This is the fourth question from the Udacity CS212 final exam. I will update this article with my solution in 1 week. Please do not comment anything during the final exam week !!
"""
UNIT 4: Search
Your task is to maneuver a car in a crowded parking lot. This is a kind of
puzzle, which can be represented with a diagram like this:
| | | | | | | |
| G G . . . Y |
| P . . B . Y |
| P * * B . Y @
| P . . B . . |
| O . . . A A |
| O . S S S . |
| | | | | | | |
A '|' represents a wall around the parking lot, a '.' represents an empty square,
and a letter or asterisk represents a car. '@' marks a goal square.
Note that there are long (3 spot) and short (2 spot) cars.
Your task is to get the car that is represented by '**' out of the parking lot
(on to a goal square). Cars can move only in the direction they are pointing.
In this diagram, the cars GG, AA, SSS, and ** are pointed right-left,
so they can move any number of squares right or left, as long as they don't
bump into another car or wall. In this diagram, GG could move 1, 2, or 3 spots
to the right; AA could move 1, 2, or 3 spots to the left, and ** cannot move
at all. In the up-down direction, BBB can move one up or down, YYY can move
one down, and PPP and OO cannot move.
You should solve this puzzle (and ones like it) using search. You will be
given an initial state like this diagram and a goal location for the ** car;
in this puzzle the goal is the '.' empty spot in the wall on the right side.
You should return a path -- an alternation of states and actions -- that leads
to a state where the car overlaps the goal.
An action is a move by one car in one direction (by any number of spaces).
For example, here is a successor state where the AA car moves 3 to the left:
| | | | | | | |
| G G . . . Y |
| P . . B . Y |
| P * * B . Y @
| P . . B . . |
| O A A . . . |
| O . . . . . |
| | | | | | | |
And then after BBB moves 2 down and YYY moves 3 down, we can solve the puzzle
by moving ** 4 spaces to the right:
| | | | | | | |
| G G . . . . |
| P . . . . . |
| P . . . . * *
| P . . B . Y |
| O A A B . Y |
| O . . B . Y |
| | | | | | | |
You will write the function
solve_parking_puzzle(start, N=N)
where 'start' is the initial state of the puzzle and 'N' is the length of a side
of the square that encloses the pieces (including the walls, so N=8 here).
We will represent the grid with integer indexes. Here we see the
non-wall index numbers (with the goal at index 31):
| | | | | | | |
| 9 10 11 12 13 14 |
| 17 18 19 20 21 22 |
| 25 26 27 28 29 30 31
| 33 34 35 36 37 38 |
| 41 42 43 44 45 46 |
| 49 50 51 52 53 54 |
| | | | | | | |
The wall in the upper left has index 0 and the one in the lower right has 63.
We represent a state of the problem with one big tuple of (object, locations)
pairs, where each pair is a tuple and the locations are a tuple. Here is the
initial state for the problem above in this format:
"""
puzzle1 = (
('@', (31,)),
('*', (26, 27)),
('G', (9, 10)),
('Y', (14, 22, 30)),
('P', (17, 25, 33)),
('O', (41, 49)),
('B', (20, 28, 36)),
('A', (45, 46)),
('|', (0, 1, 2, 3, 4, 5, 6, 7, 8, 15, 16, 23, 24, 32, 39,
40, 47, 48, 55, 56, 57, 58, 59, 60, 61, 62, 63)))
# A solution to this puzzle is as follows:
# path = solve_parking_puzzle(puzzle1, N=8)
# path_actions(path) == [('A', -3), ('B', 16), ('Y', 24), ('*', 4)]
# That is, move car 'A' 3 spaces left, then 'B' 2 down, then 'Y' 3 down,
# and finally '*' moves 4 spaces right to the goal.
# Your task is to define solve_parking_puzzle:
N = 8
def solve_parking_puzzle(start, N=N):
"""Solve the puzzle described by the starting position (a tuple
of (object, locations) pairs). Return a path of [state, action, ...]
alternating items; an action is a pair (object, distance_moved),
such as ('B', 16) to move 'B' two squares down on the N=8 grid."""
# But it would also be nice to have a simpler format to describe puzzles,
# and a way to visualize states.
# You will do that by defining the following two functions:
def locs(start, n, incr=1):
"Return a tuple of n locations, starting at start and incrementing by incr."
def grid(cars, N=N):
"""Return a tuple of (object, locations) pairs -- the format expected for
this puzzle. This function includes a wall pair, ('|', (0, ...)) to
indicate there are walls all around the NxN grid, except at the goal
location, which is the middle of the right-hand wall; there is a goal
pair, like ('@', (31,)), to indicate this. The variable 'cars' is a
tuple of pairs like ('*', (26, 27)). The return result is a big tuple
of the 'cars' pairs along with the walls and goal pairs."""
def show(state, N=N):
"Print a representation of a state as an NxN grid."
# Initialize and fill in the board.
board = ['.'] * N**2
for (c, squares) in state:
for s in squares:
board[s] = c
# Now print it out
for i,s in enumerate(board):
print s,
if i % N == N - 1: print
# Here we see the grid and locs functions in use:
puzzle1 = grid((
('*', locs(26, 2)),
('G', locs(9, 2)),
('Y', locs(14, 3, N)),
('P', locs(17, 3, N)),
('O', locs(41, 2, N)),
('B', locs(20, 3, N)),
('A', locs(45, 2))))
puzzle2 = grid((
('*', locs(26, 2)),
('B', locs(20, 3, N)),
('P', locs(33, 3)),
('O', locs(41, 2, N)),
('Y', locs(51, 3))))
puzzle3 = grid((
('*', locs(25, 2)),
('B', locs(19, 3, N)),
('P', locs(36, 3)),
('O', locs(45, 2, N)),
('Y', locs(49, 3))))
# Here are the shortest_path_search and path_actions functions from the unit.
# You may use these if you want, but you don't have to.
def shortest_path_search(start, successors, is_goal):
"""Find the shortest path from start state to a state
such that is_goal(state) is true."""
if is_goal(start):
return [start]
explored = set() # set of states we have visited
frontier = [ [start] ] # ordered list of paths we have blazed
while frontier:
path = frontier.pop(0)
s = path[-1]
for (state, action) in successors(s).items():
if state not in explored:
explored.add(state)
path2 = path + [action, state]
if is_goal(state):
return path2
else:
frontier.append(path2)
return []
def path_actions(path):
"Return a list of actions in this path."
return path[1::2]
My solution:
"""
UNIT 4: Search
Your task is to maneuver a car in a crowded parking lot. This is a kind of
puzzle, which can be represented with a diagram like this:
| | | | | | | |
| G G . . . Y |
| P . . B . Y |
| P * * B . Y @
| P . . B . . |
| O . . . A A |
| O . S S S . |
| | | | | | | |
A '|' represents a wall around the parking lot, a '.' represents an empty square,
and a letter or asterisk represents a car. '@' marks a goal square.
Note that there are long (3 spot) and short (2 spot) cars.
Your task is to get the car that is represented by '**' out of the parking lot
(on to a goal square). Cars can move only in the direction they are pointing.
In this diagram, the cars GG, AA, SSS, and ** are pointed right-left,
so they can move any number of squares right or left, as long as they don't
bump into another car or wall. In this diagram, GG could move 1, 2, or 3 spots
to the right; AA could move 1, 2, or 3 spots to the left, and ** cannot move
at all. In the up-down direction, BBB can move one up or down, YYY can move
one down, and PPP and OO cannot move.
You should solve this puzzle (and ones like it) using search. You will be
given an initial state like this diagram and a goal location for the ** car;
in this puzzle the goal is the '.' empty spot in the wall on the right side.
You should return a path -- an alternation of states and actions -- that leads
to a state where the car overlaps the goal.
An action is a move by one car in one direction (by any number of spaces).
For example, here is a successor state where the AA car moves 3 to the left:
| | | | | | | |
| G G . . . Y |
| P . . B . Y |
| P * * B . Y @
| P . . B . . |
| O A A . . . |
| O . . . . . |
| | | | | | | |
And then after BBB moves 2 down and YYY moves 3 down, we can solve the puzzle
by moving ** 4 spaces to the right:
| | | | | | | |
| G G . . . . |
| P . . . . . |
| P . . . . * *
| P . . B . Y |
| O A A B . Y |
| O . . B . Y |
| | | | | | | |
You will write the function
solve_parking_puzzle(start, N=N)
where 'start' is the initial state of the puzzle and 'N' is the length of a side
of the square that encloses the pieces (including the walls, so N=8 here).
We will represent the grid with integer indexes. Here we see the
non-wall index numbers (with the goal at index 31):
| | | | | | | |
| 9 10 11 12 13 14 |
| 17 18 19 20 21 22 |
| 25 26 27 28 29 30 31
| 33 34 35 36 37 38 |
| 41 42 43 44 45 46 |
| 49 50 51 52 53 54 |
| | | | | | | |
The wall in the upper left has index 0 and the one in the lower right has 63.
We represent a state of the problem with one big tuple of (object, locations)
pairs, where each pair is a tuple and the locations are a tuple. Here is the
initial state for the problem above in this format:
"""
puzzle0 = (
('@', (31,)),
('*', (26, 27)),
('G', (9, 10)),
('Y', (14, 22, 30)),
('P', (17, 25, 33)),
('O', (41, 49)),
('B', (20, 28, 36)),
('A', (45, 46)),
('|', (0, 1, 2, 3, 4, 5, 6, 7, 8, 15, 16, 23, 24, 32, 39,
40, 47, 48, 55, 56, 57, 58, 59, 60, 61, 62, 63)))
# A solution to this puzzle is as follows:
# path = solve_parking_puzzle(puzzle1, N=8)
# path_actions(path) == [('A', -3), ('B', 16), ('Y', 24), ('*', 4)]
# That is, move car 'A' 3 spaces left, then 'B' 2 down, then 'Y' 3 down,
# and finally '*' moves 4 spaces right to the goal.
# Your task is to define solve_parking_puzzle:
N = 8
def solve_parking_puzzle(start, N=N):
"""Solve the puzzle described by the starting position (a tuple
of (object, locations) pairs). Return a path of [state, action, ...]
alternating items; an action is a pair (object, distance_moved),
such as ('B', 16) to move 'B' two squares down on the N=8 grid."""
return shortest_path_search(start, csuccessors, isGoal, N)
def csuccessors(state, N, goal):
if isGoal(state, goal):
return {}
cars = [car for car in 'ABCDEFGHIJKLMNOPQRSTUVWXYZ*']
successors = []
currentStateDict = getStateDict(state, N)
successors += [successor for symbol, coords in state if symbol in cars for successor in getSuccessors(symbol, coords, state, currentStateDict, N)]
#print successors
return dict(successors)
def getStateDict(state, N):
dict = {}
for i in range(1, N*N + 1):
dict[i] = '.' #mark as freespot
for symbol, coords in state:
for coord in coords:
dict[coord] = symbol #override used spots
return dict
def drivesHorizontally(coords):
#we check if first two coordinates are adjacent
return True if coords[0] == coords[1] - 1 else False
def getSuccessors(car, coords, state, dict, N):
#[('A', -3), ('B', 16), ('Y', 24), ('*', 4)]
# That is, move car 'A' 3 spaces left, then 'B' 2 down, then 'Y' 3 down,
# and finally '*' moves 4 spaces right to the goal.
result = []
if drivesHorizontally(coords):
#move left
i = 1
while (coords[0] - i) in dict and dict[coords[0] - i] in ['.', '@']:
#print car + ' moves left'
result.append((getNewState(state, car, -i), (car, -i)))
i += 1
#move right
i = 1
while (coords[-1] + i) in dict and dict[coords[-1] + i] in ['.', '@']:
#print car + ' moves right'
result.append((getNewState(state, car, i), (car, i)))
i += 1
else:
#move up
i = 1
while (coords[0] -i * N) in dict and dict[coords[0] -i * N] in ['.', '@']:
#print car + ' moves up'
result.append((getNewState(state, car, -i * N), (car, -i * N)))
i += 1
#move down
i = 1
while (coords[-1] + i * N) in dict and dict[coords[-1] + i * N] in ['.', '@']:
#print car + ' moves down'
result.append((getNewState(state, car, i * N), (car, i * N)))
i += 1
return result
def getNewState(oldState, car, movedBy):
newState = []
for symbol, coords in oldState:
if symbol == car:
newState.append((symbol, addToVector(coords, movedBy)))
else:
newState.append((symbol, coords))
return tuple(newState)
def isGoal(state, goal):
#The solution state should contain both the car and the goal ('*', (30, 31)), ('@', (31,))
# ('@', (31,)),
# ('*', (26, 27)),
for symbol, coords in state:
if symbol == '*':
if goal in coords:
return True
return False
def getGoal(state):
#return the goal coordinate
for symbol, coords in state:
if symbol == '@':
return coords[0]
# But it would also be nice to have a simpler format to describe puzzles,
# and a way to visualize states.
# You will do that by defining the following two functions:
def locs(start, n, incr=1):
"Return a tuple of n locations, starting at start and incrementing by incr."
return tuple([start] + [start + incr*i for i in range(1,n)])
def getGoalCoordinate(N):
if N%2 == 0:
return N-1 + (N/2 -1)*N
else:
return N-1 + (N-1)/2 *N
def getWallCoords(N):
goal = getGoalCoordinate(N)
for i in range(1,N +1):
if i == 1:
for x in range(N):
yield x
elif i == N:
start = (N-1) * N
for x in range(start, start +N):
yield x
else:
start = (i-1) * N
end = start + N - 1
yield start
if end != goal:
yield end
def grid(cars, N=N):
"""Return a tuple of (object, locations) pairs -- the format expected for
this puzzle. This function includes a wall pair, ('|', (0, ...)) to
indicate there are walls all around the NxN grid, except at the goal
location, which is the middle of the right-hand wall; there is a goal
pair, like ('@', (31,)), to indicate this. The variable 'cars' is a
tuple of pairs like ('*', (26, 27)). The return result is a big tuple
of the 'cars' pairs along with the walls and goal pairs."""
return tuple([('@', (getGoalCoordinate(N),)), ('|', tuple(getWallCoords(N)))] + [car for car in cars])
def show(state, N=N):
"Print a representation of a state as an NxN grid."
# Initialize and fill in the board.
board = ['.'] * N**2
for (c, squares) in state:
for s in squares:
board[s] = c
# Now print it out
for i,s in enumerate(board):
print s,
if i % N == N - 1: print
# Here we see the grid and locs functions in use:
puzzle1 = grid((
('*', locs(26, 2)),
('G', locs(9, 2)),
('Y', locs(14, 3, N)),
('P', locs(17, 3, N)),
('O', locs(41, 2, N)),
('B', locs(20, 3, N)),
('A', locs(45, 2))))
puzzle2 = grid((
('*', locs(26, 2)),
('B', locs(20, 3, N)),
('P', locs(33, 3)),
('O', locs(41, 2, N)),
('Y', locs(51, 3))))
puzzle3 = grid((
('*', locs(25, 2)),
('B', locs(19, 3, N)),
('P', locs(36, 3)),
('O', locs(45, 2, N)),
('Y', locs(49, 3))))
# Here are the shortest_path_search and path_actions functions from the unit.
# You may use these if you want, but you don't have to.
def shortest_path_search(start, successors, is_goal, N):
"""Find the shortest path from start state to a state
such that is_goal(state) is true."""
goal = getGoal(start)
if is_goal(start, goal):
return [start]
explored = set() # set of states we have visited
frontier = [ [start] ] # ordered list of paths we have blazed
while frontier:
path = frontier.pop(0)
s = path[-1]
for (state, action) in successors(s,N, goal).items():
if state not in explored:
explored.add(state)
path2 = path + [action, state]
#print action
#print show(state,N)
if is_goal(state, goal):
return path2
else:
frontier.append(path2)
return []
def path_actions(path):
"Return a list of actions in this path."
return path[1::2]
def addToVector(v1, n):
v2 = tuple((n for i in range(len(v1))))
return add_vectors(v1,v2)
def add_vectors(v1, v2):
return tuple(x + y for x,y in zip(v1,v2))
def showAndSolve(puzzle):
show(puzzle, N)
solution = solve_parking_puzzle(puzzle, N)
print path_actions(solution)
