Cyrus Beck Algorithm

February 2011 Bachelor of Computer Application (BCA) Semester 5 BC0052 Theory of figurer science 4 Credits Assignment Set 1 (60 Marks) Ques1. carry that the relationis a ? b(mod m ) equivalence relation. Ans. Ques2. Ans. Ques3. Prove by the mode of contradiction in borders that is not a rational heel. Ans. A rational number is of the form p/q where , and p, q are not having any allday factors. As marrowe that is a rational number. So it can be create verbally as If p is even, then it can be written as p = 2k. and so 4 = 2 . and then q is even. This is a contradiction to our assumption that p and q have no common factors. Therefore rational number. Ques.4. Prove by numeric instalment is not a Ans. (i) Base Step: Let n = 0. and so the sum on the left over(p) is zero, since there is nothing to add. The reflect image on the right is also zero. If n=1, left berth = Right side = =1 = = 1. thereof the get out is legitimate for n = 1. (i i) Induction Hypothesis: Assume that the matter to be true for n = m because Adding the ? th term i.e. 2 to some(prenominal) sides of the above equation, we flummox, = = = = Therefore the result is true for n= m+1. Hence by mathematical induction the prone result is true for all positive integers n. Ques5.

Prove that The sum of the degrees of the vertices of a graph G is twice the number of edges Ans. (The induction is by induction on the number of edges e). Case-(i): view e = 1. Suppose f is the edge in G with f = uv. Then d(v) = 1, d(u) = 1. Therefore = 2 x (number of edges) Hence by induction we get that the sum of the degrees of the vertices of the gra ph G is twice the poem of edges. Ques6. Pro! ve that T is a head there is wizard and only one highroad amid every straddle of vertices Ans. Part 1: Suppose T is a tree. Then T is a connected graph and contains no circuits. Since T is connected, there exists at least one path amongst every pair of vertices in T. Suppose that between devil vertices a and b of T, there are two...If you desire to get a full essay, order it on our website: OrderEssay.net

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