CS 3240 – Chapter 9

 Turing Machines  From Alan Turing, 1936  Turns out to be the LAST machine  The only machine you'll ever need  The Church-Turing Thesis  All algorithms have a Turing Machine equivalent ▪ Some TMs are not algorithms, however 6/13/20162CS Chapter 9

 We have shown that a n b n c n is not context free  By the Pumping Lemma  What if we add an extra stack to a PDA? 6/13/20163CS Chapter 9

input, pop1/pop2, push1/push2 6/13/20164CS Chapter 9

 a n is accepted by an FA  a n b n is accepted by a PDA  a n b n c n is accepted by a 2PDA  What about a n b n c n d n ? 6/13/20165CS Chapter 9

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 An FA is a machine with no auxiliary memory  A PDA is like an FA but with unlimited, restricted memory  LIFO access  A Turing Machine is like an FA but with unlimited, unrestricted memory  We can move to any memory cell  We can both read and write there 6/13/20167CS Chapter 9

LanguageMachineGrammar RegularFinite AutomatonRegular Expression, Regular Grammar Context-FreePushdown AutomatonContext-Free Grammar Recursively Enumerable Turing MachineUnrestricted Phrase- Structure Grammar 8CS Introduction

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 Control Unit (Deterministic)  states (1 start state; 1 or more or halt states)  transitions: input,output,move (L or R)  I/O Tape (2-way, infinite)  pre-populated with (finite) input  output left on tape after processing  special blank symbol ☐ ▪ not part of input alphabet ▪ used as a separator and/or a boundary 6/13/201611CS Chapter 9

 Contents left on tape is ignored  Typically has only one accepting state  with no out-edges  Any string that causes the TM to halt in an accepting state is in the TM’s language  otherwise it is not in the language  (an explicit reject state may also be used) 6/13/201612CS Chapter 9

 Pseudocode:  Replace the first a with an X  Move right to first b, replace with Y  Move to right first c, replace with Z  Move left to right-most X  Repeat the above for steps until no more a’s  When no more a’s, skip over all Y’s and Z’s  fail if anything other than a Y or Z is encountered  Halt and accept or reject 6/13/201613CS Chapter 9

aabbcc Xabbcc XaYbcc XaYbZc XXYbZc XXYYZc XXYYZZ XXYYZZ ☐ XXYYZZ ☐☐ (accept) 6/13/201614CS Chapter 9

6/13/201615CS Chapter 9 Trace aabbcc

(q 0 )aabbcc X(q 1 )abbcc Xa(q 1 )bbcc XaY(q 2 )bcc XaYb(q 2 )cc XaY(q 3 )bZc Xa(q 3 )YbZc X(q 3 )aYbZc (q 3 )XaYbZc X(q 0 )aYbZc XX(q 1 )YbZc XXY(q 1 )bZc XXYY(q 2 )Zc XXYYZ(q 2 )c XXYY(q 3 )ZZ XXY(q 3 )YZZ XX(q 3 )YYZZ X(q 3 )XYYZZ XX(q 0 )YYZZ XXY(q 4 )YZZ XXYY(q 4 )ZZ XXYYZ(q 4 )Z XXYYZZ(q 4 ) ☐ XXYYZZ ☐ (q 5 ) ☐ (accept) 6/13/201616CS Chapter 9

 A sequence of configurations in a TM leading to a halt state  i.e., it doesn’t crash  If the TM is an acceptor, it explicitly accepts  If the TM represents some function, f, it leaves f(x) on the tape, for each valid input x 6/13/2016CS Chapter 917

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 What would be a reasonable strategy? 6/13/2016CS Chapter 919

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 Suppose f: D → R is a function from domain D to range R  If there is a TM that transforms every d in D to f(d)  That is, it reads d from the tape, does its work, halts, and leaves f(d) on the tape…  Then f is Turing-computable  This is the formal definition of an algorithm 6/13/201621CS Chapter 9

 Must first decide how to encode numbers on a TM tape  Then we must separate the numbers, so we can distinguish between them  Plan:  represent the numbers in unary notation  then move the 1’s from one number to the other! ▪ i.e., just remove the separator, moving the 1’s together ▪ we’ll use zero as a separator ▪ see next slide… 6/13/201622CS Chapter 9

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 Using unary notation, as usual 6/13/2016CS Chapter 925

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6/13/201627CS Chapter 9 Trace ababb

 TMs typically have one start and one or two halt states (but can have many):  one if the TM is a function  two if the TM is an acceptor  You can “call” a TM as a routine by setting up appropriate “linkage”  return to the desired state in the calling machine  make sure the tape workspace is in an acceptable configuration for your instructions 6/13/201628CS Chapter 9

 Tape configuration:  start: [x]0[y]0  finish: [x]0[y]0[xy]  First, modify copy to accommodate a 0-delimiters:  0[y]0… → 0[y]0…[y]  and finish by positioning at left of y  Then, mark a 1 in x with a, copy y to end, and repeat; then restore x’s a’s to 1’s 6/13/201629CS Chapter 9

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See Example 9.11 for an explanation of Comparer. See xplusyTM.py 6/13/201632CS Chapter 9

 T = (Q, ∑, Γ, δ, q 0, H)  Q = set of states  ∑ = input alphabet  Γ = tape alphabet  includes ∑ and blank symbol ( ☐ )  q 0 = start state  H = one or more halt states  no out-arrows  each typically represents a different meaning ▪ e.g., final vs. non-final 6/13/2016CS Chapter 933

6/13/2016CS Chapter 934 δ(q 0,1) = (q 0,1,R) δ(q 0,0) = (q 1,1,R) δ(q 1,1) = (q 1,1,R) δ(q 1, ☐ ) = (q 2, ☐,L) δ(q 2,1) = (q h, ☐,R)

δ ☐ 01 q0q0 (q 1,1,R)(q 0,1,R) q1q1 (q 2, ☐,L) (q 1,1,R) q2q2 (q h, ☐,R) 6/13/2016CS Chapter 935 δ(q 0,1) = (q 0,1,R) δ(q 0,0) = (q 1,1,R) δ(q 1,1) = (q 1,1,R) δ(q 1, ☐ ) = (q 2, ☐,L) δ(q 2,1) = (q h, ☐,R)

 q 0,1,q 0,1,R  q 0,0,q 1,1,R  q 1,1,q 1,1,R  q 1,_,q 2,_,L  q 2,1,q h,_,R  (_ = ☐ ; Use in Program 4) 6/13/2016CS Chapter 936

 Input:  A TM ▪ start state ▪ halt states ▪ quintuple format: state1,input,state2,output,direction  An input string (initial tape contents)  Output:  A trace of the actions of the machine ▪ with the current state positioned at the read-write head  Final state and tape contents 6/13/2016CS Chapter 937

 TMs can make multiple passes over the data  in multiple directions  And can process an arbitrary amount of auxiliary data in addition to the original input  They can give more meaningful output  vs. just yes/no  Interesting fact:  TMs came first! 6/13/2016CS Chapter 938

 TMs suffer from a vulnerability that FAs and PDAs do not:  They are not guaranteed to halt  They may hang (loop forever)!  Depends on the nature of the computation  More on this later 6/13/2016CS Chapter 939

 “Any computation that can be carried out by mechanical means can be performed by some TM”  Anything that can be done on any digital computer can be done by a TM  it’s the “lowest-level” programming language  No one has yet found a solvable problem for which a TM cannot be written  No alternative computation model invented is more powerful then a TM 6/13/201640CS Chapter 9

 The Turing Machine is the model for most programming as we know it  Imperative Programming  Defines a machine architecture  Characterized by reading and writing memory  Instructions and data share the same memory  There are other models of computation  Lambda Calculus (Functional paradigm; CS 4450)  Post Systems, Markov Systems, Structured Prog… 6/13/2016CS Chapter 941