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Solving the Digital Root Algorithm using JavaScript

isabel k. on May 17, 2020

One of my favorite algorithms is finding the digital root of any given integer. A digital root is a single-digit sum that is reached when you itera...

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Kobe Raypole • May 17 '20

Very cool! I'm a Ruby enthusiast so I tried it out in that.

Your Iterative strategy was great but I think using recursion could help reduce some of the complexity and make for some cleaner code.

def d_root(num)
  return num if num < 10

  d_root(num.to_s.split('').reduce(0) {|acc, digit| acc + digit.to_i}) 
end

Ruby gives us some great helpers for integers so we can simplify this even more.

def d_root(num)
  return num if num < 10

  d_root(num.digits.sum) 
end

Nice and simple!

Duy K. Bui • May 17 '20

Can I cheat?

function d_root(num) {
  return (num % 9) || 9;
}

Pavitra Golchha • May 17 '20

I am really curious how do you know this?

Duy K. Bui • May 17 '20

I learnt it first as a math trick for children in Vietnam: to know if something is divisible by 9, you calculate its digitial root and see if the digital root is divisible by 9 (same goes for 3).

When I grew up, I was curious and tried to prove that mathematically, which is how I ended up with digitalRoot(num) = num % 9 (or 9 if divisible by 9) :D

Felix Terkhorn • May 17 '20

mathworld.wolfram.com/DigitalRoot....

Looking good in this corner of the thread!

Django Janny • May 18 '20

What about d_root(0) ?

Athul Anil Kumar • May 18 '20

Amazing intuition!

Kushal Niroula • May 18 '20

Are we sure this is a recursive approach? I feel like this is an iterative solution as we just run a block until a condition is met. The function is not called multiple times.

function dRoot(num) {
  const parts = num.toString().split("");
  const sum = parts.reduce((aggr,part) => parseInt(part) + parseInt(aggr));
  if(sum > 9) return dRoot(sum);
  return sum;
}

I think this qualifies as recursive. But the iterative approach in the article is much more performant I think. Nice article.

Phil • May 17 '20

Thank you! I like those tiny little programming "snacks" :)

shortest (recursion) solution I could find:

const dr = r => r < 10 ? r : dr([...r+''].reduce((b,c) => +b+(+c)));

dr(129); // 3

Raymond Price • May 17 '20

Cool demo, but that's not actually recursion, that's just iteration. An actual recursive solution, if you wanted to do that for some strange reason, would look more like

function digitalRoot(number) {
   let arr = number.toString().split("");
   let reducer = (a,b) => parseInt(a) + parseInt(b);
   let sum = arr.reduce(reducer)

   if (sum > 9) {
       return digitalRoot(sum);
   } else {
       return sum;
   }
}

Also, sidenote, this function does break on very large numbers. That's not OP's fault, that's because Javascript can't reasonably represent things like 1234567891234567891234567891 as an integer. digitalRoot("1234567891234567891234567891") works just fine, though