Implementing the FizzBuzz Algorithm in Python

Implementation Overview

In the project sneiderrincon/pf-l3-fizzbuzz-individual, we are implementing the classic FizzBuzz algorithm. This exercise serves as a fundamental logic challenge, requiring us to iterate through numbers and apply conditional logic based on divisibility.

The Core Logic

To solve FizzBuzz, we need to check if a number is divisible by 3, 5, or both. Think of this like a traffic controller: instead of just letting cars pass, we check their license plates against specific rules before directing them to a destination lane.

def run_fizzbuzz(n):
    for i in range(1, n + 1):
        if i % 3 == 0 and i % 5 == 0:
            print("FizzBuzz")
        elif i % 3 == 0:
            print("Fizz")
        elif i % 5 == 0:
            print("Buzz")
        else:
            print(i)

This function iterates from 1 to n. By using the modulo operator (%), we determine the remainder of a division. If the remainder is zero, the number is evenly divisible. We check the condition for both (divisible by 15) first, because otherwise, a number like 15 would be caught by the 3 or 5 check before we could label it "FizzBuzz".

Why This Matters

Understanding modular arithmetic and conditional prioritization is essential for building more complex decision-making systems in your applications. It teaches developers how to structure logic so that specific, narrower cases are handled before broader, generic ones.

Actionable Takeaway

Try refactoring your next conditional chain to handle the most restrictive cases first. This "top-down" approach to validation keeps your code readable and prevents common logic bugs.

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