Let's say I have a board like this
. . x . . .
. . . . . .
. . x . . x
x is used box and '.' are free. I need to put triomines to fill all the area, so there will be no free cells. Triomines are L-shaped, and I mark same triomino with the same number.
So the solution could be like this:
1 1 x 3 5 5
1 2 3 3 4 5
2 2 x 4 4 x
What the possible backtracking python algorithm might be?
algorithm is quite simple, first we get a list of the cells available in this board configuration, basically, list all the possible cells minus the forbidden ones.
then we make a solution step by iterating over the available cell list and trying to fit a triomino piece into this particular cell position, using 4 possible orientations (available cell is the corner, we have 4 orientations because of rotation).
if the piece fits, increase the step, remove occupied cells from the list and try solving again, until there are no cells available -- that means the whole board is covered.
#!/usr/bin/env python
solution = {}
orientations = [(1,1),(1,-1),(-1,1),(-1,-1)] # 4 possible orientations
def solve( avl, step ) :
if not len(avl) :
print 'done!'
return True
for i,j in avl :
for oi,oj in orientations :
if (i+oi,j) in avl and (i,j+oj) in avl :
occupied = [(i,j),(i+oi,j),(i,j+oj)]
# remove occupied from available, increase step and keep solving
if solve( [a for a in avl if a not in occupied], step + 1 ) :
for occ in occupied :
solution[occ] = step
return True
# initial conditions for this configuration
#
# . . x . . .
# . . . . . .
# . . x . . x
X_SIZE, Y_SIZE = 6, 3
forbidden_cells = [(0,2),(2,2),(2,5)]
if __name__ == '__main__' :
available = []
# fill the available cells list
for i in range(Y_SIZE) :
for j in range(X_SIZE) :
if (i,j) not in forbidden_cells :
available.append( (i,j) )
# print the original problem
for i in range(Y_SIZE) :
for j in range(X_SIZE) :
print '.' if (i,j) in available else 'x',
print
# solve away!
if solve( available, 1 ) :
for i in range(Y_SIZE) :
for j in range(X_SIZE) :
print solution[(i,j)] if (i,j) in available else 'x',
print
else :
print 'sorry, no solution found'
The output is:
$ ./triomines.py
. . x . . .
. . . . . .
. . x . . x
done!
1 1 x 3 2 2
1 4 3 3 5 2
4 4 x 5 5 x
$