Question

Please answer only problem 2. Accurate answers with work shown will receive a 100% rating ASAP. Thank you!

Let G = (V, E) be a graph. We say that a subset S of the vertices V is an independent set if there is no edge in G joining two vertices in S. For example, given a proper colouring of the vertices of G, each colour class (i.e. the set of vertices that have some fixed colour) forms an independent set, by the definition of proper colourings. Let a(G) denote the size of the largest independent set in G. It is known that determining a(G) given an input graph G is a very difficult problem, even if the input graph is restricted to planar graphs. This means that most computer scientists expect that an efficient algorithm for finding a(G) does not exist. So the best we can hope to do is to give some bounds on a(G). In this homework we will look at some bounds on a(G) for planar graphs. Problem 1. (5 points) Let G = (V, E) be a planar graph. Show that a(G) 2 IV (hint: use the four colour theorem) A better general lower bound is not possible due to the example K4. Problem 2. (10 points) Let G = (V, E) be a connected planar graph with minimum degree 3. Show that a(G) <ŽIV- Some hints: 1. You may assume that G has a "triangulation”, a planar embedding such that every face has size 3. This is because adding edges to a graph only makes it harder to find a large independent set (you should be able to formalize this). - E + F= 2, where F is the number of faces 2. You will probably need to use Euler's formula: V in a planar embedding. 3. To get some intuition about this problem, think about why the minimum degree condition is necessary. What happens for connected planar graphs with minimum degree 1 or 2? Also, try to find some examples of graphs (that satisfy the hypothesis) where the inequality is "tight”, as in a(G) = |VI- These examples may help you find a general argument.

Answer 1

P. Ab ! ☺ given the data :- 1. let si be an indepondent set with isl = Q(G). for any edge e in G, both the end. vertices of e. can not lie in . s. so at least one end of e belongs to s Hence s is a vertor eurer of Gr. 151 =p-a(G) since B(G) is the coordinality of a minimum leater cover, BCGU < 1541-9-2 (6) => 3 (G) +BCG) =] = > 0 Next let & be a minimum verter courrier, - i. lxl = B(G) since x contains at least one and vertar of every edge, no two verties in te are aduent.

P.No :- Hence tc is an independent sot. -:p - BCG) = l«l < 2 (G) Fix is independent) I aCG) is madlimuna independent = 2(G) + B(a) zp - from 0 € We have (G) +B(61)= p since G is a bi-partite graph. then by minumax theram we get. BCG) = d'(o) - @ from ③ & ④ we get a(G) + d'(G) =