# suppose that a connected planar simple graph has 40 edges. if a plane drawing of this graph has 15 faces. how many vertices does this graph have?

### Mohammed

Guys, does anyone know the answer?

get suppose that a connected planar simple graph has 40 edges. if a plane drawing of this graph has 15 faces. how many vertices does this graph have? from screen.

## Suppose that a planar graph has $k$ connected components, $e$ edges, and $v$ vertices. Also suppose that$\dots$

Question: Suppose that a planar graph has $k$ connected components, $e$ edges, and $v$ vertices. Also suppose that the plane is divided into $r$ regions by a planar representation of the graph. Fin...

Suppose that a planar graph hask k

connected components,

e e edges, and v v

vertices. Also suppose that

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Asked 8 years, 1 month ago

Modified 8 years, 1 month ago

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**Question:**

Suppose that a planar graph has

k k

connected components,

e e edges, and v v

vertices. Also suppose that the plane is divided into

r r

regions by a planar representation of the graph. Find a formula for

r r in terms of e,v, e,v, and k k .

**Attempt:**

Let r i , v i , e i ri,vi,ei

be the number of regions, edges, vertices in the

i th ith

connected component, thus

r i = e i − v i +2 ri=ei−vi+2

Suppose we ignore the outer most regions for now, our new equation is,

r ′ i = e i − v i +2−1 ri′=ei−vi+2−1 = r ′ i = e i − v i +1 =ri′=ei−vi+1

Adding them all, we get,

where r ′ 1 + r ′ 2 +… r ′ k = r ′′ r1′+r2′+…rk′=r″ ; e 1 + e 2 +… e k =e e1+e2+…ek=e ; v 1 + v 2 +… v k =v v1+v2+…vk=v , therefore r ′ 1 + r ′ 2 +⋯+ r ′ k = r ′′ =e−v+k

r1′+r2′+⋯+rk′=r″=e−v+k

Since r ′′ r″

is deprived of any outer regions, we add one to reintroduce the outer region, thus

r= r ′′ +1 r=r″+1 , r=e−v+k+1 r=e−v+k+1

Is that attempt correct? It seems too ad-hoc for me (I've never done or seen this pattern of solving problem before).

graph-theory Share

asked Aug 16, 2014 at 18:28

Joey Andres 1431 1 silver badge 5 5 bronze badges

Looks good to me. Your proof is an actual hidden induction by

k k

, after you found the right formula, it might be easier to prove it directly by induction. –

N. S.

Aug 16, 2014 at 18:40

Sorry, could you explain how you got the first expression for

r i ri ? – user770533

Dec 22, 2021 at 17:36

Add a comment

## 1 Answer

1

Your proof seems valid to me. This is exactly the approach I was going to take.

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answered Aug 16, 2014 at 18:37

Ragnar 6,16515 15 silver badges 21 21 bronze badges

Sorry, from where does it follow the first expression for

r i ri in his proof? – user770533

Dec 22, 2021 at 17:39

see here: en.wikipedia.org/wiki/Planar_graph#Euler's_formula –

Ragnar

Dec 23, 2021 at 3:19

Add a comment

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स्रोत : **math.stackexchange.com**

## Planar Graphs

Skip to main content

Discrete Mathematics:An Open Introduction, 3rd edition

Oscar Levin 🔗

## 4.3 Planar Graphs

🔗

### Investigate!

When a connected graph can be drawn without any edges crossing, it is called planar. When a planar graph is drawn in this way, it divides the plane into regions called faces.

Draw, if possible, two different planar graphs with the same number of vertices, edges, and faces.

Draw, if possible, two different planar graphs with the same number of vertices and edges, but a different number of faces.

🔗

When is it possible to draw a graph so that none of the edges cross? If this is possible, we say the graph is planar (since you can draw it on the plane).

🔗

Notice that the definition of planar includes the phrase “it is possible to.” This means that even if a graph does not look like it is planar, it still might be. Perhaps you can redraw it in a way in which no edges cross. For example, this is a planar graph:

🔗

That is because we can redraw it like this:

🔗

The graphs are the same, so if one is planar, the other must be too. However, the original drawing of the graph was not a planar representation of the graph.

🔗

When a planar graph is drawn without edges crossing, the edges and vertices of the graph divide the plane into regions. We will call each region a face. The graph above has 3 faces (yes, we do include the “outside” region as a face). The number of faces does not change no matter how you draw the graph (as long as you do so without the edges crossing), so it makes sense to ascribe the number of faces as a property of the planar graph.

🔗

WARNING: you can only count faces when the graph is drawn in a planar way. For example, consider these two representations of the same graph:

🔗

If you try to count faces using the graph on the left, you might say there are 5 faces (including the outside). But drawing the graph with a planar representation shows that in fact there are only 4 faces.

🔗

There is a connection between the number of vertices (

v

), the number of edges (

e

) and the number of faces (

f

) in any connected planar graph. This relationship is called Euler's formula.

🔗

### Euler's Formula for Planar Graphs.

For any connected planar graph with

v vertices, e edges and f faces, we have . v−e+f=2. 🔗

Why is Euler's formula true? One way to convince yourself of its validity is to draw a planar graph step by step. Start with the graph :

P2: 🔗

Any connected graph (besides just a single isolated vertex) must contain this subgraph. Now build up to your graph by adding edges and vertices. Each step will consist of either adding a new vertex connected by a new edge to part of your graph (so creating a new “spike”) or by connecting two vertices already in the graph with a new edge (completing a circuit).

🔗

What do these “moves” do? When adding the spike, the number of edges increases by 1, the number of vertices increases by one, and the number of faces remains the same. But this means that

v−e+f

does not change. Completing a circuit adds one edge, adds one face, and keeps the number of vertices the same. So again,

v−e+f does not change. 🔗

Since we can build any graph using a combination of these two moves, and doing so never changes the quantity ,

v−e+f,

that quantity will be the same for all graphs. But notice that our starting graph

P2 has , v=2, e=1 and , f=1, so . v−e+f=2.

This argument is essentially a proof by induction. A good exercise would be to rewrite it as a formal induction proof.

🔗

### Non-planar Graphs

🔗

### Investigate!

For the complete graphs ,

Kn,

we would like to be able to say something about the number of vertices, edges, and (if the graph is planar) faces. Let's first consider :

K3:

How many vertices does

K3

have? How many edges?

If K3

is planar, how many faces should it have?

Repeat parts (1) and (2) for ,

K4, , K5, and . K23.

What about complete bipartite graphs? How many vertices, edges, and faces (if it were planar) does

K7,4

have? For which values of

m and n are Kn and Km,n planar? 🔗

Not all graphs are planar. If there are too many edges and too few vertices, then some of the edges will need to intersect. The smallest graph where this happens is .

K5. 🔗

If you try to redraw this without edges crossing, you quickly get into trouble. There seems to be one edge too many. In fact, we can prove that no matter how you draw it,

K5

will always have edges crossing.

🔗

### Theorem 4.3.1.

K5 is not planar. 🔗

### Proof.

The proof is by contradiction. So assume that

K5

is planar. Then the graph must satisfy Euler's formula for planar graphs.

K5

has 5 vertices and 10 edges, so we get

, 5−10+f=2,

which says that if the graph is drawn without any edges crossing, there would be

f=7 faces.

Now consider how many edges surround each face. Each face must be surrounded by at least 3 edges. Let

स्रोत : **discrete.openmathbooks.org**

## A connected planar graph with 15 vertices divides the plane into 12 regions. How many edges does the graph have?

Answer: The answer should be a simple application of the Euler’s formula for connected planar graphs. Euler's Formula Let G be a connected planar graph, and let n, m and f denote, respectively, the numbers of vertices, edges, and faces (regions) in a plane drawing of G. Then n - m + f = 2. In...

A connected planar graph with 15 vertices divides the plane into 12 regions. How many edges does the graph have?

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Sort Michal Forišek

Ph. D. in theoretical Computer ScienceUpvoted by

Igor Markov

, long-time researcher in applied graph theory and

Alon Amit

, Ph.D. in Graph Theory.Author has 1.2K answers and 9.9M answer views4y

Related

How do I find a graph with 6 vertices, 12 edges, and does not contain K4 as a subgraph?

The complement of your graph should have 6 vertices, only 3 edges, and there cannot be an independent set of size 4 (i.e., four vertices with no edges between them).

Clearly, each graph with three disjoint edges has the desired property - regardless of how you pick the four vertices, you have to pick both endpoints of (at least) one of the edges.

Using this point of view it can also easily be shown that the above graphs are the only graphs with this property. There are only four other possible layouts of three edges, and for each of those we can easily find four vertices with no edges between th

Alon Amit

Ph.D. in Graph Theory.Author has 7.5K answers and 117M answer views5y

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What is the maximum number of vertices in a complete and planar graph?

Four points can be placed in the plane and all of them can be connected by paths so that the paths don't go through each other.

You can draw it like this:

Or even like this:

If you have five points, however, then you can't do that. It doesn't matter where you place the points, it doesn't matter if you use straight lines or squiggly wiggly paths, you can't draw all of the ten paths that connect them in pairs in the plane. This is what people mean when they say that

K 5 K5 is not planar. But why?

When you draw any graph in the plane, you'll have some number of points or “vertices”,

v v

, some number of pat

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A connected, planar graph has 10 edges. It does not divide the plane into smaller regions. What are the number of its vertices?

If it doesn't divide the plane into smaller regions, it can't have any cycles. Therefore it must be a tree, since it is connected. And so the number of vertices is 11 (1 more than the number of edges).

Vikram

is just another Qurious QuoranAuthor has 253 answers and 786.2K answer views5y

The answer should be a simple application of the Euler’s formula for connected planar graphs.

Euler's Formula

Let G be a connected planar graph, and let n, m and f denote, respectively, the numbers of vertices, edges, and faces (regions) in a plane drawing of G. Then

n - m + f = 2.

In case you need an introduction to the concept behind the Euler’s formula, the link below should be sufficient.

Graph Theory

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Derrin Gottlieb

BA in Mathematics, New College of Florida (Graduated 1987)Author has 85 answers and 111.1K answer views2y

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How many vertices does a graph with 15 edges have if a plane drawing of this graph has 7 faces?

If the graph is planar then Euler’s polyhedron formula applies:

number of vertices - number of edges + number of faces = 2. Using letters:

V - E + F = 2.

Here E is 15 and F is 7:

V - 15 + 7 = 2. V - 8 = 2 V = 10.

So the graph has 10 vertices.

I hope this is helpful.

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If a planar graph A with seven (7) vertices divides the plane into 8 regions, how many edges must graph A have?

Apply Euler’s formula for planar graphs, which is V - E + F = 2, where V is the number of vertices, E is the number of edges, and F is the number of faces. We know that V = 7, and that F = 8 (faces are another term for regions). Then

7 - E + 8 = 2 15 - E = 2 E = 13

Therefore Graph A has 13 edges.

Amitabha Tripathi

have been teaching Graph Theory for almost three decadesAuthor has 4.3K answers and 9.1M answer views2y

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A planar graph has 12 vertices, 7 faces, and 2 components. How many edges must this graph have?

Euler’s formula for Planar graph - Wikipedia

G is n−e+f=1+c(G) n−e+f=1+c(G) , where n n , e e , f f

denote the number of vertices, edges, faces, respectively of

G G and c(G) c(G)

denotes the number of components of

Guys, does anyone know the answer?