# 5 Best Ways to Find the Super Digit of a Number in Python

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π‘ Problem Formulation: The task is to calculate the “super digit” of a given number. In mathematics, the super digit of a number is the single-digit result obtained by recursively summing the digits until only one digit remains. For example, if the input is `9875`, the super digit is `2` because `9+8+7+5 = 29` and `2+9 = 11` and finally `1+1 = 2`.

## Method 1: Iterative Approach

This iterative approach simply repeats the process of summing the digits of a number until a single digit is left. The function specification involves a while loop that continues this process until the condition of having one digit is met.

Here’s an example:

```def super_digit_iterative(n):
while len(str(n)) > 1:
n = sum(int(digit) for digit in str(n))
return n

print(super_digit_iterative(9875))```

Output: `2`

This method takes a number, converts it into a string to iterate over each digit, sums the digits, and continues the process until only a single digit remains.

## Method 2: Recursive Approach

In the recursive approach, a function calls itself with the sum of the digits of the given number, continuing until the base case of a single digit number is reached.

Here’s an example:

```def super_digit_recursive(n):
if n < 10:
return n
else:
return super_digit_recursive(sum(int(digit) for digit in str(n)))

print(super_digit_recursive(9875))```

Output: `2`

This snippet defines a recursive function that adds the digits of the number until it is a single digit. It is a classic example of a recursive function with a base case and a recursive case.

## Method 3: Using Divmod

Divmod is a built-in Python function that returns the quotient and the remainder when dividing two numbers. It’s used here to break down the number digit by digit in an iterative super digit finding algorithm.

Here’s an example:

```def super_digit_divmod(n):
while n >= 10:
n = sum(divmod(n, 10))
return n

print(super_digit_divmod(9875))```

Output: `2`

The function employs divmod to decompose the number into individual digits and sums them up in an iterative manner, until a single-digit is achieved.

## Method 4: Using Mathematics

Mathematically, the super digit can be found using a congruence formula that takes advantage of modulo operation properties, specifically for 9. This method avoids loops and recursive calls.

Here’s an example:

```def super_digit_math(n):
if n == 0:
return 0
if n % 9 == 0:
return 9
else:
return n % 9

print(super_digit_math(9875))```

Output: `2`

This method makes use of the congruence property (all numbers have the same remainder when divided by 9 as their sum of digits) to find the super digit in constant time, without explicit digit separation or recursion.

## Bonus One-Liner Method 5: Using a Single Line Python Expression

The single line Python expression condenses the calculation of a super digit into a one-liner by combining recursion and the digit summation inside the Python interpreter’s evaluation of arithmetic expression.

Here’s an example:

```super_digit_oneliner = lambda n: n if n < 10 else super_digit_oneliner(sum(int(digit) for digit in str(n)))
print(super_digit_oneliner(9875))```

Output: `2`

This one-liner defines a lambda function with a recursive call embedded within it. Lambda functions are a concise way to write functions in one line and often used for simple tasks.

## Summary/Discussion

• Method 1: Iterative Approach. Straightforward and intuitive. Potentially slower for very large numbers due to iterative processing.
• Method 2: Recursive Approach. Elegant use of recursion. May cause a stack overflow for large input numbers due to the recursive calls.
• Method 3: Using Divmod. Makes use of built-in divmod function for digit extraction. Slower for large numbers but more legible than arithmetic hacks.
• Method 4: Using Mathematics. Fast and efficient, works in constant time. However, less straightforward and requires understanding of modulo arithmetic.
• Method 5: One-Liner. Very concise. However, may be less readable and also susceptible to stack overflow for very large numbers due to recursion.