Loops Practice Questions

The gate function processes an integer n by extracting digits from left to right and constructing a newnum based on alternating digits.

1. Initialization: For n = 14362, newnum = 0, turn = 0, t = 1.
2. First while loop: t becomes the smallest power of 10 greater than n.
   • t starts at 1, then becomes 10, 100, 1000, 10000, 100000.
   • Since n = 14362 is not greater than or equal to t = 100000, the loop terminates.
3. Adjustment: t is then divided by 10, making t = 10000.
4. Second while loop: This loop iterates as long as t > 0.
   • Iteration 1 (t = 10000): d = 14362 / 10000 = 1. n = 14362 % 10000 = 4362. t = 10000 / 10 = 1000. turn is 0 (false), so newnum remains 0. turn = (0 + 1) % 2 = 1.
   • Iteration 2 (t = 1000): d = 4362 / 1000 = 4. n = 4362 % 1000 = 362. t = 1000 / 10 = 100. turn is 1 (true), so newnum = 10 * 0 + 4 = 4. turn = (1 + 1) % 2 = 0.
   • Iteration 3 (t = 100): d = 362 / 100 = 3. n = 362 % 100 = 62. t = 100 / 10 = 10. turn is 0 (false), so newnum remains 4. turn = (0 + 1) % 2 = 1.
   • Iteration 4 (t = 10): d = 62 / 10 = 6. n = 62 % 10 = 2. t = 10 / 10 = 1. turn is 1 (true), so newnum = 10 * 4 + 6 = 46. turn = (1 + 1) % 2 = 0.
   • Iteration 5 (t = 1): d = 2 / 1 = 2. n = 2 % 1 = 0. t = 1 / 10 = 0. turn is 0 (false), so newnum remains 46. turn = (0 + 1) % 2 = 1.
5. Loop Termination: t is now 0, so the loop terminates.
6. Return Value: The function returns newnum, which is 46.

[CORRECT_OPTION: 46]

Let's analyze the C function f(int x, int y):

```c
int f (int x, int y)  {
    for (int i = 0; i < y; i++ )  {
        x = x + x + y;
        // Equivalent to x = 2*x + y;
    }
    return x;
}

The loop runs `y` times. In each iteration, `x` is updated as `x = 2*x + y`.
Let $x_0$ be the initial value of `x`.
After 1 iteration: $x_1 = 2x_0 + y$
After 2 iterations: $x_2 = 2x_1 + y = 2(2x_0+y) + y = 4x_0 + 2y + y = 4x_0 + 3y$
After 3 iterations: $x_3 = 2x_2 + y = 2(4x_0+3y) + y = 8x_0 + 6y + y = 8x_0 + 7y$
In general, after $i$ iterations, $x_i = 2^i x_0 + (2^i - 1)y$.
Since the loop runs $y$ times, the final value of $x$ (returned by the function) will be $x_y = 2^y x_0 + (2^y - 1)y$.

Let's test the options:
(a) If $x_0 = 20, y = 10$:
$x_{10} = 2^{10}(20) + (2^{10}-1)(10) = 1024 \times  20 + (1023) \times  10 = 20480 + 10230 = 30710$.
Is $30710 > 2^{20} = 1048576$? No, $30710$ is much smaller. So (a) is FALSE.

(b) If $x_0 = 20, y = 20$:
$x_{20} = 2^{20}(20) + (2^{20}-1)(20) = 20 \times  2^{20} + 20 \times  2^{20} - 20 = 40 \times  2^{20} - 20$.
Is $40 \times  2^{20} - 20 > 2^{20}$? Yes, clearly. So (b) is TRUE.

(c) If $x_0 = 20, y = 10$: (Same as a)
$x_{10} = 30710$.
Is $30710 < 2^{10} = 1024$? No, $30710$ is much larger. So (c) is FALSE.

(d) If $x_0 = 10, y = 20$:
$x_{20} = 2^{20}(10) + (2^{20}-1)(20) = 10 \times  2^{20} + 20 \times  2^{20} - 20 = 30 \times  2^{20} - 20$.
Is $30 \times  2^{20} - 20 > 2^{20}$? Yes, clearly. So (d) is TRUE.

Therefore, statements (b) and (d) are TRUE.

The function fun is called with n = 2^40. Let's trace the execution.

The first for loop runs as long as i > 1, with i being halved in each iteration.

• i starts at $2^{40}$ and takes values $2^{40}, 2^{39}, \dots, 2^1$.
• The loop executes 40 times before i becomes 1 and the loop terminates.
• In each iteration, j is incremented. So, after this loop, j will be 40.

The second for loop runs as long as j > 1, with j being halved.

• j starts at 40. The loop iterates for j values of 40, 20, 10, 5, and 2.
• When j is 2, j/2 becomes 1. The loop condition j > 1 fails for the next iteration.
• The loop body sum++ executes 5 times.
• Therefore, the final value of sum returned is 5.

The program performs division using the relationship x = y*q + r, where q is the quotient and r is the remainder. This equation must hold as a loop invariant, meaning it must be true before the loop begins. The loop terminates when r < y.

To satisfy the invariant x = y*q + r at the start, we can initialize the variables appropriately. If we set q = 0 and r = x, the equation becomes x = y*0 + x, which simplifies to x = x. This initialization is correct. Additionally, for the division to be meaningful and for the loop to terminate, the divisor y must be greater than 0. Thus, the required pre-conditions are (q == 0) && (r == x) && (y > 0).

The function finds the maximum value by using two pointers, a starting from index 0 and b from n-1. In each iteration of the while loop, it compares p[a] and p[b] and moves the pointer corresponding to the smaller element inward. This process effectively discards the smaller of the two boundary elements. The loop should continue until both pointers converge on the same element, which will be the maximum value. The condition b != a ensures the loop terminates precisely when a and b point to the same index.

The question asks for a loop invariant, which is a condition that holds true at the beginning of every loop iteration. Let's test the condition $X^Y = \text{res} \times a^b$.

1. Initialization: Before the loop, res = 1, a = X, b = Y. The condition is $X^Y = 1 \times X^Y$, which is true.
2. Maintenance (b is even): res is unchanged, a becomes $a^2$, b becomes $b/2$. The expression is $\text{res} \times (a^2)^{b/2} = \text{res} \times a^b$. The invariant is maintained.
3. Maintenance (b is odd): res becomes res*a, a is unchanged, b becomes b-1. The expression is $(\text{res} \times a) \times a^{b-1} = \text{res} \times a^b$. The invariant is maintained.
   Since the condition is true at the start and is preserved through each step, it is the correct loop invariant.

The provided pseudo code implements the algorithm for integer division. The variable r is initialized with the dividend x, and the variable q (the quotient) is initialized to 0.

The while loop repeatedly subtracts the divisor y from r as long as r is greater than or equal to y. For each successful subtraction, the quotient q is incremented.

The loop terminates when r < y, at which point r holds the remainder of the division. The relationship between these variables is defined by the division algorithm: $x = q \cdot y + r$, with the condition that the remainder r is less than the divisor y.

The function MyX calculates the sum of all elements in the array E and stores it in Y during the first loop.
The nested loops then iterate through all possible subarrays of E. For each subarray, Z calculates its sum.
If the current subarray sum Z is greater than the current maximum sum stored in Y, Y is updated to Z.
This process effectively tracks the maximum sum encountered among all possible subarrays, including the full array sum initially stored in Y.
Therefore, the function returns the maximum possible sum of elements in any sub-array of array E.

The total number of multiplications is determined by how many times the innermost loop D = D * 3 executes.
This corresponds to the number of combinations of choosing 3 distinct indices i, j, k such that 1 <= i <= j < k <= n.
Let's analyze the loop bounds:

• The i loop runs from 1 to n.
• The j loop runs from i to n.
• The k loop runs from j+1 to n.

The number of times the innermost statement D = D * 3 is executed is given by the sum:
$\sum_{i=1}^{n} \sum_{j=i}^{n} \sum_{k=j+1}^{n} 1$
This sum represents the number of ways to choose three distinct integers $i$, $j'$, $k$ from $1, \ldots, n$ such that $i \le j' < k$. If we rename $j'$ to $j$ for clarity, the condition is $1 \le i \le j < k \le n$. This is equivalent to choosing 3 distinct numbers from $n$ numbers and then arranging them. This is $\binom{n}{3}$.
$\binom{n}{3} = \frac{n(n-1)(n-2)}{3 \times 2 \times 1} = \frac{n(n-1)(n-2)}{6}$
This is one-sixth of the product of the three consecutive integers $n$, $(n-1)$, and $(n-2)$.

The solution provided in the PDF shows:
$\sum_{i=1}^{n} \sum_{j=i}^{n} \sum_{k=j+1}^{n} 1 = \frac{n(n-1)(n-2)}{6}$.

Let $k$ be the number of times the loop body executes.
Initially, $j=1 = 2^0$.
In each iteration, $j$ is multiplied by 2. So, $j$ takes values $2^0, 2^1, 2^2, \dots, 2^{k-1}$ for the $k$ successful iterations.
For these $k$ iterations, the condition $j \le n$ must be true. Thus, the last value of $j$ for which the loop condition is true is $2^{k-1}$, meaning $2^{k-1} \le n$.
After the $k$-th iteration, $j$ becomes $2^k$. At this point, the loop condition $2^k \le n$ must be false, which implies $2^k > n$.

Combining these, we have the inequality:
$2^{k-1} \le n < 2^k$
Taking the base-2 logarithm of all parts of the inequality:
$\log_2(2^{k-1}) \le \log_2 n < \log_2(2^k)$
$k-1 \le \log_2 n < k$
This inequality indicates that $k-1$ is the largest integer less than or equal to $\log_2 n$. By definition, this is the floor function:
$k-1 = \left \lfloor \log_2 n \right \rfloor$
Solving for $k$:
$k = \left \lfloor \log_2 n \right \rfloor + 1$
The number of comparisons made in the execution of the loop, interpreted as the number of times the loop condition evaluates to true (i.e., the number of loop iterations), is $k$.

The while loop runs as long as $i < n$ and $j < m$, terminating when $i=n$ or $j=m$. At termination, if $i=n$, all $n$ elements from array $a$ and $j$ elements from array $b$ have been copied to $c$, so the total number of elements copied is $k = n + j$. Similarly, if $j=m$, all $m$ elements from array $b$ and $i$ elements from array $a$ have been copied to $c$, so $k = i + m$. Condition I claims $k = n + j - 1$, and Condition II claims $k = m + i - 1$. Since $k$ must be $n + j$ or $i + m$ respectively, both claims regarding $k$ are incorrect. Therefore, neither condition (I) nor (II) holds.

In C89/C90, if a function is used without an explicit declaration, the compiler assumes an implicit declaration of int fro();. The statement fro(I = 0, I < N, I++); is therefore parsed as a valid function call where three expressions (I = 0, I < N, and I++) are passed as separate arguments to the function fro. Since this structure adheres to standard C syntax rules for function calls, the compiler generates no lexical or syntactic errors while creating the object-module, deferring any potential linking issues to the later stage.

The C program implements the Euclidean algorithm for finding the greatest common divisor (GCD) of two integers $x$ and $y$ using repeated subtraction. The core logic utilizes the mathematical property that $\gcd(m, n) = \gcd(m - n, n)$ if $m > n$, and $\gcd(m, n) = \gcd(m, n - m)$ if $n > m$. The while loop repeatedly performs these subtractions until $m$ becomes equal to $n$. At the point of termination, the value of $n$ (or $m$) represents the largest positive integer that divides both original inputs $x$ and $y$.

The algorithm implements the Babylonian method (also known as Heron's method) to find the square root of a number $m$.
When the loop terminates, the condition $x \approx y$ holds, meaning the values have converged to a limit $L$.
At this limit, substituting $y = \frac{m}{x}$ into the update equation $x = \frac{x + y}{2}$ yields:
$L = \frac{L + m/L}{2}$
Solving for $L$: $2L = L + \frac{m}{L} \implies L = \frac{m}{L} \implies L^2 = m$.
Therefore, $L = \sqrt{m} = m^{1/2}$.

Initially, the integer is $n = d_{1}d_{2}...d_{m}$ and $rev = 0$. In each iteration $i$, the operation $n \% 10$ extracts the current last digit of $n$ (which is $d_{m-i+1}$), and $rev = rev * 10 + n \% 10$ appends this digit to $rev$. The operation $n = n / 10$ removes this digit from $n$. After $i$ iterations, $n$ contains the remaining prefix $d_{1}d_{2}...d_{m-i}$, while $rev$ contains the digits $d_{m}, d_{m-1}, ..., d_{m-i+1}$ in reversed order. Thus, the loop invariant is $n = d_{1}d_{2}...d_{m-i}$ and $rev = d_{m}d_{m-1}...d_{m-i+1}$.

The function computes the partial sum of the Taylor series expansion for the exponential function. Initializing $s=1$ and $p=1$, the loop iteratively calculates each subsequent term $p_i = p_{i-1} \cdot \frac{x}{i} = \frac{x^i}{i!}$ and adds it to the sum $s$.
The resulting value is $s = \sum_{i=0}^{y-1} \frac{x^i}{i!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^{y-1}}{(y-1)!}$.
The Taylor series for $e^x$ is given by the infinite sum $e^x = \sum_{n=0}^{\infty } \frac{x^n}{n!}$.
As $y \to \infty$, the partial sum computed by the function converges to the definition of $e^x$.