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On this page
  • Challenge: is a string a palindrome?
  • Recursive algorithm for computing :
  1. math
  2. Algorithms (khan academy)

Recursion

PreviousInsertion sortNextSolve Hanoi recursively

Last updated 6 years ago

The factorial function is defined for all positive integers, along with 000. What value should 0!0!0! have? It's the product of all integers greater than or equal to 111 and less than or equal to 000. But there are no such integers. Therefore, we define 0! to equal the identity for multiplication, which is 111. (Defining 0!=10! = 10!=1 meshes nicely with the formula for choosing kkk things out of nnn things. Suppose that we want to know how many ways there are to choose nnn things out of nnnthings. That's easy, because there is only one way: choose all nnn things. So now we know that, using our formula, n!/(n!⋅(n−n)!)n! / (n! \cdot (n-n)!)n!/(n!⋅(n−n)!)should equal 111. But (n−n)!(n-n)!(n−n)! is 0!0!0!, so now we know that n!/(n!⋅0!)n! / (n! \cdot 0!)n!/(n!⋅0!) should equal 111. If we cancel out the n!n!in both the numerator and denominator, we see that 1/(0!)1/(0!)1/(0!) should equal 111, and it does because 0!0!0! equals 111.)

So now we have a way to think about n!n!n!. It equals 111 when n=0n = 0n=0 and it equals 1⋅2â‹Ŋ(n−1)⋅n1\cdot 2 \cdots (n-1) \cdot n1⋅2â‹Ŋ(n−1)⋅n when nnn is positive.

var factorial = function(n) {
    var result = 1;

    for (var i = 1; i <= n; i++) {
        result *= i;
    }

    return result;
};

println("The value of 5! should be " + 5*4*3*2*1);
println("The value of 5! is " + factorial(5));
println("The value of 0! should be 1");
println("The value of 0! is " + factorial(0));

Program.assertEqual(factorial(5), 120);
Program.assertEqual(factorial(0), 1);

Here is the basic idea behind recursive algorithms:

To solve a problem, solve a subproblem that is a smaller instance of the same problem, and then use the solution to that smaller instance to solve the original problem.

var factorial = function(n) {
	// base case: 
	if (n === 0) {
	    return 1;
	}
	// recursive case:
	return factorial(n-1) * n;
}; 

println("The value of 0! is " + factorial(0) + ".");
println("The value of 5! is " + factorial(5) + ".");

Program.assertEqual(factorial(0), 1);
Program.assertEqual(factorial(5), 120);

We can distill the idea of recursion into two simple rules:

  1. Each recursive call should be on a smaller instance of the same problem, that is, a smaller subproblem.

  2. The recursive calls must eventually reach a base case, which is solved without further recursion.

Challenge: is a string a palindrome?

// Returns the first character of the string str
var firstCharacter = function(str) {
    return str.slice(0, 1);
};

// Returns the last character of a string str
var lastCharacter = function(str) {
    return str.slice(-1);
};

// Returns the string that results from removing the first
//  and last characters from str
var middleCharacters = function(str) {
    return str.slice(1, -1);
};

var isPalindrome = function(str) {
    // base case #1
    if (str.length <= 1){
        return true;
    }
    // base case #2
    if (firstCharacter(str) === lastCharacter(str)){
        return true;
    } else { 
        return false;
    }
    // recursive case
    return isPalindrome(middleCharacters(str));
};

var checkPalindrome = function(str) {
    println("Is this word a palindrome? " + str);
    println(isPalindrome(str));
};

checkPalindrome("a");
Program.assertEqual(isPalindrome("a"), true);
checkPalindrome("motor");
Program.assertEqual(isPalindrome("motor"), false);
checkPalindrome("rotor");
Program.assertEqual(isPalindrome("rotor"), true);
checkPalindrome("moon");
var isEven = function(n) {
    return n % 2 === 0;
};

var isOdd = function(n) {
    return !isEven(n);
};

var power = function(x, n) {
    println("Computing " + x + " raised to power " + n + ".");
    // base case
    if (n === 0){
        return 1;
    }
    // recursive case: n is odd
    if (isOdd(n)){
        return power(x, n-1) * x;
    }
    // recursive case: n is even
    if (isEven(n)) {
        var y = power(x, n/2);
        return y*y;
    }
    // recursive case: n is negative
    if (n < 0) {
        return 1 / power(x, -n);r
    }
};

var displayPower = function(x, n) {
    println(x + " to the " + n + " is " + power(x, n));
};

displayPower(3, 0);
Program.assertEqual(power(3, 0), 1);
displayPower(3, 1);
Program.assertEqual(power(3, 1), 3);
displayPower(3, 2);
Program.assertEqual(power(3, 2), 9);
displayPower(3, -1);
Program.assertEqual(power(3, -1), 1/3);

Recursive algorithm for computing xnx^nxn:

The base case is when n=0n = 0n=0 and x0=1x^0 = 1x0=1.

If nnn is positive and even, recursively compute y=xn/2y = x^{n/2}y=xn/2, and then xn=y⋅yx^n = y \cdot yxn=y⋅y. Notice that you can get away with making just one recursive call in this case, computing xn/2x^{n/2}xn/2 just once, and then you multiply the result of this recursive call by itself.

If nnn is positive and odd, recursively compute xn−1x^{n-1}xn−1, so that the exponent either is 000 or is positive and even. Then, xn=xn−1⋅xx^n = x^{n-1} \cdot xxn=xn−1⋅x.

If nnn is negative, recursively compute x−nx^{-n}x−n, so that the exponent becomes positive. Then, xn=1/x−nx^n = 1 / x^{-n}xn=1/x−n.

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