Recursion and Dynamic Programming

Factorial (Recursive)

factorial(0) = 1, otherwise factorial(n) = n * factorial(n - 1). The smallest example of recursion with a single base case.

Algorithm

Basic Implementation

basic.cs 
using System;

class Program {
	static int Factorial(int n) {
		if (n == 0) {
			return 1;
		}
		return n * Factorial(n - 1);
	}

	static void Main() {
		int result = Factorial(5);
		Console.WriteLine(result);
	}
}

The pinned run is factorial(5). The diagrams separate the descent, the base case, and the return values so the stack does not feel invisible.

Step 1 - Descend to the base case

Each call waits for one smaller call until f(0) returns 1.

Call tree for factorial(5): f(5) waits on f(4), down to f(0).f(5)waitsf(4)waitsf(3)waitsf(2)waitsf(1)waitsf(0)base = 1

Step 2 - Base value starts the unwind

The first finished frame is f(0) = 1; f(1) can now compute 1 * 1.

Call stack just before unwind begins.top -> bottomknown returnf(0)1f(1)waitingf(2)waitingf(3)waitingf(4)waitingf(5)waiting

Step 3 - Unwind returns 120

Each frame multiplies its n by the completed smaller result.

Return chain for factorial(5).framecalculationreturnsf(0)base1f(1)1 * 11f(2)2 * 12f(3)3 * 26f(4)4 * 624f(5)5 * 24120

Complexity

  • Time: O(n)
  • Space: O(n) call stack

Implementation notes

  • The checked-in method is static int Factorial(int n), so every argument and return value is a copied int. The base case is exactly n == 0, returning the literal 1.
  • Each recursive call uses a normal CLR call-stack frame; there is no heap allocation or container state in the recursive path, so GC is not part of the replayed behavior.
  • The multiplication is plain n * Factorial(n - 1) with no checked block. For the fixture Factorial(5), overflow is not visible; the trace focuses on frame descent and unwind values such as 5 * 24 = 120.
base case `if (n == 0) return 1;`
recursive call `n * Factorial(n - 1)`