answer : (2+10-1)C10 * (2+15-1)C15 * (2+14-1)C14 = 2640.

14 votes

Let $G$ be any connected, weighted, undirected graph.

- $G$ has a unique minimum spanning tree, if no two edges of $G$ have the same weight.
- $G$ has a unique minimum spanning tree, if, for every cut of $G$, there is a unique minimum-weight edge crossing the cut.

Which of the following statements is/are TRUE?

- I only
- II only
- Both I and II
- Neither I nor II

1

number of ways to distribute k similar objects to n distinct set is : (n + k - 1)Ck

answer : (2+10-1)C10 * (2+15-1)C15 * (2+14-1)C14 = 2640.

answer : (2+10-1)C10 * (2+15-1)C15 * (2+14-1)C14 = 2640.

3

i am not getting what you want to deliver,

in the question they are saying,

if all edge weights are unique, then G has unique minimum spanning tree. ---- TRUE

Converse :- If G has unique minimum spanning tree, then all edge weights are unique. ----- False

in the question they are saying,

if all edge weights are unique, then G has unique minimum spanning tree. ---- TRUE

Converse :- If G has unique minimum spanning tree, then all edge weights are unique. ----- False

3

why the following is not a counter example for second statement:

The cut has minimum unique weight but multiple MSTs.

0

the line is

"if for every cut of G ,there is a unique minimum weight edge crossing the cut,=> then G has a unique minimum spanning tree"

"if for every cut of G ,there is a unique minimum weight edge crossing the cut,=> then G has a unique minimum spanning tree"

0

Cut set is the set of edges whose removal disconnect the connected graph..so do you not think that cut is also same as cut set and it is min cut set??

0

I donot think it is telling about cut set. Because, if it is a cut set then can cut more than one edge, which is violation of spanning tree

right?

0

Actually they are telling to remove minimum edge from a graph, (which they described as cut set)

And after that removal, is it forming a MST? As minimum edge edge is unique , so it forming a unique spanning tree

0

Here they did not mention cut set..cut canbe bridge also..but if we can find atleast one counter example then is not it that statement can not be true..

Please let me know if my understanding is correct

Please let me know if my understanding is correct

4

They mean for any cut of the graph. Now what is a cut. Any partition of vertices of graph forms a cut. And for every cut if we choose the min edge passing that cut, then that gives us the mst of graph. Now they hv asked if this min for each cut is unique which implies we have a unique mst, so it's true.

13 votes

Best answer

1. If edge weights are distinct then there exist unique MST.

2. If for every cut of a graph there is a unique light edge crossing the cut then the graph has a unique minimum spanning tree but converse may not be true.

__Proof by contradiction:__

**Lemma:*** *If an edge $e$ is contained in some minimum spanning tree, then it is a minimum cost edge crossing some cut of the graph.

Assume MST is not unique and there exist two MST's $T_1$ and $T_2$

Suppose $e_1∈T_1$ but $e_1 \notin T_2$, if we remove $e_1$ from $T_1$, then we will have disconnected graph with two set of vertices $V_1$ and $V_2$. According to lemma $e_1$ is a minimum cost edge in the cut between $V_1$ and $V_2$.

Suppose $e_2∈T_2$ but $e_2 \notin T_1$, if we remove $e_2$ from $T_2$, then we will have disconnected graph with two set of vertices $V_1$ and $V_2$. According to lemma $e_2$ is a minimum cost edge in the cut between $V_1$ and $V_2$.

Because the minimum cost edge is unique implies $e_1$ and $e_2$ is the same edge. $e_1 \in T_2$ and $e_2 \in T_1$. We have chosen $e_1$ at random, of all edges in $T_1$, also in $T_2$ and same for $e_2$. As a result, the MST is unique.

Why converse is not true always?

https://stanford.edu/~rezab/discrete/Midterm/midtermsoln.pdf

So both statements in the question are TRUE.

Answer is (C).

6 votes

Statement 1 : True

Statement 2 : True. Explanation below.

For this I have an intuitive approach. This is not a formal mathematical proof but it occurred to me after thinking for quite some time.

Please point it out if you think something is wrong.

4 votes

Statement 1 is true. Proof can be found easily.

The biggest confusion in Statement 2 is what is a *cut*? Is it cut set, or cut vertex, or cute edge?

Actually, a cut is anything that creates a partition, or "cut" in the set of vertices $V$ of a graph.

A

cut$C=(S,T)$ is a partition of $V$ of a graph $G=(V,E)$ into twosubsetsandS.T

Notice "cut" is actually a set of two subsets.

One might say that *cut* is a *cut set*.

So, if for **every ***cut set *of a graph, there exists a unique minimum weight edge; then graph has a unique MST. True.

Why? Because the whole graph can be seen as the collection of these *cut sets *and from each cut set we can only pick the unique minimum weight edge.

Option C; both the statements are true.

Let

Gbe a connected, undirected graph. Acut inwhose removal...Gis asetof edges

These wordings are taken from GATE 1999, Question Number 5.

2 votes

If for every cut of a graph there is a unique minimum weight edge crossing the cut,

then the graph has a unique minimum spanning tree.

This statement is true and can be proved ea sily by contradiction.

then the graph has a unique minimum spanning tree.

This statement is true and can be proved ea sily by contradiction.

1 vote

Explanation:

Given G be a connected, weighted and undirected graph,

I. TRUE: G Graph is unique, no two edges of the graph is same.

Step-1: Using Kruskal's algorithm, arrange each weights in ascending order.

17, 18, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30

Step-2:

Step-3: 17 + 18 + 20 + 21 + 22 + 23 + 26 = 147

Step-4: Here, all the elements are distinct. So, the possible MCST is 1.

II. TRUE: As per the above graph, if we are cut the edge, that should the be the minimum edge.

Because we are already given, all minimum edge weights if graph is distinct.