Fibonacci Using Recursion in DSA Javascript - Build from Scratch

Fibonacci Using Recursion

📖 Scenario: You want to find the nth number in the Fibonacci sequence. The Fibonacci sequence starts with 0 and 1, and each next number is the sum of the two before it.This sequence is like counting rabbits: each pair produces another pair, so the numbers grow step by step.

🎯 Goal: Build a small program that uses a recursive function to find the Fibonacci number at position n.You will first set the position number, then write the recursive function, and finally print the result.

📋 What You'll Learn

Create a variable n with the value 7

Create a recursive function fibonacci that takes one number parameter num

The function should return 0 if num is 0

The function should return 1 if num is 1

Otherwise, the function should return the sum of fibonacci(num - 1) and fibonacci(num - 2)

Print the result of fibonacci(n)

💡 Why This Matters

🌍 Real World

Fibonacci numbers appear in nature, computer algorithms, and financial models. Understanding recursion helps solve problems that break down into smaller similar problems.

💼 Career

Many programming jobs require understanding recursion and sequence calculations, useful in algorithm design, coding interviews, and problem solving.

1

Set the Fibonacci position

Create a variable called n and set it to 7.

DSA Javascript

# Your code here

Hint

Use const n = 7; to store the position number.

2

Write the recursive Fibonacci function

Create a function called fibonacci that takes one parameter num. Inside the function, return 0 if num is 0, and return 1 if num is 1.

DSA Javascript

const n = 7;
// Your code here

Hint

Use if statements to check if num is 0 or 1 and return the correct value.

3

Add the recursive call to calculate Fibonacci

Inside the fibonacci function, after the if checks, return the sum of fibonacci(num - 1) and fibonacci(num - 2).

DSA Javascript

const n = 7;

function fibonacci(num) {
    if (num === 0) {
        return 0;
    }
    if (num === 1) {
        return 1;
    }
    // Your code here
}

Hint

Use return fibonacci(num - 1) + fibonacci(num - 2); to get the sum of the two previous Fibonacci numbers.

4

Print the Fibonacci number at position n

Write a console.log statement to print the result of fibonacci(n).

DSA Javascript

const n = 7;

function fibonacci(num) {
    if (num === 0) {
        return 0;
    }
    if (num === 1) {
        return 1;
    }
    return fibonacci(num - 1) + fibonacci(num - 2);
}
// Your code here

Hint

Use console.log(fibonacci(n)); to show the Fibonacci number for n.

Practice

1. What does a recursive Fibonacci function do when called with fib(4)?

easy

A. Throws a syntax error because recursion is not allowed

B. Returns 4 immediately without any calculation

C. Runs an infinite loop without returning

D. Calculates the 4th Fibonacci number by calling itself with smaller numbers

Solution

1. Step 1: Understand recursive calls for fib(4)

   The function calls fib(3) and fib(2) to calculate fib(4).

2. Step 2: Combine results of smaller calls

   fib(3) and fib(2) are calculated recursively and their sum is returned as fib(4).

3. Final Answer:

   Calculates the 4th Fibonacci number by calling itself with smaller numbers -> Option D

4. Quick Check:

   Recursive calls break problem into smaller parts [OK]

Hint: Recursion breaks problem into smaller calls [OK]

Common Mistakes:

• Thinking recursion returns input directly
• Assuming recursion causes infinite loop always
• Confusing recursion with iteration

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

easy

A. if (n = 0) return 0;

B. if (n <= 1) return n;

C. if n == 1 then return 1

D. if (n > 1) return n;

Solution

1. Step 1: Identify correct JavaScript syntax for base case

   The base case must check if n is 0 or 1 and return n using proper JavaScript syntax.

2. Step 2: Validate each option

   if (n <= 1) return n; uses correct syntax: if (n <= 1) return n; Options A and B have syntax errors or wrong operators, D is logically incorrect.

3. Final Answer:

   if (n <= 1) return n; -> Option B

4. Quick Check:

   Base case uses if (n <= 1) return n; [OK]

Hint: Base case returns n when n is 0 or 1 [OK]

Common Mistakes:

• Using assignment (=) instead of comparison (== or ===)
• Missing parentheses in if condition
• Incorrect logical condition for base case

3. What is the output of the following code?

function fib(n) {
  if (n <= 1) return n;
  return fib(n - 1) + fib(n - 2);
}
console.log(fib(5));

medium

A. 5

B. 8

C. 3

D. 13

Solution

1. Step 1: Trace fib(5) recursive calls

   fib(5) = fib(4) + fib(3), fib(4) = fib(3) + fib(2), fib(3) = fib(2) + fib(1), and so on until base cases.

2. Step 2: Calculate values bottom-up

   fib(0)=0, fib(1)=1, fib(2)=1, fib(3)=2, fib(4)=3, fib(5)=5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Fibonacci(5) = 5 [OK]

Hint: Remember Fibonacci sequence: 0,1,1,2,3,5... [OK]

Common Mistakes:

• Confusing index with Fibonacci value
• Miscounting recursive calls
• Assuming fib(5) is 8 or 13 incorrectly

4. Identify the error in this recursive Fibonacci function:

function fib(n) {
  if (n == 0) return 0;
  if (n == 1) return 1;
  return fib(n - 1) - fib(n - 2);
}

medium

A. The recursive step uses subtraction instead of addition

B. The base cases are incorrect

C. The function will cause infinite recursion

D. The function syntax is invalid

Solution

1. Step 1: Check base cases correctness

   Base cases for n==0 and n==1 correctly return 0 and 1 respectively.

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 A

4. Quick Check:

   Fibonacci sums previous two, not subtracts [OK]

Hint: Fibonacci adds previous two numbers, not subtracts [OK]

Common Mistakes:

• Using wrong operator (+ vs -)
• Confusing base cases
• Assuming syntax error when logic is wrong

5. How can you optimize the recursive Fibonacci function to avoid repeated calculations?

hard

A. Remove base cases to speed up recursion

B. Increase the recursion depth limit

C. Use memoization to store and reuse computed values

D. Replace recursion with a while loop only

Solution

1. Step 1: Understand repeated calculations in recursion

   Recursive Fibonacci recalculates same values many times, causing inefficiency.

2. Step 2: Apply memoization to store results

   Memoization saves results of fib(n) calls in a cache to reuse and avoid repeated work.

3. Final Answer:

   Use memoization to store and reuse computed values -> Option C

4. Quick Check:

   Memoization improves recursion efficiency [OK]

Hint: Store results to avoid repeated recursion calls [OK]

Common Mistakes:

• Thinking increasing recursion limit helps
• Believing loops are only solution
• Removing base cases causing infinite recursion