[FEATURE]: Add Digital Root Algorithm to JavaScript/Maths and its recursive implementation #1747

Motivation

Digital root algorithm is a very simple but powerful tool used in various fields such as checksum calculations (e.g., ISBN validation), cyclic redundancy checks, and understanding number properties in modular arithmetic. Plus, it offers a great way for beginners to learn and understand recursion problems (if its recursive implementation is also included)

Adding this feature helps people learn both the mathematical approach (constant time complexity) and recursive thinking. It would be useful for anyone interested in algorithms and number theory.

Examples

1. For n = 15:

   • Calculation:

     digital_root(15) = 1 + (15 - 1) % 9
                      = 1 + 14 % 9
                      = 1 + 5
                      = 6
     

   • Output: 6
   • Explanation: The digital root of 15 is 6 because the sum of its digits 1 + 5 = 6.

2. For n = 123:

   • Calculation:

     digital_root(123) = 1 + (123 - 1) % 9
                       = 1 + 122 % 9
                       = 1 + 5
                       = 6
     

   • Output: 6
   • Explanation: The digital root of 123 is 6 because the sum of its digits 1 + 2 + 3 = 6.

Possible workarounds

No response

Additional information

Algorithm overview

The digital root of a number is the value obtained by repeatedly summing the digits of the number until a single digit is reached.
This issue proposes two implementations

1. Best-case constant time complexity algorithm
   This calculates the digital root with constant time complexity using modulo arithmetic.

export function digitalRoot(num) {
  if (num < 0) num = -num;
  return num === 0 ? num : 1 + ((num - 1) % 9);
}

2. Recursive Implementation
   This has been proposed as a learning tool rather than for practical application. This recursive solution can help beginners in understanding recursion, and help them delve further into algorithms.

export function digitalRoot(num) {
  if (num < 0) num = -num;
  if (num < 10) return num;
  const sum = (num % 10) + digitalRoot(Math.floor(num / 10));
  return sum >= 9 ? sum - 9 : sum;
}

Also, if kindly assign this to me if this is accepted for implementation