Ilya Nevolin

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# Road to Genius: superior #61

Each day I solve several coding challenges and puzzles from Codr's ranked mode. The goal is to reach genius rank, along the way I explain how I solve them. You do not need any programming background to get started, and you will learn a ton of new and interesting things as you go.

``````function TNQ(n) {
let res = 0;
const dfs = (n, row, cols, pie, na) => {
if (row >= n) {
res++;
return;
}
let bits = ~(cols | pie | na) & (1 << n) - ðŸ’°;
while (bits) {
let ðŸ’š = bits & -bits;
bits = bits & bits - 1;
dfs(n, row + 1, cols | ðŸš€, (pie | p) << 1, (na | p) >> 1);
}
};
dfs(n, 0, 0, 0, 0);
return res;
}
let A = TNQ(9);

// ðŸ’° = ? (number)
// ðŸš€ = ? (identifier)
// ðŸ’š = ? (identifier)
// such that A = 352 (number)
``````

I don't think we've encountered the function `TNQ` yet. I have no clue what it does nor how it works. But from the looks of it, it's involving a lot of bit operations. Let's get started.

The first bug appears here:

``````let bits = ~(cols | pie | na) & (1 << n) - ðŸ’°;
``````

I have no clue what ðŸ’° should be; the answers we can choose from are 0, 1 and 9. Subtracting zero from any number makes no sense, subtracting 9 is a bit odd as well, the most probable answer would be 1.

The second bug ðŸ’š is a variable name declaration, the only undeclared variable below that line is `p`.

The final bug appears on this line:

``````dfs(n, row + 1, cols | ðŸš€, (pie | p) << 1, (na | p) >> 1);
``````

The bug is a right-hand variable in an or-operation; this could be anything. But if you look at its neighboring arguments (to the right), both of them are using `p` as the right-hand variable in the or-operation as well. So let's try that out:

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