1. COSC 3340: Introduction to Theory of Computation
University of Houston
Dr. Verma
Lecture 23
Lecture 23
UofH - COSC Dr. Verma

2. Definition of Nondeterministic TM
A Nondeterministic Turing Machine M = (K, , , s)
K, , and s are as for standard Turing machine
  (K X ) X ((K  {h}) X (  {L, R}))
Configurations and the relation |─M and |─M* are defined in the natural way but now |─M need not be single-valued:
One configuration may yield several others in one step.
NTMs will only be constructed as language acceptors
Only one accepting computation required for input to be accepted.
Lecture 23
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3. Example: L = {w  {a, b}*: w contains an occurrence of the substring abaab#
b, #
b, #
a, #
b, # #
Lecture 23
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4. Example: L = {w  {a, b}*: w contains an occurrence of the substring abaab}
The machine receives input in the form #w#, where w  {a,b}*.
It first scans from the right to left and nondeterministically guesses a point in w to start checking for the specified pattern.
It checks to see that the next five symbols are, from right to left, b, a, a, b, a.
If this pattern is detected, the machine halts and therefore accepts the input string.
If a violation of this pattern is found, the machine goes into an infinite loop.
Lecture 23
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5. Example: L = {In: n is composite}#
RIR IR # #
Given input #In#
Nondeterministically choose two numbers p, q > 1 and transform the input into #In #Ip #Iq #. See TM for writing Ip above.
Use the multiplication machine to transform the result of Step 1 into #In #Ip*q #.
Check to see that In = Ip*q;
this is easy, since only the lengths of these strings need to be compared. Halt if the lengths are equal; otherwise continue computing forever in some trivial fashion.
Lecture 23
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6. Simulation Idea for NTM
DTM will explore the computation graph in BFS (breadth-first order ) fashion.
Enumerate all possible computation starting from all 1-step computations, then 2-step computations, …
Initial configurations NTM 1 2 b
NTM are designed as language
acceptors only
. . . 1 1 b
... b
... 1 b
... 2 2 2 .
Lecture 23
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halt & accept

7. Simulation Idea for NTM…
input tape 1 □ □ D … x x # 1 x □
simulation tape … 1 2 3 3 2 3 1 2 1 1 3 □
address tape
Deterministic TM D simulating nondeterministic TM N
Lecture 23
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8. Simulation Idea for NTM N
Initially tape 1 contains the input w, and tapes 2 and 3 are empty.
Copy tape 1 onto tape 2.
Use tape 2 to simulate N with input w on one branch of its nondeterministic computation.
Consult tape 3 to determine which choice to make.
If no more symbol remain on tape 3 or if choice invalid, abort and go to next step.
Go to next step if a reject configuration is encountered
Replace the string on tape 3 with the lexicographically next string. Simulate the next branch of N’s computation by going to stage 2.
Lecture 23
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9. Church-Turing Thesis The thesis states that:
The Turing Machine is the formal equivalent of the informal concept of algorithm.
It is NOT a theorem, since it cannot be proved. (Algorithm is an informal concept).
Recall, algorithm -- a sequence of precise steps, each of which can be executed in a finite amount of time with finite resources.
Precise means no intelligence or judgment required, can be executed literally.
[Historical Note: Stated by Alonzo Church and Alan Turing – Turing was a student of Church]
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10. Evidence for the Thesis
All reasonable formal models so far have turned out to be equivalent to TM.
Reasonable -- means no infinite resources or power.
Examples of other formal models:
Post Correspondence systems
Random Access Machine model
Unrestricted grammars.
Partial Recursive Functions, etc.
Lecture 23
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11. Universal Turing Machine
We shall assume that there are fixed countably infinite sets
K = {q1, q2, q3, …}
and
 = {a1, a2, a3, …}
such that for every TM, the state set is a finite subset of K and the tape alphabet is a finite subset of .
Lecture 23
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12. Universal Turing Machine (contd.)
Let c be another symbol
We encode TM over the two-symbol alphabet {c, I}.
Let M = (K, , , s) be a TM where now K  K and   .
Thus K = {qi1, qi2, …, qik} and
 = {aj1, aj2, …, ajk} where
i1 < i2 < … < ik and j1 < j2 < … < jk
s = qim for some m, 1  m  k 
() qi
Ii+1 h I L R II ai
Ii+2
Lecture 23
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13. Universal Turing Machine (contd.)
Next we define kl strings Spr the transition function where 1  p  k and 1  r  l
Let (qip, ajr) = (q’, b), where
q’  K  {h}
b    {L, R}
Let Spr = cw1cw2cw3cw4c, where
w1 = (qip)
w2 = (ajr)
w3 = (q’)
w4 = (b)
Lecture 23
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14. Universal Turing Machine (contd.)
A Universal TM uses the encoding (M) of other machines as programs to direct its operation.
Therefore, U takes two arguments
(M) – description of machine M
(w) – encoding of input string w
and performs whatever M does on w.
Lecture 23
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15. Universal Turing Machine (contd.)
If M halts on input w, then U halts on an input consisting of the encoding of M followed by the encoding of w.
If U halts on an input consisting of the encoding of M followed by the encoding of w, then M halts on input w
Lecture 23
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16. Universal TM as a 3-tape TM
Let U’ = 3-tape TM
U’ uses it three tapes as follows:
The first tape contains the encoding of the tape of M
The second tape contains the encoding of M itself
The third tape contains the encoding of the state of M at the current point in the simulated computation.
Lecture 23
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17. Universal TM as a 3-tape TM U’
U’ is started with some string #(M)(w) on its first tape and the other two tapes are blank.
First U’ moves (M) onto the second tape and shift (w) down to the left end of the first tape, preceding it by #c(#) and following it by (#)c. Thus at this point the first tape contains #(#w#).
From (M), U’ extracts the encoding of the initial state of M, which it copies onto the third tape.
Lecture 23
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18. Universal TM as a 3-tape TM U’ (contd.)
U’ sets about simulating the steps of the computation of M
U’ keeps the head on the second and third tapes at the left
Since M starts off scanning the last blank of #w#, U’ should begin its simulation by moving its head on the first tape to the last c on the tape.
Lecture 23
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19. Universal TM as a 3-tape TM U’ (contd.)
Now U’ finds on its second tape the block of the form ccIicIjcIkcIlcc such that
Ii is the sting of I’s ending at the current head position on the third tape
Ij is that on the first tape, and changes the first tape appropriately.
Finally, U’ puts Ik on the third tape and checks whether the third tape now contains (h).
If it does not, then U’ simulates another computation step by M
If it does, then U’ itself halts.
If I_l is lambda(L) or lambda(R), this simply involves moving the head a few symbols to the left or right, but if it is lambda(a) then it
may be necessary to shift the right-hand part of the string on the first tape.
Lecture 23
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