Q04. What will be the output of the following pseudo code? 1. Integer j 2. Integer arr[4] = {3, 2, 3, 1} 3. arr[2] = (arr[0] + 2) & arr[1] 4. for (each j from 5 to 6) 5. arr[j mod 3] = (arr[1] + arr[2]) + arr[0] 6. arr[j mod 4] = (arr[2] + arr[3]) & arr[2] 7. End for 8. Print arr[2] + arr[3] Note: mod finds the remainder after the division of one number by another. For example, the expression "5 mod 2" would evaluate to 1 because 5 divided by 2 leaves a quotient of 2 and a remainder of 1. &: bitwise AND - The bitwise AND operator (&) compares each bit of the first operand to the corresponding bit of the second operand. If both bits are 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0. Options: A. 8 B. 14 C. 5 D. 0
Answer (1)
Let's analyze the given pseudo code step by step to find the output:
Initialization
We have an integer array, arr[4], initialized with values {3, 2, 3, 1}.
First Assignment
The statement arr[2] = (arr[0] + 2) & arr[1] updates arr[2].
arr[0] + 2 evaluates to 3 + 2 = 5.
In binary, 5 is 101 and 2 is 010.
Using the bitwise AND (&), we get: 101 & 010 = 000 which is 0.
So, arr[2] is now 0. The array becomes {3, 2, 0, 1}.
Loop Execution
The for loop runs with j from 5 to 6.
Iteration 1 (j = 5):
Calculate index: j mod 3 = 5 mod 3 = 2.
arr[2] = (arr[1] + arr[2]) + arr[0] = (2 + 0) + 3 = 5.
The array becomes {3, 2, 5, 1}.
Calculate index: j mod 4 = 5 mod 4 = 1.
arr[1] = (arr[2] + arr[3]) & arr[2] = (5 + 1) & 5 = 6 & 5.
In binary, 6 is 110 and 5 is 101.
Using the bitwise AND, 110 & 101 = 100 which is 4.
So, arr[1] becomes 4. The array becomes {3, 4, 5, 1}.
Iteration 2 (j = 6):
Calculate index: j mod 3 = 6 mod 3 = 0.
arr[0] = (arr[1] + arr[2]) + arr[0] = (4 + 5) + 3 = 12.
The array becomes {12, 4, 5, 1}.
Calculate index: j mod 4 = 6 mod 4 = 2.
arr[2] = (arr[2] + arr[3]) & arr[2] = (5 + 1) & 5 = 6 & 5.
Using the bitwise AND, 6 & 5 = 4.
So, arr[2] becomes 4. The array becomes {12, 4, 4, 1}.
Final Output Calculation
Finally, the pseudo code prints arr[2] + arr[3].
arr[2] + arr[3] = 4 + 1 = 5.
Thus, the output of the given pseudo code is 5, which corresponds to option C.
