How can I change and keep the value of a variable inside a recursive function without passing it as a parameter?

Question

function newtonRaphson(x, count){
  while(count > 0){
    var previousValue = 0
    if(previousValue === 0){
      previousValue = x/2
    } 
    var y = (previousValue + (x/previousValue))/2 
    previousValue = y
    console.log(previousValue, count)
    return newtonRaphson(x, count - 1)
  }  
  return y
}

I have this function that is supposed to perform the Newton Raphson method, it would look like this:

x = 20
count = 4
previousValue(0) = 10
y(1) = (10 + (20/10))/2 = 6
previousValue = 6
y(2) = (6 + (20/6))/2 = 4.666
previousValue = 4.666
y(3) = (4.666 + (20/4.666))/2 = 4.476
previousValue = 4.476
y(4) = (4.476+ (20/4.476))/2 = 4.472

My question is, is there a way I can maintain previousValue between iterations without passing it to the function? Because x needs to be passed no matter what, as well as count, but I also need to tell the function what the previousValue is, so I'm at a loss

Maybe you could wrap it in another function and make it local to the wrapper function? — mykaf, Mar 25 '22 at 17:45
I have tried and failed, the global variable is a good solution, don't know if it's possible without it though. — Dumbasswithaname, Mar 25 '22 at 17:54
Sorry I wasn't specific, I tried to use a wrapper function, but just like before the value of previousValue keeps resetting every iteration — Dumbasswithaname, Mar 25 '22 at 17:57
Passing it as an argument is the right thing to do in a recursive function. Not sure why you're trying to avoid that. — Bergi, Mar 25 '22 at 17:57
Please show your work on your wrapper function. We can't tell what's going on without seeing it. But @Bergi is right; passing it is the way to go. What's preventing that from working? — mykaf, Mar 25 '22 at 18:09

score 2 · Accepted Answer · answered Mar 27 '22 at 01:08

This code is confused.

First off, while it's called newtonRaphson, it seems to be only for a very specific case of the Newton-Raphson method, that of finding square roots using the Babylonian Method. So ideally it should have a better name.

It tries to simultaneously solve the problem recursively and with a while-loop. We need to choose one or the other. Here we choose a recursive version.

Then we note that these lines make no real sense:

    var previousValue = 0
    if(previousValue === 0){
      previousValue = x/2
    }

This is a ridiculously round-about way of writing

    var previousValue = x/2

You ask about doing this without passing the previous value to the recursive function. You could, just by tracking it at a higher scope. (Please, please, not the global scope!) It might look like this:

const sqrt = (x, count) => {
  let prev = x / 2
  const _sqrt = (count) => {
    if (count <= 0) return prev
    prev = (prev + (x / prev)) / 2
    return _sqrt (count - 1)
  }
  return _sqrt (count)
}

console .log (sqrt (25, 1)) //=> 7.25
console .log (sqrt (25, 2)) //=> 5.349137931034482
console .log (sqrt (25, 3)) //=> 5.011394106532552
console .log (sqrt (25, 4)) //=> 5.000012953048684
console .log (sqrt (25, 5)) //=> 5.000000000016778
console .log (sqrt (25, 6)) //=> 5

But I would not recommend this at all. Recursion is simplest when you pass the necessary variables into the function. And this is not hard to do, either with a defaulted parameter, like this:

const sqrt = (x, count, prev = x / 2) =>
  count <= 0
    ? prev
    : sqrt (x, count - 1, (prev + (x / prev)) / 2)

or with a public wrapper around an internal recursive helper function, like this:

const _sqrt = (x, count, prev) =>
  count <= 0
    ? prev
    : _sqrt (x, count - 1, (prev + (x / prev)) / 2)

const sqrt = (x, count) => 
  _sqrt (x, count, x / 2)

Of the two, I usually prefer the defaulted parameter, but there are some potential problems with them, especially if you are not in control of how your function is called. Then the public/internal split makes sense.

How can I change and keep the value of a variable inside a recursive function without passing it as a parameter?

1 Answers1