TIFR CSE 2019 | Part B | Question: 12

Question

Let $G=(V,E)$ be a directed graph with $n(\geq 2)$ vertices, including a special vertex $r$. Each edge $e \in E$ has a strictly positive edge weight $w(e)$. An arborescence in $G$ rooted at $r$ is a subgraph $H$ of $G$ in which every vertex $u \in V \backslash \{r\}$ has a directed path to the special vertex $r$. The weight of an arborescence $H$ is the sum of the weights of the edges in $H$.

Let $H^*$ be a minimum arborescence rooted at $r$, and $w^*$ the weight of $H^*$. Which of the following is $NOT$ always true?

• A. $\displaystyle w^* \geq \sum_{u \in V \backslash \{r\}} \min_{(u,v) \in E} w((u,v))$
• B. $\displaystyle w^* \geq \sum_{u \in V \backslash \{r\}} \min_{(v,u) \in E} w((v,u))$
• C. $H^*$ has exactly $n-1$ edges
• D. $H^*$ is acyclic
• E. $w^*$ is less than the weight of the minimum weight directed Hamiltonian cycle in $G$, when $G$ has a directed Hamiltonian cycle

Comments

Struggled for 15 minutes eliminated few options I was getting B. Idk how took few examples. Pls, share your approach!

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Here it is saying a graph given G

and it has a special vertex r

Now, H is subgraph of graph G and every vertex of H which has a directed edge to r

Now I thought G is a graph like this

Now as H is any subgraph of G, So, A) and B) false , there is always a subgraph which contains minimum weight of the graph

C) false because H may not contain exactly n-1 edges. 

E) False , because H and G can be both same graph (i.e. improper subgraph)

One case definitely there , where it is false

Now D) H is cyclic or not totally depend on ur designing of graph

So, it is not always true (See in the above diagram, if H=G, then It contains a cycle)

 

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@srestha You have to read and understand the question CORRECTLY before solving. Otherwise you are spending time and getting negatives. 

Answer 1

For these questions it is better to get some counter examples by trying small graphs. Consider the following graph:

Now, as per the question, an arborescence is a subgraph with all incoming edges to the special vertex $r.$ So, for the above graph it is just the edge $a-r$ which is of weight $2.$

Now, option A says that if we sum up all the outgoing edge weights (minimum weight in case multiple outgoing edges exist) of all vertices excluding $r$, the sum will be greater than or equal to the weight of the minimum arborescence. This is true because all the edges in the minimum arborescence will be included here as the outgoing edges of the other vertices. 

Now, option B is false. Because instead of outgoing edges we are adding the weights of incoming edges. And these edges are not part of the arborescence. For example, in the above graph, only one vertex $a \in V \backslash \{r\}.$

$ \sum_{u \in V \backslash \{r\}} \min_{(v,u) \in E} w((v,u)) = w(r,a) = 5 >  w^*.$

Options C and D are straightforward TRUE as for minimum arborescence we only need to consider the incoming edges to $r$ from $n-1$ other vertices thus making it acyclic too. 

For option E, suppose a graph has a directed Hamiltonian cycle. In an arborescence every vertex is connected and we need "one more" edge to make it a cycle. But this may or may not be a Hamiltonian cycle as a Hamiltonian cycle requires that in the path from the start and end of any vertex, no vertex is repeated. But if we take any directed Hamiltonian cycle, every vertex has a directed path to every other vertex (special vertex $r$ can be any of the given vertices) and so it is an arborescence. What will happen if we remove an edge $(u,v)$ from such a minimum weighted directed Hamiltonian cycle?. Then we no longer have a path to vertex $v$ but all vertices still have a path to vertex $u$ and so $u$ can be the special vertex making it still an arborescence. Since the weight of this edge $(u,v)$ is guaranteed to be positive, this means the weight of this arborescence must be STRICTLY less than the weight of the minimum Hamiltonian cycle. Thus option E is TRUE. 

So, Correct answer: B.

Comments

yes, @Sourajit25 is correct. Instead of path I was considering direct connectivity. Have corrected now. 

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@Arjun Sir, i like to suggest one small correction in above answer, in place of word "edge",  "path" should be used. Please see below picture.

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Thanks for that. Fixed now.