Question

Let G = (V, E) be a finite graph. We will use a few definitions for the statement of this problem. The Tutte polynomial is defined as the polynomial in 2 variables, 2 and y, given by: Definition 1 Tg(x,y) = (x - 1)*(A)-k(E)(y - 1)*(A)+|A1-1V1 ACE where for A CE, k(A) is the number of connected components of the graph (V, A). For this problem we will need the following definition: Definition 2 (Acyclic Graph) A graph is called acyclic if it contains no cyclic subgraphs. And at some point the following theorem may be useful, which you can use cite proving: Theorem 1 If G=(V, E) is a finite connected graph, then it is a tree if and only if |VI = |El +1.

Answer 1

Given that, ACE TG COC, y) = E COC-UJK(A) -K(E) K(A)+(Al-v1 (4-1) het G = (v, E) be a Fioite graph for any subset of ASE, K(A) K(E) where deletion of edges either increases the no. of Components or keeps it remain same Suppose |AL</ul, that means there are either (IVI-IAL-1) leaves in the graph (V, A) or (1v1- LA) number of components to (V,A) .: K(A)+AL-1V120 We know that G = (VE) & Eg = (1,2) Ez=(2,3) Ey =(3,1) Es= {(1,2),(a, 3? E6 = {(1,3), (2,3)} Ez = {(1,3), (1,2) E8 = E

TG (x, y) = Es K(A)-K(E) È (2-1) K(A)+(Al-Iv! U-1) AZEI Es KCA)+1A) - 3 RCA)-1 E (x-1) (4-1) A=E1 cocesa (4-1) +606-10'04-1° (1+1+1) +(1+1+1)+ (56-1°(4-1) = [(2–132+ 3(2-1) + 14-1) +3] (3) Ta(2.a) = [(a−13++ 3 (2-1) + (2-1)+3] 8 TG (22) 1+3(1)+1+3] TG (22) = TG (2,2) = 2 3 37 TG (2, 2) = al El [:where le) = 3 Hence shown (4) Given, TG (132) counts the number of Subgraphs H=(W; F) of G with W = V and having the same number of Components as G TG (1,2) = [,-1)2 + 3(1-1)+(2-1) +3

- 4 ) TG (1, 2) = [0+ 3103+1+3] TG (1,2)= [143 TG (1,2) lebono f Subgraphs which I components has edge set E5, E6, Ez and Eg. (5) TQ (2,1) = (2-1843(2-1)+(1-1)+3] TG (2, 1) = (1+361) +0+3] TG C2, 1) = (1+3+3]. TG (2,1) Number of acylic subgraphs of G that contain all vertices is Es, Ei , E2, E₃, E4, E5, E6 7