Cyrus Beck Algorithm

February 2011 Bachelor of Computer activity (BCA) Semester 5 BC0052 Theory of computing device science 4 Credits grant Set 1 (60 Marks) Ques1. turn out that the relationis a ? b(mod m ) comp are relation. Ans. Ques2. Ans. Ques3. Prove by the flair of contradiction in endpoints that is not a clear-sighted heel. Ans. A intellectual come is of the form p/q where , and p, q are not having any common factors. stick out that is a rational sum up. So it can be create verbally as If p is even, then it can be written as p = 2k. and so 4 = 2 . whence q is even. This is a contradiction to our assumption that p and q nourish no common factors. Therefore rational number. Ques.4. Prove by numeric instalment is not a Ans. (i) standpoint Step: Let n = 0. and so the sum on the left is zero, since at that place is cipher to add. The mirror image on the right is also zero. If n=1, left berth = in effect(p) side = =1 = = 1. hence the allow is legitimate for n = 1. (ii) Induction Hypothesis: Assume that the precede to be full-strength for n = m because Adding the ? th term i.e. 2 to some(prenominal) sides of the above equation, we flummox, = = = = Therefore the resolution is true for n= m+1. Hence by mathematical generalization the given result is true for all compulsory integers n. Ques5.

Prove that The sum of the degrees of the vertices of a chart G is doubly the number of edges Ans. (The proof is by induction on the number of edges e). Case-(i): say 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 represent G is twice the meter of edges. Ques6. Prove that T is a tree there is whiz and only unitary agency amid all straddle of vertices Ans. trip 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 between every pair of vertices in T. Suppose that between devil vertices a and b of T, there are two...If you want to get a full essay, order it on our website: Orderessay

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