N queens math (TheAlgorithms#2175)

* add new file for another solution to the n queens problem

* Add the code for the algorithm, add comments and add at the top a general explanation

* Update backtracking/n_queens_math.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update backtracking/n_queens_math.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update backtracking/n_queens_math.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update backtracking/n_queens_math.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* No newline at the end of the file

* Type hints

* whitespaces fixed

* Fixed whitespaces

* Add type hints

* CodeSpell fixed

* update

* All changes made except changing the board variable to local

* Add doctest

* Update

* Update

* Update

* Update n_queens_math.py

Co-authored-by: Christian Clauss <cclauss@me.com>

Author: DavidBanda

backtracking/n_queens_math.py

@@ -0,0 +1,165 @@
+r"""
+Problem:
+
+The n queens problem is of placing N queens on a N * N chess board such that no queen
+can attack any other queens placed on that chess board.  This means that one queen
+cannot have any other queen on its horizontal, vertical and diagonal lines.
+
+Solution:
+
+To solve this problem we will use simple math. First we know the queen can move in all
+the possible ways, we can simplify it in this: vertical, horizontal, diagonal left and
+ diagonal right.
+
+We can visualize it like this:
+
+left diagonal = \
+right diagonal = /
+
+On a chessboard vertical movement could be the rows and horizontal movement could be
+the columns.
+
+In programming we can use an array, and in this array each index could be the rows and
+each value in the array could be the column. For example:
+
+    . Q . .     We have this chessboard with one queen in each column and each queen
+    . . . Q     can't attack to each other.
+    Q . . .     The array for this example would look like this: [1, 3, 0, 2]
+    . . Q .
+
+So if we use an array and we verify that each value in the array is different to each
+other we know that at least the queens can't attack each other in horizontal and
+vertical.
+
+At this point we have that halfway completed and we will treat the chessboard as a
+Cartesian plane.  Hereinafter we are going to remember basic math, so in the school we
+learned this formula:
+
+    Slope of a line:
+
+           y2 - y1
+     m = ----------
+          x2 - x1
+
+This formula allow us to get the slope. For the angles 45º (right diagonal) and 135º
+(left diagonal) this formula gives us m = 1, and m = -1 respectively.
+
+See::
+https://www.enotes.com/homework-help/write-equation-line-that-hits-origin-45-degree-1474860
+
+Then we have this another formula:
+
+Slope intercept:
+
+y = mx + b
+
+b is where the line crosses the Y axis (to get more information see:
+https://www.mathsisfun.com/y_intercept.html), if we change the formula to solve for b
+we would have:
+
+y - mx = b
+
+And like we already have the m values for the angles 45º and 135º, this formula would
+look like this:
+
+45º: y - (1)x = b
+45º: y - x = b
+
+135º: y - (-1)x = b
+135º: y + x = b
+
+y = row
+x = column
+
+Applying this two formulas we can check if a queen in some position is being attacked
+for another one or vice versa.
+
+"""
+from typing import List
+
+
+def depth_first_search(
+    possible_board: List[int],
+    diagonal_right_collisions: List[int],
+    diagonal_left_collisions: List[int],
+    boards: List[List[str]],
+    n: int,
+) -> None:
+    """
+    >>> boards = []
+    >>> depth_first_search([], [], [], boards, 4)
+    >>> for board in boards:
+    ...     print(board)
+    ['. Q . . ', '. . . Q ', 'Q . . . ', '. . Q . ']
+    ['. . Q . ', 'Q . . . ', '. . . Q ', '. Q . . ']
+    """
+
+    """ Get next row in the current board (possible_board) to fill it with a queen """
+    row = len(possible_board)
+
+    """
+    If row is equal to the size of the board it means there are a queen in each row in
+    the current board (possible_board)
+    """
+    if row == n:
+        """
+        We convert the variable possible_board that looks like this: [1, 3, 0, 2] to
+        this: ['. Q . . ', '. . . Q ', 'Q . . . ', '. . Q . ']
+        """
+        possible_board = [". " * i + "Q " + ". " * (n - 1 - i) for i in possible_board]
+        boards.append(possible_board)
+        return
+
+    """ We iterate each column in the row to find all possible results in each row """
+    for col in range(n):
+
+        """
+        We apply that we learned previously. First we check that in the current board
+        (possible_board) there are not other same value because if there is it means
+        that there are a collision in vertical. Then we apply the two formulas we
+        learned before:
+
+        45º: y - x = b or 45: row - col = b
+        135º: y + x = b or row + col = b.
+
+        And we verify if the results of this two formulas not exist in their variables
+        respectively.  (diagonal_right_collisions, diagonal_left_collisions)
+
+        If any or these are True it means there is a collision so we continue to the
+        next value in the for loop.
+        """
+        if (
+            col in possible_board
+            or row - col in diagonal_right_collisions
+            or row + col in diagonal_left_collisions
+        ):
+            continue
+
+        """ If it is False we call dfs function again and we update the inputs """
+        depth_first_search(
+            possible_board + [col],
+            diagonal_right_collisions + [row - col],
+            diagonal_left_collisions + [row + col],
+            boards,
+            n,
+        )
+
+
+def n_queens_solution(n: int) -> None:
+    boards = []
+    depth_first_search([], [], [], boards, n)
+
+    """ Print all the boards """
+    for board in boards:
+        for column in board:
+            print(column)
+        print("")
+
+    print(len(boards), "solutions were found.")
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()
+    n_queens_solution(4)