PREVIOUS YEAR SOLVED PAPERS - GATE 2020
- Option : C
- Explanation : Rule 1: Rule 1: It is Rule 1: L-attributed by definition. Rule 2: Rule 2: It is not L-attributed because this expression (X. Rule 2: i = A.i + Y.s) fails to satisfy the definition of L-attributed.
| CODE SECTION P |
| CODE SECTION Q |
- A
It ensures that all processes execute CODE SECTION P mutually exclusively.
- B
It ensures that at most n–1 processes are in CODE SECTION P at any time.
- C
It ensures that no process executes CODE SECTION Q before every process has finished CODE SECTION P.
- D
It ensures that at most two processes are in CODE SECTION Q at any time.
- Option : C
- Explanation : Subject Operating Systems a = 1 b = 0 count = 0 Code section P: For process 1 Wait (a) ⇒ a = 0 Count ++ ⇒ count = 1 If (count = = n) ⇒ false signal (a) ⇒ a = 0 wait (b) process blocked As shown above, all processes execute wait (b) before, signal (b) in if condition. Because, ‘if’ condition becomes True only for process Pn(nth process) [∵ if (count = = n)] So, before Pn process finishes the code section, P, no other process executes code section Q. Hence, no process executes code section Q before every process has finished code section P.
- Option : D
- Explanation : G = (V, E) T = MST Weighted edge ( μ, v) is added to T. To verify it is still a MST or not, we need to compare with all the vertices. So, it would be O(V).
- Option : A
- Explanation : L1 is regular and L2 is CFL.L1: {ω xyx| ω, x, y ϵ (0 + 1)*}We can write the regular expression for this,Let, x = 0L11 = ω 0 Y 0 ⇒ ( 0 + 1)+ 0(0 + 1)+0x = 1L12 = ω 1 Y 1 ⇒ we can generate all the strings and since a regular expression can be written for 4, we can say 4 is regular.L2 is CFL:L2 = {xy| x,y ϵ (a + b)*, |x| = |y|, x ≠ y}L2 generates strings of even length, which can be done by PDA, therefore L2 is CFL.
- Option : D
- Explanation : Given P : 155
Q : 85
R : 110
S : 30
T : 115
Current head position = 100
SSTF = algorithm
From the above sequence, we can say that option D is correct.
