Checkerboard pattern without x and y (purely based on index) - algorithm

Question

Is there a way to generate a checkerboard pattern without using nested for loops and without using x and y?
I'm sure this has already been done before, but I couldn't find it in the ~15mins I was looking for it.
Currently I have this function that generates the pattern first extracting x and y:
fn get_bg_color_of(idx: usize) -> &'static str {
const BG_BLACK: &str = "\u{001b}[48;5;126m";
const BG_WHITE: &str = "\u{001b}[48;5;145m";
let x = idx % Board::WIDTH;
let y = idx / Board::WIDTH;
let is_even_row = y % 2 == 0;
let is_even_column = x % 2 == 0;
if is_even_row && is_even_column || !is_even_row && !is_even_column {
return BG_WHITE;
}
BG_BLACK
}
Is there a simpler way to do this? If yes, please also explain how and why, I like to know what's happening in my code :)

Answer 1

If WIDTH is even, then you need to separate x and y. You can write that shorter, though:
fn get_bg_color_of(idx: usize) -> &'static str {
const BG_BLACK: &str = "\u{001b}[48;5;126m";
const BG_WHITE: &str = "\u{001b}[48;5;145m";
if ( (idx + (idx/Board::WIDTH)) % 2 == 0 ) {
return BG_WHITE;
}
return BG_BLACK;
}
Note that this doesn't work if WIDTH is odd. In that case, you can just do:
if ( idx % 2 == 0 ) {
return BG_WHITE;
}
}
If you need to handle both cases, then:
if ( ((idx%Board::WIDTH) + (idx/Board::WIDTH)) % 2 == 0 ) {
return BG_WHITE;
}