Sliding puzzle [A* Search Algorithm]¶
Time: O((MxN)x(MxN)!); Space: O((MxN)x(MxN)!); hard
On a 2x3 board, there are 5 tiles represented by the integers 1 through 5, and an empty square represented by 0.
A move consists of choosing 0 and a 4-directionally adjacent number and swapping it.
The state of the board is solved if and only if the board is [[1,2,3],[4,5,0]].
Given a puzzle board, return the least number of moves required so that the state of the board is solved. If it is impossible for the state of the board to be solved, return -1.
Example 1:
Input: board = [[1,2,3],[4,0,5]]
Output: 1
Explanation:
Swap the 0 and the 5 in one move.
Example 2:
Input: board = [[1,2,3],[5,4,0]]
Output: -1
Explanation:
No number of moves will make the board solved.
Example 3:
Input: board = [[4,1,2],[5,0,3]]
Output: 5
Explanation:
5 is the smallest number of moves that solves the board.
An example path:
After move 0: [[4,1,2],[5,0,3]]
After move 1: [[4,1,2],[0,5,3]]
After move 2: [[0,1,2],[4,5,3]]
After move 3: [[1,0,2],[4,5,3]]
After move 4: [[1,2,0],[4,5,3]]
After move 5: [[1,2,3],[4,5,0]]
Example 4:
Input: board = [[3,2,4],[1,5,0]]
Output: 14
Notes:
board will be a 2 x 3 array as described above.
board[i][j] will be a permutation of [0, 1, 2, 3, 4, 5].
Hint:
Perform a breadth-first-search (BFS), where the nodes are the puzzle boards and edges are if two puzzle boards can be transformed into one another with one move.
[3]:
import heapq
import itertools
class Solution1(object):
"""
A* Search Algorithm
"""
def slidingPuzzle(self, board):
"""
:type board: List[List[int]]
:rtype: int
"""
def dot(p1, p2):
return p1[0]*p2[0] + p1[1]*p2[1]
def heuristic_estimate(board, R, C, expected):
result = 0
for i in range(R):
for j in range(C):
val = board[C*i + j]
if val == 0:
continue
r, c = expected[val]
result += abs(r-i) + abs(c-j)
return result
R, C = len(board), len(board[0])
begin = tuple(itertools.chain(*board))
end = tuple(list(range(1, R*C)) + [0])
expected = {(C*i+j+1) % (R*C) : (i, j)
for i in range(R) for j in range(C)}
min_steps = heuristic_estimate(begin, R, C, expected)
closer, detour = [(begin.index(0), begin)], []
lookup = set()
while True:
if not closer:
if not detour:
return -1
min_steps += 2
closer, detour = detour, closer
zero, board = closer.pop()
if board == end:
return min_steps
if board not in lookup:
lookup.add(board)
r, c = divmod(zero, C)
for direction in ((-1, 0), (1, 0), (0, -1), (0, 1)):
i, j = r+direction[0], c+direction[1]
if 0 <= i < R and 0 <= j < C:
new_zero = i*C+j
tmp = list(board)
tmp[zero], tmp[new_zero] = tmp[new_zero], tmp[zero]
new_board = tuple(tmp)
r2, c2 = expected[board[new_zero]]
r1, c1 = divmod(zero, C)
r0, c0 = divmod(new_zero, C)
is_closer = dot((r1-r0, c1-c0), (r2-r0, c2-c0)) > 0
(closer if is_closer else detour).append((new_zero, new_board))
return min_steps
[4]:
s = Solution1()
board = [[1,2,3],[4,0,5]]
assert s.slidingPuzzle(board) == 1
board = [[1,2,3],[5,4,0]]
assert s.slidingPuzzle(board) == -1
board = [[4,1,2],[5,0,3]]
assert s.slidingPuzzle(board) == 5
board = [[3,2,4],[1,5,0]]
assert s.slidingPuzzle(board) == 14
[9]:
import heapq
import itertools
class Solution2(object):
"""
A* Search Algorithm
Time: O((m * n) * (m * n)! * log((m * n)!))
Space: O((m * n) * (m * n)!)
"""
def slidingPuzzle(self, board):
"""
:type board: List[List[int]]
:rtype: int
"""
def heuristic_estimate(board, R, C, expected):
result = 0
for i in range(R):
for j in range(C):
val = board[C*i + j]
if val == 0:
continue
r, c = expected[val]
result += abs(r-i) + abs(c-j)
return result
R, C = len(board), len(board[0])
begin = tuple(itertools.chain(*board))
end = tuple(list(range(1, R*C)) + [0])
end_wrong = tuple(list(range(1, R*C-2)) + [R*C-1, R*C-2, 0])
expected = {(C*i+j+1) % (R*C) : (i, j)
for i in range(R) for j in range(C)}
min_heap = [(0, 0, begin.index(0), begin)]
lookup = {begin: 0}
while min_heap:
f, g, zero, board = heapq.heappop(min_heap)
if board == end:
return g
if board == end_wrong:
return -1
if f > lookup[board]:
continue
r, c = divmod(zero, C)
for direction in ((-1, 0), (1, 0), (0, -1), (0, 1)):
i, j = r+direction[0], c+direction[1]
if 0 <= i < R and 0 <= j < C:
new_zero = C*i+j
tmp = list(board)
tmp[zero], tmp[new_zero] = tmp[new_zero], tmp[zero]
new_board = tuple(tmp)
f = g+1+heuristic_estimate(new_board, R, C, expected)
if f < lookup.get(new_board, float("inf")):
lookup[new_board] = f
heapq.heappush(min_heap, (f, g+1, new_zero, new_board))
return -1
[10]:
s = Solution2()
board = [[1,2,3],[4,0,5]]
assert s.slidingPuzzle(board) == 1
board = [[1,2,3],[5,4,0]]
assert s.slidingPuzzle(board) == -1
board = [[4,1,2],[5,0,3]]
assert s.slidingPuzzle(board) == 5
board = [[3,2,4],[1,5,0]]
assert s.slidingPuzzle(board) == 14