Fizz Buzz
==========
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:
x := 1..=100
~y<[string]> := 1..=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:
ix3 := (x % 3) == 0
ix5 := (x % 5) == 0
Then we can use these indices to replace the values in the vector `y`:
y[ix3] = "✨"
y[ix5] = "🐝"
y[ix3 && ix5] = "✨🐝"
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:
```mech
y
```