CPMagical CP Library

Problem solution · Python

Sliding Puzzle

Sliding Puzzle: a Python solution using heap or priority queue. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Heap or priority queue
Source
Kamyu LeetCode Solutions
Length
112 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Heap or priority queue

For Sliding Puzzle, the implementation repeatedly takes the currently best candidate from a heap while inserting newly available choices.

  1. Define the priority key and whether the smallest or largest item should lead.
  2. Push each candidate when it becomes eligible.
  3. Discard stale entries when necessary and process the best live candidate.

Code notes

  • 112 lines of Python from the credited upstream file sliding-puzzle.py.
  • The implementation visibly relies on sequence storage, ordered lookup, work queue.
  • No explicit loop blocks detected.

Complexity

Count heap pushes and pops; each normally contributes a logarithmic factor in the heap size.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSliding Puzzle · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O((m * n) * (m * n)!)# Space: O((m * n) * (m * n)!) import heapqimport itertools  # A* Search Algorithmclass Solution(object):    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 xrange(R):                for j in xrange(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(range(1, R*C) + [0])        expected = {(C*i+j+1) % (R*C) : (i, j)                    for i in xrange(R) for j in xrange(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  # Time:  O((m * n) * (m * n)! * log((m * n)!))# Space: O((m * n) * (m * n)!)# A* Search Algorithmclass Solution2(object):    def slidingPuzzle(self, board):        """        :type board: List[List[int]]        :rtype: int        """        def heuristic_estimate(board, R, C, expected):            result = 0            for i in xrange(R):                for j in xrange(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(range(1, R*C) + [0])        end_wrong = tuple(range(1, R*C-2) + [R*C-1, R*C-2, 0])        expected = {(C*i+j+1) % (R*C) : (i, j)                    for i in xrange(R) for j in xrange(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

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗