# suppose a graph has vertices of degree 1,2, 4, 5 and 6. how many edges does the graph have?

### Mohammed

Guys, does anyone know the answer?

get suppose a graph has vertices of degree 1,2, 4, 5 and 6. how many edges does the graph have? from screen.

## A graph with degrees 0 2 2 4 4 4?

Given a graph with 6 vertices of degrees 0 2 2 4 4 4, in what ways may it be drawn? Simple and connected or some combination of? Obviously it can't be connected due to the vertex with degree 0, but

## A graph with degrees 0 2 2 4 4 4?

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Given a graph with 6 vertices of degrees 0 2 2 4 4 4, in what ways may it be drawn? Simple and connected or some combination of?

Obviously it can't be connected due to the vertex with degree 0, but what can I do with the remaining 5 vertices? The graph would have 8 edges since

∑deg( v i ) 2 =8 ∑deg(vi)2=8 . I noticed that K 5 K5

has 10 edges which means that each vertex has deg=4. You take away any edge and you end up with 2 edges having deg=3. At this point the only way to get desired set of degrees is to remove an edge between the 2 vertices having deg=3. But we already removed the only edge between them, so the graph of the remaining 5 vertices is either not simple or not connected.

So at this point I drew a graph where each vertex has a loop and 3 vertices are connected in a triangle, giving the desired set of degrees and number of edges.

Is this the only solution and **is there a better way to approach the problem other than trial and error in drawing the graph?**

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edited Mar 30, 2012 at 7:50

asked Mar 30, 2012 at 7:10

Robert S. Barnes 2,2973 3 gold badges 32 32 silver badges 45 45 bronze badges

You correctly deduced that the graph cannot be simple. I'm no expert, but often when you allow your graphs to have loops, you also allow double edges (=two edges from A to B)... –

Jyrki Lahtonen

Mar 30, 2012 at 7:14

Yes, I could have also drawn double edges instead of loops on the 3 vertices connected in a triangle. Any other ideas for solutions? –

Robert S. Barnes

Mar 30, 2012 at 7:33

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## 2 Answers

5

You can use the Havel-Hakimi theorem (which is applicable to simple graphs).

Using that theorem your sequence

4,4,4,2,2,0 4,4,4,2,2,0 becomes 3,3,3,1,1,0 3,3,3,1,1,0

after one step of application. The sum of those degrees is odd, therefore no such (simple) graph exists.

If you allow for loops and multiple edges, to each vertex of even degree, you can add multiple self loops.

Since the number of vertices of odd degree must be even, you can pair them off and then add multiple self loops to each.

So, as long as the degree sequence has an even number of odd degrees (which is a necessary condition), you can find a non-simple graph (with loops) with that degree sequence, and is quite uninteresting.

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edited Apr 13, 2017 at 12:21

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answered Mar 30, 2012 at 8:06

Aryabhata 79.8k8 8 gold badges 180 180 silver badges 265 265 bronze badges

I'm pretty much a novice in graph theory, could you explain a bit more in depth? –

Robert S. Barnes

Mar 30, 2012 at 10:40

@RobertS.Barnes: Which part? –

Aryabhata

Mar 30, 2012 at 14:34

The Havel-Hakimi Theorem. I looked over the link, but I'm not sure what the algorithm is or it's consequences. –

Robert S. Barnes

Mar 30, 2012 at 15:08

@RobertS.Barnes: The theorem gives you a way of transforming a giving degree sequence, into another sequence, in which you reduce the maximum degree (and possibly the number of nodes) such that the original sequence is graphical if and only if the transformed sequence is graphical. So if you keep transforming, you should end up with all

0 0 ultimately. – Aryabhata

Mar 30, 2012 at 15:36

Havel-Hakimi is for simple graphs, you wrote this, but OP asked for general graphs. Also adding loops is simplest possible way to do this, and OP mentioned this in his question, and seems he looks for more elegant way. –

Saeed

Mar 30, 2012 at 21:54

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3

I will assume that loops and multiple edges are not allowed.

Since you have one vertex of degree 0, the graph would consist of one isolated vertex and the remaining vertices would form a graph with degree sequence

(2,2,4,4,4) (2,2,4,4,4) .

Now you have 5 vertices and 3 of them have degree 4, so these 3 vertices must be connected to all remaining 4 degrees. This implies that each vertex of this 5-vertex subgraph must have degree at least 3 (it is adjacent at least to the 3 vertices having degree 4). This is a contradiction.

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answered Mar 30, 2012 at 8:58

Martin Sleziak 50.3k18 18 gold badges 171 171 silver badges 346 346 bronze badges They are allowed. – Robert S. Barnes

Mar 30, 2012 at 10:33

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### Linked

22

What condition need to be imposed on Havel-Hakimi theorem to check for connected graph?

### Related

6

Homework - Proof: Is this particular graph Hamiltonian?

5

Graph Theory: Simple Graph

2

Show a simple, no loop, 3 connected graph for which

min(deg(v))> edge connectivity >vertex- connectivity

min(deg(v))> edge connectivity >vertex- connectivity

स्रोत : **math.stackexchange.com**

## Answers to questions

## Answers to questions

### Question 1

Unfortunately no. You can do this by trial and error on Figure 3. But it will help to alter Figure 3 slightly in the following way. Put a dot on every land mass and join two dots by a line for each bridge that connects them. So we get Figure A1 below. And then it would help to think a little more deeply than trial and error.

Figure A1: Königsberg reduced to dots and lines

Suppose that you could do the round trip walk. And suppose that you started at land mass A. Then you would have to go out from A on a bridge. So you start by using one bridge. At any time later that you came back to A you would use one bridge going in to A and one edge going out. So far you have used an odd number of bridges. Eventually you'll go back to A and use one final bridge to make the bridge count even. BUT! No land mass is attached to an even number of bridges. SO! There is no round trip walk.

### Question 2

Apply the argument of Q1 to show that every land mass, except for the initial and final one, has to be attached to an even number of bridges. So can you see why the answer here is `no' too?

### Question 3

It turns out that, over time, Königsberg in Prussia has become Kaliningrad, Russia. In the process of time, the seven bridges have been reduced to five. If the current position of the bridges is as shown in Figure A2, then the walk (

A,B,C,D,B,A A,B,C,D,B,A ) can be done.

Figure A2: A walk with five bridges

How did you go with six bridges? Can you only find a walk that starts on one part of the city and ends up somewhere else? Why?

### Question 4

Yes, put in two landmasses

X X and Y Y between C C and D D and then A,B,C,D,X,Y,BA A,B,C,D,X,Y,BA

will get you round. But can you be more imaginative than that?

### Question 5

Well it just depends on how you deploy your bridges. Sometimes there is a round trip walk and sometimes there ain't. What is the important feature that will guarantee a walk?

### Question 6

The key point is the even-ness of the number of bridges at each land mass. Is it possible to prove that if the number of bridges at each land mass is even, then there is a round trip walk? Is it possible to show that if there is a round trip walk, then the number of bridges at each land mass is even? So is it true that there is a round trip walk if and only if the number of bridges at each land mass is even? Can you prove this? Beware the lure of a simple result. Make sure that you carefully word your answer. (See Euler's Tour Theorem.)

### Question 7

The answer is no, but we'll get back to this later when we've thought about 'planarity' and

K 3 K3

- whatever they are.

### Question 8

Putting one dot on the page should give a graph with one vertex, so A is not a possibility. How could we have more than one graph though? Perhaps we could have differently shaped vertices? No, we will assume that however you draw the vertices they are the same vertex (though we usually draw them as small circles). So what else could we do? There is the possibility of edges. Could we draw an edge from a vertex to a vertex? There seems to be nothing in the rules to stop that. Such things are called ** loops**. We show one in Figure A3.

Figure A3: A graph with a loop

Those of you who wanted to include loops would have answered D to Q1. If you are against loops, though, you will have answered B.

### Question 9

Because we are not allowing loops, it looks at the start as if there might only be two graphs on two vertices. But then look at Figure A4 below. Are these graphs the same or different?

Figure A4: Graphs with different looking lines

### Question 10

After some thought I'd guess that there are two schools of thought. Those who like C and those who like D. The problem again is 'when are two graphs the same?'. We will go for C, but why?

### Question 11

There are 11 graphs on 4 vertices. We show them in Figure A5. But we show them systematically there. What system did we use? Why is the sixth one 'alone' with no partners?

Figure A5: The simple graphs on four vertice

### Question 12

Another way to do this is: is

a↔v,b↔u,c↔w a↔v,b↔u,c↔w , and d↔x d↔x

. This sends the edges

ab↔vu=uv,ac↔vw,bd↔ux

ab↔vu=uv,ac↔vw,bd↔ux

.

### Question 13

Doing this simply, interchange the positions of

c c and d d

in Figure A6 and then rotate the graph .

Figure A6: An isomorphic pair of graphs - from Z to

Π Π

Using a bijection we have

a↔x,b↔u,c↔v

## Suppose a connected graph has 17 edges, 1 vertex of degree 2, 2 vertices of degree 3, 3 vertex of degrees 4, and all others of degree 7. How many vertices does the graph have?

Answer (1 of 3): 17 edges means 34 edge-vertex incidences. 1 degree 2 vertex accounts for 2 incidences. That leaves 32. 2 degree 3 vertices accounts for 6 incidences. That leaves 26. 3 degree 4 vertices accounts for 12 incidences. That leaves 14. The remaining degree 7 vertices account for th...

Suppose a connected graph has 17 edges, 1 vertex of degree 2, 2 vertices of degree 3, 3 vertex of degrees 4, and all others of degree 7. How many vertices does the graph have?

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Sort Michael Wish

M.S. in Mathematics & Physics (college major), Naval Postgraduate School (Graduated 2022)Upvoted by

Steve Powell

, MA Mathematics, New College, University of Oxford (1975)Updated 11mo

The sum of the degrees of any connected graph must equal twice the number of edges. So if the graph has

17 17

edges, the degree of the graph is

34 34

. Sum and multiply the vertices and their degrees:

1⋅2+2⋅3+3⋅4+7⋅n=34 1⋅2+2⋅3+3⋅4+7⋅n=34

where n is the number of vertices we don't know, but is degree

7 7

. A little algebra and you get

7⋅n=14 7⋅n=14 , so n=2 n=2

. But we're not done, that just means there are two vertices that are of degree

7 7

. To get the total number of vertices we need to add to what's already given:

1+2+3+2=8 1+2+3+2=8 . There are 8 8

total vertices in the graph.

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17 edges means 34 edge-vertex incidences.

1 degree 2 vertex accounts for 2 incidences. That leaves 32.

2 degree 3 vertices accounts for 6 incidences. That leaves 26.

3 degree 4 vertices accounts for 12 incidences. That leaves 14.

The remaining degree 7 vertices account for the remaining 14 incidences 7 at a time, which means there are exactly two of them.

Therefore, assuming such a graph exists, it has 1+2+3+2=8 vertices. And given that it has more than enough edges to connect it, we can be fairly sure it does. Go ahead and draw a graph with this degree distribution if you like.

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A simple graph has 9 edges, 3 vertices of degree 4 and all others of degree 2. How many vertices does the graph have?

These are very straightforward questions. If you had taken the trouble of opening your textbook, you would not be asking this question. You must make an effort to try problems yourself first. You gain very little having your questions answered without first trying.

The first theorem of Graph Theory says that the sum of all vertex degrees equals twice the number of edges. So if there are

n n

vertices in the graph, then

(3⋅4)+2(n−3)=2⋅9=18 (3⋅4)+2(n−3)=2⋅9=18 . This gives n=6 n=6 . ■ ◼ Sayan Banerjee

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Rest is easy,

2×4+3+2+(n−4)×1=2×(n−1)

2×4+3+2+(n−4)×1=2×(n−1)

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Thus, The tree has 11 vertices (and 10 edges).

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Among any four vertices of the graph on n>3 vertices, one is connected by edges with all the others. What is the least amount of edges that the graph contains?

Guys, does anyone know the answer?