Mastering the Maze: Is it Solvable in Just One Line?
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Chapter 1: Understanding the Maze
In this section, we delve into the concept of mazes and how to determine their solvability. A maze is represented as a list of strings, each containing four distinct symbols:
- S: Starting point
- X: Goal
- -: Open path
- #: Wall
To win, the player must navigate from S to X, moving only in four directions—up, down, left, or right—one step at a time. The player can traverse only the open paths (-) and cannot walk through walls (#).
Here are two examples of mazes:
maze1 = [
'S-#--',
'--#--',
'-#---',
'----X'
]
maze2 = [
'S-#--',
'--#--',
'-#---',
'#---X'
]
In this scenario, maze1 is solvable because a path exists from S to X, while maze2 is not solvable due to the walls obstructing the path.
This video explores the challenge of determining whether a maze can be solved in a single line of code.
Section 1.1: The Challenge
Your task is to create a function called is_solvable(maze) that accepts a maze and returns True if it can be solved or False if it cannot. Before jumping into a one-liner solution, consider implementing a multi-line version first.
Here’s a multi-line implementation to start with:
def valid_coordinate(new_row, new_col, maze):
if new_row < 0 or new_col < 0:
return Falsetry:
symbol = maze[new_row][new_col]
return symbol in ['-', 'X']
except:
return False
def is_solvable(maze):
seen = set()
queue = [(0, 0)]
while len(queue) > 0:
print(queue)
row, col = queue.pop(0)
seen.add((row, col))
# Check if at the goal
if maze[row][col] == 'X':
return True
# Add unseen neighbors to the queue
for a, b in [(0, 1), (1, 0), (0, -1), (-1, 0)]:
new_row = row + a
new_col = col + b
if not valid_coordinate(new_row, new_col, maze):
continueif (new_row, new_col) not in seen:
queue.append((new_row, new_col))return False
This implementation checks all possible paths to find whether X can be reached from S.
Section 1.2: From Multi-line to One-liner
Transitioning to a one-liner is not straightforward with our current setup. However, utilizing recursion allows for a more compact implementation.
Here’s a recursive version of the function:
def is_solvable(maze, seen=set(), r=0, c=0):
seen.update([(r, c), maze[r][c]])
for dr, dc in [(0, 1), (1, 0), (-1, 0), (0, -1)]:
nr = dr + r
nc = dc + c
if nr >= 0 and nc >= 0 and (nr, nc) not in seen and maze[nr][nc] != '#':
if maze[nr][nc] == 'X' or is_solvable(maze, seen, nr, nc):
return Truereturn False
Now, let's condense it into a one-liner, focusing on simplifying variable names and employing list comprehensions:
def is_solvable(M, S=set(), r=0, c=0):
S.update([(r, c), M[r][c]]);
return S[-1] == 'X' or any(is_solvable(M, S=S, r=dr+r, c=dc+c) for dr, dc in [(0, 1), (1, 0), (-1, 0), (0, -1)] if r + dr >= 0 and c + dc >= 0)
This allows us to determine if the maze is solvable in a single line.
The second video discusses finding Fibonacci numbers in one line, showcasing how concise coding can be approached.
Conclusion
Through this exploration, we've transformed a complex problem into a succinct solution. Mastering such techniques not only sharpens coding skills but also enhances problem-solving efficiency. If you found this guide helpful, consider sharing your thoughts or supporting the content.