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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. $2^m, \: \text{ where } m=\frac{n(n-1)}{2}$
edited | 1.3k views
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n! ?

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 Problem statement meaning. Your resultant graph should be acyclic.

Lets say you 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 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 these structures are Isomorphic...But still, there are $3!$ ways to reach such structure.

let there are 3 vertices,

Howmany ways we can arrage them ? it is 3! = 6

i) a,b,c

ii) a,c,b

iii) b,a,c

iv) b,c,a

v) c,a,b

vi) c,b,a

let take each order, and check howmany ways you make it as directed edges ( remember there are 3 edges in K3 ) with it

ex :- b,a,c ===> we make b ---> a ----> c, note that remaining edge must be b ---> c, there is no other possibility for it,due to  if you make it c ---> 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 is an edge u --> v, then u should be before v )

in how many ways can you choose a direction for the edges so that there are no directed cycles? ==> No.of permutations of vertices = n !

edited
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Nice explanation.Thanks :-)
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It will help to solve it ....

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For isomorphism

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 $n*(n-1)*(n-2)*...*1 = n!$.
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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.

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.

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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

For vertex 1 , we have 6 choices , similarly we have 4 different choices for every vertex, which makes 24 ways. which is 4! for K4 .

Kindly Correct me if I'm wrong somewhere. Thank U

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.

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