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.
PREVIOUS YEAR SOLVED PAPERS MCQ - GATE 2020
32
Each of a set of Q.28 n processes executes the following code using two semaphores a and b initialized to 1 and 0, respectively. Assume that count is a shared variable initialized to 0 and not used in CODE SECTION P.
wait(a); count=count+1;
if (count==n) signal(b);
signal(a); wait(b); signal(b);
What does the code achieve?
| 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.
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.
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).
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.
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.
From the above sequence, we can say that option D is correct.