CS50AI - Crossword - Solution
If you have doubts refer this.Don't try to copy and answer.
Name: CS50's Introduction to Artificial Intelligence with Python
University: Harvard University
URL: https://cs50.harvard.edu/ai/2020/
My GitHub: https://github.com/Dinesh-Kumar-E
Automatic CS50 Submit Tool: https://github.com/Dinesh-Kumar-E/CS50-Submit
See: https://cs50.harvard.edu/college/2021/fall/syllabus/#academic-honesty
generate.py
import sys
import copy
from crossword import *
class CrosswordCreator():
def __init__(self, crossword):
"""
Create new CSP crossword generate.
"""
self.crossword = crossword
self.domains = {
var: self.crossword.words.copy()
for var in self.crossword.variables
}
def letter_grid(self, assignment):
"""
Return 2D array representing a given assignment.
"""
letters = [
[None for _ in range(self.crossword.width)]
for _ in range(self.crossword.height)
]
for variable, word in assignment.items():
direction = variable.direction
for k in range(len(word)):
i = variable.i + (k if direction == Variable.DOWN else 0)
j = variable.j + (k if direction == Variable.ACROSS else 0)
letters[i][j] = word[k]
return letters
def print(self, assignment):
"""
Print crossword assignment to the terminal.
"""
letters = self.letter_grid(assignment)
for i in range(self.crossword.height):
for j in range(self.crossword.width):
if self.crossword.structure[i][j]:
print(letters[i][j] or " ", end="")
else:
print("█", end="")
print()
def save(self, assignment, filename):
"""
Save crossword assignment to an image file.
"""
from PIL import Image, ImageDraw, ImageFont
cell_size = 100
cell_border = 2
interior_size = cell_size - 2 * cell_border
letters = self.letter_grid(assignment)
# Create a blank canvas
img = Image.new(
"RGBA",
(self.crossword.width * cell_size,
self.crossword.height * cell_size),
"black"
)
font = ImageFont.truetype("assets/fonts/OpenSans-Regular.ttf", 80)
draw = ImageDraw.Draw(img)
for i in range(self.crossword.height):
for j in range(self.crossword.width):
rect = [
(j * cell_size + cell_border,
i * cell_size + cell_border),
((j + 1) * cell_size - cell_border,
(i + 1) * cell_size - cell_border)
]
if self.crossword.structure[i][j]:
draw.rectangle(rect, fill="white")
if letters[i][j]:
w, h = draw.textsize(letters[i][j], font=font)
draw.text(
(rect[0][0] + ((interior_size - w) / 2),
rect[0][1] + ((interior_size - h) / 2) - 10),
letters[i][j], fill="black", font=font
)
img.save(filename)
def solve(self):
"""
Enforce node and arc consistency, and then solve the CSP.
"""
self.enforce_node_consistency()
self.ac3()
return self.backtrack(dict())
def enforce_node_consistency(self):
"""
Update `self.domains` such that each variable is node-consistent.
(Remove any values that are inconsistent with a variable's unary
constraints; in this case, the length of the word.)
"""
# do the deep copy of domains
domain_copy = copy.deepcopy(self.domains)
# iterate through domains copy
for variable in domain_copy:
# getting the variable length
length = variable.length
# iterate through words in domain
for word in domain_copy[variable]:
if len(word) != length:
# if length of the word doesn't fit variable, delete it from
# the original domain (not copy)
self.domains[variable].remove(word)
def revise(self, x, y):
"""
Make variable `x` arc consistent with variable `y`.
To do so, remove values from `self.domains[x]` for which there is no
possible corresponding value for `y` in `self.domains[y]`.
Return True if a revision was made to the domain of `x`; return
False if no revision was made.
"""
# getting x and y ovelapping cells, unpack cords to variables
xoverlap, yoverlap = self.crossword.overlaps[x, y]
# make variable describing if revision was made
revision_made = False
# making domains copy
domains_copy = copy.deepcopy(self.domains)
# if overlap occurs
if xoverlap:
# iterate through words in x's domain
for xword in domains_copy[x]:
matched_value = False
# iterate through words in y's domain
for yword in self.domains[y]:
# if x's word and y's word have same letter in overlapping position
if xword[xoverlap] == yword[yoverlap]:
matched_value = True
break # no need to check rest of y's words for that x
if matched_value:
continue # if x and y was matched, proceed with another x
else:
self.domains[x].remove(xword) # no matching y's word to x, removing word from domain
revision_made = True
# return bolean if revision was made
return revision_made
def ac3(self, arcs=None):
"""
Update `self.domains` such that each variable is arc consistent.
If `arcs` is None, begin with initial list of all arcs in the problem.
Otherwise, use `arcs` as the initial list of arcs to make consistent.
Return True if arc consistency is enforced and no domains are empty;
return False if one or more domains end up empty.
"""
if not arcs:
# no arcs provided, start with an initial queue of all of the arcs in the problem
queue = []
# populating queue
for variable1 in self.domains:
for variable2 in self.crossword.neighbors(variable1):
if self.crossword.overlaps[variable1, variable2] is not None:
queue.append((variable1, variable2))
while len(queue) > 0:
x, y = queue.pop(0)
if self.revise(x, y):
if len(self.domains[x]) == 0:
return False
for neighbour in self.crossword.neighbors(x):
if neighbour != y:
queue.append((neighbour, x))
return True
def assignment_complete(self, assignment):
"""
Return True if `assignment` is complete (i.e., assigns a value to each
crossword variable); return False otherwise.
"""
for variable in self.domains:
if variable not in assignment:
return False
return True
def consistent(self, assignment):
"""
Return True if `assignment` is consistent (i.e., words fit in crossword
puzzle without conflicting characters); return False otherwise.
"""
# all values are distinct, every value is the correct length,
# and there are no conflicts between neighboring variables.
# check if all values are distinct
words = [*assignment.values()]
if len(words) != len(set(words)):
return False
# check if every value is the correct length
for variable in assignment:
if variable.length != len(assignment[variable]):
return False
# check if there are any conflicts between neighbouring variables
for variable in assignment:
for neighbour in self.crossword.neighbors(variable):
if neighbour in assignment:
x, y = self.crossword.overlaps[variable, neighbour]
if assignment[variable][x] != assignment[neighbour][y]:
return False
# all cases checked, no conflicts, can return True
return True
def order_domain_values(self, var, assignment):
"""
Return a list of values in the domain of `var`, in order by
the number of values they rule out for neighboring variables.
The first value in the list, for example, should be the one
that rules out the fewest values among the neighbors of `var`.
"""
# make temporary dict for holding values
word_dict = {}
# getting neighbours of var
neighbours = self.crossword.neighbors(var)
# iterating through var's words
for word in self.domains[var]:
eliminated = 0
for neighbour in neighbours:
# don't count if neighbor has already assigned value
if neighbour in assignment:
continue
else:
# calculate overlap between two variables
xoverlap, yoverlap = self.crossword.overlaps[var, neighbour]
for neighbour_word in self.domains[neighbour]:
# iterate through neighbour's words, check for eliminate ones
if word[xoverlap] != neighbour_word[yoverlap]:
eliminated += 1
# add eliminated neighbour's words to temporary dict
word_dict[word] = eliminated
# sort variables dictionary by number of eliminated neighbour values
sorted_dict = {k: v for k, v in sorted(word_dict.items(), key=lambda item: item[1])}
return [*sorted_dict]
def select_unassigned_variable(self, assignment):
"""
Return an unassigned variable not already part of `assignment`.
Choose the variable with the minimum number of remaining values
in its domain. If there is a tie, choose the variable with the highest
degree. If there is a tie, any of the tied variables are acceptable
return values.
"""
choice_dict = {}
# iterating through variables in domains
for variable in self.domains:
# iterating through variables in assignment
if variable not in assignment:
# if variable is not yet in assigment, add it to temp dict
choice_dict[variable] = self.domains[variable]
# make list of variables sorted by number of remaining values
sorted_list = [v for v, k in sorted(choice_dict.items(), key=lambda item:len(item[1]))]
# return variable with the minimum number of remaining values
return sorted_list[0]
def backtrack(self, assignment):
"""
Using Backtracking Search, take as input a partial assignment for the
crossword and return a complete assignment if possible to do so.
`assignment` is a mapping from variables (keys) to words (values).
If no assignment is possible, return None.
"""
# if assignment is already ready
if len(assignment) == len(self.domains):
return assignment
# selecting one of unassigned variables
variable = self.select_unassigned_variable(assignment)
# iterating through words in that variable
for value in self.domains[variable]:
# making assignment copy, with updated variable value
assignment_copy = assignment.copy()
assignment_copy[variable] = value
# checking for consistency, getting result of that new assignment backtrack
if self.consistent(assignment_copy):
result = self.backtrack(assignment_copy)
if result is not None:
return result
return None
def main():
# Check usage
if len(sys.argv) not in [3, 4]:
sys.exit("Usage: python generate.py structure words [output]")
# Parse command-line arguments
structure = sys.argv[1]
words = sys.argv[2]
output = sys.argv[3] if len(sys.argv) == 4 else None
# Generate crossword
crossword = Crossword(structure, words)
creator = CrosswordCreator(crossword)
assignment = creator.solve()
# Print result
if assignment is None:
print("No solution.")
else:
creator.print(assignment)
if output:
creator.save(assignment, output)
if __name__ == "__main__":
main()
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