Question

5. Let G is a simple planar graph containing no triangles. (i) Using Euler's formula, show that G contains a vertex of degree at most 3. (ii) Use induction to deduce that G is 4-colorable-(v).

Answer 1

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AS FOR VIVEN DATA...

Let G be a simple planar graph containing no triangles. (i) Using Euler’s formula, show that G contains a vertex of degree at most 3. (ii) Use induction to deduce that G is 4-colorable.

EXPLANATION :-

For the first part we can use Euler's F−E+V=2, where F is the number of faces (counting the face that includes ∞), E is the number of edges, and V

the number of vertices.

If there are no triangles then each face is enclosed by at least 4

edges. So 4F/2<E, i.e. F<E/2. From this we get V>E/2+2. If all vertices have order ≥4 then 4V/2<E, i.e. V<E/2. From this we get 0>2. So, we cannot have no triangles and at the same time all vertices with degree ≥4

.

For the second part take a vertex that has degree ≤3

. Color it and its neighbors with different colors A,B,C,D. We can do this only because it has degree ≤3. We can delete this vertex and the edges insident on it. If we color the rest of the graph there is always a way to color the deleted vertex. Notice that the graph with this vertex deleted still has no triangles, but less vertices. Apply induction on the number of vertices now.

ii>

a triangle-free planar graph always has a vertex with degree 3 or less. any graph on 4 (or less) vertices is 4-colourable

Lets say , every triangle-free planar graph on n vertices is 4-colourable. Now take any triangle-free planar graph on (n+1) vertices.

If the graph has a vertex with degree three or less then remove it, the remaining graph can be 4 coloured by the induction hypothesis. The removed vertex had 3 or less neighbours only so we can colour it using the fourth colour.

There is always a vertex having degree 3 or less. We can suppose that the graph is connected and has at least 4 vertices.

If it is not connected than take any connected component of the graph. Every face has at least 4 edges. Hence the number of edges is at least 2f. v-e+f=2 implies v-e+e/2 is at least 2. So, on one hand e<=2v-4.

On the other hand the sum of the degrees is 2e. If every vertex had degree 4 or larger then 2e would be at least 4v which is not possible since e<=2v-4.

Hence we could deduce that G is 4 colourable.

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