PREVIOUS YEAR SOLVED PAPERS - GATE 2017 Shift 2
- Option : A
- Explanation : Given: code1 00000 code2 01011 code3 10101 code4 11110 Hamming distance between code 1 and code 2 is 3 Hamming distance between code 1 and code 3 is 3 Hamming distance between code 1 and code 4 is 4 Hamming distance between code 2 and code 3 is 4 Hamming distance between code 2 and code 4 is 3 Hamming distance between code 3 and code 4 is 3 So, as per Hamming code, minimum Hamming distance of all code words is considered as Hamming distance i.e., 3 (p). Now, the max number of erroneous bits that can be corrected by the Hamming code is 2d + 1 So, So option A is correct.
- Option : B
- Explanation :
Hence, system is in safe state , because P1 and P3 anyone can execute first and after them P2 can be executed, no deadlocked option B is correct.PID Current allocation Max Need Available Needed P1 3 3 2 0 P2 1 6 - 5 P3 3 5 - 2
- Option : A
- Explanation : Conflict conditions RW WR WW ∴ 5 conflicts T1 - T2 a b c d r1(X) w1(X) r1(Y) w1(Y) r2(Y) w2(Y) r2(Z) r2(Z) 1 2 3 4 Constraints: a<b<c<d 1<2<3<4 d<1 (or) 2<c only 1 way Total = 70 - (12 + 5) = 53 Therefore total no of, 53 + 1 = 54
39. If w, x, y, z are Boolean variables, then which one of the following is INCORRECT?
- Option : C
- Explanation : A)LHS: wx + w(x + y) + x(x+y) = x + wy
RHS
⇒ wx + wx + wy + x + xy
⇒ wx + wy + x + xy [∵ xx = x]
⇒ x(w+y+1) + wy [∵1 + x = 1]
⇒ x + wy
⇒ L.H.S = R.H.S
L.H.S ≠ R.H.S
D) L.H.S.:(w + y)(wxy + wyz) = wxy + wyz
(w + y)(wxy + wyz)
wxy + wyz + wxy + wyz
wxy + wyz
L.H.S = R.H.S
- Option : A
- Explanation : Assume starting address of string is 100 to make the whole expression is easy to understand.
Note: Whenever we have characters in the arithmetic expressions, we can replace those with their ASCII values
strlen (100 + x + 11 - x - 1)[assume x has the ASCII value of I]
⇒Strlen(110)
is printed
It gives address of second last character in the string. So it prints length 2.
