N Queens in Python using backtracking
I write this solution to the popular N Queens problem using backtracking algorithm. I am relatively new to Python. I would like to know what are the ways to refactor this code and also code style of Python in general.
def isSafe (board, row, col):
# check left row
for y in range(col):
if board[row][y] == 1:
return False
# check diagonal left top
for x, y in zip(range(row, -1, -1), range(col, -1, -1)):
if board[x][y] == 1:
return False
# check diagonal left bottom
for x, y in zip(range(row, N, 1), range(col, -1, -1)):
if board[x][y] == 1:
return False
return True
def generateSolution(board, col):
# terminating condition
# all columns covered
global N
if col >= N:
return True
# loop over all the rows
for i in range(N):
if isSafe(board, i, col) == True:
board[i][col] = 1
# recursively place other queens
if generateSolution(board, col + 1) == True:
return True
# unmark queen spot
board[i][col] = 0
# backtrack
return False
N = int(input())
startCol = 0
board = [[0 for i in range(N)] for j in range(N)]
# print(board)
if generateSolution(board, startCol) == False:
print("No Solution Exists")
else:
print("Solution exists")
print(board)
python algorithm python-3.x backtracking n-queens
New contributor
add a comment |
I write this solution to the popular N Queens problem using backtracking algorithm. I am relatively new to Python. I would like to know what are the ways to refactor this code and also code style of Python in general.
def isSafe (board, row, col):
# check left row
for y in range(col):
if board[row][y] == 1:
return False
# check diagonal left top
for x, y in zip(range(row, -1, -1), range(col, -1, -1)):
if board[x][y] == 1:
return False
# check diagonal left bottom
for x, y in zip(range(row, N, 1), range(col, -1, -1)):
if board[x][y] == 1:
return False
return True
def generateSolution(board, col):
# terminating condition
# all columns covered
global N
if col >= N:
return True
# loop over all the rows
for i in range(N):
if isSafe(board, i, col) == True:
board[i][col] = 1
# recursively place other queens
if generateSolution(board, col + 1) == True:
return True
# unmark queen spot
board[i][col] = 0
# backtrack
return False
N = int(input())
startCol = 0
board = [[0 for i in range(N)] for j in range(N)]
# print(board)
if generateSolution(board, startCol) == False:
print("No Solution Exists")
else:
print("Solution exists")
print(board)
python algorithm python-3.x backtracking n-queens
New contributor
add a comment |
I write this solution to the popular N Queens problem using backtracking algorithm. I am relatively new to Python. I would like to know what are the ways to refactor this code and also code style of Python in general.
def isSafe (board, row, col):
# check left row
for y in range(col):
if board[row][y] == 1:
return False
# check diagonal left top
for x, y in zip(range(row, -1, -1), range(col, -1, -1)):
if board[x][y] == 1:
return False
# check diagonal left bottom
for x, y in zip(range(row, N, 1), range(col, -1, -1)):
if board[x][y] == 1:
return False
return True
def generateSolution(board, col):
# terminating condition
# all columns covered
global N
if col >= N:
return True
# loop over all the rows
for i in range(N):
if isSafe(board, i, col) == True:
board[i][col] = 1
# recursively place other queens
if generateSolution(board, col + 1) == True:
return True
# unmark queen spot
board[i][col] = 0
# backtrack
return False
N = int(input())
startCol = 0
board = [[0 for i in range(N)] for j in range(N)]
# print(board)
if generateSolution(board, startCol) == False:
print("No Solution Exists")
else:
print("Solution exists")
print(board)
python algorithm python-3.x backtracking n-queens
New contributor
I write this solution to the popular N Queens problem using backtracking algorithm. I am relatively new to Python. I would like to know what are the ways to refactor this code and also code style of Python in general.
def isSafe (board, row, col):
# check left row
for y in range(col):
if board[row][y] == 1:
return False
# check diagonal left top
for x, y in zip(range(row, -1, -1), range(col, -1, -1)):
if board[x][y] == 1:
return False
# check diagonal left bottom
for x, y in zip(range(row, N, 1), range(col, -1, -1)):
if board[x][y] == 1:
return False
return True
def generateSolution(board, col):
# terminating condition
# all columns covered
global N
if col >= N:
return True
# loop over all the rows
for i in range(N):
if isSafe(board, i, col) == True:
board[i][col] = 1
# recursively place other queens
if generateSolution(board, col + 1) == True:
return True
# unmark queen spot
board[i][col] = 0
# backtrack
return False
N = int(input())
startCol = 0
board = [[0 for i in range(N)] for j in range(N)]
# print(board)
if generateSolution(board, startCol) == False:
print("No Solution Exists")
else:
print("Solution exists")
print(board)
python algorithm python-3.x backtracking n-queens
python algorithm python-3.x backtracking n-queens
New contributor
New contributor
edited 8 mins ago
Jamal♦
30.2k11116226
30.2k11116226
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asked 14 mins ago
Abrar
1205
1205
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