Basic JavaScript - Replace Loops using Recursion

[mcmichaeltyler93]

I’m not understanding [quote=“mcmichaeltyler93, post:3, topic:555703”]
multiply(arr, n - 1) * arr[n - 1]
[/quote]

[JeremyLT]

@JeremyLT

@JeremyLT

@JeremyLT

@mcmichaeltyler93

multiply(arr, n - 1) * arr[n - 1]

This is the product of the first n-1 elements multiplied by the nth element.

Using this formula instead, if I multiply the product of the first 3 numbers, 6, by the 4th number, 3, I get 18, which is the same thing as ‘the product of the first 4 numbers’.

Did this part make sense?

Could you please actually answer if this explanation makes any sense.

[mcmichaeltyler93]

But N is the input that is being passed in…

[mcmichaeltyler93]

Yes, it does. I just don’t know how one got to that conclusion.

[JeremyLT]

You aren’t being asked to reinvent math here. I’m trying to get you to agree that multiply([2, 3, 1, 3, 4], 4) === 18 === 6 * 3 === multiply([2, 3, 1, 3, 4], 3) * [2, 3, 1, 3, 4][3].

The entire idea behind recursion is to make a ‘smaller’ version of the problem that takes less work to solve. To do that, you need a way to make the answer with the result of a function call with ‘smaller’ inputs.

[mcmichaeltyler93]

2 * 3 === 6 * 1 === 6 * 3 === 18 * 4 === 72. Or would it be 2 * 3 * 1 * 3 since there is 3 after the comma?

[JeremyLT]

I am not sure where those numbers are coming from? You are using = in between expressions that are not the same.

[mcmichaeltyler93]

@JeremyLT

multiply([2, 3, 1, 3, 4], 3)

multiply([2, 3, 1, 3, 4], 3)

[JeremyLT]

@mcmichaeltyler93

2 * 3 === 6 * 1 === 6 * 3 === 18 * 4 === 72

These values are not equivalent. 6 is not the same thing as 72.

@mcmichaeltyler93

multiply([2, 3, 1, 3, 4], 3)

The value of this expression is not 72. Also, it really helps if you use words instead of copy-pasting code at me.

[mcmichaeltyler93]

OMG i get it!!!

[mcmichaeltyler93]

Thanks for being patient with me (: I always get it eventually