Undecidability

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Undecidability
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Church-Turing Thesis

• A Turing machine that halts on all inputs is the
  precise formal notion corresponding to the
  intuitive notion of an algorithm
• An algorithm means a precisely defined set of
  instructions
• This thesis cannot be formally proven
• See this link.

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• Consequence of Church-Turing thesis If a
  problem cannot be solved by a Turing machine then
  it cannot be solved by a human using a precisely
  defined sequence of instructions, either
• Knowing a problem is undecidable helps avoid
  wasted work
• Can extend Church-Turing thesis to
  semidecidability (e.g. theorem proving)

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Universal Turing machine

• A UTM is a general purpose computer
• Can simulate any other Turing machine
• How simple can it be?
• U Given description T of machine T and x of
  input to T, simulates T on x and halts if and
  only if T halts on x.
• Need to represent arbitrary TM in fixed alphabet
  see this link.

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• Details of U are uninteresting. But try to
  understand how one or more UTM work.
• Examples
• Hopcroft and Ullman 40 state UTM
• A 4 state 6 symbol UTM
• Minsky 7 state 4 symbol UTM
• Wolframs Universal Turing Machines
• Details for Hopcroft and Ullman machine

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Unsolvability of Halting Problem

• A Turing machine can be represented by a sequence
  of 4-tuples (state, symbol, new state, direction
  or symbol) or 5-tuples of the form (state,
  symbol, new state, direction, new symbol)
• Thus a TM can be represented by a string of
  symbols
• This string can be encoded in binary

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• Thus a TM can be represented as a binary string
• Not all binary strings represent TM if a binary
  string does not represent a TM we can say it
  represents a TM that always halts right away
• A binary string can be interpreted as a binary
  integer j
• Thus each binary integer j corresponds to a TM
  (call it Tj )

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• The integer j can also be represented in unary
• This permits j to be read by a TM having only one
  symbol other than blank in its alphabet

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• Every TM can be represented as Tj for some j
• Let ? be the problem whose base set is the
  positive integers, and such that the right answer
  for an integer j is yes if Tj does not halt on
  input j, no otherwise
• Claim ? is not partially decidable.
• Proof By contradiction.
• Suppose ? is partially decidable. Then there is
  a Turing machine M that always gives the right
  answer and halts on input j if Tj does not halt
  on input j, and M loops otherwise.

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• But then M must be Ti for some i.
• Then for any j, M halts on input j if and only if
  Tj does not halt on input j.
• In particular, for j i, this is true.
• Thus M halts on input i if and only if Ti does
  not halt on input i.
• But M is Ti .
• Thus Ti halts on input i if and only if Ti does
  not halt on input i.
• This is a contradiction.
• Thus M does not exist and ? is not partially
  decidable.

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• Note that self-reference is essential for this
  proof.
• The problem occurs when Tj reads input j.
• Thus Tj is reading a description of itself.
• Self awareness is part of consciousness.
• Does this mean that consciousness is in some way
  releted to undecidability?

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• Corollary ? is not decidable either.
• Thus the halting problem is not decidable.
• The halting problem H The base set is the set
  of pairs (T,x) where T is a Turing machine and x
  is the input to T.
• The right answer is yes if T halts on input x
  and no otherwise.
• If H were decidable then ? would be too. So H is
  undecidable.
• H is partially decidable (use a universal Turing
  machine).

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• ?H is not partially decidable (if it were then H
  would be decidable).
• H is not decidable on any computing device.
• Dont waste time trying to solve it.
• Illustration In a certain village the barber
  shaves everyone who does not shave himself, and
  only those people who do not shave themselves.
• Who shaves the barber?

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• Unsolvability of H results from TM running on
  their own descriptions. A kind of
  self-awareness. Is this significant?

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Other Unsolvable Problems

• Show other problems unsolvable by showing that if
  you could solve them, you could solve the halting
  problem.
• Many problems are known to be unsolvable.
• Examples
• Blank tape halting problem
• Hilberts tenth problem
• Theorem proving in first-order logic

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• But many of these unsolvable problems are
  partially decidable.
• A proof that the blank tape halting problem is
  undecidable

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Unsolvability of BTHP

• Let M have start state s, blank B, and these
  5-tuples
• (s,0,t,0,R) (s,1,t,1,R) (s,B,y,B,R)
• (t,0,s,0,R) (t,1,s,1,R) (t,B,n,B,R)
• M tests if a binary string has even length.
• Let M have start state a, blank symbol B, all 5
  tuples of M, and these 5-tuples
• (a,B,b,0,R) (b,B,c,1,R) (c,B,d,1,L) (d,1,s,1,L)
• Then M started on blank tape writes 011, moves
  left to the 0, and transfers control to M.

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• Thus M halts on input 011 if and only if M halts
  on blank tape.
• If we could solve the blank tape halting problem
  then we could use this to tell whether M halts on
  input 011.
• The same idea works in general If we could
  solve BTHP we could solve the general halting
  problem.
• Because the halting problem is unsolvable, BTHP
  is unsolvable.

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Busy Beaver Problem (step counting version)

• BB(n) is the largest number of steps taken by any
  n state TM with a binary alphabet, starting on
  blank tape, that eventually halts
• Suppose there were 5 TM having 3 states. Suppose
  M1 took 3 steps and halted, M2 took 50 steps and
  halted, M3 looped, M4 took 7 steps and halted,
  and M5 looped. Then BB(3) 50.

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• Now BB(n) is not computable and no computable
  function is always larger.
• Proof If we could compute BB(n) we could solve
  the halting problem. (How?)
• But with a few extra bits for each n we could
  compute BB(n).
• Just say which TM runs the longest and halts.
• Thus with a few extra bits we could solve the
  halting problem. These bits will never be known
  unless by revelation.
• BB(n) grows very fast, faster than any computable
  function.

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• The problem is classifying Turing computations
  that do not obviously halt or loop.
• This will forever be impossible to do unless some
  computing device is more powerful than a Turing
  machine. The behavior of Turing machines is too
  subtle for us to fully understand.
• See this link
• There is no easy way to distinguish the beginning
  of a halting computation from the beginning of a
  looping computation.

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• If there were an easy way to look at the
  beginning of a computation and tell that it was
  going to halt, then one would quickly know that
  the other computations were looping.
• If there were an easy way to look at the
  beginning of a computation and tell that it was
  going to loop, then one would quickly know that
  the other computations would halt.
• Neither is possible. Halting and looping
  computations are very hard to distinguish.

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Known bounds on BB(n)

• The function Sigma(n) denotes the maximal number
  of tape marks which a Turing Machine (TM) with n
  internal states and a two-way infinite tape can
  produce onto an initially empty tape and then
  halt. The function S(n) denotes the maximal
  number of steps (shifts) which such a TM can do
  (it needs not produce many tape marks). The
  following table gives some known values

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• n Sigma(n) S(n)
• 1 1 1
• 2 4 6
• 3 6 21
• 4 13 107
• 5 gt 4098 gt 47,176,870
• 6 gt 1.2910865 gt 3101730

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• Some researchers believe that Sigma(5) will never
  be known exactly. There are too many Turing
  machines to consider and their behavior is too
  subtle.

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• Following are transition tables and values of S
  and Sigma for a few Turing machines.
• Try simulating them for a few steps to understand
  how they work!
• The difficulty of understanding such machines
  helps us to see why the halting problem is
  unsolvable.
• No matter how advanced we become, there will be
  infinitely many machines that are beyond our
  understanding.
• We will not know whether they eventually halt.
  Our math is not strong enough to say.

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Record 5-state Turing Machines

• The following table lists for each input state
  and symbol the output state, symbol and tape
  movement. The initial state is A, the halting
  state is H. The blank symbol is 0 and the tape
  is 2-way infinite.
• No A0 A1 B0 B1 C0 C1 D0 D1 E0
  E1 ones steps
• 1 B1L C1R C1L B1L D1L E0R A1R D1R H1L A0R 4098
  47176870
• 2 B1L A1L C1R B1R A1L D1R A1L E1R H1R C0R 4098
  11798826
• 3 B1L D0R C1R D1L A1R C1R H1L E1L A1L B0L 4097
  23554764
• 4 B1L A1L C1R D1L A1R C1R H1L E0L C1R B1L 4097
  11798796
• 5 B1L A1L C1R D0L A1R C1R H1L E1L E0R B1L 4096
  11804910
• 6 B1L A1L C1R D0L A1R C1R H1L E1L C1R B1L 4096
  11804896
• 7 B1L D1L C1R E0R A0L B0R E1L H1L C1R C1L 1471
  2358064

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Record 6-state Turing machines

• This is a list of 6-state TMs producing many
  ones, found by J.Buntrock and H.Marxen by
  scanning around 10 of all 6-state TMs with a
  program from June 2000 to October 2000. The
  blank symbol is 0 and the tape is 2-way infinite.

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• Name a
• ones 17485734
• steps 95547257425490
• ones gt 1.7107 steps gt 9.51013
• A B1R C0L
• B A1L A0R
• C D0L Z1R
• D E1R D1L
• E F0L E0L
• F F1R B0L

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• Name b
• ones 36109969
• steps 1137975066814256
• ones gt 3.6107
• steps gt 1.11015
• A B1R C0L
• B A1L E0R
• C D1L F0L
• D B0L C1L
• E B1R A1R
• F B1R Z1R

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• Name c
• ones 36109970
• steps 758650111948072
• ones gt 3.6107
• steps gt 7.51014
• A B1R C0L
• B A1L E0R
• C D1L F1L
• D B0L C1L
• E B1R A1R
• F C0L Z1R

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• Name q
• ones gt 6.410462
• steps gt 6.110925
• A B1R B0L
• B C0R B1L
• C D1R A0L
• D E1L F1L
• E A1L D0L
• F Z1R E1L

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• Name r
• ones gt 1.210865
• steps gt 3.0101730
• A B1R F0L
• B C0R D0R
• C D1L E1R
• D E0L D0L
• E A0R C1R
• F A1L Z1R

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Simulation Results

• Check these links for the results of simulations
  of machines 1,2,3,a,r
• Busy Beaver Simulation 1
• Busy Beaver Simulation 2
• Busy Beaver Simulation 3
• Busy Beaver Simulation a
• Busy Beaver Simulation r (the record holder)

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Chaitins Omega

• A real number between 0 and 1
• If we could compute it we could solve the halting
  problem.
• Omega is random. Writing n bits of omega
  requires an n bit program. Omega has no simple
  patterns.
• See this link and this link.

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• ? ?p?P2-p
• where P is the set of all halting Turing machines
  and p is the length in bits of the encoding of
  p.
• Let Pn be the set of TM that halt in n steps or
  less.
• ?n ?p?Pn2-p
• Then ?n approaches ? as n approaches ?.

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• ?n can be computed to any accuracy.
• To solve the halting problem using ?
• If M halts in n steps then answer yes.
• If ? - ?n lt 2-M then answer no.
• Do this for n 1, 2, 3, until an answer is
  given.
• Eventually either a yes or no answer will be
  given.
• Thus ? will never be known. It is unknowable
  unless it is given by revelation.

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Partial solutions to the halting problem

• A loop tester is a Turing machine M that on input
  i, halts if Ti loops on input i, and M loops
  otherwise.
• A partial loop tester is a Turing machine M such
  that if M halts on input i, then Ti loops on
  input i.
• No loop tester exists. But partial loop testers
  exist.

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Improving partial loop testers

• Suppose M is a partial loop tester. Let M be Tj.
  What does M do on input j?
• If M halts on input j then Tj loops on input j.
  But Tj is M. Contradiction.
• Therefore M loops on input j. But M does not
  know that it loops on input j.
• M cannot solve this instance of the loop testing
  problem. But we can solve it.
• Thus we excel every partial loop tester.

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• Let b(M) (better M) be M modified so that it
  halts on input j.
• Thus b(M) knows that M halts on input j.
• It is easy to construct Turing machine b(M) from
  M.
• b(M) is also a partial loop tester.
• But b(M) does not know that b(M) halts on input
  k where b(M) is Tk.
• Consider b(b(M)). This machine knows that b(M)
  halts on input k. Et cetera.

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• In this way we construct b(M), b2(M), b3(M), et
  cetera. Each halts more often than the previous
  one.
• Let b?(M) be a Turing machine that halts on input
  i if any of M, b(M), b2(M), et cetera halt on
  input i. This improves all bn(M).
• Such a machine can be constructed by simulating
  M, b(M), b2(M), in parallel.
• Then one can improve b?(M) to get b(b?(M)),
  b(b(b?(M))) et cetera.
• Notation b?1(M), b?2(M) .

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• One obtains in the same way b 2? (M), b3?(M),
  b4?(M), , and eventually b??(M).
• This can be written b?2(M).
• Similarly one obtains b?3(M) b?4(M) b?5(M) .
• This goes on for a long time!
• Related to the ordinals.

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A paradox of consciousness

• Suppose H is a Turing machine encoding human
  thought about looping of Turing machines.
• Suppose H halts on input i if a human with lots
  of time can eventually deduce mathematically that
  Ti loops on input i. H loops otherwise.
• Then H is a partial loop tester. Therefore H
  loops on input j where H is Tj.

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• Because H loops on input j, this means that
  humans cannot deduce that Tj loops on input j.
• Tj is H, so humans cannot deduce that H loops on
  input j.
• But we just showed that H loops on input j. We
  are humans. Contradiction.
• Possibilities
• H is not a partial loop tester human logic is
  faulty.
• Humans can never know H.
• Human thought cannot be represented by a Turing
  machine.

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• All possibilities are disturbing.
• The first means that we cant think correctly
  even when reasoning mathematically.
• The second means that we can never know how we
  think.
• The third means that our thought in some way
  exceeds the capacity of a Turing machine.
• Note that not much human logic is needed to
  reason that H loops on input j.
• Some people think that our consciousness, or
  self-awareness, permits us to excel Turing
  machines.

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• Another possibility H is a partial loop tester
  but we can never know that H is a partial loop
  tester.
• This implies that human mathematical thought
  employs reasoning that human thought cannot
  justify.
• What would be an example of this?