Develop Reference

Recursive approach to Persistent bugger problem returns undefined

I've been trying to solve the following problem in codewars using recursion:
Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.
For example (Input --> Output):
39 --> 3 (because 3*9 = 27, 2*7 = 14, 1*4 = 4 and 4 has only one digit)
999 --> 4 (because 9*9*9 = 729, 7*2*9 = 126, 1*2*6 = 12, and finally 1*2 = 2)
4 --> 0 (because 4 is already a one-digit number)
Here's what I've tried:
var numOfIterations = 0;
function persistence(num) {
//code me
var i;
var digits=[];
var result = 1;
if (num.toString().length==1) {
return numOfIterations;
} else {
numOfIterations++;
digits = Array.from(String(num), Number);
for (i=0;i<digits.size;i++) {
result=result*digits[i];
}
persistence(result);
}
}
But for some reason, instead of returning the number of iterations, it returns undefined. I've been told that I'm not using recursion correctly, but I just can't find the problem.

Other answers have explained what's wrong with your code. I just want to point out a simpler implementation:
const multiplyDigits = (n) =>
n < 10 ? n : (n % 10) * multiplyDigits (n / 10 | 0);
const persistence = (n) =>
n < 10 ? 0 : 1 + persistence (multiplyDigits (n));
[39, 999, 4] .forEach (t => console .log (`${t}:\t${persistence (t)}`));
multiplyDigits does just what it says, recursively multiplying the final digit by the number left when you remove that last digit (Think of | 0 as like Math .floor), and stopping when n is a single digit.
persistence checks to see if we're already a single digit, and if so, returns zero. If not, we add one to the value we get when we recur on the multiple of the digits.

I've been told that I'm not using recursion correctly
You're recursing, but you're not returning the result of that recursion. Imagine for a moment just this structure:
function someFunc() {
if (someCondition) {
return 1;
} else {
anotherFunc();
}
}
If someCondition is false, what does someFunc() return? Nothing. So it's result is undefined.
Regardless of any recursion, at its simplest if you want to return a result from a function then you need to return it:
function persistence(num) {
//...
if (num.toString().length==1) {
//...
} else {
//...
return persistence(result); // <--- here
}
}

As #David wrote in his answer, you were missing the return of the recursive call to itself.
Plus you were using digits.size instead of digits.length.
Anyway consider that one single digit being zero will collpse the game because that's enough to set the result to zero despite how many digits the number is made of.
To deal with the reset of numOfIterations, at first I tried using function.caller to discriminate between recursive call and direct call and set the variable accordingly. Since that method is deprecated as shown here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/caller
I opted for the optional argument iteration that gets set to zero as default, to keep track of that value while it goes down the call stack. This solution still fulfills the fact that the caller doesn't need to know a new interface for the function to work.
//var numOfIterations = 0;
function persistence(num, iteration=0) {
/*
Commented strategy using the function.caller
working but deprecated so I can't recommend anymore
used optional argument iteration instead
//gets the name of the caller scope
let callerName = persistence.caller?.name;
//if it's different from the name of this function
if (callerName !== 'persistence')
//reset the numOfIterations
numOfIterations = 0;
*/
var digits=[];
if (num.toString().length==1){
return iteration;
} else {
var result = 1;
digits = Array.from(String(num), Number);
for (let i=0;i<digits.length;i++) {
result = result * digits[i];
}
return persistence(result, iteration+1);
}
}
console.log( persistence(39) ); //-> 3
console.log( persistence(999 ) ); //-> 4
console.log( persistence(4) ); //-> 0

You can do something like this
const persistenceTailRecursive = (num, iterations = 0) => {
const str = '' + num;
if(str.length === 1){
return iterations;
}
return persistenceTailRecursive(str.split('').reduce((res, a) => res * parseInt(a), 1), iterations + 1)
}
const persistence = (num) => {
const str = '' + num;
if(str.length === 1){
return 0;
}
return 1 + persistence(str.split('').reduce((res, a) => res * parseInt(a), 1))
}
console.log(persistenceTailRecursive(93))
console.log(persistenceTailRecursive(999))
console.log(persistence(93))
console.log(persistence(999))
There are 2 versions
1 tailRecursive call the same method with the exact signature (preventing stackoverflow in some languages like scala)
2 basic the result is calculated at the end

reduce sum of digits recursivel down to a one digit number

I'm trying to solve a challenge (Digit Degree) on Code signal where the task is to find the the number of times we need to replace this number with the sum of its digits until we get to a one digit number. I.e. of the incoming number is 5 it's already a one digit number so the outcome should be 0. If the number is 100 the sum of its digits is 1 which is one digit so the outcome should be 1 and so on...
I'm doing a recursive solution like this:
let count = 0;
function digitDegree(n) {
if (n < 10) {
console.log(count);
return count;
};
const arr = n.toString().split('').map(Number);
const sumOfDigits = arr.reduce((acc, curr) => acc + curr);
count++;
digitDegree(sumOfDigits);
}
At the second loop and forth I'm getting null as output even if the console log shows the correct value. Where does it go wrong?
I saw that the question was up before but the answer was quite mathematical. Would this approach be ok or would it be considered bad practise?

You need a return statement for getting the result of the recursion.
return digitDegree(sumOfDigits);
A shorter approach would remove explicit conversions in advance and remove temporary arrays.
Then take another parameter for count and omit a global variable.
function digitDegree(n, count = 0) {
if (n < 10) return count;
return digitDegree(
n .toString()
.split('')
.reduce((acc, curr) => acc + +curr, 0),
count + 1
);
}
console.log(digitDegree(9999));

To return a value, update the last line in your function to return digitDegree(sumOfDigits)
function digitDegree(n) {
if (n < 10) {
console.log(count);
return count;
};
const arr = n.toString().split('').map(Number);
const sumOfDigits = arr.reduce((acc, curr) => acc + curr);
count++;
// add return
return digitDegree(sumOfDigits);
}
Pitfalls with your current approach:
It's impure.
digitDegree(100) returns 1 the first time you run it but returns 2 when you run it again. This is because count was declared outside the function (making it global)
let count = 0
function digitDegree(n) {
if (n < 10) {
return count;
};
const arr = n.toString().split('').map(Number);
const sumOfDigits = arr.reduce((acc, curr) => acc + curr);
count++;
return digitDegree(sumOfDigits);
}
// wrong - output should not change
console.log(digitDegree(100)) //=> 1
console.log(digitDegree(100)) //=> 2
console.log(digitDegree(100)) //=> 3
Make your function pure
A pure function is a specific kind of value-producing function that not only has no side effects but also doesn’t rely on side effects from other code—for example, it doesn’t read global bindings whose value might change.
A pure function has the pleasant property that, when called with the same arguments, it always produces the same value (and doesn’t do anything else)
Source
Suggestions:
Pass count as an argument
function digitDegree(n,count=0) {
if (n < 10) {
return count;
};
const arr = n.toString().split('').map(Number);
const sumOfDigits = arr.reduce((acc, curr) => acc + curr);
count++;
return digitDegree(sumOfDigits, count);
}
// correct
console.log(digitDegree(100)) //=> 1
console.log(digitDegree(100)) //=> 1
console.log(digitDegree(100)) //=> 1
Wrap your recursion function inside another function
function recursionWrapper(num){
let count = 0;
function digitDegree(n) {
if (n < 10) {
return count;
};
const arr = n.toString().split('').map(Number);
const sumOfDigits = arr.reduce((acc, curr) => acc + curr);
count++;
// notice how we don't need a return in this approach
digitDegree(sumOfDigits)
}
digitDegree(num)
return count
}
// correct
console.log(recursionWrapper(100)) //=> 1
console.log(recursionWrapper(100)) //=> 1
console.log(recursionWrapper(100)) //=> 1
Further reading:
How to deal with dirty side effects in your pure functional JavaScript
Javascript and Functional Programming — Pt. 3: Pure Functions

I need to extract every nth char of a string in Javascript

Ive been reading everything online but its not exactly what I need
var x = 'a1b2c3d4e5'
I need something to get me to
using 1 the answer should be abcde
using 2 the answer should be 12345
using 3 the answer should be b3e
the idea behind it if using 1 it grabs 1 skips 1
the idea behind it if using 2 it grabs 2 skips 2
the idea behind it if using 3 it grabs 3 skips 3
I dont want to use a for loop as it is way to long especially when your x is longer than 300000 chars.
is there a regex I can use or a function that Im not aware of?
update
I'm trying to some how implement your answers but when I use 1 that's when I face the problem. I did mention trying to stay away from for-loops the reason is resources on the server. The more clients connect the slower everything becomes. So far array.filter seem a lot quicker.
As soon as I've found it I'll accept the answer.

As others point out, it's not like regular expressions are magic; there would still be an underlying looping mechanism. Don't worry though, when it comes to loops, 300,000 is nothing -
console.time('while')
let x = 0
while (x++ < 300000)
x += 1
console.timeEnd('while')
// while: 5.135 ms
console.log(x)
// 300000
Make a big string, who cares? 300,000 is nothing -
// 10 chars repeated 30,000 times
const s =
'abcdefghij'.repeat(30000)
console.time('while string')
let x = 0
let interval = 2
let values = []
while (x < s.length)
{ values.push(s[x])
x += interval
}
let result = values.join('')
console.timeEnd('while string')
// while string: 31.990ms
console.log(result)
console.log(result.length)
// acegiacegiacegiacegiacegiacegiacegiacegia...
// 150000
Or use an interval of 3 -
const s =
'abcdefghij'.repeat(30000)
console.time('while string')
let x = 0
let interval = 3
let values = []
while (x < s.length)
{ values.push(s[x])
x += interval
}
let result = values.join('')
console.timeEnd('while string')
// while string: 25.055ms
console.log(result)
console.log(result.length)
// adgjcfibehadgjcfibehadgjcfibehadgjcfibe...
// 100000
Using a larger interval obviously results in fewer loops, so the total execution time is lower. The resulting string is shorter too.
const s =
'abcdefghij'.repeat(30000)
console.time('while string')
let x = 0
let interval = 25 // big interval
let values = []
while (x < s.length)
{ values.push(s[x])
x += interval
}
let result = values.join('')
console.timeEnd('while string')
// while string: 6.130
console.log(result)
console.log(result.length)
// afafafafafafafafafafafafafafafafafafafafafafa...
// 12000
You can achieve functional style and stack-safe speed simultaneously -
const { loop, recur } = require('./lib')
const everyNth = (s, n) =>
loop
( (acc = '', x = 0) =>
x >= s.length
? acc
: recur(acc + s[x], x + n)
)
const s = 'abcdefghij'.repeat(30000)
console.time('loop/recur')
const result = everyNth(s, 2)
console.timeEnd('loop/recur')
// loop/recur: 31.615 ms
console.log(result)
console.log(result.length)
// acegiacegiacegiacegiacegiacegiacegia ...
// 150000
The two are easily implemented -
const recur = (...values) =>
({ recur, values })
const loop = f =>
{ let acc = f()
while (acc && acc.recur === recur)
acc = f(...acc.values)
return acc
}
// ...
module.exports =
{ loop, recur, ... }
And unlike the [...str].filter(...) solutions which will always iterate through every element, our custom loop is much more flexible and receives speed benefit when a higher interval n is used -
console.time('loop/recur')
const result = everyNth(s, 25)
console.timeEnd('loop/recur')
// loop/recur: 5.770ms
console.log(result)
console.log(result.length)
// afafafafafafafafafafafafafafa...
// 12000
const recur = (...values) =>
({ recur, values })
const loop = f =>
{ let acc = f()
while (acc && acc.recur === recur)
acc = f(...acc.values)
return acc
}
const everyNth = (s, n) =>
loop
( (acc = '', x = 0) =>
x >= s.length
? acc
: recur(acc + s[x], x + n)
)
const s = 'abcdefghij'.repeat(30000)
console.time('loop/recur')
const result = everyNth(s, 2)
console.timeEnd('loop/recur')
// loop/recur: 31.615 ms
console.log(result)
console.log(result.length)
// acegiacegiacegiacegiacegiacegiacegia ...
// 150000

Since I'm not an expert of regex, I'd use some fancy es6 functions to filter your chars.
var x = 'a1b2c3d4e5'
var n = 2;
var result = [...x].filter((char, index) => index % n == 0);
console.log(result);
Note that because 0 % 2 will also return 0, this will always return the first char. You can filter the first char by adding another simple check.
var result = [...x].filter((char, index) => index > 0 && index % n == 0);

As a variant:
function getNth(str, nth) {
return [...str].filter((_, i) => (i + 1) % nth === 0).join('');
}
console.log(getNth('a1b2c3d4e5', 2)); // 12345
console.log(getNth('a1b2c3d4e5', 3)); // b3e

What I'd suggest, to avoid having to iterate over the entire array, is to step straight into the known nth's.
Here's a couple of flavors:
function nthCharSubstr(str, nth) {
let res = "";
for (let i = nth - 1; i < str.length; i += nth) {
res += string[i];
}
return res;
}
More ES6-y:
const nthCharSubstr = (str, nth) =>
[...Array(parseInt(str.length / nth)).keys()] // find out the resulting number of characters and create and array with the exact length
.map(i => nth + i * nth - 1) // each item in the array now represents the resulting character's index
.reduce((res, i) => res + str[i], ""); // pull out each exact character and group them in a final string
This solution considers this comment as being valid.

Disqualifying consecutive characters in an Alphabetical Order. Sol [ASCII Format]

I was given this problem by my friend. The question asks to remove all the alphabetically consecutive characters from the string input given.
So I did it using Javascript, I need expert help if I performed it precisely.
I thought using Array.prototype.reduce will be the best way, do we have other possible ways?
/**
* #author Abhishek Mittal <abhishekmittaloffice#gmail.com>
* #description function can deal with both any types followed in a consecutive manner in ASCII Chart.
* #param {string} str
*/
function improvise(str) {
// Backup for original input.
const bck = str || '';
const bckArr = bck.split('').map( e => e.charCodeAt(0)); // converting the alphabets into its ASCII for simplicity and reducing the mayhem.
let result = bckArr.reduce( (acc, n) => {
// Setting up flag
let flag1 = n - acc.rand[acc.rand.length - 1];
let flag2 = n - acc.temp;
if(flag1 === 1 || flag2 === 1) {
(flag2 !== NaN && flag2 !== 1) ? acc.rand.pop() : null; // update the latest value with according to the case.
acc.temp = n
}else{
acc.rand.push(n); // updating the random to latest state.
acc.temp = null;
}
return acc;
}, {rand: [], temp: null} /* setting up accumulative situation of the desired result */)
result = result.rand.map(e => String.fromCharCode(e)).join('')
return result ? result : '' ;
}
function init() {
const str = "ab145c";
const final = improvise(str);
console.log(final)
}
init();
Well, the output is coming out to be correct.
Input: ab145c
Output: 1c

There's no way to solve this using any remotely reasonable regular expression, unfortunately.
I think it would be a lot clearer to use .filter, and check whether either the next character or the previous character is consecutive:
const code = char => char
? char.charCodeAt(0)
: -2; // will not be === to any other codes after addition or subtraction
function improvise(str) {
return [...str]
.filter((char, i) => {
const thisCode = code(char);
return (
thisCode !== code(str[i - 1]) + 1
&& thisCode !== code(str[i + 1]) - 1
);
})
.join('');
}
console.log(improvise('ab145c'));
(alternatively, you could check only whether the next character is consecutive, but then you'd have to check the validity of the last character in the string as well)
If you need continuously replace characters until no consecutive characters remain, then keep calling improvise:
const code = char => char
? char.charCodeAt(0)
: -2; // will not be === to any other codes after addition or subtraction
function improvise(str) {
return [...str]
.filter((char, i) => {
const thisCode = code(char);
return (
thisCode !== code(str[i - 1]) + 1
&& thisCode !== code(str[i + 1]) - 1
);
})
.join('');
}
let result = 'hishakmitalaaaaabbbbbbcccccclmnojqyz';
let same = false;
while (!same) {
const newResult = improvise(result);
if (newResult !== result) {
result = newResult;
console.log(result);
} else {
same = true;
}
}
console.log('FINAL:', result);

Well, that's great code but it doesn't gives the exact solution to the problem, see:
INPUT: hishakmitalaaaaabbbbbbcccccclmnojqyz
i got
Output: shakmitalaaaabbbbcccccjq
you see 'ab' & 'bc' remains intact, we need to increase the number of loops maybe to check those, can increase complexity as well.
But, my solution doesn't as well gives me the answer I desire e.g. for the above string I should get
ishakmitalaacccjq
but rather my solution gives
shakmitalcccccjq
hope you understand the question. We need a change in the way we traverse throughout the string.

How do I replace while loops with a functional programming alternative without tail call optimization?

I am experimenting with a more functional style in my JavaScript; therefore, I have replaced for loops with utility functions such as map and reduce. However, I have not found a functional replacement for while loops since tail call optimization is generally not available for JavaScript. (From what I understand ES6 prevents tail calls from overflowing the stack but does not optimize their performance.)
I explain what I have tried below, but the TLDR is:
If I don't have tail call optimization, what is the functional way to implement while loops?
What I have tried:
Creating a "while" utility function:
function while(func, test, data) {
const newData = func(data);
if(test(newData)) {
return newData;
} else {
return while(func, test, newData);
}
}
Since tail call optimization isn't available I could rewrite this as:
function while(func, test, data) {
let newData = *copy the data somehow*
while(test(newData)) {
newData = func(newData);
}
return newData;
}
However at this point it feels like I have made my code more complicated/confusing for anyone else who uses it, since I have to lug around a custom utility function. The only practical advantage that I see is that it forces me to make the loop pure; but it seems like it would be more straightforward to just use a regular while loop and make sure that I keep everything pure.
I also tried to figure out a way to create a generator function that mimics the effects of recursion/looping and then iterate over it using a utility function like find or reduce. However, I haven't figured out an readable way to do that yet.
Finally, replacing for loops with utility functions makes it more apparent what I am trying to accomplish (e.g. do a thing to each element, check if each element passes a test, etc.). However, it seems to me that a while loop already expresses what I am trying to accomplish (e.g. iterate until we find a prime number, iterate until the answer is sufficiently optimized, etc.).
So after all this, my overall question is: If I need a while loop, I am programming in a functional style, and I don't have access to tail call optimization, then what is the best strategy.

An example in JavaScript
Here's an example using JavaScript. Currently, most browsers do not support tail call optimisation and therefore the following snippet will fail
const repeat = n => f => x =>
n === 0 ? x : repeat (n - 1) (f) (f(x))
console.log(repeat(1e3) (x => x + 1) (0)) // 1000
console.log(repeat(1e5) (x => x + 1) (0)) // Error: Uncaught RangeError: Maximum call stack size exceeded
Trampolines
We can work around this limitation by changing the way we write repeat, but only slightly. Instead of returning a value directly or immediately recurring, we will return one of our two trampoline types, Bounce or Done. Then we will use our trampoline function to handle the loop.
// trampoline
const Bounce = (f,x) => ({ isBounce: true, f, x })
const Done = x => ({ isBounce: false, x })
const trampoline = ({ isBounce, f, x }) => {
while (isBounce)
({ isBounce, f, x } = f(x))
return x
}
// our revised repeat function, now stack-safe
const repeat = n => f => x =>
n === 0 ? Done(x) : Bounce(repeat (n - 1) (f), f(x))
// apply trampoline to the result of an ordinary call repeat
let result = trampoline(repeat(1e6) (x => x + 1) (0))
// no more stack overflow
console.log(result) // 1000000
Currying slows things down a little bit too, but we can remedy that using an auxiliary function for the recursion. This is nice too because it hides the trampoline implementation detail and does not expect the caller to bounce the return value. This runs about twice as fast as the above repeat
// aux helper hides trampoline implementation detail
// runs about 2x as fast
const repeat = n => f => x => {
const aux = (n, x) =>
n === 0 ? Done(x) : Bounce(x => aux (n - 1, x), f (x))
return trampoline (aux (n, x))
}
Clojure-style loop/recur
Trampolines are nice and all but it's kind of annoying to have to have to worry about calling trampoline on your function's return value. We saw the alternative was to use an auxiliary helper, but c'mon that's kind of annoying, too. I'm sure some of you aren't too keen about the Bounce and Done wrappers, too.
Clojure creates a specialised trampoline interface that uses a pair of functions, loop and recur – this tandem pair lends itself to a remarkably elegant expression of a program
Oh and it's really fast, too
const recur = (...values) =>
({ recur, values })
const loop = run =>
{ let r = run ()
while (r && r.recur === recur)
r = run (...r.values)
return r
}
const repeat = n => f => x =>
loop
( (m = n, r = x) =>
m === 0
? r
: recur (m - 1, f (r))
)
console.time ('loop/recur')
console.log (repeat (1e6) (x => x + 1) (0)) // 1000000
console.timeEnd ('loop/recur') // 24 ms
Initially this style will feel foreign, but over time I am finding it to be the most consistent while producing durable programs. Comments below help ease you into the expression-rich syntax -
const repeat = n => f => x =>
loop // begin a loop with
( ( m = n // local loop var m: counter, init with n
, r = x // local loop var r: result, init with x
) =>
m === 0 // terminating condition
? r // return result
: recur // otherwise recur with
( m - 1 // next m value
, f (r) // next r value
)
)
The continuation monad
This is one of my favourite topics tho, so we're gonna see what this looks like with the continuation monad. Reusing loop and recur, we implement a stack-safe cont that can sequence operations using chain and run operation sequences using runCont. For repeat, this is senseless (and slow), but it's cool to see the mechanics of cont at work in this simple example -
const identity = x =>
x
const recur = (...values) =>
({ recur, values })
const loop = run =>
{ let r = run ()
while (r && r.recur === recur)
r = run (...r.values)
return r
}
// cont : 'a -> 'a cont
const cont = x =>
k => recur (k, x)
// chain : ('a -> 'b cont) -> 'a cont -> 'b cont
const chain = f => mx =>
k => recur (mx, x => recur (f (x), k))
// runCont : ('a -> 'b) -> a cont -> 'b
const runCont = f => mx =>
loop ((r = mx, k = f) => r (k))
const repeat = n => f => x =>
{ const aux = n => x =>
n === 0 // terminating condition
? cont (x) // base case, continue with x
: chain // otherise
(aux (n - 1)) // sequence next operation on
(cont (f (x))) // continuation of f(x)
return runCont // run continuation
(identity) // identity; pass-thru
(aux (n) (x)) // the continuation returned by aux
}
console.time ('cont monad')
console.log (repeat (1e6) (x => x + 1) (0)) // 1000000
console.timeEnd ('cont monad') // 451 ms
Y combinator
The Y combinator is my spirit combinator; this answer would be incomplete without giving it some place amongst the other techniques. Most implementations of the Y combinator however, are not stack-safe and will overflow if the user-supplied function recurs too many times. Since this answer is all about preserving stack-safe behaviour, of course we'll implement Y in a safe way – again, relying on our trusty trampoline.
Y demonstrates the ability to extend easy-to-use, stack-safe, synchronous infinite recursion without cluttering up your function.
const bounce = f => (...xs) =>
({ isBounce: true, f, xs })
const trampoline = t => {
while (t && t.isBounce)
t = t.f(...t.xs)
return t
}
// stack-safe Y combinator
const Y = f => {
const safeY = f =>
bounce((...xs) => f (safeY (f), ...xs))
return (...xs) =>
trampoline (safeY (f) (...xs))
}
// recur safely to your heart's content
const repeat = Y ((recur, n, f, x) =>
n === 0
? x
: recur (n - 1, f, f (x)))
console.log(repeat (1e5, x => x + 1, 0)) // 10000
Practicality with while loop
But let's be honest: that's a lot of ceremony when we're overlooking one of the more obvious potential solutions: use a for or while loop, but hide it behind a functional interface
For all intents and purposes, this repeat function works identically to the ones provided above – except this one is about one or two gadzillion times faster (with exception to the loop/recur solution). Heck, it's arguably a lot easier to read too.
Granted, this function is perhaps a contrived example – not all recursive functions can be converted to a for or while loop so easily, but in such a scenario where it's possible, it's probably best to just do it like this. Save the trampolines and continuations for heavy lifting when a simple loop won't do.
const repeat = n => f => x => {
let m = n
while (true) {
if (m === 0)
return x
else
(m = m - 1, x = f (x))
}
}
const gadzillionTimes = repeat(1e8)
const add1 = x => x + 1
const result = gadzillionTimes (add1) (0)
console.log(result) // 100000000
setTimeout is NOT a solution to the stack overflow problem
OK, so it does work, but only paradoxically. If your dataset is small, you don't need setTimeout because there will be no stack overflow. If your dataset is large and you use setTimeout as a safe recursion mechanism, not only do you make it impossible to synchronously return a value from your function, it will be so F slow you won't even want to use your function
Some people have found an interview Q&A prep site that encourages this dreadful strategy
What our repeat would look like using setTimeout – notice it's also defined in continuation passing style – ie, we must call repeat with a callback (k) to get the final value
// do NOT implement recursion using setTimeout
const repeat = n => f => x => k =>
n === 0
? k (x)
: setTimeout (x => repeat (n - 1) (f) (x) (k), 0, f (x))
// be patient, this one takes about 5 seconds, even for just 1000 recursions
repeat (1e3) (x => x + 1) (0) (console.log)
// comment the next line out for absolute madness
// 10,000 recursions will take ~1 MINUTE to complete
// paradoxically, direct recursion can compute this in a few milliseconds
// setTimeout is NOT a fix for the problem
// -----------------------------------------------------------------------------
// repeat (1e4) (x => x + 1) (0) (console.log)
I can't stress enough how bad this is. Even 1e5 takes so long to run that I gave up trying to measure it. I'm not including this in the benchmarks below because it's just too slow to even be considered a viable approach.
Promises
Promises have the ability to chain computations and are stack safe. However, achieving a stack-safe repeat using Promises means we'll have to give up our synchronous return value, the same way we did using setTimeout. I'm providing this as a "solution" because it does solve the problem, unlike setTimeout, but in a way that's very simple compared to the trampoline or continuation monad. As you might imagine though, the performance is somewhat bad, but nowhere near as bad as the setTimeout example above
Worth noting in this solution, the Promise implementation detail is completely hidden from the caller. A single continuation is provided as a 4th argument and its called when the computation is complete.
const repeat = n => f => x => k =>
n === 0
? Promise.resolve(x).then(k)
: Promise.resolve(f(x)).then(x => repeat (n - 1) (f) (x) (k))
// be patient ...
repeat (1e6) (x => x + 1) (0) (x => console.log('done', x))
Benchmarks
Seriously, the while loop is a lot faster - like almost 100x faster (when comparing best to worst – but not including async answers: setTimeout and Promise)
// sync
// -----------------------------------------------------------------------------
// repeat implemented with basic trampoline
console.time('A')
console.log(tramprepeat(1e6) (x => x + 1) (0))
console.timeEnd('A')
// 1000000
// A 114 ms
// repeat implemented with basic trampoline and aux helper
console.time('B')
console.log(auxrepeat(1e6) (x => x + 1) (0))
console.timeEnd('B')
// 1000000
// B 64 ms
// repeat implemented with cont monad
console.time('C')
console.log(contrepeat(1e6) (x => x + 1) (0))
console.timeEnd('C')
// 1000000
// C 33 ms
// repeat implemented with Y
console.time('Y')
console.log(yrepeat(1e6) (x => x + 1) (0))
console.timeEnd('Y')
// 1000000
// Y 544 ms
// repeat implemented with while loop
console.time('D')
console.log(whilerepeat(1e6) (x => x + 1) (0))
console.timeEnd('D')
// 1000000
// D 4 ms
// async
// -----------------------------------------------------------------------------
// repeat implemented with Promise
console.time('E')
promiserepeat(1e6) (x => x + 1) (0) (console.log)
console.timeEnd('E')
// 1000000
// E 2224 ms
// repeat implemented with setTimeout; FAILED
console.time('F')
timeoutrepeat(1e6) (x => x + 1) (0) (console.log)
console.timeEnd('F')
// ...
// too slow; didn't finish after 3 minutes
Stone Age JavaScript
The above techniques are demonstrated using newer ES6 syntaxes, but you could implement a trampoline in the earliest possible version of JavaScript – it only requires while and first class functions
Below, we use stone age javascript to demonstrate infinite recursion is possible and performant without necessarily sacrificing a synchronous return value – 100,000,000 recursions in under 6 seconds - this is a dramatic difference compared to setTimeout which can only 1,000 recursions in the same amount of time.
function trampoline (t) {
while (t && t.isBounce)
t = t.f (t.x);
return t.x;
}
function bounce (f, x) {
return { isBounce: true, f: f, x: x };
}
function done (x) {
return { isBounce: false, x: x };
}
function repeat (n, f, x) {
function aux (n, x) {
if (n === 0)
return done (x);
else
return bounce (function (x) { return aux (n - 1, x); }, f (x));
}
return trampoline (aux (n, x));
}
console.time('JS1 100K');
console.log (repeat (1e5, function (x) { return x + 1 }, 0));
console.timeEnd('JS1 100K');
// 100000
// JS1 100K: 15ms
console.time('JS1 100M');
console.log (repeat (1e8, function (x) { return x + 1 }, 0));
console.timeEnd('JS1 100M');
// 100000000
// JS1 100K: 5999ms
Non-blocking infinite recursion using stone age JavaScript
If, for some reason, you want non-blocking (asynchronous) infinite recursion, we can rely on setTimeout to defer a single frame at the start of the computation. This program also uses stone age javascript and computes 100,000,000 recursions in under 8 seconds, but this time in a non-blocking way.
This demonstrates that there's nothing special about having a non-blocking requirement. A while loop and first-class functions are still the only fundamental requirement to achieve stack-safe recursion without sacrificing performance
In a modern program, given Promises, we would substitute the setTimeout call for a single Promise.
function donek (k, x) {
return { isBounce: false, k: k, x: x };
}
function bouncek (f, x) {
return { isBounce: true, f: f, x: x };
}
function trampolinek (t) {
// setTimeout is called ONCE at the start of the computation
// NOT once per recursion
return setTimeout(function () {
while (t && t.isBounce) {
t = t.f (t.x);
}
return t.k (t.x);
}, 0);
}
// stack-safe infinite recursion, non-blocking, 100,000,000 recursions in under 8 seconds
// now repeatk expects a 4th-argument callback which is called with the asynchronously computed result
function repeatk (n, f, x, k) {
function aux (n, x) {
if (n === 0)
return donek (k, x);
else
return bouncek (function (x) { return aux (n - 1, x); }, f (x));
}
return trampolinek (aux (n, x));
}
console.log('non-blocking line 1')
console.time('non-blocking JS1')
repeatk (1e8, function (x) { return x + 1; }, 0, function (result) {
console.log('non-blocking line 3', result)
console.timeEnd('non-blocking JS1')
})
console.log('non-blocking line 2')
// non-blocking line 1
// non-blocking line 2
// [ synchronous program stops here ]
// [ below this line, asynchronous program continues ]
// non-blocking line 3 100000000
// non-blocking JS1: 7762ms

A better loop/recur pattern
There are two things that I dislike about the loop/recur pattern described in the accepted answer. Note that I actually like the idea behind the loop/recur pattern. However, I dislike the way it has been implemented. So, let's first look at the way I would have implemented it.
// Recur :: a -> Result a b
const Recur = (...args) => ({ recur: true, args });
// Return :: b -> Result a b
const Return = value => ({ recur: false, value });
// loop :: (a -> Result a b) -> a -> b
const loop = func => (...args) => {
let result = func(...args);
while (result.recur) result = func(...result.args);
return result.value;
};
// repeat :: (Int, a -> a, a) -> a
const repeat = loop((n, f, x) => n === 0 ? Return(x) : Recur(n - 1, f, f(x)));
console.time("loop/recur/return");
console.log(repeat(1e6, x => x + 1, 0));
console.timeEnd("loop/recur/return");
Compare this with the loop/recur pattern described in the aforementioned answer.
// recur :: a -> Recur a
const recur = (...args) => ({ recur, args });
// loop :: (a? -> Recur a ∪ b) -> b
const loop = func => {
let result = func();
while (result && result.recur === recur) result = func(...result.args);
return result;
};
// repeat :: (Int, a -> a, a) -> a
const repeat = (n, f, x) => loop((m = n, r = x) => m === 0 ? r : recur(m - 1, f(r)));
console.time("loop/recur");
console.log(repeat(1e6, x => x + 1, 0));
console.timeEnd("loop/recur");
If you notice, the type signature of the second loop function uses default parameters (i.e. a?) and untagged unions (i.e. Recur a ∪ b). Both of these features are at odds with the functional programming paradigm.
Problem with default parameters
The loop/recur pattern in the aforementioned answer uses default parameters to supply the initial arguments of the function. I think this is an abuse of default parameters. You can just as easily supply initial arguments using my version of loop.
// repeat :: (Int, a -> a, a) -> a
const repeat = (n, f, x) => loop((n, x) => n === 0 ? Return(x) : Recur(n - 1, f(x)))(n, x);
// or more readable
const repeat = (n, f, x) => {
const repeatF = loop((n, x) => n === 0 ? Return(x) : Recur(n - 1, f(x)));
return repeatF(n, x);
};
Futhermore, it allows eta conversion when the all the arguments are passed through.
// repeat :: (Int, a -> a, a) -> a
const repeat = (n, f, x) => loop((n, f, x) => n === 0 ? Return(x) : Recur(n - 1, f, f(x)))(n, f, x);
// can be η-converted to
const repeat = loop((n, f, x) => n === 0 ? Return(x) : Recur(n - 1, f, f(x)));
Using the version of loop with default parameters does not allow eta conversion. In addition, it forces you to rename parameters because you can't write (n = n, x = x) => ... in JavaScript.
Problem with untagged unions
Untagged unions are bad because they erase important information, i.e. information of where the data came from. For example, because my Result type is tagged I can distinguish Return(Recur(0)) from Recur(0).
On the other hand, because the right-hand side variant of Recur a ∪ b is untagged, if b is specialized to Recur a, i.e. if the type is specialized to Recur a ∪ Recur a, then it's impossible to determine whether the Recur a came from the left-hand side or the right-hand side.
One criticism might be that b will never be specialized to Recur a, and hence it doesn't matter that b is untagged. Here's a simple counterexample to that criticism.
// recur :: a -> Recur a
const recur = (...args) => ({ recur, args });
// loop :: (a? -> Recur a ∪ b) -> b
const loop = func => {
let result = func();
while (result && result.recur === recur) result = func(...result.args);
return result;
};
// repeat :: (Int, a -> a, a) -> a
const repeat = (n, f, x) => loop((m = n, r = x) => m === 0 ? r : recur(m - 1, f(r)));
// infinite loop
console.log(repeat(1, x => recur(1, x), "wow, such hack, much loop"));
// unreachable code
console.log("repeat wasn't hacked");
Compare this with my version of repeat which is bulletproof.
// Recur :: a -> Result a b
const Recur = (...args) => ({ recur: true, args });
// Return :: b -> Result a b
const Return = value => ({ recur: false, value });
// loop :: (a -> Result a b) -> a -> b
const loop = func => (...args) => {
let result = func(...args);
while (result.recur) result = func(...result.args);
return result.value;
};
// repeat :: (Int, a -> a, a) -> a
const repeat = loop((n, f, x) => n === 0 ? Return(x) : Recur(n - 1, f, f(x)));
// finite loop
console.log(repeat(1, x => Recur(1, x), "wow, such hack, much loop"));
// reachable code
console.log("repeat wasn't hacked");
Thus, untagged unions are unsafe. However, even if we were careful to avoid the pitfalls of untagged unions I would still prefer tagged unions because the tags provide useful information when reading and debugging the program. IMHO, the tags make the program more understandable and easier to debug.
Conclusion
To quote the Zen of Python.
Explicit is better than implicit.
Default parameters and untagged unions are bad because they are implicit, and can lead to ambiguities.
The Trampoline monad
Now, I'd like to switch gears and talk about monads. The accepted answer demonstrates a stack-safe continuation monad. However, if you only need to create a monadic stack-safe recursive function then you don't need the full power of the continuation monad. You can use the Trampoline monad.
The Trampoline monad is a more powerful cousin of the Loop monad, which is just the loop function converted into a monad. So, let's start by understanding the Loop monad. Then we'll see the main problem of the Loop monad and how the Trampoline monad can be used to fix that problem.
// Recur :: a -> Result a b
const Recur = (...args) => ({ recur: true, args });
// Return :: b -> Result a b
const Return = value => ({ recur: false, value });
// Loop :: (a -> Result a b) -> a -> Loop b
const Loop = func => (...args) => ({ func, args });
// runLoop :: Loop a -> a
const runLoop = ({ func, args }) => {
let result = func(...args);
while (result.recur) result = func(...result.args);
return result.value;
};
// pure :: a -> Loop a
const pure = Loop(Return);
// bind :: (Loop a, a -> Loop b) -> Loop b
const bind = (loop, next) => Loop(({ first, loop: { func, args } }) => {
const result = func(...args);
if (result.recur) return Recur({ first, loop: { func, args: result.args } });
if (first) return Recur({ first: false, loop: next(result.value) });
return result;
})({ first: true, loop });
// ack :: (Int, Int) -> Loop Int
const ack = (m, n) => {
if (m === 0) return pure(n + 1);
if (n === 0) return ack(m - 1, 1);
return bind(ack(m, n - 1), n => ack(m - 1, n));
};
console.log(runLoop(ack(3, 4)));
Note that loop has been split into a Loop and a runLoop function. The data structure returned by Loop is a monad, and the pure and bind functions implement its monadic interface. We use the pure and bind functions to write a straightforward implementation of the Ackermann function.
Unfortunately, the ack function is not stack safe because it recursively calls itself until it reaches a pure value. Instead, we would like ack to return a Recur like data structure for its inductive cases. However, Recur values are of type Result instead of Loop. This problem is solved by the Trampoline monad.
// Bounce :: (a -> Trampoline b) -> a -> Trampoline b
const Bounce = func => (...args) => ({ bounce: true, func, args });
// Return :: a -> Trampoline a
const Return = value => ({ bounce: false, value });
// trampoline :: Trampoline a -> a
const trampoline = result => {
while (result.bounce) result = result.func(...result.args);
return result.value;
};
// pure :: a -> Trampoline a
const pure = Return;
// bind :: (Trampoline a, a -> Trampoline b) -> Trampoline b
const bind = (first, next) => first.bounce ?
Bounce(args => bind(first.func(...args), next))(first.args) :
next(first.value);
// ack :: (Int, Int) -> Trampoline Int
const ack = Bounce((m, n) => {
if (m === 0) return pure(n + 1);
if (n === 0) return ack(m - 1, 1);
return bind(ack(m, n - 1), n => ack(m - 1, n));
});
console.log(trampoline(ack(3, 4)));
The Trampoline data type is a combination of Loop and Result. The Loop and Recur data constructors have been combined into a single Bounce data constructor. The runLoop function has been simplified and renamed to trampoline. The pure and bind functions have also been simplified. In fact, pure is just Return. Finally, we apply Bounce to the original implementation of the ack function.
Another advantage of Trampoline is that it can be used to define stack-safe mutually recursive functions. For example, here's an implementation of the Hofstadter Female and Male sequence functions.
// Bounce :: (a -> Trampoline b) -> a -> Trampoline b
const Bounce = func => (...args) => ({ bounce: true, func, args });
// Return :: a -> Trampoline a
const Return = value => ({ bounce: false, value });
// trampoline :: Trampoline a -> a
const trampoline = result => {
while (result.bounce) result = result.func(...result.args);
return result.value;
};
// pure :: a -> Trampoline a
const pure = Return;
// bind :: (Trampoline a, a -> Trampoline b) -> Trampoline b
const bind = (first, next) => first.bounce ?
Bounce(args => bind(first.func(...args), next))(first.args) :
next(first.value);
// female :: Int -> Trampoline Int
const female = Bounce(n => n === 0 ? pure(1) :
bind(female(n - 1), f =>
bind(male(f), m =>
pure(n - m))));
// male :: Int -> Trampoline Int
const male = Bounce(n => n === 0 ? pure(0) :
bind(male(n - 1), m =>
bind(female(m), f =>
pure(n - f))));
console.log(Array.from({ length: 21 }, (_, n) => trampoline(female(n))).join(" "));
console.log(Array.from({ length: 21 }, (_, n) => trampoline(male(n))).join(" "));
The major pain point of writing monadic code is callback hell. However, this can be solved using generators.
// Bounce :: (a -> Trampoline b) -> a -> Trampoline b
const Bounce = func => (...args) => ({ bounce: true, func, args });
// Return :: a -> Trampoline a
const Return = value => ({ bounce: false, value });
// trampoline :: Trampoline a -> a
const trampoline = result => {
while (result.bounce) result = result.func(...result.args);
return result.value;
};
// pure :: a -> Trampoline a
const pure = Return;
// bind :: (Trampoline a, a -> Trampoline b) -> Trampoline b
const bind = (first, next) => first.bounce ?
Bounce(args => bind(first.func(...args), next))(first.args) :
next(first.value);
// bounce :: (a -> Generator (Trampoline b)) -> a -> Trampoline b
const bounce = func => Bounce((...args) => {
const gen = func(...args);
const next = data => {
const { value, done } = gen.next(data);
return done ? value : bind(value, next);
};
return next(undefined);
});
// female :: Int -> Trampoline Int
const female = bounce(function* (n) {
return pure(n ? n - (yield male(yield female(n - 1))) : 1);
});
// male :: Int -> Trampoline Int
const male = bounce(function* (n) {
return pure(n ? n - (yield female(yield male(n - 1))) : 0);
});
console.log(Array.from({ length: 21 }, (_, n) => trampoline(female(n))).join(" "));
console.log(Array.from({ length: 21 }, (_, n) => trampoline(male(n))).join(" "));
Finally, mutually recursive functions also demonstrate the advantage of having a separate trampoline function. It allows us to call a function returning a Trampoline value without actually running it. This allows us to build bigger Trampoline values, and then run the entire computation when required.
Conclusion
If you're want to write indirectly or mutually recursive stack-safe functions, or monadic stack-safe functions then use the Trampoline monad. If you want to write non-monadic directly recursive stack-safe functions then use the loop/recur/return pattern.

Programming in the sense of the functional paradigm means that we are guided by types to express our algorithms.
To transform a tail recursive function into a stack-safe version we have to consider two cases:
base case
recursive case
We have to make a choice and this goes well with tagged unions. However, Javascript doesn't have such a data type so either we have to create one or fall back to Object encodings.
Object Encoded
// simulate a tagged union with two Object types
const Loop = x =>
({value: x, done: false});
const Done = x =>
({value: x, done: true});
// trampoline
const tailRec = f => (...args) => {
let step = Loop(args);
do {
step = f(Loop, Done, step.value);
} while (!step.done);
return step.value;
};
// stack-safe function
const repeat = n => f => x =>
tailRec((Loop, Done, [m, y]) => m === 0
? Done(y)
: Loop([m - 1, f(y)])) (n, x);
// run...
const inc = n =>
n + 1;
console.time();
console.log(repeat(1e6) (inc) (0));
console.timeEnd();
Function Encoded
Alternatively, we can create a real tagged union with a function encoding. Now our style is much closer to mature functional languages:
// type/data constructor
const Type = Tcons => (tag, Dcons) => {
const t = new Tcons();
t.run = cases => Dcons(cases);
t.tag = tag;
return t;
};
// tagged union specific for the case
const Step = Type(function Step() {});
const Done = x =>
Step("Done", cases => cases.Done(x));
const Loop = args =>
Step("Loop", cases => cases.Loop(args));
// trampoline
const tailRec = f => (...args) => {
let step = Loop(args);
do {
step = f(step);
} while (step.tag === "Loop");
return step.run({Done: id});
};
// stack-safe function
const repeat = n => f => x =>
tailRec(step => step.run({
Loop: ([m, y]) => m === 0 ? Done(y) : Loop([m - 1, f(y)]),
Done: y => Done(y)
})) (n, x);
// run...
const inc = n => n + 1;
const id = x => x;
console.log(repeat(1e6) (inc) (0));

See also unfold which (from Ramda docs)
Builds a list from a seed value. Accepts an iterator function, which
returns either false to stop iteration or an array of length 2
containing the value to add to the resulting list and the seed to be
used in the next call to the iterator function.
var r = n => f => x => x > n ? false : [x, f(x)];
var repeatUntilGreaterThan = n => f => R.unfold(r(n)(f), 1);
console.log(repeatUntilGreaterThan(10)(x => x + 1));
<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.22.1/ramda.min.js"></script>

I've been thinking about this question a lot. Recently I came across the need for a functional while loop.
It seems to me like the only thing this question really wants is a way to inline a while loop. There IS a way to do that using a closure.
"some string "+(a=>{
while(comparison){
// run code
}
return result;
})(somearray)+" some more"
Alternatively, if what you want is something that chains off an array, than you can use the reduce method.
somearray.reduce((r,o,i,a)=>{
while(comparison){
// run code
}
a.splice(1); // This would ensure only one call.
return result;
},[])+" some more"
None of this actually turns our while loop at its core into a function. But it does allow us the use of an inline loop. And I just wanted to share this with anyone who this might help.