Question

Consider the following pseudocode.

Create empty stack S
Set x = 0, flag = 0, sum = 0
Push x onto S
while (S is not empty){
    if (flag equals 0){
        Set x = x + 1
        Push x onto S
    }
    if (x equals 8):
        Set flag = 1
    if (flag equals 1){
        x = Pop(S)
        if (x is odd):
            Pop(S)
        Set sum = sum + x
    }
}
Output sum

The value of \( sum \) output by a program executing the above pseudocode is:

Hint:

When analyzing pseudocode with stack operations, always keep track of the stack contents and the conditions that trigger operations like pop. In this case, the alternation between odd and even values affects how the stack is manipulated.

Answer 1

Analyzing the Pseudocode.
Initially, \( x = 0 \), \( flag = 0 \), and \( sum = 0 \).
First, \( x = 0 \) is pushed onto the stack.
The loop starts with \( flag = 0 \), and the value of \( x \) is incremented by 1 on each iteration until \( x = 8 \). So the stack contains: \( [0, 1, 2, 3, 4, 5, 6, 7, 8] \).
When \( x = 8 \), \( flag \) is set to 1.
Now, \( flag = 1 \), and we start popping values from the stack.
First, \( x = 8 \) is popped from the stack.
Since \( x = 8 \) is even, we don't pop again and add \( x \) to the sum. So, \( sum = 0 + 8 = 8 \).
Next, \( x = 7 \) is popped from the stack.
Since \( x = 7 \) is odd, we pop the next value, \( x = 6 \), and add \( x \) to the sum. So, \( sum = 8 + 6 = 14 \).
Next, \( x = 5 \) is popped from the stack.
Since \( x = 5 \) is odd, we pop the next value, \( x = 4 \), and add \( x \) to the sum. So, \( sum = 14 + 4 = 18 \).
Next, \( x = 3 \) is popped from the stack.
Since \( x = 3 \) is odd, we pop the next value, \( x = 2 \), and add \( x \) to the sum. So, \( sum = 18 + 2 = 20 \).
Finally, \( x = 1 \) is popped from the stack.
Since \( x = 1 \) is odd, we pop the next value, \( x = 0 \), and add \( x \) to the sum. So, \( sum = 20 + 4 = 24 \).
Step 2: Conclusion.
The final value of \( sum \) after the program executes is \( \boxed{24} \).