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Recursion is a technique where a function calls itself to solve a problem. It breaks down a large problem into smaller, more manageable subproblems.
// Recursive function to solve a problem
function recursiveSolve(problem):
if baseCase(problem):
return baseResult
smallerProblem = breakIntoSubproblem(problem)
result = recursiveSolve(smallerProblem)
return combine(result)
Recursion provides a natural and elegant solution to many problems, especially those that can be defined in terms of smaller subproblems. It is essential to ensure a valid base case to avoid infinite recursion and stack overflow.
Problem Statement:
Find the factorial of a number n, where factorial is defined as:
0! = 1
n! = n × (n - 1) × (n - 2) × ... × 1
for n > 0
For example:
factorial(5) = 5 × 4 × 3 × 2 × 1 = 120
This problem can be elegantly solved using recursion because it naturally fits the definition of factorial: n! = n × (n - 1)!
factorial(n)
.n == 0
, return 1
(base case).n × factorial(n - 1)
.// Recursive factorial
function factorial(n):
if n == 0:
return 1
return n * factorial(n - 1)
This method solves the problem by repeatedly breaking it into smaller subproblems until it reaches the simplest case (0!). The result is built by multiplying the results of the smaller problems.
The recursive factorial problem is a foundational example to understand recursion. It introduces the concepts of base cases, recursive calls, and the call stack. As the problem size increases, understanding stack depth and function return values becomes crucial.
Problem Statement:
Find the nth Fibonacci number, where the Fibonacci sequence is defined as:
F(0) = 0
F(1) = 1
F(n) = F(n - 1) + F(n - 2)
for n ≥ 2
This means every number in the sequence is the sum of the two previous numbers. Example sequence:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
Recursion is a technique where a function calls itself to solve smaller instances of the same problem. In Fibonacci, we can define the solution recursively because each number depends on the previous two.
n
is 0, return 0.n
is 1, return 1.fib(n - 1) + fib(n - 2)
.// Simple recursive function to find nth Fibonacci number
function fib(n):
if n == 0:
return 0
if n == 1:
return 1
return fib(n - 1) + fib(n - 2)
fib(4)
= fib(3) + fib(2)
= (fib(2) + fib(1)) + (fib(1) + fib(0))
= ((fib(1) + fib(0)) + 1) + (1 + 0)
= ((1 + 0) + 1) + (1 + 0)
= 2 + 1 = 3
This recursive solution mirrors the mathematical definition of Fibonacci. It’s intuitive and easy to implement.
n
.Note: While recursion is elegant, it’s not the most efficient way to solve Fibonacci due to overlapping subproblems. But it is a great way to understand how recursion works!
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