Question

Task 3:N ns Brute For In mathematics, the notion of permutation relates to the act of arranging all the members of a set into some sequence or order, or if the set is already ordered, rearranging (reordering) its elements, a process called permuting. The number of permutations on a set of n elements is given by n! (Read as n factorial). For example, there are 2!2 x 1- 2 permutations of 11,2), 2,1) and 3!-3x2x16 permutations of (1,2,3), 1,3,2], 13,1,2), 13,2,1), (2,3,1], 2,1,3) A brute force algorithm for N-queens problem goes as follow:s . Generate all possible permutations of 1,2,.. , Ni,... , N. Set a counter to 0 . For each permutation place the queens on the board . Test if this configuration is valid; if so, increment the counter. Implement the brute force algorithm for 4-queens problem For example, your program may output Enter the size of N: 4 Number of solutions: 2 00 Q0 Q000 000Q 0Q00 0Q00 000Q Q000 00 Q0

python code，please！

Answer 1

import itertools

def isValidPermutation(permutation): # checks whether the given permutation is valid N queens placement or not
canPlace = [] # keeps track of whether a queen can be placed on the board of not
N = len(permutation)
for i in range(N):
row = []
for j in range(N):
row.append(True)
canPlace.append(row)
for i in range(N): # iterate through all rows
if canPlace[i][permutation[i]] is False:
return False
for j in range(i + 1, N): # iterate through all rows below current row
canPlace[j][permutation[i]] = False # set the cell directly below the current column to be False
if permutation[i] - (j - i) >= 0: # set the cell on the left diagonal to be False
canPlace[j][permutation[i] - (j - i)] = False
if permutation[i] + (j - i) < N: # set the cell on the right diagonal to be False
canPlace[j][permutation[i] + (j-i)] = False
return True

def printBoard(permutation): # prints the board for the given permutation
N = len(permutation)
for i in range(N):
for j in range(N):
if j == permutation[i]:
print 'Q',
else:
print '0',
print ''

N = input("Enter the size of N: ")
permutations = itertools.permutations(range(N)) # get all the permutations
valid_permutations = []
valid_count = 0
for permutation in permutations:
if isValidPermutation(permutation):
valid_count += 1
valid_permutations.append(permutation)
print 'Number of solutions:', valid_count
print ''
for permutation in valid_permutations:
printBoard(permutation)
print ''

Enter the size of N: 4 Number of solutions: 2 0 Q 0 0 0 0 Q 0 0 0 Q 0 Q 0 0 0 0 0 0 Q 0 Q 0 0

import itertools def isValidPermutation(permutation): # checks whether the given permutation is valid N queens placement or not # keeps track of whether a queen can be placed on the board of not canPlace[] N len(permutation) for i in range(N): row] for j in range(N): row.append (True) canPlace.append ( row) for i in range(N) if canPlace[i] [permutation [i]] is False: return False # iterate through aïï rows for j in range(i +1, N) # iterate through all rows below current row canPlace[j] [permutation [i]] = False # set the cell directly below the current column to be False permutation [i] - (j - 1) > canPlace [jl [permutation[i] -(j -i) False if permutation[i] + (j ) N: canPlace[jl [permutation [i] + (j-i)1 -False # set the cell on the left diagonal to be False # set the cell on the right diagonal to be False return True ef printBoard(permutation): # prints the boa rd for the given permutation N len(permutation) for i in range (N): for j in range(N) if j permutation [1]: print 'Q' else: print '0' print N input ("Enter the size of N:") permutations itertools.permutations (range(N)) valid_permutations [l valid_count for permutation in permutations: if isValidPermutation(permutation): # get all the permutations valid_count 1 valid_permutations.append (permutation) print 'Number of solutions: '", valid_count print ' for permutation in valid_permutations: printBoard(permutation) print'