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test_sample.doctest - simple test for rush hour solution finder
>>> from x import find_solutions
>>> def get_moves(board):
... """Return a list or printable solutions (steps)."""
... return [
... ','.join(move for move, board in solution)
... for solution in find_solutions(board)]
>>> board = '''
... ....AA
... ..BBCC
... rr..EF
... GGHHEF
... ...IEF
... ...IJJ
... '''.strip().split()
>>> solutions = get_moves(board)
>>> len(solutions[0].split(','))
13
>>> 'AL4,BL2,CL2,EU2,FU2,HR2,IU1,JL4,ID1,HL2,ED3,FD3,rR4' in solutions
True
With the current algorithm it is very hard to determine whether a (manually)
precomputed solution will be found because there are a lot of orders in which
steps can be taken and since we try to avoid finding moves that result in the
exact same board configuration we are bound to miss some.
>>> board = '''
... ..OOOP
... .....P
... rrAB.P
... Q.ABCC
... Q.....
... QRRR..
... '''.strip().split()
>>> solutions = get_moves(board)
>>> len(solutions[0].split(','))
6
For example, the first one is not found with the current implementation but
the second (which is basically the exact same solution) is:
>>> 'OL2,BU2,CL1,PD3,AD1,rR4' in solutions
False
>>> 'OL2,AD1,BU2,CL1,PD3,rR4' in solutions
True
Another simple example.
>>> board = '''
... ..OABB
... ..OA.C
... ..OrrC
... ..PPPD
... .....D
... ......
... '''.strip().split()
>>> solutions = get_moves(board)
>>> len(solutions[0].split(','))
9
>>> 'DD1,PR1,OD3,rL2,AD1,BL3,CU1,AU1,rR3' in solutions
False
>>> 'DD1,PR1,OD3,rL2,AD1,BL3,AU1,CU1,rR3' in solutions
True
This is the hardest example from the Rush Hour game.
>>> board = '''
... OOO..P
... .....P
... ..ArrP
... ..ABCC
... D.EBFF
... D.EQQQ
... '''.strip().split()
>>> solutions = get_moves(board)
>>> len(solutions[0].split(','))
29
The algorithm should also deal fine with larger or smaller boards, non-square
boards and longer cars.
>>> board = '''
... ....DAAAAA
... ..rrD..B..
... .......B..
... ....CCCCC.
... '''.strip().split()
>>> solutions = get_moves(board)
>>> len(solutions[0].split(','))
7
The algorithm will return nothing (empty iterator) for impossible boards.
>>> board = '''
... ....B.
... ..AAB.
... rr..B.
... ..ECCC
... ..E.D.
... FFF.D.
... '''.strip().split()
>>> len(get_moves(board))
0
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