Turing g Machines Reading: Chapter 8
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Turing Machines are…
Very powerful (abstract) machines that could simulate any modern day computer (although very, very slowly!) Why design such a machine?
For every input, answer YES or NO
If a p problem cannot be “solved” even using g a TM, then it implies that the problem is undecidable
Computability vs. Decidability
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A Turing Machine (TM)
This is like the CPU & program counter t
M = (Q, ∑, , , q0,B,F) Finite control Tape head
Infinite tape p with tape p symbols y …
B B B X1 X2
X3
Tape is the memory
…
Xi
…
Xn
B B …
Input & output tape symbols B: blank symbol (special symbol reserved to indicate data boundary) 3
Transition function
You can also use: for R for L
One move (denoted by ) in a TM does the following:
(q,X) = (p,Y,D)
q
X / Y,D
p
q is the current state X is the current tape symbol pointed by tape head State changes from q to p After the move:
X is replaced with symbol Y If D=“L”, the tape head moves “left” by one position. Alternatively, if D=“R” D R the tape head moves “right” by one position. 4
ID of a TM
Instantaneous Description or ID :
X1X2…Xi1qXiXi+1…Xn means:
q is the current state Tape head is pointing g to Xi X1X2…Xi1XiXi+1…Xn are the current tape symbols
( Xi) = ((p,Y,R) (q,X Y R) iis same as:
X1…Xi1qXi…Xn  X1…Xi1YpXi+1…Xn
(q Xi) = (p,Y,L) (q,X (p Y L) is same as:
X1…Xi1qXi…Xn  X1…pXi1YXi+1…Xn 5
Way to check for Membership
Is a string w accepted by a TM?
Initial condition:
The (whole) input string w is present in TM, preceded and followed by infinite blank symbols
Final acceptance:
Accept w if TM enters final state and halts If TM halts and not final state, then reject
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Example: L =
Strategy:
n n {0 1
 n≥1}
w = 000111 …
…
B B 0
0
0
1
1
1
B B
…
…
B B X X 0
Y Y 1
B B
…
B B
…
… …
B B X 0
0
1
1
1
B B
…
B B X X X Y Y 1
… …
B B X 0
0
… Y 1
1
B B
…
Y 1
1
B B
…
B B X X X Y Y Y B B
…
B B X X X Y Y Y B B
…
… …
B B X X 0
Accept
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TM for
n n {0 1
 n≥1}
Y / Y,R 0 / 0,R q0
0 / X,R XR
2.
q1 1 / Y,L
Y / Y,R YR X / X,R Y / Y,R
q3 B / B,R q4
1.
q2 Y / Y,L 0 / 0,L
3.
4.
Mark next unread 0 with X and move right Move to the right all the way to the first unread 1 1, and mark it with Y Move back (to the left) all the way to the last marked X, and then move one position to the right If the next position is 0, then goto step g p 1. Else move all the way to the right to ensure there are no excess 1s. If not move right to the next blank symbol and stop & accept. 8
*state diagram representation preferred
TM for
n n {0 1
 n≥1}
Next Tape Symbol Curr. State
0
1
X
Y
B
q0
(q1,X,R)


(q3,Y,R)

q1
(q1,0,R)
(q2,Y,L)

(q1,Y,R)

q2
(q2,0,L)

(q0,X,R)
(q2,Y,L)

q3



(q3,Y,R)
(q4,B,R)
*q4





Table representation of the state diagram
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TMs for calculations
TMs can also be used for calculating values
Like arithmetic computations Eg addition, Eg., addition subtraction subtraction, multiplication multiplication, etc.
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Example 2: monus subtraction “m  n” = max{mn,0} ...B 0mn B.. (if m>n) 0m10n ...BB…B.. (otherwise) Give sta ate diagra am
1.
2.
For every 0 on the left (mark X), mark off a 0 on the right (mark Y) Repeat process, until one of the following happens: 1. // No more 0s remaining on the left of 1 Answer is 0, so flip all excess 0s on the right of 1 to Bs ( d th (and the 1 ititself) lf) and dh halt lt 2. //No more 0s remaining on the right of 1 Answer is mn, so simply halt after making 1 to B
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Give sta ate diagra am
Example 3: Multiplication
0m10n1 (input),
Pseudocode: 1.
2.
3.
0mn1 (output)
Move tape head back & forth such that for every 0 seen in 0m, write n 0s to the right of the last delimiting 1 Once written, that zero is changed to B to get marked k d as fifinished i h d After completing on all m 0s, make the remaining g n 0s and 1s also as Bs 12
Calculations vs. Languages A “calculation” is one that takes an input and outputs a value (or values) A “language” is a set of strings that meet certain criteria
The “language” for a certain calculation is the set of strings of the form “”, where the output corresponds to a valid calculated value for the input E.g., The language Ladd for the addition operation
Membership question == verifying a solution e.g., is “” a member of Ladd ?
“” “” … “” … 13
Language g g of the Turing g Machines
Regular (DFA)
Contextfree (PDA)
Recurs sively Enume erable
Recursive Enumerable (RE) language
Con ntext sen nsitive
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Variations of Turing Machines
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TMs with storage
E.g., TM for 01* + 10*Current Current state
storage
Tape symbol
Next state
1
New N Storage symbol
• ([q0,B],a) = ([q1,a], a, R) Tape head
B B 0
Storage symbol
Transition function :
q
[q,a]:
Generic description Will work for both a=0 and a=1
1
1
1
1
where q is current state, a is the symbol in storage
B B
• ([q1,a],a) = ([q1,a], a, R) …
• ([q1,a],B) = ([q2,B], B, R)
Are the standard TMs equivalent to TMs with storage?
Yes 16
Standard TMs are equivalent to TMs with storage  Proof Claim: Every TM w/ storage can be simulated by a TM w/o storage as follows: For every [state, symbol] combination in the TM w/ storage:
Create a new state in the TM w/o storage Define transitions induced by TM w/ storage
Since there are only finite number of states and symbols in the TM with storage storage, the number of states in the TM without storage will also be finite 17
Multitrack Turing Machines
TM with multiple tracks, but just one unified tape head control
One tape head to read k symbols from the k tracks at one step.
…
…
Track 2
…
…
…
…
T k1 Track
Track k
…
… 18
MultiTrack TMs
TM with multiple “tracks” but just one E.g., g TM for {{wcw  w {{0,1}* } } head
but w/o modifying original input string
BEFORE
control t l
AFTER
Tape head
control t l Tape head
… B B 0 1 0 c 0 1 0 B … Track 1 …
B B 0 1 0 c 0 1 0 B
… Track 1
… B B B B B B B B B B … Track 2 …
B B X X X c Y Y Y B
… Track 2
Second track mainly used as a scratch space for marking
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Multitrack TMs are equivalent q to basic (singletrack) TMs
Let M be a singletrack TM
Let M’ be a multitrack TM (k tracks)
M = (Q, ∑, , , q0,B,F)
M = (Q’ M’ (Q , ∑ ’, ’, ’, q q’0,B,F B F’)) ’(qi,) = (qj, , L/R)
Claims:
For every M, there is an M’ s.t. L(M)=L(M’).
(proof trivial here) 20
Multitrack TM ==> TM (proof)
For every M’, there is an M s.t. L(M’)=L(M).
Main idea: M = (Q, (Q ∑, ∑ , , q0,[B,B,…],F) [B B ] F) Create one composite symbol to represent Where: every combination of Q = Q’ y ∑ = ∑ ‘ x ∑ ‘ x … (k times for ktrack) k track) k symbols = ’ x ’ x … (k times for ktrack) q0 = q’0 F = F’ (qi,[a1,a2,…ak]) = ’(qi, )
Multitrack TMs are just a different way to p singletrack g TMs,, and is a matter of represent design convenience. 21
Multitape Turing Machines
TM with multiple tapes, each tape with a separate head
Each head can move independently of the others control k separate heads …
…
Tape 2
…
…
…
…
…
Tape 1
Tape k
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On how a Multitape p TM would operate
Initially: The input is in tape #1 surrounded by blanks All other tapes contain only blanks st The tape head for tape #1 points to the 1 symbol of the input The heads for all other tapes p p point at an arbitrary y cell (doesn’t matter because they are all blanks anyway) A move: Is a function (current state, the symbols pointed by all the heads) After each move, each tape head can move independently ( f or right)) off one another (left 23
Multitape TMs Basic TMs
Theorem: Every language accepted by a ktape p TM is also accepted p by y a singletape g p TM Proof by construction:
Construct a singletape TM with 2k tracks, where each tape of the ktape TM is simulated by 2 tracks of basic TM k out the 2k tracks simulate the k input tapes The other k out of the 2k tracks keep track of the k tape head positions 24
Multitape TMs Basic TMs …
To simulate one move of the ktape TM:
Move from the leftmost marker to the rightmost marker (k markers) and in the process, gather all the input symbols into storage Then, take the action same as done by the ktape TM (rewrite tape symbols & move L/R using the markers)
control
Track 1
…
Track 2
…
Track 3 …
Track 4
storage
A1
A2
…
Ai
…
… …
… …
x x B1
B2
…
Bi
…
Bj
… … 25
Nondeterministic TMs Deterministic TMs
Nondeterministic TMs
A TM can have nondeterministic moves:
(q,X) = { (q1,Y1,D1), (q2,Y2,D2), … }
Simulation using a multitape deterministic TM: Control Input tape ID1
Marker tape
*
ID2
ID3
*
*
ID4 *
Scratch tape
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Summary
TMs == Recursively Enumerable languages TMs can be used as both:
B i TM is Basic i equivalent i l t to t all ll the th below: b l 1. 2 2. 3. 4.
Language recognizers Calculators/computers TM + storage Multitrack Multi track TM Multitape TM Nondeterministic TM
TMs are like universal computing machines with unbounded storage 27