csr2011 june16 14_00_kari
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Snakes and Cellular Automata
Reductions and Inseparability Results
Jarkko Kari
Department of Mathematics, University of Turku, Finland
TUCS (Turku Centre for Computer Science)
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Wang tiles
A Wang tile is a unit size square tile with colored edges.
A tile set T is a finite collection of such tiles:
A B C D
A valid tiling is an assignment
Z2 −→ T
of tiles on infinite square lattice so that the abutting edges of
adjacent tiles have the same color.
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For example, with copies of the given four tiles we can
properly tile a 5× 5 square. . .
A
B
C
D
C
A
C
B
D
C
B
D
A
C
C
B
D
C
A
C
B
A
C
D
C
. . . and since the colors on the borders match this square can
be repeated to form a valid periodic tiling of the plane.
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A decision problem:
The domino problem: Given a set T of Wang tiles, does T
admit a valid tiling of the plane?
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An obvious attempt to solve the domino problem:
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An obvious attempt to solve the domino problem:
• Test if the 2× 2 square can be tiled. If not, answer “no”.
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An obvious attempt to solve the domino problem:
• Test if the 2× 2 square can be tiled. If not, answer “no”.
• Test if the 3× 3 square can be tiled. If not, answer “no”.
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An obvious attempt to solve the domino problem:
• Test if the 2× 2 square can be tiled. If not, answer “no”.
• Test if the 3× 3 square can be tiled. If not, answer “no”.
• Test if the 4× 4 square can be tiled. If not, answer “no”.
• . . .
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An obvious attempt to solve the domino problem:
• Test if the 2× 2 square can be tiled. If not, answer “no”.
• Test if the 3× 3 square can be tiled. If not, answer “no”.
• Test if the 4× 4 square can be tiled. If not, answer “no”.
• . . .
If T does not admit a tiling, then eventually we encounter a
number n such that T does not tile the n× n square, and we
get the correct“no”-answer.
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An obvious attempt to solve the domino problem:
• Test if the 2× 2 square can be tiled. If not, answer “no”.
• Test if the 3× 3 square can be tiled. If not, answer “no”.
• Test if the 4× 4 square can be tiled. If not, answer “no”.
• . . .
If T does not admit a tiling, then eventually we encounter a
number n such that T does not tile the n× n square, and we
get the correct“no”-answer.
A big problem: If T admits tiling then the process never
stops. We never get the positive answer. This is only a
semi-algorithm for non-tileability.
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Second attempt: For every n = 2, 3, 4 . . ., construct all
possible tilings of the n× n square.
• If the n× n square can not be tiled, answer “no”.
• If the n× n square can be tiled in a way that can be
repeated periodically to cover the plane, answer “yes”.
Now we get the correct “no” -answer if no tiling exists, and the
correct “yes” -answer if a periodic tiling exists.
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Second attempt: For every n = 2, 3, 4 . . ., construct all
possible tilings of the n× n square.
• If the n× n square can not be tiled, answer “no”.
• If the n× n square can be tiled in a way that can be
repeated periodically to cover the plane, answer “yes”.
Now we get the correct “no” -answer if no tiling exists, and the
correct “yes” -answer if a periodic tiling exists.
But: there exist aperiodic tile sets that tile the plane only in
non-periodic ways. On such cases our second attempt fails to
return any answer.
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In fact, no general procedure exists:
Theorem (R. Berger 1966): The domino problem is
undecidable.
As part of his proof, R. Berger also constructed the first
known aperiodic tile set.
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We also need the following variant of the domino problem:
The finite domino problem: Given a set T of Wang tiles,
containing a specific blank tile b, does T admit a valid tiling of
the plane such that all but a finite, non-zero number of tiles
are blank:
A
B C
bC
b b b
b
b
bb
b
b
b
b
D
D
b
b
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Theorem. The finite domino problem is undecidable.
The proof is easy, using H.Wang’s original technique to
simulate Turing machines on tiles.
Remark:
• The negative instances of the domino problem
• The positive instances of the finite domino problem
are semi-decidable.
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Outline of the talk
1. Wang tiles and the Domino problem
2. Two applications
(A) Self-assembly
• Termination problem
• Snake tiling problem
• Loop tiling problem
• Recursive inseparability
(B) Cellular automata
• Reversibility and surjectivity
• Inseparability
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Self-assembly
• Self-assembly = the process by which objects
autonomously come together to form complex structures
• Examples
– Atoms bind by chemical bonds to form molecules
– Molecules may form crystals or macromolecules
– Cells interact to form organisms
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Motivation for self-assembly
• Nanotechnology: miniaturization in medicine,
electronics, engineering, material science, manufacturing
– Top-down techniques: nanolithography
– Bottom-up techniques: self-assembly. Design
molecules that self-assemble to form the target
structure.
• Computation using self-assembly: In 1994 L.Adleman
demonstrated that the self-assembly process of DNA
molecules can be used to solve computational problems
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Theoretical model based on Wang tiles
A simplified theoretical model to study algorithms and
complexity related to self-assembly (L. Adleman, E. Winfree).
A B C D
Each Wang tile is a “molecule” that may participate in the
assembly. Tiles are placed on the plane at integer lattice
positions indexed by Z2. The tiles are not rotated.
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Two adjacent tiles stick if they have the same color at the
abutting edges:
DA
In the model,
• We have a finite tile set of different tiles. [Finite number of
different “molecules” in use.]
• We have an unlimited supply of copies of the tiles
available. [We do not control the precise number of
molecules participating in the self-assembly process.]
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Structures self-assemble with tiles from T by incrementally
adding single tiles that stick:
A
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Structures self-assemble with tiles from T by incrementally
adding single tiles that stick:
A
C
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Structures self-assemble with tiles from T by incrementally
adding single tiles that stick:
A
C B
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Structures self-assemble with tiles from T by incrementally
adding single tiles that stick:
A
C
C B
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Structures self-assemble with tiles from T by incrementally
adding single tiles that stick:
A
C
C B
B
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Structures self-assemble with tiles from T by incrementally
adding single tiles that stick:
AB
C
C B
B
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Structures self-assemble with tiles from T by incrementally
adding single tiles that stick:
AB
C
C B
B
A
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Structures self-assemble with tiles from T by incrementally
adding single tiles that stick:
AB
C
D
C B
B
A
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There are two natural choices of matching rules:
• Under strong sticking rules, each new tile must stick to
all its neighbors that have been assembled before it.
AB
C
D
C B
B
A
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There are two natural choices of matching rules:
• Under strong sticking rules, each new tile must stick to
all its neighbors that have been assembled before it.
AB
C
D
C B
B
A
C
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There are two natural choices of matching rules:
• Under strong sticking rules, each new tile must stick to
all its neighbors that have been assembled before it.
• Under weak sticking rules, it is enough for a tile to stick
to one previous tile.
AB
C
D
C B
B
A
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There are two natural choices of matching rules:
• Under strong sticking rules, each new tile must stick to
all its neighbors that have been assembled before it.
• Under weak sticking rules, it is enough for a tile to stick
to one previous tile.
AB
C
D
C B
B
A
B
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There are two natural choices of matching rules:
• Under strong sticking rules, each new tile must stick to
all its neighbors that have been assembled before it.
• Under weak sticking rules, it is enough for a tile to stick
to one previous tile.
A more general approach: assign a strength value to each
color and say that a tile sticks if the sum of the strengths
between the tile and its neighbors exceeds some given
threshold.
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The tiling model is a simplification of real self-assembly
systems. However, it is already complex enough to admit
universal computation.
If some questions concerning the tiling model are
(computationally) difficult, then surely the same questions
about real self-assembly systems are at least as difficult.
=⇒The tiling model gives Lower bounds on
computational problems on self-asssembly.
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Several interesting algorithmic questions arise:
• Does a given tile set assemble a unique shape ?
• Complexity questions: How many tiles are needed in a set
that self-assembles into a given shape ?
• Can the given tile set self-assemble without limit,
i.e., is unbounded growth possible ?
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The termination problem
A pattern is a terminal assembly if no more tiles can be
added.
A tile set admits unbounded growth if an infinite sequence
of tile additions without ever reaching a terminal assembly is
possible.
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The termination problem
A pattern is a terminal assembly if no more tiles can be
added.
A tile set admits unbounded growth if an infinite sequence
of tile additions without ever reaching a terminal assembly is
possible.
=⇒ Decision problem. . .
The termination problem: Does a given tile set admit
unbounded growth (under strong/weak sticking rules) ?
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Easy to see:
• an unbounded assembly is possible if and only if the tiles
admit a tiling of a (bi-infinite) snake.
Snake: an injective function
s : Z −→ Z2
such that s(i) and s(i + 1) are neighbors for all i.
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• If the assembly uses strong sticking rules then the snake
must be tiled so that all neighboring tiles of the snake
must match. We call this a strong snake.
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• If the assembly uses strong sticking rules then the snake
must be tiled so that all neighboring tiles of the snake
must match. We call this a strong snake.
• If the assembly uses weak sticking rules then it is enough
that the consecutive tiles along the snake stick. This is
a weak snake.
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=⇒ Decision problems. . .
Snake tiling problems: Does a given finite set of Wang tiles
admit a strong/weak snake?
These are equivalent to the termination problems in
self-assembly (under the strong/weak sticking rules,
respectively).
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=⇒ Decision problems. . .
Snake tiling problems: Does a given finite set of Wang tiles
admit a strong/weak snake?
These are equivalent to the termination problems in
self-assembly (under the strong/weak sticking rules,
respectively).
The Snake tiling problem was actually asked by Y. Etzion-
Petruschka, D.Harel and D.Myers already in 1994, long before
the self-assembly motivation of the problem.
Our answer: The questions are undecidable. We can
reduce the domino problem into them.
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Tile set Snakes
In the reduction of the Domino problem into the Snake
tiling problem, a specific tile set Snakes is used.
Tiles of Snakes have an arrow printed on them. The arrow is
horizontal or vertical and it points to one of the four neighbors
of the tile:
Such tiles with arrows are called directed tiles.
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A directed snake is a snake, tiled with directed tiles, in
which the direction of each tile points to the next tile along
the snake.
Ok
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A directed snake is a snake, tiled with directed tiles, in
which the direction of each tile points to the next tile along
the snake.
Not Ok
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A directed snake is a snake, tiled with directed tiles, in
which the direction of each tile points to the next tile along
the snake.
Not Ok
Two variants: Strong and weak directed snakes – but we
only consider the strong variant.
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The directed tile set Snakes has the following plane-filling
property:
(PFP) A strong directed snake formed by Snakes is
necessarily a plane-filling path.
[More precisely: for every positive integer n there exists an
n× n square all of whose positions are visited by the path.]
Snakes also has the property that it admits a valid tiling.
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The strong directed snakes by Snakes have the shape of the
well known plane-filling Hilbert-curve
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The strong directed snakes by Snakes have the shape of the
well known plane-filling Hilbert-curve
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The strong directed snakes by Snakes have the shape of the
well known plane-filling Hilbert-curve
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The strong directed snakes by Snakes have the shape of the
well known plane-filling Hilbert-curve
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The strong directed snakes by Snakes have the shape of the
well known plane-filling Hilbert-curve
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The strong directed snakes by Snakes have the shape of the
well known plane-filling Hilbert-curve
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The strong directed snakes by Snakes have the shape of the
well known plane-filling Hilbert-curve
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The strong directed snakes by Snakes have the shape of the
well known plane-filling Hilbert-curve
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Directed snake tiling problem: Does a given set of
directed tiles admit a strong directed snake ?
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Directed snake tiling problem: Does a given set of
directed tiles admit a strong directed snake ?
Proof of undecidability: We reduce the Domino problem,
using the tile set Snakes.
For any given Wang tile set T , construct the directed tile set
D = T × Snakes.
• Matching condition is that both layers stick.
• Snakes -component gives the direction.
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Claim: The sandwich tiles in D admit a strong directed snake
if and only if T admits a tiling of the plane
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Claim: The sandwich tiles in D admit a strong directed snake
if and only if T admits a tiling of the plane
(⇐=) If T tiles the plane then D tiles the plane. Any path
that follows the arrows is a strong directed snake.
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Claim: The sandwich tiles in D admit a strong directed snake
if and only if T admits a tiling of the plane
(=⇒) Suppose D admits a strong directed snake s.
• Plane-filling property of Snakes =⇒ s covers arbitrarily
large squares.
• No tiling errors in the T component along s =⇒ T can tile
arbitrarily large squares =⇒ T admits a tiling of the plane.
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Claim: The sandwich tiles in D admit a strong directed snake
if and only if T admits a tiling of the plane
=⇒ Directed snake tiling problem is undecidable.
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(Undirected) snake tiling problems
Next we reduce the Directed snake tiling problem to the
(undirected, strong and weak) Snake tiling problems.
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(Undirected) snake tiling problems
Next we reduce the Directed snake tiling problem to the
(undirected, strong and weak) Snake tiling problems.
Let D be any given set of directed tiles, called macro-tiles.
We effectively construct a set T of Wang tiles, called
mini-tiles, such that
• If D admits a directed strong snake then T admits
(undirected) strong and weak snakes, and
• If D does not admit a directed strong snake then T does
not admit (undirected) strong or weak snakes.
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(Undirected) snake tiling problems
A motif construction is used to convert
• Directed −→ Undirected
• Strong sticking −→ weak sticking
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For each macro-tile d ∈ D we build a sequence of undirected
mini-tiles. The mini-tiles are colored with unique colors that
force them to form a motif: a finite snake that goes around the
the four edges of the macro-tile d:
macro-tile d Corresponding motif of mini-tiles
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Since the entry direction into a tile is not specified by the
arrow, we need three motifs for each d ∈ D, one for every
possible entry direction:
Left entry Bottom entry Right entry
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Each motif matches other motifs only at the two ”free” ends.
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Gluing motifs from their free ends forms snakes that exactly
correspond to paths of macro-tiles specified by arrows:
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Gluing motifs from their free ends forms snakes that exactly
correspond to paths of macro-tiles specified by arrows:
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Gluing motifs from their free ends forms snakes that exactly
correspond to paths of macro-tiles specified by arrows:
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So far the motifs have no constraints corresponding to edge
coloring of the macro-tiles. To simulate the colors, the motifs
are bent to form one bump or dent on each side of the motif.
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The N and E side of each motif contains one bump, and the S
and W side contains one dent:
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The exact position of the bump or the dent encodes the color
of the edge in the corresponding macro-tile. Each color
corresponds to a unique position.
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The exact position of the bump or the dent encodes the color
of the edge in the corresponding macro-tile. Each color
corresponds to a unique position.
If the colors of two adjacent macro-tiles match then the bump
exactly fits inside the dent:
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The exact position of the bump or the dent encodes the color
of the edge in the corresponding macro-tile. Each color
corresponds to a unique position.
But if the colors do not match then the two motifs would
overlap, which is not possible:
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Claim: Macro-tiles in D admit a strong directed snake if and
only if the mini-tiles in T admit a (strong/weak) infinite snake.
=⇒ (Strong and weak) snake tiling problems are
undecidable, as are the termination problems
of self-assembly.
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Loops
With a similar argument we can prove the undecidability of
the Loop tiling problems.
Loop: a finite snake whose ends meet. A loop of length n is a
one-to-one function
s : {1, 2, . . . , n} −→ Z2
such that s(i) and s(i + 1 mod n) are neighbors for all
i = 1, 2, . . . , n.
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We tile loops with Wang tiles:
• Strong loop: all neighboring tiles on the loop stick.
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We tile loops with Wang tiles:
• Strong loop: all neighboring tiles on the loop stick.
• Weak loop: only consecutive tiles are required to stick.
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We tile loops with Wang tiles:
• Strong loop: all neighboring tiles on the loop stick.
• Weak loop: only consecutive tiles are required to stick.
Or we tile them with directed tiles:
• Strong directed loop: all neighboring tiles on the loop
stick, and the loop follows the arrows on the tiles.
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We tile loops with Wang tiles:
• Strong loop: all neighboring tiles on the loop stick.
• Weak loop: only consecutive tiles are required to stick.
Or we tile them with directed tiles:
• Strong directed loop: all neighboring tiles on the loop
stick, and the loop follows the arrows on the tiles.
• Weak directed loop. . .
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=⇒ Decision problems. . .
Loop tiling problems: Does a given set of (directed) Wang
tiles admit a strong/weak (directed) loop ?
All variants are undecidable.
Remark:
• The negative instances of the snake tiling problems
• The positive instances of the loop tiling problems
are semi-decidable.
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Tile set Loops
To prove the Loop tiling problems undecidable, we reduce
the Finite domino problem using a specific set Loops of
directed tiles.
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The tiles in Loops are partitioned into three parts:
(A) Normal tiles (B) Blue tiles (C) Green tiles
Loops has a rectangle-covering property:
(RCP) Any strong directed loop formed by Loops
completely covers some rectangle whose boundary tiles
are green and there is a single blue tile in the interior.
Loops also has the property that it admits valid tilings
containing arbitrarily long loops.
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We reduce the finite domino problem to the directed loop
tiling problem, using the tile set Loops:
For any given Wang tile set T , construct the directed tile set
D ⊆ T × Loops.
Loops
with the constraints that in (t, l) ∈ D
• if l is green then t is blank, and
• if l is blue then t is not blank.
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Claim: The sandwich tiles in D admit a strong directed loop
if and only if T admits a tiling of the plane such that all but a
finite, non-zero number of tiles are blank.
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Claim: The sandwich tiles in D admit a strong directed loop
if and only if T admits a tiling of the plane such that all but a
finite, non-zero number of tiles are blank.
(⇐=)
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Claim: The sandwich tiles in D admit a strong directed loop
if and only if T admits a tiling of the plane such that all but a
finite, non-zero number of tiles are blank.
(⇐=)
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Claim: The sandwich tiles in D admit a strong directed loop
if and only if T admits a tiling of the plane such that all but a
finite, non-zero number of tiles are blank.
(=⇒)
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Claim: The sandwich tiles in D admit a strong directed loop
if and only if T admits a tiling of the plane such that all but a
finite, non-zero number of tiles are blank.
(=⇒)
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Claim: The sandwich tiles in D admit a strong directed loop
if and only if T admits a tiling of the plane such that all but a
finite, non-zero number of tiles are blank.
(=⇒)
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Claim: The sandwich tiles in D admit a strong directed loop
if and only if T admits a tiling of the plane such that all but a
finite, non-zero number of tiles are blank.
Any algorithm to solve the directed loop tiling problem
yields an algorithm to solve the finite domino problem
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To change
• directed −→ undirected,
• strong −→ weak,
we use the same motif construction, obtaining the
undecidability of all variants of the loop tiling problem.
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Recursive inseparability
Disjoint sets A,B ⊆ Σ∗ are recursively inseparable if there
is no recursive R such that A ⊆ R and B ∩R = ∅.
A B
R Σ \ R*
Recursively inseparable RE sets are known to exist.
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The undecidability proofs for snakes and loops can be
combined to show the rec. inseparability of the following two
classes of tile sets:
A: Tile sets T that do not admit any infinite snakes or any
loops (not even weak ones)
B: Tile sets T that admit arbitrarily long (even strong) loops
(=⇒ admit also infinite snakes).
No loops,No snakes
Both loopsand snakes
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To prove rec. inseparability:
• Merge Snakes and Loops into a single tile set Comb
with both PFP and RCP.
• Fix two rec. inseparable RE sets A,B ⊆ Σ∗.
• For any given x ∈ Σ∗ effectively construct tile sets TA and
TB such that
x ∈ A⇐⇒ TA admits a finite tiling,
x ∈ B ⇐⇒ TB admits no infinite tiling.
• Construct three layer sandwiches
Comb
T
T
A
B
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Outline of the talk
1. Wang tiles and the Domino problem
2. Two applications
(A) Self-assembly
• Termination problem
• Snake tiling problem
• Loop tiling problem
• Recursive inseparability
(B) Cellular automata
• Reversibility and surjectivity
• Inseparability
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Cellular automata (CA)
Cellular automata are among the oldest models of natural
computing. They are investigated
• in physics as discrete models of physical systems,
• in computer science as models of massively parallel
computation under the realistic constraints of locality and
uniformity,
• in mathematics as endomorphisms of the full shift in the
context of symbolic dynamics.
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Example: the Game-of-Life by John Conway.
• Infinite square grid whose squares (=cells) are colored
black (=alive) or white (=dead).
• At each discrete time step each cell counts the number of
living cells surrounding it, and based on this number
determines its new state.
• All cells change their state simultaneously.
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The local update rule asks each cell to check the present states
of the eight surrounding cells.
• If the cell is alive then it stays alive (survives) iff it has
two or three live neighbors. Otherwise it dies of loneliness
or overcrowding.
• If the cell is dead then it becomes alive iff it has exactly
three living neighbors.
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The local update rule asks each cell to check the present states
of the eight surrounding cells.
• If the cell is alive then it stays alive (survives) iff it has
two or three live neighbors. Otherwise it dies of loneliness
or overcrowding.
• If the cell is dead then it becomes alive iff it has exactly
three living neighbors.
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The local update rule asks each cell to check the present states
of the eight surrounding cells.
• If the cell is alive then it stays alive (survives) iff it has
two or three live neighbors. Otherwise it dies of loneliness
or overcrowding.
• If the cell is dead then it becomes alive iff it has exactly
three living neighbors.
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The local update rule asks each cell to check the present states
of the eight surrounding cells.
• If the cell is alive then it stays alive (survives) iff it has
two or three live neighbors. Otherwise it dies of loneliness
or overcrowding.
• If the cell is dead then it becomes alive iff it has exactly
three living neighbors.
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More generally: A two-dimensional cellular automaton
(2D CA) over a finite state set S is determined by a local
update rule that gives the new state of each cell from the old
states of its neighbors.
The update rule is applied synchronously at all cells.
The resulting function
G : SZ2
−→ SZ2
is the CA function.
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Reversible CA
A CA is called
• injective if G is one-to-one,
• surjective if G is onto,
• bijective if G is both one-to-one and onto.
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Reversible CA
A CA is called
• injective if G is one-to-one,
• surjective if G is onto,
• bijective if G is both one-to-one and onto.
A CA G is a reversible (RCA) if there is another CA
function F that is its inverse, i.e.
G ◦ F = F ◦G = identity function.
RCA G and F are called the inverse automata of each other.
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A reversible CA must be bijective but also the converse is true:
Theorem (Curtis-Hedlund-Lyndon): A cellular
automaton G is reversible if and only if it is bijective.
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A reversible CA must be bijective but also the converse is true:
Theorem (Curtis-Hedlund-Lyndon): A cellular
automaton G is reversible if and only if it is bijective.
In bijective CA, each cell can determine its previous
state by looking at the current states in some bounded
neighborhood around it.
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Easy to see:
G injective =⇒ G surjective
so
G injective ⇐⇒ G bijective ⇐⇒ G reversible
Injective= bijective= reversible
Surjective
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Decision problems. . .
The reversibility problem: Is a given CA reversible ?
The surjectivity problem: Is a given CA surjective ?
Both problems are undecidable for 2D CA.
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Decision problems. . .
The reversibility problem: Is a given CA reversible ?
The surjectivity problem: Is a given CA surjective ?
Both problems are undecidable for 2D CA.
Remark:
• The positive instances of the reversibility problem
• The negative instances of the surjectivity problem
are semi-decidable.
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Even better: The classes of
A: Reversible 2D CA
B: Non-surjective 2D CA
are recursively inseparable
Non-surjective2D CA Reversible
2D CA
Proof: direct reduction from the snake and loop tiling
problems.
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The reductions: For a given Wang tile set T , construct a 2D
CA with the state set
T × {←, ↑,→, ↓} × {0, 1}.
0
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The reductions: For a given Wang tile set T , construct a 2D
CA with the state set
T × {←, ↑,→, ↓} × {0, 1}.
0
The neighbor in the direction of the arrow is the follower.
A cell is active iff the tile matches in color with the follower.
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The reductions: For a given Wang tile set T , construct a 2D
CA with the state set
T × {←, ↑,→, ↓} × {0, 1}.
0
1
Local rule:
• Cell inactive (=no match) =⇒ no change in the cell.
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The reductions: For a given Wang tile set T , construct a 2D
CA with the state set
T × {←, ↑,→, ↓} × {0, 1}.
0
1
Local rule:
• Cell inactive (=no match) =⇒ no change in the cell.
• Cell active (=colors match) =⇒ the bit component is
XORed with the bit of follower.
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The reductions: For a given Wang tile set T , construct a 2D
CA with the state set
T × {←, ↑,→, ↓} × {0, 1}.
1
1
Local rule:
• Cell inactive (=no match) =⇒ no change in the cell.
• Cell active (=colors match) =⇒ the bit component is
XORed with the bit of follower.
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Claim 1: If T admits a weak loop, the CA is not surjective.
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Claim 1: If T admits a weak loop, the CA is not surjective.
Proof: Assume a weak loop.
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Claim 1: If T admits a weak loop, the CA is not surjective.
Proof. Assume a weak loop:
0 10
00
0
0
0Pattern with no pre-image (single cell of the loop carries bit 1).
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Claim 2: If the CA is not reversible then T admits a loop or a
snake
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Claim 2: If the CA is not reversible then T admits a loop or a
snake
Proof. Assume two different configurations have the same
image:
0 1
They differ in the bit components only.
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Claim 2: If the CA is not reversible then T admits a loop or a
snake
Proof. Assume two different configurations have the same
image:
0a
1a
The cell must be active, and the bits in the follower must be
different.
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Claim 2: If the CA is not reversible then T admits a loop or a
snake
Proof. Assume two different configurations have the same
image:
0a
b
1a
b
The reasoning can be repeated on the follower. It must be
active, and the bits in its follower are different.
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Claim 2: If the CA is not reversible then T admits a loop or a
snake
Proof. Assume two different configurations have the same
image:
0a
b
1a
b
Repeat the reasoning over-and-over =⇒ A weak snake or a
loop.
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So we have
• T admits loops and snakes =⇒ CA is not surjective.
• T has no loops or snakes =⇒ CA is reversible.
Theorem: Reversibility and non-surjectivity are recursively
inseparable among 2D CA.
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Conclusions
Any property of 2D CA that is between reversibility and
surjectivity is undecidable. Some examples of such properties:
• Openness,
• Closingness,
• Density of periodic orbits.
Density of periodic orbits is conjectured to be equivalent to
surjectivity. We have no idea on the solution of this long
standing open problem but, in any case, we know now that it
is undecidable if a given 2D CA has dense periodic orbits.
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Conclusions
It would be interesting to refine the inseparability result by
shrinking the classes. For example, are
• periodic CA
• eventually periodic, but not periodic CA
recursively inseparable ?
Non-surjective2D CA Reversible
2D CA
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Thank You