PREVIOUS YEAR SOLVED PAPERS - GATE 2018
Which one of the following is correct for the given parse tree?
- Option : A
- Explanation : here $ is higher precedence than # $ is left associative because in the sub expression b $ c $ d, b $ c will be evaluated first as per given tree. As per the given tree structure right # if higher precedence than left #. Hence it is right associative.
| Producer | Consumer |
| do{ wait(p); wait(mutex); //Add item to buffer signal(mutex); signal(Q); }while (1); | do{ wait(R); wait(mutex); //Consume item from buffer signal (mutex); signal (); }while (1); |
Which one of the following assignments to P, Q, R and S will yield the correct solution?
- Option : C
- Explanation : Full = N, empty = 0, mutex = 1 Initially buffer will be empty, so consumer should not start first, so options B, D are eliminated. With option A consumer will never consume the item, so it is wrong. Option ‘C’ is the correct answer which proper functionality of produce and consumer.
- Option : D
- Explanation : T3T3 because from first T3 bbaac is taken from second T3 abc is taken longest possible prefix. Hence T3T3 token output.
- Option : C
- Explanation : Binary code: (b7b6b5b4b3 ⋅ b2b1b0)2 (31.5)10 = (11111.1)2 (0.875)10 = (0.111)2 (12.100)10 = (01100.000110.....)2 ↓ Only 3 bits of fraction space available So can't be stored. (3.001)10 = (00011.000000..........)2 It is also not accurate storage.
- Option : D
- Explanation : P : Set of rational number → countable
Q : Set of functions from {0, 1} to N → N
0 can be assigned in N ways
1 can be assigned in N ways
There are functions, cross product of countable set in countable.
R : Set of functions from N to {0, 1}
Each of thus boxes can be assigned to 0 or 1 so each such function is a binary number with infinite number of bits.
