Question

In problem 3, you will write a function, findvertically (), which will determine whether or not a given word exists in a word search puzzle vertically. In word search puzzles, words can be written upwards or downwards. For example, "BEAK" appears at [1] [3] written downwards, while "BET" appears at [2] [2] written upwards: [["C", "A", "T", "X", "R"], ["D", "T", "E", "B", "L"], ["A" "R" "B", "E", "Z"], ["X", "O", "E", "A", "U"], ["J", "S", "O", "K", "W"]] Your function must look through each column for the word. If it is found, it must return the row and column at which the word starts and the direction in which the word is written. Details for function findvertically () Parameter Variables (Input) • puzzle (list): The word search puzzle (a table-a list of lists—with the same number of rows as columns). • word (string): The word to find written vertically in the puzzle (all words will be uppercase). Processing • For each column in puzzle: o If you find word written downwards (top-to-bottom, like "BEAK" is in the example above), stop searching. o If you find word written uowards (bottom-to-top, like "BET" is in the example above), stop searching. o Otherwise, move to the next column in puzzle.

Answer 1

def findVertically(puzzle, word):

global DIR_U, DIR_D

# If the length of word is greater than
# the puzzle no need to check
if len(word) == 0 or len(word) > len(puzzle):
return None
# Taking the transpose of the puzzle so that,
# we can process the words horizontally.
# Each row in puzzle becomes column in puzzle_transpose.
# Each column in puzzle becomes row in puzzle_transpose.
puzzle_transpose = [list(i) for i in zip(*puzzle)]
    
for row in puzzle_transpose:

# Row in puzzle_transpose is column of puzzle,
# hence the row index of puzzle_transpose gives column index in puzzle
# Column in puzzle_transpose is row of puzzle,
# hence the column index of puzzle_transpose gives row index in puzzle.
col_index = puzzle_transpose.index(row)
    
# Find word in row of puzzle_transpose in down direction.
row_str = ''.join(row)
row_index = row_str.find(word)
# Find word in row of puzzle_transpose in up direction by,
# searching for the reverse of word.
rev_row_index = row_str.find(word[::-1])

if row_index >= 0:
return {"row": row_index, "column":col_index, "direction": DIR_D}
elif rev_row_index >= 0:
# Correcting the index as the direction is from bottom to up.
return {"row": len(puzzle) - rev_row_index, "column":col_index, "direction": DIR_U}

return None

Please refer to the below screenshot of the above code if, there are any indent issues.

The output of the above script is as follows:

def findvertically(puzzle, word): # If the length of word is greater than # the puzzle no need to check if len(word) == 0 or len(word) > len(puzzle): return None # Taking the transpose of the puzzle so that, # we can process the words horizontally. # Each row in puzzle becomes column in puzzle transpose. # Each column in puzzle becomes row in puzzle_transpose. puzzle_transpose = [list(i) for i in zip(*puzzle)] for row in puzzle_transpose: # Row in puzzle transpose is column of puzzle, # hence the row index of puzzle_transpose gives column index in puzzle # Column in puzzle transpose is row of puzzle, # hence the column index of puzzle_transpose gives row index in puzzle. col_index = puzzle_transpose. index (row) # Find word in row of puzzle_transpose in down direction. row_str = ".join(row) row_index = row_str.find(word) # Find word in row of puzzle_transpose in up direction by, # searching for the reverse of word. rev_row_index = row_str.find(word[::-1]) if row index >= 0; return {"row": row_index, "column":col_index, "direction": "DIR_D"} elif rev_row_index >= 0: # Correcting the index as the direction is from bottom to up. return {"row": len(puzzle) - rev_row_index, "column":col_index, "direction": "DIR_U"} return None

puzzle = ["0", "W", "E", "R"], "W", "0", "R", "D"], ["z", "R", "C", "v"], ["V", "N", "N", "M"]] print(findvertically (puzzle, word="WORN")) print(findvertically (puzzle, word="ROW')) print(findvertically(puzzle, word="LIKE")) {'row': 0, 'column': 1, 'direction': 'DIR_D'} {'row': 4, 'column': 1, 'direction': 'DIR_U'} None

If you have any queries please free to comment.