Do we always need to take some graphs and check options by hit and trial in graph theory or they can be done using some axioms.?

thor
asked
in Graph Theory
Dec 28, 2016

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## 2 Answers

1 vote

We can solve this type of problem by assuming graph which holds true for their constraint given,

For statement 1 and 2 , graph contain six vertices and have degree = 2 ( for each node )

simple graph , graph which have no multiple edges and self loop is known as simple graph but it can be connected or not.

each node have two degree means , each node is connected with two different nodes , because simple graph doesn't have multiple edges or self loop.

Then graph have to be cyclic like [ A - B - C - D - E - F - A ] , all have degree equals to two.

Hence statement one is true.

Second, euler circuit : start from any node , traverse each edge exactly once and come back to the starting node is know as euler circuit, which is true for this graph, hence statement 2 is correct.

Statement 3 : you can surely draw a graph, which have no self loop and no multiple edges, and have degree equals to three , and it is disconnected.

Hence s1 and s2 is correct , s3 is not.

For statement 1 and 2 , graph contain six vertices and have degree = 2 ( for each node )

simple graph , graph which have no multiple edges and self loop is known as simple graph but it can be connected or not.

each node have two degree means , each node is connected with two different nodes , because simple graph doesn't have multiple edges or self loop.

Then graph have to be cyclic like [ A - B - C - D - E - F - A ] , all have degree equals to two.

Hence statement one is true.

Second, euler circuit : start from any node , traverse each edge exactly once and come back to the starting node is know as euler circuit, which is true for this graph, hence statement 2 is correct.

Statement 3 : you can surely draw a graph, which have no self loop and no multiple edges, and have degree equals to three , and it is disconnected.

Hence s1 and s2 is correct , s3 is not.

### 7 Comments

S1: Two complete graph of 3 vertices. Each vertex having degree 2. Graph is disconnected.

S2: Same graph as above. Since it's disconnected, no Euler circuit will be there.

S3: A graph on set of 8 vertices, consisting of two components. Each component being a complete graph of 4 vertices.

Now tell me, what's wrong with this.

S2: Same graph as above. Since it's disconnected, no Euler circuit will be there.

S3: A graph on set of 8 vertices, consisting of two components. Each component being a complete graph of 4 vertices.

Now tell me, what's wrong with this.

3

Yep, all options are true or false on different - different simple graph...

Option A : Onlys S1 is true, (false ) because statement 2 and 3 can also be false.

Option B : S1 and S2 are true , yep , possibble , they together be true or false

Option C : Only S3 is true , false , s1 and s2 can be true

Option D : S2 and S3 are true , its not necessary they together be true, both are independent of each other

Hence option B is correct.

Option A : Onlys S1 is true, (false ) because statement 2 and 3 can also be false.

Option B : S1 and S2 are true , yep , possibble , they together be true or false

Option C : Only S3 is true , false , s1 and s2 can be true

Option D : S2 and S3 are true , its not necessary they together be true, both are independent of each other

Hence option B is correct.

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