# Tiling Solver¶

Tiling a n-dimensional box into non-intersecting n-dimensional polyominoes.

This uses dancing links code which is in Sage. Dancing links were originally introduced by Donald Knuth in 2000 [Knuth1]. In particular, Knuth used dancing links to solve tilings of a region by 2d pentaminoes. Here we extend the method to any dimension.

In particular, the sage.games.quantumino module is based on the Tiling Solver and allows to solve the 3d Quantumino puzzle.

This module defines two classes:

• sage.combinat.tiling.Polyomino class, to represent polyominoes in arbitrary dimension. The goal of this class is to return all the rotated, reflected and/or translated copies of a polyomino that are contained in a certain box.
• sage.combinat.tiling.TilingSolver class, to solve the general problem of tiling a rectangular $$n$$-dimensional box with a set of $$n$$-dimensional polyominoes. One can specify if rotations and reflections are allowed or not and if pieces can be reused or not. This class convert the tiling data into rows of a matrix that are passed to the DLX solver. It also allows to compute the number of solutions.

AUTHOR:

• Sebastien Labbe, June 2011

EXAMPLES:

## 2d Easy Example¶

Here is a 2d example. Let us try to fill the $$3 \times 2$$ rectangle with a $$1 \times 2$$ rectangle and a $$2 \times 2$$ square. Obviously, there are two solutions:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0), (0,1)])
sage: q = Polyomino([(0,0), (0,1), (1,0), (1,1)])
sage: T = TilingSolver([p,q], box=[3,2])
sage: it = T.solve()
sage: next(it)
[Polyomino: [(0, 0), (0, 1), (1, 0), (1, 1)], Color: gray, Polyomino: [(2, 0), (2, 1)], Color: gray]
sage: next(it)
[Polyomino: [(1, 0), (1, 1), (2, 0), (2, 1)], Color: gray, Polyomino: [(0, 0), (0, 1)], Color: gray]
sage: next(it)
Traceback (most recent call last):
...
StopIteration
sage: T.number_of_solutions()
2


## 1d Easy Example¶

Here is an easy one dimensional example where we try to tile a stick of length 6 with three sticks of length 1, 2 and 3. There are six solutions:

sage: p = Polyomino([[0]])
sage: q = Polyomino([[0],[1]])
sage: r = Polyomino([[0],[1],[2]])
sage: T = TilingSolver([p,q,r], box=[6])
sage: len(T.rows())
15
sage: it = T.solve()
sage: next(it)
[Polyomino: [(0,)], Color: gray, Polyomino: [(1,), (2,)], Color: gray, Polyomino: [(3,), (4,), (5,)], Color: gray]
sage: next(it)
[Polyomino: [(0,)], Color: gray, Polyomino: [(1,), (2,), (3,)], Color: gray, Polyomino: [(4,), (5,)], Color: gray]
sage: T.number_of_solutions()
6


## 2d Puzzle allowing reflections¶

The following is a puzzle owned by Florent Hivert:

sage: from sage.combinat.tiling import Polyomino, TilingSolver
sage: L = []
sage: L.append(Polyomino([(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)], 'yellow'))
sage: L.append(Polyomino([(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2)], "black"))
sage: L.append(Polyomino([(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,3)], "gray"))
sage: L.append(Polyomino([(0,0),(0,1),(0,2),(0,3),(1,0),(1,3)],"cyan"))
sage: L.append(Polyomino([(0,0),(0,1),(0,2),(0,3),(1,0),(1,1)],"red"))
sage: L.append(Polyomino([(0,0),(0,1),(0,2),(0,3),(1,1),(1,2)],"blue"))
sage: L.append(Polyomino([(0,0),(0,1),(0,2),(0,3),(1,1),(1,3)],"green"))
sage: L.append(Polyomino([(0,1),(0,2),(0,3),(1,0),(1,1),(1,3)],"magenta"))
sage: L.append(Polyomino([(0,1),(0,2),(0,3),(1,0),(1,1),(1,2)],"orange"))
sage: L.append(Polyomino([(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)],"pink"))


By default, rotations are allowed and reflections are not. In this case, there are no solution:

sage: T = TilingSolver(L, (8,8))
sage: T.number_of_solutions()                       # long time (2.5 s)
0


If reflections are allowed, there are solutions. Solve the puzzle and show one solution:

sage: T = TilingSolver(L, (8,8), reflection=True)
sage: solution = next(T.solve())
sage: G = sum([piece.show2d() for piece in solution], Graphics())
sage: G.show(aspect_ratio=1, axes=False)


Compute the number of solutions:

sage: T.number_of_solutions()                         # long time (2.6s)
328


Create a animation of all the solutions:

sage: a = T.animate()                            # not tested
sage: a                                          # not tested
Animation with 328 frames


## 3d Puzzle¶

The same thing done in 3d without allowing reflections this time:

sage: from sage.combinat.tiling import Polyomino, TilingSolver
sage: L = []
sage: L.append(Polyomino([(0,0,0),(0,1,0),(0,2,0),(0,3,0),(1,0,0),(1,1,0),(1,2,0),(1,3,0)]))
sage: L.append(Polyomino([(0,0,0),(0,1,0),(0,2,0),(0,3,0),(1,0,0),(1,1,0),(1,2,0)]))
sage: L.append(Polyomino([(0,0,0),(0,1,0),(0,2,0),(0,3,0),(1,0,0),(1,1,0),(1,3,0)]))
sage: L.append(Polyomino([(0,0,0),(0,1,0),(0,2,0),(0,3,0),(1,0,0),(1,3,0)]))
sage: L.append(Polyomino([(0,0,0),(0,1,0),(0,2,0),(0,3,0),(1,0,0),(1,1,0)]))
sage: L.append(Polyomino([(0,0,0),(0,1,0),(0,2,0),(0,3,0),(1,1,0),(1,2,0)]))
sage: L.append(Polyomino([(0,0,0),(0,1,0),(0,2,0),(0,3,0),(1,1,0),(1,3,0)]))
sage: L.append(Polyomino([(0,1,0),(0,2,0),(0,3,0),(1,0,0),(1,1,0),(1,3,0)]))
sage: L.append(Polyomino([(0,1,0),(0,2,0),(0,3,0),(1,0,0),(1,1,0),(1,2,0)]))
sage: L.append(Polyomino([(0,0,0),(0,1,0),(0,2,0),(1,0,0),(1,1,0),(1,2,0)]))


Solve the puzzle and show one solution:

sage: T = TilingSolver(L, (8,8,1))
sage: solution = next(T.solve())
sage: G = sum([piece.show3d(size=0.85) for piece in solution], Graphics())
sage: G.show(aspect_ratio=1, viewer='tachyon')


Let us compute the number of solutions:

sage: T.number_of_solutions()                              # long time (3s)
328


## Donald Knuth example : the Y pentamino¶

Donald Knuth [Knuth1] considered the problem of packing 45 Y pentaminoes into a $$15 \times 15$$ square:

sage: from sage.combinat.tiling import Polyomino, TilingSolver
sage: y = Polyomino([(0,0),(1,0),(2,0),(3,0),(2,1)])
sage: T = TilingSolver([y], box=(5,10), reusable=True, reflection=True)
sage: T.number_of_solutions()
10
sage: solution = next(T.solve())
sage: G = sum([piece.show2d() for piece in solution], Graphics())
sage: G.show(aspect_ratio=1)

sage: T = TilingSolver([y], box=(15,15), reusable=True, reflection=True)
sage: T.number_of_solutions()                      #not tested
212


## 5d Easy Example¶

Here is a 5d example. Let us try to fill the $$2 \times 2 \times 2 \times 2 \times 2$$ rectangle with reusable $$1 \times 1 \times 1 \times 1 \times 1$$ rectangles. Obviously, there is one solution:

sage: from sage.combinat.tiling import Polyomino, TilingSolver
sage: p = Polyomino([(0,0,0,0,0)])
sage: T = TilingSolver([p], box=(2,2,2,2,2), reusable=True)
sage: rows = T.rows()                               # long time (3s)
sage: rows                                          # long time (fast)
[[0], [1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30], [31]]
sage: T.number_of_solutions()                       # long time (fast)
1


REFERENCES:

 [Knuth1] (1, 2) Knuth, Donald (2000). “Dancing links”. Arxiv cs/0011047.
class sage.combinat.tiling.Polyomino(coords, color='gray')

Return the polyomino defined by a set of coordinates.

The polyomino is the union of the unit square (or cube, or n-cube) centered at those coordinates. Such an object should be connected, but the code do not make this assumption.

INPUT:

• coords - iterable of tuple
• color - string (optional, default: 'gray'), the color

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: Polyomino([(0,0,0), (0,1,0), (1,1,0), (1,1,1)], color='blue')
Polyomino: [(0, 0, 0), (0, 1, 0), (1, 1, 0), (1, 1, 1)], Color: blue

boundary()

Return the boundary of a 2d polyomino.

INPUT:

• self - a 2d polyomino

OUTPUT:

• list of edges (an edge is a pair of adjacent 2d coordinates)

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0), (1,0), (0,1), (1,1)])
sage: p.boundary()
[((0.5, 1.5), (1.5, 1.5)), ((-0.5, -0.5), (0.5, -0.5)), ((0.5, -0.5), (1.5, -0.5)), ((-0.5, 1.5), (0.5, 1.5)), ((-0.5, 0.5), (-0.5, 1.5)), ((-0.5, -0.5), (-0.5, 0.5)), ((1.5, 0.5), (1.5, 1.5)), ((1.5, -0.5), (1.5, 0.5))]
sage: len(_)
8
sage: p = Polyomino([(5,5)])
sage: p.boundary()
[((4.5, 5.5), (5.5, 5.5)), ((4.5, 4.5), (5.5, 4.5)), ((4.5, 4.5), (4.5, 5.5)), ((5.5, 4.5), (5.5, 5.5))]

bounding_box()

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0,0),(1,0,0),(1,1,0),(1,1,1),(1,2,0)], color='deeppink')
sage: p.bounding_box()
[[0, 0, 0], [1, 2, 1]]

canonical()

Returns the translated copy of self having minimal and positive coordinates

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0,0),(1,0,0),(1,1,0),(1,1,1),(1,2,0)], color='deeppink')
sage: p
Polyomino: [(0, 0, 0), (1, 0, 0), (1, 1, 0), (1, 1, 1), (1, 2, 0)], Color: deeppink
sage: p.canonical()
Polyomino: [(0, 0, 0), (1, 0, 0), (1, 1, 0), (1, 1, 1), (1, 2, 0)], Color: deeppink


TESTS:

sage: p
Polyomino: [(0, 0, 0), (1, 0, 0), (1, 1, 0), (1, 1, 1), (1, 2, 0)], Color: deeppink
sage: p + (3,4,5)
Polyomino: [(3, 4, 5), (4, 4, 5), (4, 5, 5), (4, 5, 6), (4, 6, 5)], Color: deeppink
sage: (p + (3,4,5)).canonical()
Polyomino: [(0, 0, 0), (1, 0, 0), (1, 1, 0), (1, 1, 1), (1, 2, 0)], Color: deeppink

canonical_orthogonals(orientation_preserving=True)

Iterator over the image of self under orthogonal transformations where the coordinates are all positive and minimal.

Note

No guarantee of uniqueness.

INPUT:

• orientation_preserving - bool (optional, default: True), whether the orientation is preserved

OUTPUT:

iterator of Polyomino

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0,0), (0,1,0), (1,1,0), (1,1,1)], color='blue')
sage: L = list(p.canonical_orthogonals())
sage: len(L)
24


They might not be all different:

sage: s = set(p.canonical_orthogonals())
sage: len(s)
12


With the non orientation-preserving:

sage: s = set(p.canonical_orthogonals(False))
sage: len(s)
24

center()

Return the center of the polyomino.

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0,0),(0,0,1)])
sage: p.center()
(0, 0, 1/2)


In 3d:

sage: p = Polyomino([(0,0,0),(1,0,0),(1,1,0),(1,1,1),(1,2,0)], color='deeppink')
sage: p.center()
(4/5, 4/5, 1/5)


In 2d:

sage: p = Polyomino([(0,0),(1,0),(1,1),(1,2)])
sage: p.center()
(3/4, 3/4)

color()

Return the color of the polyomino.

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0,0), (0,1,0), (1,1,0), (1,1,1)], color='blue')
sage: p.color()
'blue'

neighbor_edges()

Return an iterator over the pairs of neighbor coordinates of the polyomino.

Two points $$P$$ and $$Q$$ are neighbor if $$P - Q$$ has one coordinate equal to $$+1$$ or $$-1$$ and zero everywhere else.

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0,0),(0,0,1)])
sage: list(sorted(edge) for edge in p.neighbor_edges())
[[(0, 0, 0), (0, 0, 1)]]


In 3d:

sage: p = Polyomino([(0,0,0),(1,0,0),(1,1,0),(1,1,1),(1,2,0)], color='deeppink')
sage: L = sorted(sorted(edge) for edge in p.neighbor_edges())
sage: for a in L: a
[(0, 0, 0), (1, 0, 0)]
[(1, 0, 0), (1, 1, 0)]
[(1, 1, 0), (1, 1, 1)]
[(1, 1, 0), (1, 2, 0)]


In 2d:

sage: p = Polyomino([(0,0),(1,0),(1,1),(1,2)])
sage: L = sorted(sorted(edge) for edge in p.neighbor_edges())
sage: for a in L: a
[(0, 0), (1, 0)]
[(1, 0), (1, 1)]
[(1, 1), (1, 2)]

orthogonals(orientation_preserving=True)

Iterator over the images of self under orthogonal transformations.

Note

No guarantee of uniqueness.

INPUT:

• orientation_preserving - bool (optional, default: True), whether the orientation is preserved

OUTPUT:

iterator of Polyomino

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0,0), (0,1,0), (1,1,0), (1,1,1)], color='blue')
sage: L = list(p.orthogonals())
sage: len(L)
24
sage: L = list(p.orthogonals(False))
sage: len(L)
48

show2d(size=0.7, color='black', thickness=1)

Returns a 2d Graphic object representing the polyomino.

INPUT:

• self - a polyomino of dimension 2
• size - number (optional, default: 0.7), the size of each square.
• color - color (optional, default: 'black'), color of the boundary line.
• thickness - number (optional, default: 1), how thick the boundary line is.

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0),(1,0),(1,1),(1,2)], color='deeppink')
sage: p.show2d()              # long time (0.5s)
Graphics object consisting of 17 graphics primitives

show3d(size=1)

Returns a 3d Graphic object representing the polyomino.

INPUT:

• self - a polyomino of dimension 3
• size - number (optional, default: 1), the size of each 1 \times 1 \times 1 cube. This does a homothety with respect to the center of the polyomino.

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0,0), (0,1,0), (1,1,0), (1,1,1)], color='blue')
sage: p.show3d()
Graphics3d Object

translated(box)

Returns an iterator over the translated images of self inside a box.

INPUT:

• box - tuple, size of the box

OUTPUT:

iterator of 3d polyominoes

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0,0),(1,0,0),(1,1,0),(1,1,1),(1,2,0)], color='deeppink')
sage: for t in p.translated(box=(5,8,2)): t
Polyomino: [(0, 0, 0), (1, 0, 0), (1, 1, 0), (1, 1, 1), (1, 2, 0)], Color: deeppink
Polyomino: [(0, 1, 0), (1, 1, 0), (1, 2, 0), (1, 2, 1), (1, 3, 0)], Color: deeppink
Polyomino: [(0, 2, 0), (1, 2, 0), (1, 3, 0), (1, 3, 1), (1, 4, 0)], Color: deeppink
Polyomino: [(0, 3, 0), (1, 3, 0), (1, 4, 0), (1, 4, 1), (1, 5, 0)], Color: deeppink
Polyomino: [(0, 4, 0), (1, 4, 0), (1, 5, 0), (1, 5, 1), (1, 6, 0)], Color: deeppink
Polyomino: [(0, 5, 0), (1, 5, 0), (1, 6, 0), (1, 6, 1), (1, 7, 0)], Color: deeppink
Polyomino: [(1, 0, 0), (2, 0, 0), (2, 1, 0), (2, 1, 1), (2, 2, 0)], Color: deeppink
Polyomino: [(1, 1, 0), (2, 1, 0), (2, 2, 0), (2, 2, 1), (2, 3, 0)], Color: deeppink
Polyomino: [(1, 2, 0), (2, 2, 0), (2, 3, 0), (2, 3, 1), (2, 4, 0)], Color: deeppink
Polyomino: [(1, 3, 0), (2, 3, 0), (2, 4, 0), (2, 4, 1), (2, 5, 0)], Color: deeppink
Polyomino: [(1, 4, 0), (2, 4, 0), (2, 5, 0), (2, 5, 1), (2, 6, 0)], Color: deeppink
Polyomino: [(1, 5, 0), (2, 5, 0), (2, 6, 0), (2, 6, 1), (2, 7, 0)], Color: deeppink
Polyomino: [(2, 0, 0), (3, 0, 0), (3, 1, 0), (3, 1, 1), (3, 2, 0)], Color: deeppink
Polyomino: [(2, 1, 0), (3, 1, 0), (3, 2, 0), (3, 2, 1), (3, 3, 0)], Color: deeppink
Polyomino: [(2, 2, 0), (3, 2, 0), (3, 3, 0), (3, 3, 1), (3, 4, 0)], Color: deeppink
Polyomino: [(2, 3, 0), (3, 3, 0), (3, 4, 0), (3, 4, 1), (3, 5, 0)], Color: deeppink
Polyomino: [(2, 4, 0), (3, 4, 0), (3, 5, 0), (3, 5, 1), (3, 6, 0)], Color: deeppink
Polyomino: [(2, 5, 0), (3, 5, 0), (3, 6, 0), (3, 6, 1), (3, 7, 0)], Color: deeppink
Polyomino: [(3, 0, 0), (4, 0, 0), (4, 1, 0), (4, 1, 1), (4, 2, 0)], Color: deeppink
Polyomino: [(3, 1, 0), (4, 1, 0), (4, 2, 0), (4, 2, 1), (4, 3, 0)], Color: deeppink
Polyomino: [(3, 2, 0), (4, 2, 0), (4, 3, 0), (4, 3, 1), (4, 4, 0)], Color: deeppink
Polyomino: [(3, 3, 0), (4, 3, 0), (4, 4, 0), (4, 4, 1), (4, 5, 0)], Color: deeppink
Polyomino: [(3, 4, 0), (4, 4, 0), (4, 5, 0), (4, 5, 1), (4, 6, 0)], Color: deeppink
Polyomino: [(3, 5, 0), (4, 5, 0), (4, 6, 0), (4, 6, 1), (4, 7, 0)], Color: deeppink


This method is independant of the translation of the polyomino:

sage: q = Polyomino([(0,0,0), (1,0,0)])
sage: list(q.translated((2,2,1)))
[Polyomino: [(0, 0, 0), (1, 0, 0)], Color: gray, Polyomino: [(0, 1, 0), (1, 1, 0)], Color: gray]
sage: q = Polyomino([(34,7,-9), (35,7,-9)])
sage: list(q.translated((2,2,1)))
[Polyomino: [(0, 0, 0), (1, 0, 0)], Color: gray, Polyomino: [(0, 1, 0), (1, 1, 0)], Color: gray]


Inside smaller boxes:

sage: list(p.translated(box=(2,2,3)))
[]
sage: list(p.translated(box=(2,3,2)))
[Polyomino: [(0, 0, 0), (1, 0, 0), (1, 1, 0), (1, 1, 1), (1, 2, 0)], Color: deeppink]
sage: list(p.translated(box=(3,2,2)))
[]
sage: list(p.translated(box=(1,1,1)))
[]
sage: list(p.translated(box=(1,1,-1)))
[]

translated_orthogonals(box, orientation_preserving=True)

Return the translated and rotated of self that lies in the box.

INPUT:

• box - tuple of size three, size of the box
• orientation_preserving - bool (optional, default: True), whether the orientation is preserved

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino
sage: p = Polyomino([(0,0,0),(1,0,0),(1,1,0),(1,1,1),(1,2,0)], color='deeppink')
sage: L = list(p.translated_orthogonals(box=(5,8,2)))
sage: len(L)
360

sage: p = Polyomino([(0,0,0),(1,0,0),(1,1,0),(1,2,0),(1,2,1)], color='orange')
sage: L = list(p.translated_orthogonals(box=(5,8,2)))
sage: len(L)
180

sage: p = Polyomino([(0,0,0),(1,0,0),(1,1,0),(1,2,0),(1,2,1)], color='orange')
sage: L = list(p.translated_orthogonals((5,8,2), False))
sage: len(L)
360

class sage.combinat.tiling.TilingSolver(pieces, box, rotation=True, reflection=False, reusable=False)

Tiling solver

Solve the problem of tiling a rectangular box with a certain number of pieces, called polyominoes, where each polyomino must be used exactly once.

INPUT:

• pieces - iterable of Polyominoes
• box - tuple, size of the box
• rotation - bool (optional, default: True), whether to allow rotations
• reflection - bool (optional, default: False), whether to allow reflections
• reusable - bool (optional, default: False), whether to allow the pieces to be reused

EXAMPLES:

By default, rotations are allowed and reflections are not allowed:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: T
Tiling solver of 3 pieces into the box (1, 1, 6)
Rotation allowed: True
Reflection allowed: False
Reusing pieces allowed: False


Solutions are given by an iterator:

sage: it = T.solve()
sage: for p in next(it): p
Polyomino: [(0, 0, 0)], Color: gray
Polyomino: [(0, 0, 1), (0, 0, 2)], Color: gray
Polyomino: [(0, 0, 3), (0, 0, 4), (0, 0, 5)], Color: gray


Another solution:

sage: for p in next(it): p
Polyomino: [(0, 0, 0)], Color: gray
Polyomino: [(0, 0, 1), (0, 0, 2), (0, 0, 3)], Color: gray
Polyomino: [(0, 0, 4), (0, 0, 5)], Color: gray


TESTS:

sage: T = TilingSolver([p,q,r], box=(1,1,6), rotation=False, reflection=True)
Traceback (most recent call last):
...
NotImplementedError: When reflection is allowed and rotation is not allowed

animate(partial=None, stop=None, size=0.75, axes=False)

Return an animation of evolving solutions.

INPUT:

• partial - string (optional, default: None), whether to include partial (incomplete) solutions. It can be one of the following:
• None - include only complete solutions
• 'common_prefix' - common prefix between two consecutive solutions
• 'incremental' - one piece change at a time
• stop - integer (optional, default:None), number of frames
• size - number (optional, default: 0.75), the size of each 1 \times 1 square. This does a homothety with respect to the center of each polyomino.
• axes - bool (optional, default:False), whether the x and y axes are shown.

EXAMPLES:

sage: from sage.combinat.tiling import Polyomino, TilingSolver
sage: y = Polyomino([(0,0),(1,0),(2,0),(3,0),(2,1)], color='cyan')
sage: T = TilingSolver([y], box=(5,10), reusable=True, reflection=True)
sage: a = T.animate()
sage: a
Animation with 10 frames


Include partial solutions (common prefix between two consecutive solutions):

sage: a = T.animate('common_prefix')
sage: a
Animation with 19 frames


Incremental solutions (one piece removed or added at a time):

sage: a = T.animate('incremental')      # long time (2s)
sage: a                                 # long time (2s)
Animation with 123 frames

sage: a.show()                 # optional -- ImageMagick


The show function takes arguments to specify the delay between frames (measured in hundredths of a second, default value 20) and the number of iterations (default value 0, which means to iterate forever). To iterate 4 times with half a second between each frame:

sage: a.show(delay=50, iterations=4) # optional


Limit the number of frames:

sage: a = T.animate('incremental', stop=13)     # not tested
sage: a                                         # not tested
Animation with 13 frames

coord_to_int_dict()

Returns a dictionary mapping coordinates to integers.

OUTPUT:

dict

EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: A = T.coord_to_int_dict()
sage: sorted(A.iteritems())
[((0, 0, 0), 3), ((0, 0, 1), 4), ((0, 0, 2), 5), ((0, 0, 3), 6), ((0, 0, 4), 7), ((0, 0, 5), 8)]


Reusable pieces:

sage: p = Polyomino([(0,0), (0,1)])
sage: q = Polyomino([(0,0), (0,1), (1,0), (1,1)])
sage: T = TilingSolver([p,q], box=[3,2], reusable=True)
sage: B = T.coord_to_int_dict()
sage: sorted(B.iteritems())
[((0, 0), 0), ((0, 1), 1), ((1, 0), 2), ((1, 1), 3), ((2, 0), 4), ((2, 1), 5)]

dlx_common_prefix_solutions()

Return an iterator over the row indices of solutions and of partial solutions, i.e. the common prefix of two consecutive solutions.

The purpose is to illustrate the backtracking and construct an animation of the evolution of solutions.

OUPUT:

iterator

EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: list(T.dlx_solutions())
[[0, 7, 14], [0, 12, 10], [6, 13, 5], [6, 14, 2], [11, 9, 5], [11, 10, 3]]
sage: list(T.dlx_common_prefix_solutions())
[[0, 7, 14], [0], [0, 12, 10], [], [6, 13, 5], [6], [6, 14, 2], [], [11, 9, 5], [11], [11, 10, 3]]

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: y = Polyomino([(0,0),(1,0),(2,0),(3,0),(2,1)], color='yellow')
sage: T = TilingSolver([y], box=(5,10), reusable=True, reflection=True)
sage: for a in T.dlx_common_prefix_solutions(): a
[64, 83, 149, 44, 179, 62, 35, 162, 132, 101]
[64, 83, 149, 44, 179]
[64, 83, 149, 44, 179, 154, 35, 162, 132, 175]
[64, 83, 149]
[64, 83, 149, 97, 39, 162, 35, 62, 48, 106]
[64]
[64, 157, 149, 136, 179, 62, 35, 162, 132, 101]
[64, 157, 149, 136, 179]
[64, 157, 149, 136, 179, 154, 35, 162, 132, 175]
[]
[82, 119, 58, 97, 38, 87, 8, 63, 48, 107]
[82, 119, 58, 97, 38]
[82, 119, 58, 97, 38, 161, 8, 63, 140, 107]
[82, 119]
[82, 119, 150, 136, 180, 63, 8, 161, 131, 175]
[82, 119, 150]
[82, 119, 150, 171, 38, 87, 8, 63, 48, 107]
[82, 119, 150, 171, 38]
[82, 119, 150, 171, 38, 161, 8, 63, 140, 107]

dlx_incremental_solutions()

Return an iterator over the row indices of the incremental solutions.

Between two incremental solution, either one piece is added or one piece is removed.

The purpose is to illustrate the backtracking and construct an animation of the evolution of solutions.

OUPUT:

iterator

EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: list(T.dlx_solutions())
[[0, 7, 14], [0, 12, 10], [6, 13, 5], [6, 14, 2], [11, 9, 5], [11, 10, 3]]
sage: list(T.dlx_incremental_solutions())
[[0, 7, 14], [0, 7], [0], [0, 12], [0, 12, 10], [0, 12], [0], [], [6], [6, 13], [6, 13, 5], [6, 13], [6], [6, 14], [6, 14, 2], [6, 14], [6], [], [11], [11, 9], [11, 9, 5], [11, 9], [11], [11, 10], [11, 10, 3]]

sage: y = Polyomino([(0,0),(1,0),(2,0),(3,0),(2,1)], color='yellow')
sage: T = TilingSolver([y], box=(5,10), reusable=True, reflection=True)
sage: for a in T.dlx_solutions(): a
[64, 83, 149, 44, 179, 62, 35, 162, 132, 101]
[64, 83, 149, 44, 179, 154, 35, 162, 132, 175]
[64, 83, 149, 97, 39, 162, 35, 62, 48, 106]
[64, 157, 149, 136, 179, 62, 35, 162, 132, 101]
[64, 157, 149, 136, 179, 154, 35, 162, 132, 175]
[82, 119, 58, 97, 38, 87, 8, 63, 48, 107]
[82, 119, 58, 97, 38, 161, 8, 63, 140, 107]
[82, 119, 150, 136, 180, 63, 8, 161, 131, 175]
[82, 119, 150, 171, 38, 87, 8, 63, 48, 107]
[82, 119, 150, 171, 38, 161, 8, 63, 140, 107]
sage: len(list(T.dlx_incremental_solutions()))
123

dlx_solutions()

Return an iterator over the row indices of the solutions.

OUPUT:

iterator

EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: list(T.dlx_solutions())
[[0, 7, 14], [0, 12, 10], [6, 13, 5], [6, 14, 2], [11, 9, 5], [11, 10, 3]]

dlx_solver()

Return the sage DLX solver of that 3d tiling problem.

OUTPUT:

DLX Solver

EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: x = T.dlx_solver()
sage: x

int_to_coord_dict()

Returns a dictionary mapping integers to coordinates.

EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: B = T.int_to_coord_dict()
sage: sorted(B.iteritems())
[(3, (0, 0, 0)), (4, (0, 0, 1)), (5, (0, 0, 2)), (6, (0, 0, 3)), (7, (0, 0, 4)), (8, (0, 0, 5))]


Reusable pieces:

sage: from sage.combinat.tiling import Polyomino, TilingSolver
sage: p = Polyomino([(0,0), (0,1)])
sage: q = Polyomino([(0,0), (0,1), (1,0), (1,1)])
sage: T = TilingSolver([p,q], box=[3,2], reusable=True)
sage: B = T.int_to_coord_dict()
sage: sorted(B.iteritems())
[(0, (0, 0)), (1, (0, 1)), (2, (1, 0)), (3, (1, 1)), (4, (2, 0)), (5, (2, 1))]


TESTS:

The methods int_to_coord_dict and coord_to_int_dict returns dictionary that are inverse of each other:

sage: A = T.coord_to_int_dict()
sage: B = T.int_to_coord_dict()
sage: all(A[B[i]] == i for i in B)
True
sage: all(B[A[i]] == i for i in A)
True

is_suitable()

Return whether the volume of the box is equal to sum of the volume of the polyominoes and the number of rows sent to the DLX solver is larger than zero.

If these conditions are not verified, then the problem is not suitable in the sense that there are no solution.

Note

The DLX solver throws a Segmentation Fault when the number of rows is zero:

sage: from sage.combinat.matrices.dancing_links import dlx_solver
sage: rows = []
sage: x = dlx_solver(rows)
sage: x.search()        # not tested
BOOM !!!


EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: T.is_suitable()
True
sage: T = TilingSolver([p,q,r], box=(1,1,7))
sage: T.is_suitable()
False

nrows_per_piece()

Return the number of rows necessary by each piece.

OUPUT:

list

EXAMPLES:

sage: from sage.games.quantumino import QuantuminoSolver
sage: q = QuantuminoSolver(0)
sage: T = q.tiling_solver()
sage: T.nrows_per_piece()             # long time (10s)
[360, 360, 360, 360, 360, 180, 180, 672, 672, 360, 360, 180, 180, 360, 360, 180]

number_of_solutions()

Return the number of distinct solutions.

OUPUT:

integer

EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0)])
sage: q = Polyomino([(0,0), (0,1)])
sage: r = Polyomino([(0,0), (0,1), (0,2)])
sage: T = TilingSolver([p,q,r], box=(1,6))
sage: T.number_of_solutions()
6

sage: T = TilingSolver([p,q,r], box=(1,7))
sage: T.number_of_solutions()
0

pieces()

Return the list of pieces.

OUTPUT:

list of 3d polyominoes

EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: for p in T._pieces: p
Polyomino: [(0, 0, 0)], Color: gray
Polyomino: [(0, 0, 0), (0, 0, 1)], Color: gray
Polyomino: [(0, 0, 0), (0, 0, 1), (0, 0, 2)], Color: gray

rows()

Creation of the rows

EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: rows = T.rows()
sage: for row in rows: row
[0, 3]
[0, 4]
[0, 5]
[0, 6]
[0, 7]
[0, 8]
[1, 3, 4]
[1, 4, 5]
[1, 5, 6]
[1, 6, 7]
[1, 8, 7]
[2, 3, 4, 5]
[2, 4, 5, 6]
[2, 5, 6, 7]
[2, 8, 6, 7]

solve(partial=None)

Returns an iterator of list of 3d polyominoes that are an exact cover of the box.

INPUT:

• partial - string (optional, default: None), whether to include partial (incomplete) solutions. It can be one of the following:
• None - include only complete solution
• 'common_prefix' - common prefix between two consecutive solutions
• 'incremental' - one piece change at a time

OUTPUT:

iterator of list of 3d polyominoes

EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: it = T.solve()
sage: for p in next(it): p
Polyomino: [(0, 0, 0)], Color: gray
Polyomino: [(0, 0, 1), (0, 0, 2)], Color: gray
Polyomino: [(0, 0, 3), (0, 0, 4), (0, 0, 5)], Color: gray
sage: for p in next(it): p
Polyomino: [(0, 0, 0)], Color: gray
Polyomino: [(0, 0, 1), (0, 0, 2), (0, 0, 3)], Color: gray
Polyomino: [(0, 0, 4), (0, 0, 5)], Color: gray
sage: for p in next(it): p
Polyomino: [(0, 0, 0), (0, 0, 1)], Color: gray
Polyomino: [(0, 0, 2), (0, 0, 3), (0, 0, 4)], Color: gray
Polyomino: [(0, 0, 5)], Color: gray


Including the partial solutions:

sage: it = T.solve(partial='common_prefix')
sage: for p in next(it): p
Polyomino: [(0, 0, 0)], Color: gray
Polyomino: [(0, 0, 1), (0, 0, 2)], Color: gray
Polyomino: [(0, 0, 3), (0, 0, 4), (0, 0, 5)], Color: gray
sage: for p in next(it): p
Polyomino: [(0, 0, 0)], Color: gray
sage: for p in next(it): p
Polyomino: [(0, 0, 0)], Color: gray
Polyomino: [(0, 0, 1), (0, 0, 2), (0, 0, 3)], Color: gray
Polyomino: [(0, 0, 4), (0, 0, 5)], Color: gray
sage: for p in next(it): p
sage: for p in next(it): p
Polyomino: [(0, 0, 0), (0, 0, 1)], Color: gray
Polyomino: [(0, 0, 2), (0, 0, 3), (0, 0, 4)], Color: gray
Polyomino: [(0, 0, 5)], Color: gray


Colors are preserved when the polyomino can be reused:

sage: p = Polyomino([(0,0,0)], color='yellow')
sage: q = Polyomino([(0,0,0), (0,0,1)], color='yellow')
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)], color='yellow')
sage: T = TilingSolver([p,q,r], box=(1,1,6), reusable=True)
sage: it = T.solve()
sage: for p in next(it): p
Polyomino: [(0, 0, 0)], Color: yellow
Polyomino: [(0, 0, 1)], Color: yellow
Polyomino: [(0, 0, 2)], Color: yellow
Polyomino: [(0, 0, 3)], Color: yellow
Polyomino: [(0, 0, 4)], Color: yellow
Polyomino: [(0, 0, 5)], Color: yellow


TESTS:

sage: T = TilingSolver([p,q,r], box=(1,1,7))
sage: next(T.solve())
Traceback (most recent call last):
...
StopIteration

space()

Returns an iterator over all the non negative integer coordinates contained in the box.

EXAMPLES:

sage: from sage.combinat.tiling import TilingSolver, Polyomino
sage: p = Polyomino([(0,0,0)])
sage: q = Polyomino([(0,0,0), (0,0,1)])
sage: r = Polyomino([(0,0,0), (0,0,1), (0,0,2)])
sage: T = TilingSolver([p,q,r], box=(1,1,6))
sage: list(T.space())
[(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 0, 4), (0, 0, 5)]

sage.combinat.tiling.orthogonal_transformation(n, orientation_preserving=True)

Return the list of orthogonal transformation matrices in the $$n$$-dimensional vector space.

INPUT:

• n - positive integer, dimension of the space
• orientation_preserving - bool (optional, default: True), whether the orientation is preserved

OUTPUT:

list of matrices

EXAMPLES:

sage: from sage.combinat.tiling import orthogonal_transformation
sage: orthogonal_transformation(2)
[
[1 0]  [ 0  1]  [-1  0]  [ 0 -1]
[0 1], [-1  0], [ 0 -1], [ 1  0]
]
sage: orthogonal_transformation(2, orientation_preserving=False)
[
[1 0]  [ 0 -1]  [ 1  0]  [ 0  1]  [0 1]  [-1  0]  [ 0 -1]  [-1  0]
[0 1], [-1  0], [ 0 -1], [-1  0], [1 0], [ 0 -1], [ 1  0], [ 0  1]
]
sage: orthogonal_transformation(3)
[
[1 0 0]  [ 1  0  0]  [-1  0  0]  [-1  0  0]  [0 0 1]  [ 0  0 -1]
[0 1 0]  [ 0 -1  0]  [ 0  1  0]  [ 0 -1  0]  [1 0 0]  [ 1  0  0]
[0 0 1], [ 0  0 -1], [ 0  0 -1], [ 0  0  1], [0 1 0], [ 0 -1  0],

[ 0  0 -1]  [ 0  0  1]  [0 1 0]  [ 0 -1  0]  [ 0  1  0]  [ 0 -1  0]
[-1  0  0]  [-1  0  0]  [0 0 1]  [ 0  0 -1]  [ 0  0 -1]  [ 0  0  1]
[ 0  1  0], [ 0 -1  0], [1 0 0], [ 1  0  0], [-1  0  0], [-1  0  0],

[ 0  1  0]  [ 0 -1  0]  [ 0  1  0]  [ 0 -1  0]  [ 1  0  0]  [ 1  0  0]
[ 1  0  0]  [ 1  0  0]  [-1  0  0]  [-1  0  0]  [ 0  0 -1]  [ 0  0  1]
[ 0  0 -1], [ 0  0  1], [ 0  0  1], [ 0  0 -1], [ 0  1  0], [ 0 -1  0],

[-1  0  0]  [-1  0  0]  [ 0  0 -1]  [ 0  0  1]  [ 0  0  1]  [ 0  0 -1]
[ 0  0  1]  [ 0  0 -1]  [ 0  1  0]  [ 0 -1  0]  [ 0  1  0]  [ 0 -1  0]
[ 0  1  0], [ 0 -1  0], [ 1  0  0], [ 1  0  0], [-1  0  0], [-1  0  0]
]


TESTS:

sage: orthogonal_transformation(1)
[[1]]
sage: orthogonal_transformation(0)
Traceback (most recent call last):
...
ValueError: ['B', 0] is not a valid Cartan type