Five Lectures on CA 2. Simulations Thomas Worsch Department of - - PowerPoint PPT Presentation

Five Lectures on CA

• 2. Simulations

Thomas Worsch

Department of Informatics Karlsruhe Institute of Technology http://liinwww.ira.uka.de/~thw/vl-hiroshima/

at Hiroshima University, January 2012

Outline

Turing machines Simulation of TM by CA Simulation of CA by TM

Turing machines Simulation of TM by CA Simulation of CA by TM

Definition Turing machine, one head, one tape (1)

Definition Turing machine, one head, one tape (1)

Basics:

◮ finite set of states Q ◮ finite tape alphabet S ◮ move function δ : Q × S → Q × S × D

◮ D = {−1, 0, 1}: directions for head move

−1 for moving left, 1 for moving right and 0 for no move

◮ let’s assume that δ is a total function in this lecture

◮ a blank symbol ✷ ◮ an initial state

Definition Turing machine, one head, one tape (2)

derived notions

◮ configuration of a TM

◮ c = (q, b, p) ∈ Q × SZ × Z

◮ write the move function as three separate functions:

◮ δq : Q × S → Q ◮ δs : Q × S → S ◮ δd : Q × S → D

◮ one step of a TM: ∆ : Q × SZ × Z → Q × SZ × Z ◮ For c = (q, b, p) the succesor configuration ∆(c) = (q′, b′, p′)

is defined as

◮

q′ = δq(q, b(p)),

◮

∀x ∈ Z holds b′(x) = b(x) if x = p δs(q, b(p)) if x = p

◮

p′ = p + δd(q, b(p))

Turing machines: recent developments :-)

◮ http://www.youtube.com/watch?v=E3keLeMwfHY ◮ Steam-Powered Turing Machine ◮ http://www.cs.washington.edu/general/sptm-caption.html ◮ http://www.cs.washington.edu/homes/ruzzo/

”[. . . ] principal research project involves the construction and programming of a vaguely parallel computer, consisting of 32 steam-powered Turing machines installed in the basement [. . . ] Graduate students have played an important role in the construction and operation of the engine, [...] Originally intended as a general computing engine, restrictions imposed by the Pollution Control and Noise Abatement Boards require that only algorithms running in polynomial time may be

• used. [. . . ] one of [the] students slipped on a mouldering stack of

ungraded homework exercises and fell under the write head of one

• f the machines. Now permanently embossed with a series of 1’s

and 0’s, the student is suing to have the machine dismantled.“

Turing machines Simulation of TM by CA Simulation of CA by TM

Theorem

For each TM T = (QT, ST, δT) there is a CA C = (Z, QC, {−1, 0, 1}, fC, ) simulating T without loss of time step by step.

Sketch of proof: idea

a1 a2 a3 a4 a5 q ∆T a1 a2 a4 a5 a′

3

q′

Sketch of proof: idea

a1 a2 a3 a4 a5 q ∆T a1 a2 a4 a5 a′

3

q′ FC

Sketch of proof: idea

a1 a2 a3 a4 a5 q ∆T a1 a2 a4 a5 a′

3

q′ FC

Sketch of proof: idea

a1 a2 a3 a4 a5 q ∆T a1 a2 a4 a5 a′

3

q′ a1 a2 a3 a4 a5 FC a1 a2 a4 a5 a′

3

Sketch of proof: idea

a1 a2 a3 a4 a5 q ∆T a1 a2 a4 a5 a′

3

q′ a1 a2 a3 a4 a5 q FC a1 a2 a4 a5 a′

3

q′

Sketch of proof: idea

◮ let QC = (QT ˙

∪{ }) × ST

◮ represent cT = (q, b, p) by

cC : Z → QC x → (q, b(p)) if x = p ( , b(x)) if x = p

Sketch of proof: local transition function

for the simulation to work define fC(( , al), ( , am), ( , ar)) = ( , am)

Sketch of proof: local transition function

for the simulation to work define fC(( , al), ( , am), ( , ar)) = ( , am) fC((q, al), ( , am), ( , ar)) = (q′, am) if δT(q, al) = (q′, a′

l, 1)

( , am)

• therwise

Sketch of proof: local transition function

for the simulation to work define fC(( , al), ( , am), ( , ar)) = ( , am) fC((q, al), ( , am), ( , ar)) = (q′, am) if δT(q, al) = (q′, a′

l, 1)

( , am)

• therwise

fC(( , al), (q, am), ( , ar)) = (q′, a′

m)

if δT(q, am) = (q′, a′

m, 0)

( , a′

m)

if δT(q, am) = (q′, a′

m, ±1)

Sketch of proof: local transition function

for the simulation to work define fC(( , al), ( , am), ( , ar)) = ( , am) fC((q, al), ( , am), ( , ar)) = (q′, am) if δT(q, al) = (q′, a′

l, 1)

( , am)

• therwise

fC(( , al), (q, am), ( , ar)) = (q′, a′

m)

if δT(q, am) = (q′, a′

m, 0)

( , a′

m)

if δT(q, am) = (q′, a′

m, ±1)

fC(( , al), ( , am), (q, ar)) = (q′, am) if δT(q, ar) = (q′, a′

r, −1)

( , am)

• therwise

This does not yet completely specify f . The rest is not important here.

Generalization: more tapes and heads

more complicated:

◮ TM with two tapes and one head on each of them ◮ TM with two heads on one tape ◮ TM with arbitrary numbers of tapes and heads

problem:

◮ heads can be “far away” from each other,

• e. g. copying data from one head to the other

◮ still we want to simulate that in one (or two) steps

Generalization: more tapes and heads (2)

basic idea for a solution:

Generalization: more tapes and heads (2)

basic idea for a solution:

◮ always keep the important information in one cell

• i. e. the information needed for the simulation of the next step

Generalization: more tapes and heads (2)

basic idea for a solution:

◮ always keep the important information in one cell

• i. e. the information needed for the simulation of the next step

◮ like this:

Generalization: more tapes and heads (2)

basic idea for a solution:

◮ always keep the important information in one cell

• i. e. the information needed for the simulation of the next step

◮ like this:

◮ don’t move the head(s) but the tape contents

Generalization: more tapes and heads (2)

basic idea for a solution:

◮ always keep the important information in one cell

• i. e. the information needed for the simulation of the next step

◮ like this:

◮ don’t move the head(s) but the tape contents ◮ don’t wait until everything has been shifted . . .

Generalization: more tapes and heads (2)

basic idea for a solution:

◮ always keep the important information in one cell

• i. e. the information needed for the simulation of the next step

◮ like this:

◮ don’t move the head(s) but the tape contents ◮ don’t wait until everything has been shifted . . . ◮ . . . start simulating the next step as soon as all necessary

informations are available

Shifting one tape (1)

Shifting one tape (1)

. . . q . . . . . . a b c d e f g h . . . δT(q, d) = . . . . . . (q′, x, −1) . . . q′ . . . . . . a b c

←

✷ e f g h . . . . . .

→

x . . . . . . q′ . . . . . . a b

←

✷ c x f g h . . . . . .

→

e . . . . . . q′ . . . . . . a

←

✷ b c x e g h . . . . . .

→

f . . . . . . q′ . . . . . .

←

✷ a b c x e f h . . . . . .

→

g . . .

Shifting one tape (2)

. . . q . . . . . . a b c d e f g h . . . δT(q, d) = . . . . . . (q′, x, −1) . . . q′ . . . . . . a b c

←

✷ e f g h . . . . . .

→

x . . . . . . q′ . . . . . . a b

←

✷ c x f g h . . . δT(q′, c) = . . .

→

e . . . (q′′, y, −1) . . . q′′ . . . . . . a

←

✷ b

←

✷ x e g h . . . . . .

→

y

→

f . . . . . . q′′ . . . . . .

←

✷ a

←

✷ b y e f h . . . . . .

→

x

→

g . . .

Shifting more

◮ the same technique works for more tapes and one head on

each tape

◮ if one has k heads on the same tape:

Shifting more

◮ the same technique works for more tapes and one head on

each tape

◮ if one has k heads on the same tape:

◮ simulate it by k tapes with only 1 head on each

without loss of time (Stoß, 1970) or

Shifting more

◮ the same technique works for more tapes and one head on

each tape

◮ if one has k heads on the same tape:

◮ simulate it by k tapes with only 1 head on each

without loss of time (Stoß, 1970) or

◮ use the same idea as before: store all currently visited tapes

squares in one cell . . .

Shifting more

◮ the same technique works for more tapes and one head on

each tape

◮ if one has k heads on the same tape:

◮ simulate it by k tapes with only 1 head on each

without loss of time (Stoß, 1970) or

◮ use the same idea as before: store all currently visited tapes

squares in one cell . . .

◮ as just described, this simulation now needs two CA steps for

• ne TM step

◮ there are techniques to save a factor of 2 “most of the time”

◮ but we will not go into details here . . .

Turing machines Simulation of TM by CA Simulation of CA by TM

Definition

Given a CA with quiescent state q0 and a configuration c ∈ QR the support supp(c) of c is supp(c) = {x ∈ R | c(x) = q0}

Lemma

If c is a CA configuration with finite support, then F(c) also has finite support.

Proof

Lemma

If c is a CA configuration with finite support, then F(c) also has finite support.

Proof

◮ quiescent state: f (q0, . . . , q0) = q0 ◮ If F(c)(x) = q0, then at least one neighbor y of x was not

quiescent before: c(y) = q0.

◮ But each non-quiescent cell y is only the neighbor of |N|,

i.e. finitely many cells.

◮ Since supp(c) is finite, only finitely many cells have a

non-quiescent neighbor.

◮ Therefore supp(F(c)) is finite, too.

Theorem

For each CA C = (Z, QC, {−1, 0, 1}, fC, q0) there is a TM T = (QT, ST, δT) simulating C restricted to configurations with finite support. The simulation of c → FC(c) needs time O(| supp(c)|).

Simulation of CA

◮ one CA step

a1 a2 a3 a4 a5 FC b1 b2 b3 b4 b5

◮ is simulated computing one new state after the other

Simulation of CA

◮ the new states are computed one after the other:

a1 a2 a3 a4 a5 b1 b2 q

◮ for example the TM

◮ reads a2, a3, and a4 ◮ computes b3 and writes it on the tape ◮ moves one square to the right

◮ at the end

◮ the bi are copied to the upper part and ◮ used for the simulation of the next CA step

◮ This is also the (naive) standard technique used on computers.

Summary

◮ Each TM can be simulated by a CA. ◮ Each CA using only configurations with finite support

can be simulated by a TM.