Extensions to Turing machines

1
Extensions to Turing machines

• TMs are clumsy.
• Can we make them more efficient or more powerful?
• We will consider 6 possibilities
• Multiple tapes
• Multiple heads
• Two dimensional tapes
• Random access memory
• One way tapes
• Nondeterminism

2
Multiple tapes
tape1
lt a b a a - -
head1
control
tape2
lt a b a a - -
head2
tape3
lt a b a a - -
head3

• Basic idea At each step, scan symbols of all
  heads. Then, control depends on set of symbols
  and current state the result is to replace
  symbols (move R/L) on all tapes.
• Question does this allow us to compute more
  functions?
• Answer NO.
• Theorem A function is TM-computable iff it is
  computable on a multitape TM.
• Proof Assume TM M and MTTM M.
• Show TM M is simulated by MTTM M (easy)
• Show MTTM M is simulated by TM M. (more
  interesting)

3
Multiple tapes
One encoding of two tapes.
-1 0 1 2 3
M
lt A B C D E - -
tape1
head1
-1 0 1 2 3
lt F G H I J - -
tape 2
head2
M .. (0,A,F) (0,B,G) (1,C,H)
(0,D,I)

• In M,
• location -1 contains 0 not current head
• A location -1 of tape 1
• F location -1 of tape 2
• location 0 contains 0 not current head
• B location 0 of tape 1 and posn of tape 1
  head
• G location 0 of tape 2

4
Multiple tapes
Another encoding of two tapes.
-1 0 1 2 3
M
lt A B C D E - -
tape1
head1
-1 0 1 2 3
lt F G H I J - -
tape 2
head2

• M -1 0 1 2 3
• A B C D E
• 0 1 0 0 0
• F G H I J
• 0 0 0 1 0
• 0 denotes inactive location
• 1 - denotes active location

5
Multiple tapes
Whichever encoding, consider how to simulate a
transition in M. One action in M will be matched
by several actions in M
-1 0 1 2 3
M
lt A B C D E - -
tape1
head1
-1 0 1 2 3
lt F G H I J - -
tape 2
One transition Move head tape 1 Replace symbol
on tape 2
head2
-1 0 1 2 3
M
lt A B C D E - -
tape1
head1
-1 0 1 2 3
lt F G H Z J - -
tape 2
head2

• In M
• Position head somewhere sensible right end of
  data
• Move left, memorising the symbols at all active
  locations. Move to right end.
• Move left, replacing ABC by ABC
• 010
  by 001
• and HIJ
  by HZF
• 010
  010
• Return to right end.
• If in halting state, convert contents of tape and
  halt.

6
Multiple Heads
tape
lt a b a a - -
head1
head2
head n
control

• There are n heads, but in any state, only one
  head can move.
• BUT, this doesnt add any expressive power this
  machine can be simulated by an ordinary TM
  (argument is similar to multiple tapes).

7
Two dimensional tapes
One head on a grid that expands indefinitely to
the right and down We can simulate
this machine by a TM. Number locations
diagonally in M, v v v v v gt 1
2 6 7 15 gt 3 5 8 14 gt 4
9 13 gt 10 12 gt 11 then represent as
follows v 1 v 2 3 gt 4 5 6 v 7
8 9 10 gt 11 ..
8
Random Access Memory RAM TMs

• Instead of sequential data, change model to
  incorporate random access.
• program ----- program
  counter PC
• -----
  registers R1 Ac
• 
  Rn
• control
• tape
• (RAM)
• The tape itself (memory) is still one
  dimensional, and infinite, but we index into it
  via (a finite number of) registers. Registers
  and RAM cells contain natural numbers.
• Standard TM inputs/outputs strings over S.
• RAM TM inputs/outputs values in N into RAM
  locations.

9
Turing machine

• Standard TM inputs/outputs strings over S.
• But
• RAM TM inputs/outputs values in N into RAM
  locations.
• So does this mean that the RM TM is more
  powerful?OF COURSE NOT.
• Any string in S can be represented by a numeric
  code (assign numbers to symbols, then
  concatenate). So both machines have the same
  domain.
• To show that RAM TM ? M, we construct a
  simulation.
• Dont give simulation here, but idea is
• K registers ? k3 tape TM
• tape 1 for I/O
• tape 2 for content of
  RAM tape (locn, content)
• tape 3 for working tape

10
One Way Tapes

• So far, our tape has been infinite in both
  directions. What if we restrict the tape to being
  one way infinite. Does this mean that the OWT TM
  is less powerful?OF COURSE NOT.
• Any two way tape machine M can be simulated by
  a one way tape machine M.
• Two way tape
• location -2 -1 0 1 2 3
  
• x-2 x-1 x0 x1
  x2 x3
• One way tape
• location 1 2 3 4 5 6 7 8
  
• x1 x0 x2 x-1
  x3 x-2
• Location i in M is at location 2i1 in M
  igt1
• Location i in M is at location 4-2i in M i
  lt0
• Why do we need ?
• To indicate end of tape (to wrap around).

11
countability

• The argument for one way tapes is very similar to
  a countability argument.
• Defn A set is countable if we can list the
  elements of the set (total order), and we can
  refer to the ith element of the set.
• Examples
• N 0,1,2,3, is countable
• Z -1, -2, 0,1, 2, is countable
• NxN is countable
• (m,n) lt (m,n) iff (mn)lt (mn) or if
  (mn)(mn) then mltm
• (1,1) (1,2) (1,3) (1,4)
• (2,1) (2,2) (2,3) (2,4)
• (3,1) (3,2) (3,3) (3,4)
• But R is not countable. Consider the interval
  0,1).
• Proof By contradiction (Cantors diagonisation)
• Assume there is an ordering
• 1st 0.a11a12a13

12
countability

• The argument for one way tapes is very similar to
  a countability argument.
• Defn A set is countable if we can list the
  elements of the set (total order), and we can
  refer to the ith element of the set.
• Examples
• N 0,1,2,3, is countable
• Z -1, -2, 0,1, 2, is countable
• NxN is countable
• (m,n) lt (m,n) iff (mn)lt (mn) or if
  (mn)(mn) then mltm
• (1,1) (1,2) (1,3) (1,4)
• (2,1) (2,2) (2,3) (2,4)
• (3,1) (3,2) (3,3) (3,4)
• But R is not countable. Consider the interval
  0,1).
• Proof By contradiction (Cantors diagonisation)
• Assume there is an ordering
• 1st 0.a11a12a13

13
countability

• Now change aii to bii as follows bii aii 1,
  if aii lt9
• 
  0 if aii 9
• Clearly the number 0.b11b22b33 is in the
  interval 0,1).
• But, it does not occur in the list because for
  any i, the ith digit differs from aii (it is aii
  1!) so we cannot find a position for
  0.b11b22b33 . Thus 0,1) is not countable. Nor
  is R.

14
Nondeterminism

• All machines so far have been deterministic.
  What if we allow nondeterminism? Nondeterminism
  does not affect the power of FSAs but it does
  affect PDA (something not covered here).
• What is nondeterminism for TMs? Multiple
  transitions
• (q,a) (p,b)
• (q,b) (p,b)
• (q,b) (p,b)
• So, we have nondeterministic computations
• C0
• C11 C12
  C1k
• C21 C22 .. ..
• A nondeterministic machine can produce multiple
  outputs. But is it more powerful? (OF COURSE
  NOT)

15
Nondeterminism

• Sketch of proof Show that if M semidecides L
  (if w is in L, at least one comp. halts) then M
  semidecides L (if w is in L, then M halts).
• How to simulate all computations of M? We cant
  simply run them one after another what if one
  doesnt halt (but a later one does)? We need to
  dovetail computations. Or put another way,
  instead of a depth first search, we need a
  breadth first search.
• Dovetailing
• Step 1 run first computation for 1 step.
• Step 2 run first computation for 2 steps and 2nd
  computation for 2 steps.
• Step i run I computations for I steps.
• To do the dovetailing in M, use 2 tapes to store
  the configurations from odd and even steps.
• For example, assume all configurations from the
  10th step are stored on tape 2. Then, assuming
  no computations halt, work out next
  configurations (from tape 2) and write them to
  tape 1. Erase tape 2, then start working on
  configurations from 1th step, which are all no
  stored on tape 1.
• Clearly if M halts, then so will M.
• CONCLUSION We cannot enhance a TM!