Daily Challenge #264 - Digital Root

Digital root is the recursive sum of all the digits in a number. Given n, take the sum of the digits of n. If that value has more than one digit, continue reducing in this way until a single-digit number is produced. This is only applicable to the natural numbers.

Examples

16  -->  1 + 6 = 7
942  -->  9 + 4 + 2 = 15  -->  1 + 5 = 6
132189  -->  1 + 3 + 2 + 1 + 8 + 9 = 24  -->  2 + 4 = 6
493193  -->  4 + 9 + 3 + 1 + 9 + 3 = 29  -->  2 + 9 = 11 --> 1 + 1 = 2

Tests

digital_root(456)
digital_root(16)

Good luck!

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This challenge comes from user578387 on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

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Comments

[peter279k]

Here is the Python code snippets with nest while loop:

def digital_root(n):
    result = 0
    n = str(n)

    index = 0
    while len(n) != 1:
        while index < len(n):
            result += int(n[index])
            index += 1

        index = 0
        n = str(result)
        result = 0

    return int(n)

[Priyesh Naik]

Python 3 (indent 3 is not reflecting in comment)

ip_val = input("Enter the number?")

if ip_val.isdigit():
while len(ip_val) > 1:
res = 0
for e in ip_val:
res += int(e)
ip_val = str(res)
print(ip_val)

[Ayan Banerjee]

Property of the digital root is that the result would be n % 9 and n % 9 == 0 it will be 9.

int digitalRoot(int n) {
    if (n == 0) return 0;
    if (n % 9 == 0) return 9;
    return n % 9;
}

[agtoever]

Python 3 using direct calculation. Defined as function with test cases in the tio link.

digital_root = lambda n: 0 if n == 0 else 1 + (n - 1) % 9

Try it online!

[Rafael Acioly]

why 1 + (n - 1)?

[Jean-Michel Plourde]

I tested it with just n and it works fine.

[Hayden Mankin]

It's important to remember that the % operator happens before +, this solves the edge cases around numbers divisible by 9 without needing extra if statements/ternaries.

For instance,

18 -> 1 + 8 = 9

however,

18 % 9 = 0

which does not fit the problem specification, so instead we do

(18 - 1) % 9 + 1 = 17 % 9 + 1 = 8 + 1 = 9

[Amin]

Haskell

sumDigits :: Int -> Int
sumDigits 0 = 0
sumDigits n = (mod n 10) + sumDigits (div n 10)

digitalRoot :: Int -> Int
digitalRoot n
    | n > 9     = digitalRoot (sumDigits n)
    | otherwise = n

[Hayden Mankin]

I've always liked this problem because of the n%9 trick, it's such a neat mathematical property.

I thought I'd do the obligatory n%9 solution:

let digital_root = (n) => n == 0 ? 0 : (n - 1) % 9 + 1;

console.log(digital_root(456)); // 6
console.log(digital_root(16));  // 7

As well as a one line recursive solution that doesn't use the n%9 trick:

let digital_root = (n) => (n < 10) ? n : digital_root(Array.from(n.toString()).map(Number).reduce((a, b)=>a+b));

console.log(digital_root(456)); // 6
console.log(digital_root(16));  // 7

[Jean-Michel Plourde]

Here's my solution in Python. Not bad considering I gave it only 5 minutes. I could have gone with reduce and lambda, but it's not really pythonic, not fast and the reading would be terrible. I like the short circuit behavior.

def digital_root(n):
  while True:
    n_list = [int(d) for d in str(n)]
    if len(n_list) == 1:
      return n
    n = sum(n_list)

[JMW]

n % 9 is such an interesting property! But since I had no knowledge of it, I went with a naive solution in KOTLIN which I cracked within a few minutes:

fun digitalRoot(number: Int): Int {
    return if(number < 10) {
        number
    } else {
        val digits = number.toString().split("")
        var sum = 0
        digits.forEach {
            val digit = it.toIntOrNull()
            if(digit != null) { sum += digit}
        }
        digitalRoot(sum)
    }
}

[Mirza]

Here's my solution:

def digital_root(num):
    num = str(num)

    if len(num) == 1:
        return int(num)
    else:
        temp = str(int(num[0]) + int(num[1]))
        return digital_root(temp + num[2: len(num)])

[Viclv4]

def digital_root(number):
    n = number
    while len(str(n)) != 1:
        n = sum([int(i) for i in str(n)])
    return n