PREVIOUS YEAR SOLVED PAPERS - August 2016 Paper 3
- Option : B
- Explanation : Program P( ) { x = 10; y = 3; funb (y, x, x) print x; print y; } funb (x, y, z) { y = y + 4; z = x + y + z; } Since, It is call by reference then address will be pass as argument: i.e. P( ) { x = 10; y = 3; funb (&y, &x, &x) print x; print y; } funb (x, y, z) //funb (&y, &x, &x) // { y = y + 4; //at &x 14 will be assigned // z = x + y + z; // Now &x will be assigned with 3 + 14 + 14 = 31. } There is no change in y and x is updated twice when printf called it will print x = 31 and y = 3. So, option (B) is correct.
22. The regular grammar for the language L = {anbm | n + m is even} is given by
- Option : D
- Explanation : Given, L = L = {anbm | n + m is even} For (n + m) to be even either n and m both are even or n and m both are odd. So, for n and m to be even, grammar is; S1 → aa S1| A1 A1 → bb A1| λ For n and m odd, grammar is: S2 → aaS2| aA2 A2 → bbA2| b Now, combine both; then resultant grammar is: S → S1 | S2 S1 → aa S1| A1 S2 → aaS2| aA2 A1 → bb A1| λ A2 → bb A2| b
23. Let Σ = {a, b} and language L = {aa, bb}. Then, the complement of L is
- Option : D
- Explanation : Strings generated by language L = {aa, bb} will be even length. So, complement will be universal set of stringgs - {aa, bb}. i.e. {λ, a, b, ab, ba} and strings {w ∈ {a, b}* | |w| ≥ 3}, i.e. strings of length greater than or equal to 3. So, option (D) is correct.
- Option : D
- Explanation :
- (r + s)* will generate any strings containing r or s or both. We can draw DFA for (r + s)* and it is same as (s + r)*. It is a regular expression.
- (r*)* will generate any sring containing r and its DFA can be drawn easily and it is same as r*. It is also a regular expression.
- (r* s*)* will generate any strings containing r or s or both. We can draw DFA for (r* s*)* and it is same as (r + s)*. It is a regular expression. All option are true.
So, option (D) is correct.
- Option : A
- Explanation : It is given that transmission rate of a channel is 32 kbps and ‘8’ routes from source to destination and each packet p contains 8000 bits. Total delay = routes * packets / transmission rate. i.e. 8 * 8000 b / 32 kbps = 8 * 8000 b / 32000 bps = 2 s So, option (A) is correct.
