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I'm trying to return the factorial of the provided integer. When doing it like this, e.g: 

factorialize(num) { for (var i = 1;i <= num; i++){ num*=i; } return num; } factorialize(5);

I'm getting an infinite loop. While I understand that this shouldn't give me the correct answer because my understanding of it would go something like this:

n! = 5 * 1 * 2 * 3 * 4 * 5 = 600

when really it should go:

n! = 1 * 2 * 3 * 4 * 5 = 120

But still, I don't understand why I am getting an infinite loop here?

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    make another variable to hold the result. Rather than changing the value of the parameter. Commented Apr 13, 2017 at 22:16
  • @Lixus: Please answer in the answer section (clue's in the name) Commented Apr 13, 2017 at 22:18

9 Answers 9

Reset to default
8

Lets say that the value of num variable is 2.

  • First cycle: 

    for (var i = 1; i <= 2; i++) { //true num *= i; //2 = 2 * 1 => 2 }

  • Second cycle:

    for (var i = 2; i <= 2; i++) { //true num *= i; //2 = 2 * 2 => 4 }

  • Third cycle:

    for (var i = 3; i <= 4; i++) { //true num *= i; //4 = 4 * 3 => 12 }

  • Fourth cycle:

    for (var i = 4; i <= 12; i++) { //true num *= i; //12 = 12 * 4 => 48 }

The num value increases exponentially, while the i value increases linearly.

i <= num condition will be always fulfilled, that's why you are getting an infinite loop.

enter image description here

Snippet including the chart:

var num = { x: [1, 2, 3, 4, 5], y: [2, 4, 12, 48, 240], type: 'scatter', name: 'num' }; var i = { x: [1, 2, 3, 4, 5], y: [1, 2, 3, 4, 5], type: 'scatter', name: 'i' }; var data = [num, i]; Plotly.newPlot('myDiv', data);
<script src="https://cdn.plot.ly/plotly-latest.min.js"></script> <div id="myDiv" style="width: 480px; height: 400px;">

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1 Comment

num increases exponentially, not logarithmically.
2

Your loop is chasing a moving target.

It will end when i reaches num, but you keep making num bigger; actually i will never reach num.

As you pointed out, your algorithm is wrong anyway.

function factorialize(num) { var result = 1; for (var i = 2; i <= num; i++) { result *= i; } return result; } console.log(factorialize(0)); // 1 console.log(factorialize(1)); // 1 console.log(factorialize(2)); // 2 console.log(factorialize(3)); // 6 console.log(factorialize(4)); // 24 console.log(factorialize(5)); // 120
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1

The variable that you use on the for loop it's the same variable that you use to store the multiplication.

This should work:

factorialize(num) { var len = num; for (var i = 1;i <= len; i++){ num*=i; } return num; } factorialize(5);
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0

Because you change the value of 'num' in every iteration. You should use a third variable to count the product.

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0

You are getting an infinite loop because you are always updating num. You are doing num = num * i. So it can never reach the end of the loop. i will never be greater then num.

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Taking your example:

Before run 1: num = 5; After run 1: num = 5; After run 2: num = 10; After run 3: num = 30;

As per your condition, loop would stop only when num is more than i, which would never be the case, as you're modifying it's value on every iteration by a huge difference.

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here i will never catch num

i = 1 num = 5 1 <= 5 //(num = 5*1) i = 2 num = 5 2 <= 10 //(num = 5*2) i = 3 num = 10 3 <= 30 //(num = 10*3) ...
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This is because you are dynamically updating the value of num. So, after each iteration, the number gets bigger and bigger, and that's why the value of i will never be more than the value of num (for values of num larger than 1).

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5 Comments

It will if num is <= 0 at the start of the function.
@BoundaryImpositio Yes you are right. But for value larger than 1, this will always be an infinite loop.
You said "never" :)
@BoundaryImposition Just made an edit. Thanks for noticing this! :)
No problem Omar!
0

The end condition for i<=num is always true hence the infinite loop because the argument num keeps increasing with each iteration. A better approach is

function factorialize(num) { var res = 1; if(num ===0 || num === 1){ return 1;	} for (var i = 2;i <= num; i++){ res*=i; } return res; } factorialize(5);

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