Fibonacci Using Recursion in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the base case value for n equals 0 in the Fibonacci function.

DSA Python

def fibonacci(n):
    if n == 0:
        return [1]

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

A0

B1

Cn

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 instead of 0 for n == 0

Returning n which is unnecessary here

The Fibonacci sequence starts with 0 at position 0, so the function should return 0 when n is 0.

2fill in blank

medium

Complete the code to return the base case value for n equals 1 in the Fibonacci function.

DSA Python

def fibonacci(n):
    if n == 0:
        return 0
    elif n == 1:
        return [1]

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

A0

Bn

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of 1 for n == 1

Returning n which is unnecessary here

The Fibonacci sequence has 1 at position 1, so the function should return 1 when n is 1.

3fill in blank

hard

Fix the error in the recursive call to correctly calculate the Fibonacci number.

DSA Python

def fibonacci(n):
    if n == 0:
        return 0
    elif n == 1:
        return 1
    else:
        return fibonacci(n[1]1) + fibonacci(n-2)

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

A+

B/

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' in the recursive call

Using '*' or '/' which are not valid here

The recursive call should subtract 1 from n to get the previous Fibonacci number, so the operator must be '-'.

4fill in blank

hard

Fill both blanks to complete the recursive Fibonacci function correctly.

DSA Python

def fibonacci(n):
    if n == 0:
        return 0
    elif n == 1:
        return 1
    else:
        return fibonacci(n[1]1) + fibonacci(n[2]2)

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

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' or '*' instead of '-' in recursive calls

The nth Fibonacci number is the sum of fibonacci(n-1) and fibonacci(n-2). Use the '-' operator in both blanks.

5fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps numbers 0 to 5 to their Fibonacci values.

DSA Python

fib_dict = {i: fibonacci(i) for i in range([1], [2])}

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

B0

C6

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Adding an operator in blank 1 which breaks the call

Wrong range start or end values

The dictionary comprehension calls fibonacci(i) with no change to i, so blank 1 is empty. The range starts at 0 and goes up to 6 (exclusive) to include 0 to 5.

Practice

1. What does the Fibonacci sequence represent when calculated using recursion?

easy

A. Each number is the square of the previous number

B. Each number is the product of the two preceding numbers

C. Each number is the difference of the two preceding numbers

D. Each number is the sum of the two preceding numbers

Solution

1. Step 1: Understand Fibonacci sequence definition

   The Fibonacci sequence is defined such that each number is the sum of the two numbers before it.

2. Step 2: Relate to recursion

   Recursion calculates this by calling the function for the two previous numbers and adding their results.

3. Final Answer:

   Each number is the sum of the two preceding numbers -> Option D

4. Quick Check:

   Fibonacci = sum of two previous numbers [OK]

Hint: Fibonacci adds two previous numbers, not multiply or subtract [OK]

Common Mistakes:

• Thinking Fibonacci multiplies previous numbers
• Confusing difference with sum
• Assuming Fibonacci squares previous number

2. Which of the following is the correct base case syntax for a recursive Fibonacci function in Python?

easy

A. if n == 0 or n == 1: return n

B. if n < 0: return 0

C. if n == 2: return 1

D. if n == 1: return 0

Solution

1. Step 1: Identify base cases for Fibonacci

   Fibonacci sequence starts with 0 at position 0 and 1 at position 1, so base cases are n == 0 and n == 1.

2. Step 2: Check syntax correctness

   The correct syntax returns n when n is 0 or 1, matching the base cases.

3. Final Answer:

   if n == 0 or n == 1: return n -> Option A

4. Quick Check:

   Base cases return n for 0 or 1 [OK]

Hint: Base cases return n when n is 0 or 1 [OK]

Common Mistakes:

• Using wrong base case values
• Returning 0 for n == 1
• Missing base case for n == 0

3. What is the output of the following code?

def fib(n):
    if n == 0 or n == 1:
        return n
    else:
        return fib(n-1) + fib(n-2)

print(fib(4))

medium

A. 4

B. 5

C. 3

D. 7

Solution

1. Step 1: Trace fib(4) calls

   fib(4) = fib(3) + fib(2); fib(3) = fib(2) + fib(1); fib(2) = fib(1) + fib(0)

2. Step 2: Calculate values bottom-up

   fib(0)=0, fib(1)=1; fib(2)=1+0=1; fib(3)=1+1=2; fib(4)=2+1=3

3. Final Answer:

   3 -> Option C

4. Quick Check:

   fib(4) = 3 [OK]

Hint: Calculate fib(n-1) + fib(n-2) stepwise [OK]

Common Mistakes:

• Confusing fib(4) with fib(5)
• Adding wrong values for fib(2) or fib(3)
• Forgetting base cases

4. Find the error in this recursive Fibonacci function:

def fib(n):
    if n == 0:
        return 0
    elif n == 1:
        return 1
    else:
        return fib(n-1) - fib(n-2)

medium

A. The base cases are incorrect

B. The recursive step uses subtraction instead of addition

C. The function lacks a return statement

D. The function will cause infinite recursion

Solution

1. Step 1: Check base cases

   Base cases for n == 0 and n == 1 are correct and return proper values.

2. Step 2: Analyze recursive step

   The recursive step uses subtraction (fib(n-1) - fib(n-2)) instead of addition, which is incorrect for Fibonacci.

3. Final Answer:

   The recursive step uses subtraction instead of addition -> Option B

4. Quick Check:

   Recursive step must add previous two values [OK]

Hint: Fibonacci recursion adds, never subtracts previous two [OK]

Common Mistakes:

• Using subtraction in recursive step
• Misunderstanding base cases
• Missing return keyword

5. How many total calls are made when computing fib(5) using the naive recursive Fibonacci function?

def fib(n):
    if n == 0 or n == 1:
        return n
    else:
        return fib(n-1) + fib(n-2)

hard

A. 15

B. 9

C. 7

D. 10

Solution

1. Step 1: Count calls for fib(5)

   fib(5) calls fib(4) and fib(3); fib(4) calls fib(3) and fib(2); fib(3) calls fib(2) and fib(1), etc.

2. Step 2: Calculate total calls

   Total calls = 15 (including repeated calls for fib(3), fib(2), fib(1), fib(0)) as the recursion tree expands.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Naive recursion calls grow exponentially; fib(5) = 15 calls [OK]

Hint: Count all calls in recursion tree for fib(n) [OK]

Common Mistakes:

• Counting only unique calls
• Ignoring repeated calls in recursion
• Underestimating exponential growth