Recursive Functions and the Power of Stacking

Question

Recursive Functions and the Power of Stacking

Just hoping this isn't a duplicate inquiry, as I've searched extensively but couldn't find the solution.

I'm currently experimenting with recursive functions (I'm relatively new to this), attempting to multiply each number in an array akin to calculating a factorial. However, despite having written the code below, it only returns undefined as the output.

Here is the code snippet:

 var stack = [];

function countDown(int) {
  stack.push(int);
  if (int === 1) {  
    return 1;
  }
    return countDown(int - 1);
}

function multiplyEach() {
  // Remove the last value of the stack 
  // and assign it to the variable int
  int = stack.pop();
  x = stack.length;
  // Base case
  if (x === 0 ) {
    return;
  }
  // Recursive case
  else {
    stack[int - 1] = int * stack[x - 1];
    return multiplyEach(int);
  }
}

// Call the function countDown(7)
countDown(7);
// Print out the returned value by multiplyEach()
console.log(multiplyEach());

Your insights will be greatly appreciated. Thank you!

Cheers!

javascript arrays recursion

Answer 1

Answer №1

To begin, it is important to mention that your function multiplyEach() should have a parameter named "int." The current approach being used may be more suitable for a different technique, but we will address that shortly.

Within the multiplyEach() function, there are two potential scenarios:

If the stack still contains elements, then we multiply the new top value on the stack by the previous one and proceed with another iteration of multiplyEach.
If the stack is empty, we simply return from the function.

The issue here is that we are not actually returning or preserving the final stack value for later access. The function seems to lose track of it, leaving us wondering what the output of the function would be.

In many programming languages, we would specify a return type for this function—either 'void' for no value or 'int' for returning the multiplied value. However, in JavaScript, there are no functions that do not return a value; instead, 'undefined' represents nothing in JS. When you return multiplyEach(), you essentially add another call of the function to the call stack, expecting an actual return value. In this case, the return ends up being 'return;' which JS interprets as 'return undefined;'. Unlike other languages where this might cause an error, JS handles it differently. Let's explore two possible implementations:

Using the custom stack:

// skipping to the core logic
function multiplyEach() {
    var int = stack.pop(), x = stack.length;
    stack[x - 1] *= int;
    if(x < 2)
        return;
    multiplyEach();
}

//...
multiplyEach();
console.log(stack[0]);

Introducing a parameter and utilizing a list:

function multiplyEach(index) {
    var int = list[index];
    if(index == 0)
        return int;
    return int * multiplyEach(index - 1);
}

//...
console.log(multiplyEach(list.length - 1));

Both approaches demonstrate different recursive implementations; recursion fundamentally involves a function calling itself. Tail recursion could also have been considered by having a parameter store the total instead of multiplying the return values as in option 2.

EDIT: I noticed that the base case for option 1 was misplaced, so I have corrected it. It should function properly now.

Answer 2

To begin, it is important to mention that your function multiplyEach() should have a parameter named "int." The current approach being used may be more suitable for a different technique, but we will address that shortly.

Within the multiplyEach() function, there are two potential scenarios:

If the stack still contains elements, then we multiply the new top value on the stack by the previous one and proceed with another iteration of multiplyEach.
If the stack is empty, we simply return from the function.

The issue here is that we are not actually returning or preserving the final stack value for later access. The function seems to lose track of it, leaving us wondering what the output of the function would be.

In many programming languages, we would specify a return type for this function—either 'void' for no value or 'int' for returning the multiplied value. However, in JavaScript, there are no functions that do not return a value; instead, 'undefined' represents nothing in JS. When you return multiplyEach(), you essentially add another call of the function to the call stack, expecting an actual return value. In this case, the return ends up being 'return;' which JS interprets as 'return undefined;'. Unlike other languages where this might cause an error, JS handles it differently. Let's explore two possible implementations:

Using the custom stack:

// skipping to the core logic
function multiplyEach() {
    var int = stack.pop(), x = stack.length;
    stack[x - 1] *= int;
    if(x < 2)
        return;
    multiplyEach();
}

//...
multiplyEach();
console.log(stack[0]);

Introducing a parameter and utilizing a list:

function multiplyEach(index) {
    var int = list[index];
    if(index == 0)
        return int;
    return int * multiplyEach(index - 1);
}

//...
console.log(multiplyEach(list.length - 1));

Both approaches demonstrate different recursive implementations; recursion fundamentally involves a function calling itself. Tail recursion could also have been considered by having a parameter store the total instead of multiplying the return values as in option 2.

EDIT: I noticed that the base case for option 1 was misplaced, so I have corrected it. It should function properly now.

Answer 3

Answer №2

To accomplish this recursively, you can follow these steps:

function multiplyEach(arr, size) {
  return size === 0 ? arr[size] : arr[size] * multiplyEach(arr, size - 1);
}

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9];
console.log(multiplyEach(array, array.length - 1));

Alternatively, you can use the built-in method reduce like so:

console.log([1, 2, 3, 4, 5, 6, 7, 8, 9].reduce(function(a, b) { return a * b; }));

A more concise ES6 version:

console.log([1, 2, 3, 4, 5, 6, 7, 8, 9].reduce((a, b) => a * b));

This should provide a solution to your query.

Answer 4

To accomplish this recursively, you can follow these steps:

function multiplyEach(arr, size) {
  return size === 0 ? arr[size] : arr[size] * multiplyEach(arr, size - 1);
}

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9];
console.log(multiplyEach(array, array.length - 1));

Alternatively, you can use the built-in method reduce like so:

console.log([1, 2, 3, 4, 5, 6, 7, 8, 9].reduce(function(a, b) { return a * b; }));

A more concise ES6 version:

console.log([1, 2, 3, 4, 5, 6, 7, 8, 9].reduce((a, b) => a * b));

This should provide a solution to your query.

Answer 5

Answer №3

There are situations where recursion proves to be the optimal solution for a problem, while in other cases iteration is more suitable.

Recursion is ideal for repeating an action until a specific condition is met before terminating, similar to a while loop.

Iteration works best for executing an action a set number of times and then stopping, akin to a for loop.

For instance, suppose we need to input an array and find the product of all its elements. Essentially, we need to perform N - 1 multiplications, with N representing the array's element count. Since we have a clear idea of the operations required, opting for iteration would be more efficient, as demonstrated below.

var arr = [1, 2, 3]
var result = multiplyEach(arr)
console.log(result)
// -> 6

function multiplyEach(arr) {
    var result = 1
    arr.map(function(value) {
        result *= value
    })
    return result
}

However, it's important to note that this guideline is not absolute, as there are instances where recursion offers a more elegant solution compared to iteration. Take, for example, the concise and recursive implementation of calculating factorial.

function factorial(value) {
return 1 === value ? 1 : value * factorial(value - 1)
}

In contrast, the iterative approach for calculating factorial through a while or for loop may not appear as elegant.

function factorial(value) {
    var result = 1
    while (value > 0) {
        result *= value
        value--
    }
    return result
}

function factorial(value) {
    var result = 1
    for (var i = 0; i < value; i++) {
        result *= value
    }
    return result
}

Answer 6

There are situations where recursion proves to be the optimal solution for a problem, while in other cases iteration is more suitable.

Recursion is ideal for repeating an action until a specific condition is met before terminating, similar to a while loop.

Iteration works best for executing an action a set number of times and then stopping, akin to a for loop.

For instance, suppose we need to input an array and find the product of all its elements. Essentially, we need to perform N - 1 multiplications, with N representing the array's element count. Since we have a clear idea of the operations required, opting for iteration would be more efficient, as demonstrated below.

var arr = [1, 2, 3]
var result = multiplyEach(arr)
console.log(result)
// -> 6

function multiplyEach(arr) {
    var result = 1
    arr.map(function(value) {
        result *= value
    })
    return result
}

However, it's important to note that this guideline is not absolute, as there are instances where recursion offers a more elegant solution compared to iteration. Take, for example, the concise and recursive implementation of calculating factorial.

function factorial(value) {
return 1 === value ? 1 : value * factorial(value - 1)
}

In contrast, the iterative approach for calculating factorial through a while or for loop may not appear as elegant.

function factorial(value) {
    var result = 1
    while (value > 0) {
        result *= value
        value--
    }
    return result
}

function factorial(value) {
    var result = 1
    for (var i = 0; i < value; i++) {
        result *= value
    }
    return result
}

Answer 7

Answer №4

It appears that the key issue lies in the absence of an argument within the multiplyEach() function definition, resulting in attempting to invoke it with one inside the function itself.

A revised version could look something like this:

var stack = [1, 2, 3, 4, 5, 6, 7];

function multiplyEach(arr) {
  if (arr.length == 0) {
    return 1;
  } else {
    return arr.pop() * multiplyEach(arr);
  }
}
console.log(multiplyEach(stack));

Answer 8

It appears that the key issue lies in the absence of an argument within the multiplyEach() function definition, resulting in attempting to invoke it with one inside the function itself.

A revised version could look something like this:

var stack = [1, 2, 3, 4, 5, 6, 7];

function multiplyEach(arr) {
  if (arr.length == 0) {
    return 1;
  } else {
    return arr.pop() * multiplyEach(arr);
  }
}
console.log(multiplyEach(stack));

Answer 9

Answer №5

The issue arises from the base case in the recursive function multiplyEach() returning undefined.

An easy solution can be implemented by:

var stack = [];
var multipliedStack = [];

function countDown(int) {
  stack.push(int);
  if (int === 1) {  
    return 1;
  }
    return countDown(int - 1);
}

function multiplyEach() {
  // Extract the last value from the stack 
  // and assign it to the variable int
  int = stack.pop();
  x = stack.length;
  // Base case
  if (x === 0) {
    return multipliedStack.reverse();
  }
  // Recursive case
  else {
    multipliedStack.push(int * stack[x - 1]);
    return multiplyEach(int);
  }
}

// Call the function countDown(7)
countDown(7);
// Print out the returned value of multiplyEach()
console.log(multiplyEach());

I'm not quite sure what you mean by "sort of like a factorial"? What is the intended purpose of the function?

To grasp recursion, understanding recursive base cases is essential, and I believe revisiting how push() and pop() are used in your code would also help. Recursive functions may seem daunting if you're more familiar with iteration, but persistence will lead to comprehension sooner than you think.

Answer 10

The issue arises from the base case in the recursive function multiplyEach() returning undefined.

An easy solution can be implemented by:

var stack = [];
var multipliedStack = [];

function countDown(int) {
  stack.push(int);
  if (int === 1) {  
    return 1;
  }
    return countDown(int - 1);
}

function multiplyEach() {
  // Extract the last value from the stack 
  // and assign it to the variable int
  int = stack.pop();
  x = stack.length;
  // Base case
  if (x === 0) {
    return multipliedStack.reverse();
  }
  // Recursive case
  else {
    multipliedStack.push(int * stack[x - 1]);
    return multiplyEach(int);
  }
}

// Call the function countDown(7)
countDown(7);
// Print out the returned value of multiplyEach()
console.log(multiplyEach());

I'm not quite sure what you mean by "sort of like a factorial"? What is the intended purpose of the function?

To grasp recursion, understanding recursive base cases is essential, and I believe revisiting how push() and pop() are used in your code would also help. Recursive functions may seem daunting if you're more familiar with iteration, but persistence will lead to comprehension sooner than you think.

Recursive Functions and the Power of Stacking

Answer №1

Answer №2

A more concise ES6 version:

Answer №3

Answer №4

Answer №5

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