Recursion Base Case and Recursive Case in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the base case value for factorial of 0.

DSA Rust

fn factorial(n: u32) -> u32 {
    if n == 0 {
        return [1];
    }
    n * factorial(n - 1)
}

Drag options to blanks, or click blank then click option'

A-1

B0

C1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of 1 for the base case.

Returning n which causes infinite recursion.

The factorial of 0 is defined as 1, so the base case returns 1.

2fill in blank

medium

Complete the recursive call to calculate factorial of n.

DSA Rust

fn factorial(n: u32) -> u32 {
    if n == 0 {
        return 1;
    }
    n [1] factorial(n - 1)
}

Drag options to blanks, or click blank then click option'

A+

B*

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '*' causing wrong results.

Using '-' or '/' which are incorrect for factorial.

Factorial multiplies n by factorial of n-1, so the operator is *.

3fill in blank

hard

Fix the error in the base case condition to stop recursion correctly.

DSA Rust

fn factorial(n: u32) -> u32 {
    if n [1] 0 {
        return 1;
    }
    n * factorial(n - 1)
}

Drag options to blanks, or click blank then click option'

A==

B>

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' or '<' which do not correctly identify the base case.

Using '!=' which causes infinite recursion.

The base case should check if n equals 0 to stop recursion.

4fill in blank

hard

Fill both blanks to create a recursive function that sums numbers from n down to 1.

DSA Rust

fn sum_to_n(n: u32) -> u32 {
    if n == 0 {
        return 0;
    }
    n [1] sum_to_n(n [2] 1)
}

Drag options to blanks, or click blank then click option'

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of addition.

Calling sum_to_n with n+1 causing infinite recursion.

To sum numbers, add n to sum_to_n(n-1).

5fill in blank

hard

Fill all three blanks to write a recursive function that computes the nth Fibonacci number.

DSA Rust

fn fibonacci(n: u32) -> u32 {
    if n == 0 || n == 1 {
        return n;
    }
    fibonacci([1]) [2] fibonacci([3])
}

Drag options to blanks, or click blank then click option'

An - 1

B+

Cn - 2

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition.

Using wrong indices like n+1 or n-3.

Fibonacci is sum of fibonacci(n-1) and fibonacci(n-2).

Practice

1. In a recursive function, what is the purpose of the base case?

easy

A. To increase the input size for the next call

B. To call the function again with a smaller input

C. To print the result of the recursion

D. To stop the recursion and prevent infinite calls

Solution

1. Step 1: Understand the base case role

   The base case is the condition that stops the recursion from continuing indefinitely.

2. Step 2: Differentiate from recursive case

   The recursive case calls the function again with smaller input, but the base case stops recursion.

3. Final Answer:

   To stop the recursion and prevent infinite calls -> Option D

4. Quick Check:

   Base case = stop recursion [OK]

Hint: Base case stops recursion to avoid infinite loops [OK]

Common Mistakes:

• Confusing base case with recursive case
• Thinking base case increases input size
• Assuming base case prints output

2. Which of the following is the correct syntax for a recursive function in Rust that calculates factorial?

easy

A. fn factorial(n: u32) -> u32 { loop { factorial(n - 1) } }

B. fn factorial(n: u32) -> u32 { while n > 0 { n * factorial(n - 1) } }

C. fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } }

D. fn factorial(n: u32) -> u32 { return n + factorial(n - 1); }

Solution

1. Step 1: Check base case and recursive call

   fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } } correctly uses if n == 0 { 1 } as base case and recursive multiplication.

2. Step 2: Identify syntax errors in other options

   fn factorial(n: u32) -> u32 { while n > 0 { n * factorial(n - 1) } } uses while loop incorrectly, C loops infinitely, D uses addition instead of multiplication.

3. Final Answer:

   fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } } -> Option C

4. Quick Check:

   Correct base case and recursive call = fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } } [OK]

Hint: Look for base case with if and recursive call with smaller input [OK]

Common Mistakes:

• Using loops instead of recursion
• Missing base case
• Wrong operation in recursive call

3. What is the output of this Rust code?

fn sum(n: u32) -> u32 {
    if n == 0 {
        0
    } else {
        n + sum(n - 1)
    }
}

fn main() {
    println!("{}", sum(3));
}

medium

A. 3

B. 6

C. 0

D. Stack overflow error

Solution

1. Step 1: Trace recursive calls for sum(3)

   sum(3) = 3 + sum(2), sum(2) = 2 + sum(1), sum(1) = 1 + sum(0), sum(0) = 0.

2. Step 2: Calculate total sum

   sum(3) = 3 + 2 + 1 + 0 = 6.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Sum 1 to 3 = 6 [OK]

Hint: Add numbers down to zero using recursion [OK]

Common Mistakes:

• Confusing base case value
• Forgetting to add current n
• Expecting runtime error

4. Identify the error in this recursive Rust function that calculates factorial:

fn factorial(n: u32) -> u32 {
    n * factorial(n - 1)
}

fn main() {
    println!("{}", factorial(3));
}

medium

A. Missing base case to stop recursion

B. Wrong return type for factorial

C. Incorrect function name

D. Using multiplication instead of addition

Solution

1. Step 1: Check for base case presence

   The function lacks a base case like if n == 0 { 1 } to stop recursion.

2. Step 2: Understand recursion consequences

   Without base case, recursion never stops, causing stack overflow error.

3. Final Answer:

   Missing base case to stop recursion -> Option A

4. Quick Check:

   Base case missing = Missing base case to stop recursion [OK]

Hint: Always include base case to stop recursion [OK]

Common Mistakes:

• Assuming recursion stops automatically
• Ignoring base case importance
• Confusing return types

5. You want to write a recursive function in Rust to find the nth Fibonacci number. Which of the following correctly shows the base case and recursive case?

hard

A. if n == 0 { 0 } else if n == 1 { 1 } else { fib(n - 1) + fib(n - 2) }

B. if n == 1 { 0 } else if n == 2 { 1 } else { fib(n - 1) * fib(n - 2) }

C. if n == 0 { 1 } else { fib(n - 1) + fib(n - 2) }

D. if n == 0 { 0 } else { fib(n - 1) - fib(n - 2) }

Solution

1. Step 1: Identify correct base cases for Fibonacci

   Fibonacci sequence starts with fib(0) = 0 and fib(1) = 1, so base cases must reflect that.

2. Step 2: Check recursive case correctness

   Recursive case sums previous two Fibonacci numbers: fib(n-1) + fib(n-2).

3. Final Answer:

   if n == 0 { 0 } else if n == 1 { 1 } else { fib(n - 1) + fib(n - 2) } -> Option A

4. Quick Check:

   Correct base and recursive case for Fibonacci = if n == 0 { 0 } else if n == 1 { 1 } else { fib(n - 1) + fib(n - 2) } [OK]

Hint: Fibonacci base cases: 0 and 1; recursive case sums previous two [OK]

Common Mistakes:

• Wrong base case values
• Using multiplication or subtraction instead of addition
• Missing one base case