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eight queens problem in Python

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python

8-queens problem in Python.

Hi! I only start teaching Python, so could someone explain the code written below (found in the Internet)? Some pieces of the code are complicated for me. Please, explain them. Thank you. Questions are near the code.

BOARD_SIZE = 8

def under_attack(col, queens): # (col, queens) What is their meaning? What do I need to write it this field? 
    left = right = col
    for r, c in reversed(queens): # What does reversed means in this loop? For what reson do we need r and c (their meaning is 0 by default?)?
        left, right = left-1, right+1
        if c in (left, col, right):
            return True
    return False

def solve(n):
    if n == 0: return [[]]
    smaller_solutions = solve(n-1) # For what reasons do we need to write smaller_solutions?
    return [solution+[(n,i+1)] # What is solution (is it a function or what?)? What is value of i? 
        for i in range(BOARD_SIZE)
            for solution in smaller_solutions
                if not under_attack(i+1, solution)]
for answer in solve(BOARD_SIZE): print answer

Thank you!

like image 808
Bob Avatar asked Mar 14 '26 14:03

Bob


2 Answers

Your code is wrong (cut and paste error?), but here's the gist:

You want a list of possible solutions. Each solution is a list of queens. Every queen is a tuple - a row (integer) and column (integer). For example, the solution for BOARD_SIZE=1 is [[(1,1)]] - a single solution - [(1,1)] containing a single queen - (1,1) placed on row 1 and column 1.

There are 8 smaller_solutions for BOARD_SIZE=8, and n=1 - [[(1,1)],[(1,2)],[(1,3)],[(1,4)],[(1,5)],[(1,6)],[(1,7)],[(1,8)]] - a single queen placed in every column in the first row.

You understand recursion? If not, google it NOW.

Basically, you start by adding 0 queens to a size 0 board - this has one trivial solution - no queens. Then you find the solutions that place one queen the first row of the board. Then you look for solutions which add a second queen to the 2nd row - somewhere that it's not under attack. And so on.

def solve(n):
    if n == 0: return [[]] # No RECURSION if n=0. 
    smaller_solutions = solve(n-1) # RECURSION!!!!!!!!!!!!!!
    solutions = []
    for solution in smaller_solutions:# I moved this around, so it makes more sense
        for column in range(1,BOARD_SIZE+1): # I changed this, so it makes more sense
            # try adding a new queen to row = n, column = column 
            if not under_attack(column , solution): 
                solutions.append(solution + [(n,column)])
    return solutions

That explains the general strategy, but not under_attack.

under_attack could be re-written, to make it easier to understand (for me, you, and your students):

def under_attack(column, existing_queens):
    # ASSUMES that row = len(existing_queens) + 1
    row = len(existing_queens)+1
    for queen in existing_queens:
        r,c = queen
        if r == row: return True # Check row
        if c == column: return True # Check column
        if (column-c) == (row-r): return True # Check left diagonal
        if (column-c) == -(row-r): return True # Check right diagonal
    return False

My method is a little slower, but not much.

The old under_attack is basically the same, but it speeds thing up a bit. It looks through existing_queens in reverse order (because it knows that the row position of the existing queens will keep counting down), keeping track of the left and right diagonal.

like image 93
wisty Avatar answered Mar 16 '26 03:03

wisty


BOARD_SIZE = 8

def under_attack(col, queens): # You do not need to fill in these fields. This is a helper function for the solve function.
    left = right = col
    for r, c in reversed(queens): # Reversing queens causes them to be iterated over in reverse order.
        left, right = left-1, right+1
        if c in (left, col, right):
            return True
    return False

def solve(n):
    if n == 0: return [[]]
    smaller_solutions = solve(n-1) # It appears that in solving this board, it solves all boards smaller than it in a recursive manner.
    return [solution+[(n,i+1)] # This line appears to be in error. Have you run this code and verified that it runs correctly?
        for i in range(BOARD_SIZE)
            for solution in smaller_solutions
                if not under_attack(i+1, solution)]
for answer in solve(BOARD_SIZE): print answer
like image 34
Noctis Skytower Avatar answered Mar 16 '26 02:03

Noctis Skytower



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