1024HardBacktrackingAlgorithm X

Counting the Safe Beacon Layouts

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A square yard is laid out as n rows by n columns. A beacon placed in a cell sweeps its whole row, its whole column, and both diagonals running through it, for any distance.

Place n beacons on the yard so that no beacon stands in another's sweep. Return how many arrangements manage it.

Examples

Example 1

Input: n = 2
Output: 0

On a two-by-two yard any two cells share a row, a column or a diagonal, so nothing works.

Example 2

Input: n = 6
Output: 4

Four arrangements fit on a six-by-six yard, and each is a turn or a mirror image of the others.

Example 3

Input: n = 9
Output: 352

Three hundred and fifty-two arrangements fit on a nine-by-nine yard.

Constraints

1 <= n <= 9

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def total_n_queens(n: int) -> int:

Java

public int totalNQueens(int n)

The intended approach

Time O(n!) loosely, cut down heavily by abandoning rows early.Space O(n) for the walk's depth.

There are as many beacons as rows and no two may share a row, so every row holds exactly one beacon. That turns the job into choosing a column for each row in turn. A column is open for the row in hand when no earlier row took that column and no earlier beacon lies on either diagonal through the cell. The diagonals have a tidy name: cells on the same falling diagonal all share their row less their column, and cells on the same rising diagonal all share their row plus their column. Keep three sets of bits, one per kind, and a column is open when none of its three bits is set. Placing a beacon sets those three bits and lifting it clears them again, so the walk tries every arrangement in turn and abandons a row the moment no column is open. Reaching past the last row counts one arrangement.

Pythonentry point: total_n_queens

def total_n_queens(n):
    total = 0
    def place(row, cols, falling, rising):
        nonlocal total
        if row == n:
            total += 1
            return
        for col in range(n):
            down = row - col + n
            up = row + col
            if cols >> col & 1:
                continue
            if falling >> down & 1:
                continue
            if rising >> up & 1:
                continue
            place(row + 1, cols | 1 << col, falling | 1 << down, rising | 1 << up)
    place(0, 0, 0, 0)
    return total

The obvious version

Since each row holds one beacon and no two share a column, an arrangement is just a way of ordering the columns. Run through every ordering of the columns and check each pair of beacons for a shared diagonal, counting the orderings that survive. Costs every ordering in full, with no rows abandoned early.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_total_n_queens(n):
    from itertools import permutations
    total = 0
    for order in permutations(range(n)):
        ok = True
        for a in range(n):
            for b in range(a + 1, n):
                if order[a] - order[b] == a - b or order[a] - order[b] == b - a:
                    ok = False
                    break
            if not ok:
                break
        if ok:
            total += 1
    return total

How this problem was checked

Reference solution executed against 9 test cases — 9 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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