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SPRING 2014, ASSIGNMENT
| PRIGRAM | BCA(REVISED 2007) |
| SEMESTER | 5TH |
| SUBJECT CODE & NAME | BC0052 – THEORY OF COMPUTER SCIENCE |
| CREDIT | 4 |
| BK ID | B0972 |
| MAX. MARKS | 60 |
Note: Answer all questions. Kindly note that answers for 10 marks questions should be approximately of 400 words. Each question is followed by evaluation scheme.
Q.1Define g.c.d. (m,n)
Solve recursively: (i) f(x, y) = x + y
(ii) g(x, 0) = 0, g(x, y + 1) = g(x, y) + x.
Answer:Define g.c.d. (m,n)
The greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common factor (hcf), of two or more integers (at least one of which is not zero), is the largest positive integer that divides the numbers without a remainder. For example, the GCD of 8 and 12 is 4
Q.2 Obtain a DFA to accept strings of a’s and b’s starting with the string ab.
Answer: It is clear that the string should start with ab and so, the minimum string that can be accepted by the machine is ab.
To accept the string ab, we need three states and the machine can be written as
Q.3 Prove by mathematical induction.
12 + 22 + 32+——+ n2 = (n(n+1)(2n+1))/6
Answer:- 12 + 22 + 32+——+ n2 = (n(n+1)(2n+1))/6
First of all Identify the general term and nth partial sum before beginning the problem
The general term, an, is the last term on the left hand side. an = n2
The nth partial sum, Sn, is the right hand side. Sn = n (n + 1) (2n + 1) / 6
Q.4 briefly describes Moore and Mealy machines.
Answer: -Describe Moore
An automation system in which the output depends only on the present input is called a Moore machine. Alternatively, an automaton system in which output depends both on the present input and the present state is called Mealy machine.
A Moore machine can be defined as a 6-tuple ( S, S0, Σ, Λ, T, G ) consisting of the following:
- A finite set of states ( S )
· A start state (also called initial state) S0 which is an element of (S)
· A finite set called the input alphabet ( Σ )
Q.5 If G = ({ S }, { 0,1}, { S →0S1, S →Ʌ}, S ) then find L(G),
the language generated by G.
Answer:-
Since S → ^ is a production, S ⇒ ^. This implies that ^ ∈L(G).
Now, for all n ≥ 1, we can write the following:
S ⇒ 0S1 ⇒ 00S11 … ⇒ 0nS1n ⇒ 0n1n.
Q.6 Prove that “A tree G with n vertices has (n–1) edges”
Answer:-
We prove this theorem by induction on the number vertices n.
Basic step: If n = 1, then G contains only one vertex and no edge. So the number of edges in
G is n –1 = 1 – 1 = 0.
Induction hypothesis: The statement is true for all trees with less than ‘n’ vertices. Induction step: Now
let us consider a tree with ‘n’ vertices. Let ‘ek ’ be any edge in T whose end vertices are vi and v j.
Since T is a tree, by there is no other path between vI and v
Dear students get fully solved SMU BBA Spring 2014 assignments
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