Appendix A
Tiling problems
Cyril Allauzen1 and Bruno Durand23
A.1 Introduction
In this appendix, we prove the undecidability of the following problems:
the constrained domino problem (proved undecidable by Wang): give a
tile set and a tile as input, ask whether it is possible to form a tiling of
the plane containing the given tile;
the unconstrained domino problem (Berger's Theorem): the input is a
tile set and the question is whether one can tile the plane with it;
the periodic domino problem (Berger Gurevich Koryakov): the input is
also a tile set, but the question is whether it can be used to form a periodic
tiling of the plane.
In order to prove these results, we present some recursive transformations
of Turing machines into tile sets. These constructions are not independent,
thus the reader may not understand the last one if he could not understand the
rst ones. The last construction also provides a direct proof of the recursive
inseparability result of Berger, Gurevich and Koryakov (Theorem ?? in this
book and reference [2]). Its intuitive meaning is that it is not possible to
separate with any computing device tile sets that cannot tile the plane and tile
sets that can form a periodic tiling of the plane.
We do not present here the original proofs of these theorems. They were
based on Berger's construction (see [1]): we present a simpler proof inspired
by Robinson's ideas in [3]. Both proofs are based on the construction of an
Student at the Ecole Normale Superieure de Lyon, 46 Allee d'Italie, 69364 Lyon Cedex
07, France.
2
Laboratoire de l'Informatique du Parallelisme, ENS-Lyon CNRS, 46 Allee d'Italie, 69364
Lyon Cedex 07, France.
3
This work was done during a training period of Cyril Allauzen under the direction of
Bruno Durand in summer 1995.
1
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aperiodic tile set (i.e. a tile set that can tile the plane, but not periodically).
Berger's aperiodic tile set contains more that 30000 tiles and Robinson's contains 56 tiles. The reason why these aperiodic tile sets are fundamental in these
proofs is the following: imagine that any tile sets that can tile the plane can
also form a periodic tiling. Then, one could decide the unconstrained domino
problem: form all possible n n squares of tiles. If no correct tiling is formed
then stop and answer \no". If one of these squares forms a periodic tiling of
the plane, then stop and answer \yes". Else increment n and iterate the process. If no aperiodic tile set exists then the domino problem is decidable by the
previous algorithm (and this was conjectured by Wang in [4]).
In the sequel, we rst prove the constrained domino problem, then we construct an aperiodic tile set with ad hoc properties, and then we prove the undecidability of the domino problem. We then adapt the proof in order to get the
undecidability of the periodic domino problem and the inseparability theorem.
A.2 The constrained domino problem
We consider now the constrained domino problem for which a tile set and a
particular tile are given as input. The problem is to form a tiling of the plane
containing this chosen tile. Wang proved this problem undecidable in [4]. An
other version of this problem consists of forming an almost everywhere blank
nontrivial tiling of the plane using a tile set and a blank tile (see [3]).
The proof of this theorem consists of a reduction from the halting problem
for Turing machines on an empty input. Let us consider such a machine with
a bi-innite tape; we transform it into a tile set as follows.
Let Q = fqo ; q1; : : :g be its set of states, q0 2 Q be the initial set, and
Qf Q be the set of halting states. Let S = fs0; s1; : : :g be the set of symbols,
and s0 2 S the blank symbol. Let M = fL; Rg be the possible movements
for the head of the machine. The action of the machine is determined by the
transition function : (Q ? Qf ) S ! S M Q.
The idea of the transformation is to force rows of the tilings to represent
the tape of the machine while columns will represent the evolutions of cells of
the tape during the computation. In other words, we construct a space time
diagram representing the (possibly innite) computation of the machine on an
empty tape.
More precisely, a conguration of the machine (tape, position of the head,
and state) is represented on the upper and lower edges of our tile set. This tile
set is described in Figures A.1 to A.3.
sk
sk
Figure A.1: Alphabet tile
2
qi s j
qi s j
qi
qi
sj
s
Figure A.2: Merging tiles
sj
q
sj
ql
l
qi s j
qi s j
Figure A.3: Action tiles
Remark that we use labeling and arrows to represent colors. It is an easy
exercise to prove that these representations are equivalent (i.e. it is possible to
transform a tile set with arrows into a tile set with colors without modifying
the possible tilings).
An \alphabet" (Figure A.1) tile transmits without modication the symbol
sk , and is constructed for all k. Merging tiles (Figure A.2) combine a state qi
with a symbol sj . For the sake of simplicity, we construct all these tiles but
not all of them will be able to take part of our tilings. The rst tile (resp.
the second tile) of Figure A.3 is constructed if and only if (qi; sj ) = (sk ; L; ql)
(resp. (qi; sj ) = (sk ; R; ql)).
Assume that we have a row of tiles which upper edges represent the tape of
the machine at time t. Assume also that the machine is not in a halting state.
Then, one of these tiles contains an up arrow labeled qi sj , and others contain
up arrow with a symbol label (such as sk ). Then there is only one possibility
to tile the next row: it must represent the conguration of the machine at time
t + 1.
If we assume that the machine starts on a blank tape, then we can use tiles
of Figure A.4 in order to represent its initial conguration.
s0
q 0 s0
s0
Figure A.4: Tiles involved in an initial conguration
Let us add now a blank tile to the tile set dened in Figures A.1 to A.4.
We obtain a tile set associated to the considered Turing machine. Let us select
the second tile of Figure A.4 as imposed tile. Then in order to tile the plane,
this tile must have on its left only the rst tile of Figure A.4 and on its right,
only the third tile of the same Figure. Then the only way to tile the bottom
half-plane is to use the blank tile. Thus, the upper sides of the row on which
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appears the imposed tile represent the initial conguration. Hence, one can tile
the plane with this imposed tile set if and only if the considered Turing machine
does not halt. The undecidability of the constrained domino problem is proved.
A.3 Robinson's aperiodic tile set
Figure A.5: Robinson's aperiodic tile set
Robinson's aperiodic tile set is formed by the 6 tiles of Figure A.5 and by their
images by all possible rotations and symmetries. We prove in the sequel that
this tile set can only form aperiodic tilings of the plane. These tiles are not
exactly Wang tiles since they have bumps and humps. Nevertheless, it is an
easy exercise to transform this tile set into a true set of Wang tiles.
Let us rst consider Robinson's tile set without the particular shape of
the corners. Then the rst two tiles of the second column are the same. We
represent bumps and humps by arrows: symmetrical bumps are represented
by a centered arrow, non-symmetrical ones are represented by two arrows, one
centered and the other one slightly shifted. We obtain the tiles of Figure A.6
and we add them all tiles obtained by symmetries and rotations. Then, we
transform arrows into colors and obtain a set of Wang tiles. For the sake of
simplicity we prefer to work on tiles of Figure A.6 rather than on the Wang
tiles obtained by the above transformation.
We call the rst tile of Figure A.6 \a cross" and other tiles \arms". The
direction of the pictured cross is \up-right". Arms point on the direction of their
main arrow. We represent these tiles with the help of the abbreviated symbols
of Figure A.7. The cross represents a cross-tile of any orientation while the
other one represents any arm the main arrow of which is as pictured.
On another hand, we can suppress bumps on sides of Robinson's tiles and we
obtain two polygons represented in Figure A.8 that we call respectively bumpy
and humpy.
Assume now that the whole plane is tiled by these two polygons. Then in
the neighborhood of each corner, one bumpy tile and three humpy ones must
be found. This implies that either the bumpy tile appears on one row every
4
Figure A.6: The ve basic tiles
Figure A.7: Abbreviated notations
two rows, and on this row every two cells, or it appears on one column every
two column, and on this column every two cells.
We thus obtain the periodic pattern of Figure A.9: bumpy tiles appear on
cells both coordinates of which are odd. There is another possible tiling with
these tiles which is depicted in Figure A.10. It is an easy combinatorial exercise
to prove that if a tiling contains a pattern as depicted in Figure A.9 then the
whole plane is tiled the same way, and if a pattern as depicted in Figure A.10
appears, then the whole plane is tiled the same way. More precisely, 6 tiles
placed on a 2 3 rectangle force the whole tiling.
Let us consider now a tiling of the plane with the basic tiles. Assume that
a cross appears somewhere. On its right, we nd a sequence of right arms
(possibly empty or innite), and then there is another cross. Two consecutive
crosses on a row are face-to-face or back-to-back. If they are face-to-face, then
the vertical arm between these crosses points down (see Figure A.11). If they
are back-to-back, then the vertical arm between these crosses points either up
or down: both congurations are possible (see Figure A.12).
The distance between two consecutive crosses is odd: let us assume for instance that both of them point up (see gure A.13). Then all tiles between
them have tails of arrows on their upper side {they are either vertical or horizontal arms. The tile above the left cross (resp. the right cross)is necessarily
a left arm (resp. a right arm) because this tile has a left tail (resp. right tail)
on its bottom side. Thus the upper-right tile of the cross is also a cross and by
Figure A.8: A bumpy and a humpy tile
5
Figure A.9: First solution
Figure A.10: Second solution
Figure A.11: Face-to-face crosses
Figure A.12: Back-to-back crosses
6
Figure A.13: This row is forced by two face-to-face crosses
iteration of this argument, one prove that there is one cross every two cells on
this up row. Hence the number of tiles between two crosses is odd.
Let us consider now a tiling of the plane with Robinson's tiles. We then
have the constraint that on one row every two row, one tile every two tiles
must be a cross. These particular crosses are humpy; we call them in the sequel
\1-squares".
Let us consider a 1-square. It faces another 1-square located two tiles farther. Then this row is either the bottom or the upper row of Figure A.14. A
cross must be present at the center of this square but its orientation is not
imposed. We thus obtain 4 possible 3-squares with 1-squares as corners.
Figure A.14: A 3-square
Let us consider now the central cross of a 3-square. It faces another cross.
Then this cross must be at the center of another 3-square otherwise double arms
would not coincide. We thus obtain a 7-square as illustrated in Figure A.15.
By iteration of the same construction, we can construct a (2n ? 1)-square for
all n 2 N. By Konig lemma, it is possible to tile the plane with Robinson's tile.
Let us prove now that all tilings by Robinson's tile are not periodic. We
have proved above that 1-squares must appear in all tilings. Each 1-square
uniquely determines the position of the 3-square in which it is included; there
are only 4 possible positions according to the orientation of the cross. Then,
the 3-square uniquely determines the position of the 7-square in which it is
included (4 possible positions). By iteration, for all n 2 N, the tiling contains
a 2n ? 1-square. Hence the tiling is not periodic.
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Figure A.15: A 7-square
A.4 The unconstrained domino problem
The goal of this section is to transform a Turing machine into a tile set such
that the machine does not halt on a blank tape if and only if one can tile the
plane with the tile set. The basic idea of the proof is that the computation
of the machine should be represented in a uniform manner in any tiling of the
plane. More precisely, we construct a tile set such that, given a time t, there
exists a size n such that the t rst steps of the computation of the machine
are represented in any n n square correctly tiled. Thus we have just to force
a tiling error when the halting state appears, and the theorem will be proved.
Remark that this notion of uniform distribution of computations is analogous
to the notion of quasiperiodicity: a tile set is quasi periodic if any nite pattern
that appears in the tiling appears in all suciently large squares.
Let us consider again Robinson's tiles presented in Section A.3. We proved
that in the center of all (2n+1 ? 1)-squares one must nd a cross, and that
a (2n+1 ? 1)-square is formed by four (2n ? 1)-squares. The centers of these
(2n ? 1)-squares are four crosses that delimit a square of size 2n that we call a
2n-frame. Furthermore, all crosses that are face-to-face delimit a 2k -frame.
Observe now that two of these frames intersect if and only if a corner of one
of them is the center of the other one. It is equivalent to say that a 2n-frame
intersects only one 2n+1 -frame and four 2n?1-frames (see Figure A.16).
We now modify the ve basic tiles (Figure A.6) by giving two dierent colors
for the slightly shifted arrow. This arrow can be either red or green with the
following condition: on arms, the color of horizontal arrows (resp. of vertical
arrows) must be the same. For the central tile (with two double arms), if the
vertical arrow is green, then the horizontal must be red (resp. red and green).
All tiles excepted the last one of Figure A.6 are duplicated once, hence we
obtain 9 tiles. We give to the green cross the humpy shape and we obtain a
new tile set.
8
Thus if we do not consider colors, tilings that can be formed with this tile
set are exactly the same as with Robinson's tiles. The only dierence is that
frames are colored, and if 2n-frames are red then 2n+1-frames are green and
conversely. The \humpy" constraint on the green cross imposes that 2-frames
are green and thus that a 2n-frame is green if n is odd and red if n is even.
Rouge
Vert
Figure A.16: A 4-frame
Let us consider now only red squares (that are 4n-frames). These squares
do not intersect and any tiling of the plane contains a sequence of frames of
size 4; 42; : : : ; 4n; : : :
In the sequel, we call 4n-zone the surface delimited by a 4n-frame. In this
zone, we are interested in \free" rows. These rows are those which do not
intersect a 4k -frame where k < n. In a 4n-zone, if we suppress columns and
rows that intersect a 4n?1-zone, we obtain four 4n?1-zones side-to-side. Thus if
the number of free rows is Fn?1 in a 4n?1-zone, it is Fn = 2Fn?1 in a 4n-zone.
As F1 = 3, Fn = 2n + 1. The same argument proves that there are Fn = 2n + 1
free columns in a 4n-zone. Hence there are (2n + 1)2 free cells in a 4n-zone
(located on a free row and on a free column). It is approximatively the square
root of the number of all cells of the zone.
It is easy to modify our tile set such that cells that are not free are marked
with a dierent color. In order to do that, we impose to red squares to send
vertically and horizontally an \obstruction" color outside their borders. This
transformation is very easy in its principle but tedious to prove. The reader
can do this transformation as an exercise.
Let us now consider a Turing machine. On each free tile, we superimpose
the tiles representing the space-time local behavior of the machine as represented in Figures A.1 to A.3. Tiles that are obstructed in only one direction
(vertically or horizontally) will transmit the state of the tape in the other di9
Rouge
Figure A.17: A 16-frame
10
rection (horizontally or vertically). We do not modify tiles that are obstructed
in both directions. Furthermore, we impose that the down-most free cells of a
zone are associated to the blank color of the tape (i.e. s0 ) excepted the center
one which is associated with the head in an initial state (i.e. q0 s0 ). We do not
give full details for this construction but its justication is that a copy of a
down-most tile in a red zone may only appear as a down-most tile in another
red zone. The center one is recognized because it is the only free down-most
cell which also belongs to a green frame.
Thus with this tile set, we can tile arbitrarily large squares of the plane
if and only if the Turing machine does not halt. Berger's theorem is proved.
Remark that the computation at time step t can be found in any 4t2 4t2
square of the tiling since these squares contain a 4n-zone with 4n > t2 .
A.5 The periodic problem and the inseparability result
With the previous construction, we can also prove the undecidability of the
periodic domino problem: is it possible with a tile set to tile the plane periodically? We modify the previous tile set such that it can tile the plane periodically
if and only if the associated Turing machine halts. Otherwise it can tile the
plane as in the previous proof (not periodically).
First, we transform the halting state of the machine in such a way that if
the machine enter this state, then it stays forever in this state and do nothing
on the tape. When a color qf si (where qf is a halting state) arrives on a red
frame, then this frame is transformed into a purple frame. It means that we
duplicate tiles for red frames into tiles for purple frames and that if a halting
state arrives on the frame, it is only possible to put these purple tiles. We
impose another dierence between red and purple frames: outside the zone,
edges of purple tiles are blank except for corners where the outside edges are
blue.
In other words, if the machine halts, then we have cut out suciently large
squares on which the halting state appears. These squares are blank on their
edges and blue on their corners. Then, the neighbors of such a square might
be identical squares. Thus if the machine halts, then it is possible to form a
periodic tiling on the plane with the tile set.
Conversely, if on can construct a periodic tiling of the plane, then a purple
tile (hence a purple square) must appear in it otherwise we have a tiling of the
plane with the tiles of the previous section and we proved that it cannot be
periodic. As there is a purple frame in the tiling, we can nd inside it a halting
computation of the Turing machine.
The undecidability result is proved but as an exercise, you can prove that
the tile set can form only periodic tilings if the machine halts, and can form
only aperiodic tilings if the machine does not halt.
We can strengthen this result and obtain the inseparability theorem (Theorem ?? in this book). Let us assume that the output of a Turing machine that
halts is the word between its head and the rst blank character on the tape.
11
Then we can do the previous construction with the dierence that the machine
is blocked if it enters a halting states with a `0' at the position of the head, and
with no modication otherwise. Thus, if the machine does not halt, then one
can tile the plane but only a-periodically; if the machine halts on a `0', then
the tile set cannot be used to tile the plane; otherwise, one can form a periodic
tiling of the plane. The converse of this proposition holds. Hence, if we denote
by x the tile set associated to the Turing machine x , then x (0) = 0 implies
that x cannot tile the plane, and x (0) = 1 implies that x can plane the plane
periodically.
The sets fx; x (0) = 1g and fx; x (0) = 0g are recursively inseparable:
assume that there exist a recursive set R such that fx; x (0) = 1g R and
fx; x(0) = 0g N ? R, then the recursive function f dened below with the
help of the s-n-m theorem would not have any xed point
(
0 if x 2 R;
f (x) (y ) =
1 if x 62 R:
because f (x) is everywhere dened and f (x) (0) = x (0) leads to a contradiction. Hence we have proved the recursive inseparability of the tile sets that
cannot tile the plane and those that can be used to tile the plane periodically.
12
Bibliography
[1] R. Berger. The undecidability of the domino problem. Mem. AMS, 66,
1966.
Proves the undecidability of the unconstrained domino problem introduced
by [?]. The proof has been considerably simplied by Robinson [3]. Gurevich
and Koryakov [2] have strengthened the result showing that the classes of
domino problems with no or with periodic solution respectively are recursively inseparable.
[2] Y. Gurevich and I. Koryakov. Remarks on Berger's paper on the domino
problem. , 13:319{321, 1972.
Berger's undecidability proof for the unconstrained domino problem (see
[1]) is strenghtened by showing that the classes of domino problems with no
or with periodic solution respectively are recursively inseparable.
[3] R. Robinson. Undecidability and non-periodicity for tilings of the plane. ,
12:177{209, 1971.
Simplies Berger's proof (see [1]) of the undecidability of the unconstrained
domino problem introduced by [?].
[4] H. Wang. Dominoes and the 898-case of the decision problem. In Proc.
Symp. on Mathematical Theory of Automata, pages 23{55. Brooklyn Polytechnic Institute, New York, 1962.
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