complete graph $G$ which has $n$ vertices, $v_1, v_2,\dots v_n$. To convert it into the resultant graph we have to assign direction to each edge. Now see, our resultant graph is acyclic therefore it must have a topological order. (I have not drawn all edges except $V_1$ edges.)

here every rearrangement of vertices in topological sort leads to one particular combination to choose the direction of edges.

Hence - $n!$ is answer.

Just to illustrate more, see one of the permutation out of $n!$ These two permutation shows that undirected edge between $V_1$ and $V_2$, was first chosen as $V_1 \rightarrow V_2$ and then $V_2 \rightarrow V_1$

Don't think about the labeling of vertices, If I do unlabelling of all $n!$ permutations then all structures are same. But it doesn't matter If I am arriving at the same structure, What matters is, In how many ways I can reach to that.

See this-  All the above $3$ structures are Isomorphic. But still, there are $3!$ ways to reach such structure.

Let there be $3$ vertices,

In how many ways we can arrage them? It is $3! = 6$

1. $a,b,c$
2. $a,c,b$
3. $b,a,c$
4. $b,c,a$
5. $c,a,b$
6. $c,b,a$

Let's take each order, and check in how many ways we can make it as directed edges ( remember there are 3 edges in K3 ) with it

ex :- $b,a,c:$  we make $b \to a \to c.$ Now the remaining edge must be $b \to c;$ there is no other possibility for it, as if we make $c \to b$ then it is cyclic.

So, there is only one way to assign edges, to each order !

Now, you got doubt that " There may be many assignment of direction of edges than permutations ? "

Answer for this question is, Should be NO. ( think about it, let there be an edge $u \to v,$ then $u$ should be before $v$)

In how many ways we can choose direction for the edges so that there are no directed cycles? $\to$ Number of permutations of vertices $= n !$

Correct Answer: $C$

by Boss (17.9k points)
edited
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Nice explanation.Thanks :-)
Let there are $n$ vertices $a_{1},a_{2},...,a_{n}$. If edge from any vertex direct to previously/already paved vertex it forms a cycle. To ensure it doesn't happen at every vertex there would be a choice of the edge such that it would direct the vertices not yet encountered. Therefore the number of ways turned out to be $n*(n-1)*(n-2)*...*1 = n!$.
by Loyal (6.3k points)
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He is talking abt topological structure which is mentioned in best answer ...
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Constructive prooproof
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Constructive prove.
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Think of deadlocks. A condition that needs to be satisfied for deadlocks is circular wait. To eliminate circular wait, there's some method (can't recall it's name) in which we number each process and each resource, and if a resource is allocated to a process, then the process can ask for only those resources that are numbered higher than the currently held resource. This diasbles circular wait, as we can't go in a loop — we'll always point forward.

Similarly here, label each vertex as 1,2,3,4... At each point a vertex can point to only a larger numbered vertex. (If it points to a smaller numbered, we'll get a loop).

So, $n!$

Take the example of a triangle (K3) where n = 3.

We can have total of 8 permutations - since each edge can be filled with two options ' > ' or ' < ' => 2 * 2 * 2 = 8.

We can have a cycle only in two permutations  =>  ' > > > ' and ' < < < '.

And hence, it is 8-2 = 6 when there are no directed cycles. This corresponds to only option (c), and no other option.

by Active (1.7k points)
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I have a confusion. It's may be silly.

In the question, they mean to say that the resulting graph should not contain directed cycle "with n vertices" or "it should not contain any directed cycle.Means even the directed cycle with less than n vertices?"
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If cycle length is "n", then there will be always 2 ways to form a cycle, right? '>>>...>' OR '<<<..<<'. Then the answer would have been 2^n-2, which is not in the options.
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yes, u r right , in question its saying that there would be no cycle either of n length as well as  of less than n , but k3 is the simple and small example to check which will give n!, it means its applicable for all graph either k4,k5or so on
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Soumya29 no directed cycle of any length

Draw two directed graphs, one having 3 vertices and other having 4 vertices (Draw such that there are no directed cycles)

You will notice,

In graph with 3 vertices, there is one vertex with indegree 2, one vertex with indegree 1 and one vertex with indegree 0

That is, 3*2*1 different ways to choose vertices $\Rightarrow 3! ways$

Similarly in graph with 4 vertices, there will be 1 vertex of indegree 3, one with indegree 2, one with indegree 1 and final one with indegree 0

That is 4*3*2*1 different ways $\Rightarrow 4! ways$

Similarly for n vertices n! ways
by (217 points)
+1 vote
A  cycle will be formed when the no of vertices is greater than  2. So for a complete graph with 3 vertices has 3 edges.

Lets denote f(n=3) as no of acyclic graphs possible with 3 nodes.

So  for 3 edges there are 2^3 possible directions ">" or " <" for each edge. But all "> "  or all "<" will give cycle.

Now f(3)=2^3-2=6

Since a graph is acyclic   iff  it maintains topological order. So

f(4)= 4C1*f(3) because choose any of 4 nodes and maintain topological order with remaining 3 vertices. So f(4)= 4*6= 24= 4!

Similarly f(5)= 5C1*f(4)= 5*4!=5!

So f(n)= n!.

Correct me if i have made any error.
by (133 points) 