The Fizz Buzz problem is a simple programming exercise that involves printing numbers from 1 to 100, but with a twist.

• For multiples of 3, print "Fizz" instead of the number,

• For multiples of 5 print "Buzz"

• For numbers that are multiples of both 3 and 5, print "FizzBuzz".

This is a common exercise in programming interviews to test a candidate's understanding of control flow and basic programming concepts.

In most imperative progrmaming languages you'd solve FizzBuzz like this:

for i in 1..100 {
  if (i % 3 == 0 && i % 5 == 0) {
    println("FizzBuzz");
  } else if (i % 3 == 0) {
    println("Fizz");
  } else if (i % 5 == 0) {
    println("Buzz");
  } else {
    println(i);
  }
}

One of the "tricks" of the problem is that when you implement it using a control-flow statement like if, you have to handle the "FizzBuzz" case first, otherwise you will print "Fizz" or "Buzz" instead of "FizzBuzz"; a naive implementor might code the conditions in the order they are presented in the problem statement, which would lead to incorrect results.

But in Mech we can achieve this in a more concise way. First get a vector of numbers from 1 to 100:

then replace the values based on the conditions specified using a feature called "logical indexing", which allows us to directly manipulate elements of a vector based on conditions. First we can create an index for all the multiples of 3 and 5:

Then we can use these indices to replace the values in the vector y:

The statement y[ix3] = "✨" is equivalent to saying "for all elements in y where the corresponding element in ix3 is true, set that element to '✨'".

Then we can print them all out at once: