def doBacktracking(self, sudoku, moveStack):
empties = sudoku.getEmptySquares()
if (len(empties) == 0):
self.backTrackingResult = sudoku
return True
else:
(i, j) = empties[0]
domain = sudoku.getLegalMoves(i, j)
moveStack.append(((i, j), sudoku.copy()))
for value in domain:
sudoku.setSquare(i, j, value)
self.backTrackingCounter += 1
if self.isSolvableByAC(sudoku):
self.backTrackingResult = sudoku
return True
if not sudoku.hasEmptyDomain():
if self.doBacktracking(sudoku, moveStack):
return True
sudoku = moveStack[-1][1].copy()
moveStack.pop()
return False
def doBacktracking(self, sudoku, moveStack):
empties = sudoku.getEmptySquares()
if len(empties) == 0:
self.backTrackingResult = sudoku
return True
else:
(i, j) = empties[0]
domain = sudoku.getLegalMoves(i, j)
moveStack.append(((i, j), sudoku.copy()))
for value in domain:
sudoku.setSquare(i, j, value)
self.backTrackingCounter += 1
if self.isSolvableByAC(sudoku):
self.backTrackingResult = sudoku
return True
if not sudoku.hasEmptyDomain():
if self.doBacktracking(sudoku, moveStack):
return True
sudoku = moveStack[-1][1].copy()
moveStack.pop()
return False
def feature_vector(self, sudoku):
vector = list()
#Add the level of the puzzle
vector.append(sudoku.getIntLevel())
#there are 7 features in each list that is returned
#1) the number of given squares
given_squares = 81 - len(sudoku.getEmptySquares())
vector.append(given_squares)
#4) maximum number of squares given in a set (where set is defined as row, column or 3 by 3 square)
given_in_set = self.feature_4(sudoku)
vector.append(given_in_set)
#6) maximum number of rows or columns filled in a square
max_row_or_cols = self.feature_5(sudoku)
vector.append(max_row_or_cols)
#7) maximum number of squares where a digit is in the domain.
max_in_domain = self.feature_6(sudoku)
vector.append(max_in_domain)
#2) the number of domain variables eliminated in the first 10 rules
eliminated_domains = self.arc_consistency(sudoku, 10)
size_of_domain_difference = eliminated_domains[0] - eliminated_domains[
len(eliminated_domains) - 1]
vector.append(size_of_domain_difference)
#5) maximum number of squares that can be filled by a single digit after 10 rounds of arc consistency
squares_filled = self.feature_7(sudoku)
sudoku.reset()
vector.append(squares_filled)
#3) solvable with arc consistency
self.arcConsistencyCounter = 0
solvable = self.solvableByAC(sudoku)
if solvable:
solvable = self.arcConsistencyCounter
else:
solvable = 0
vector.append(solvable)
sudoku.reset()
#8) how many times we have to backtrack to solve
self.doBacktracking(sudoku, [])
vector.append(self.backTrackingCounter)
self.backTrackingCounter = 0
return vector
def feature_vector(self, sudoku):
vector = list()
# Add the level of the puzzle
vector.append(sudoku.getIntLevel())
# there are 7 features in each list that is returned
# 1) the number of given squares
given_squares = 81 - len(sudoku.getEmptySquares())
vector.append(given_squares)
# 4) maximum number of squares given in a set (where set is defined as row, column or 3 by 3 square)
given_in_set = self.feature_4(sudoku)
vector.append(given_in_set)
# 6) maximum number of rows or columns filled in a square
max_row_or_cols = self.feature_5(sudoku)
vector.append(max_row_or_cols)
# 7) maximum number of squares where a digit is in the domain.
max_in_domain = self.feature_6(sudoku)
vector.append(max_in_domain)
# 2) the number of domain variables eliminated in the first 10 rules
eliminated_domains = self.arc_consistency(sudoku, 10)
size_of_domain_difference = eliminated_domains[0] - eliminated_domains[len(eliminated_domains) - 1]
vector.append(size_of_domain_difference)
# 5) maximum number of squares that can be filled by a single digit after 10 rounds of arc consistency
squares_filled = self.feature_7(sudoku)
sudoku.reset()
vector.append(squares_filled)
# 3) solvable with arc consistency
self.arcConsistencyCounter = 0
solvable = self.solvableByAC(sudoku)
if solvable:
solvable = self.arcConsistencyCounter
else:
solvable = 0
vector.append(solvable)
sudoku.reset()
# 8) how many times we have to backtrack to solve
self.doBacktracking(sudoku, [])
vector.append(self.backTrackingCounter)
self.backTrackingCounter = 0
return vector
