Chapter 25

The Turing-Church Thesis and Undecidable Problems

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Problem 609

Find the exact number of Turing machines of the form M=(Q,Σ,P,q0)\mathfrak {M}=\left(Q, \Sigma , P, q_{0}\right) with a predetermined set of states QQ and tape alphabet Σ\Sigma.

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Problem 610

Let the following alphabet of Turing machines be fixed: {Λ,∣}\left\{ \Lambda , \mid \right\}. For the "busy beaver" function bb

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(a)

find bb⁡(1)\operatorname {bb}(1);

(b)

prove that bb⁡(2)⩾4\operatorname {bb}(2) \geqslant 4.

Problem 611

Prove that the algorithmic reducibility relation ⩽m\leqslant_{m} is reflexive and transitive.

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Problem 612

Prove the algorithmic undecidability of the test-data completeness problem TEST\mathrm{TEST}. The problem consists of triples (π(M),x,q)(\pi (\mathfrak {M}), x, q) such that the state qq occurs in the computation of the Turing machine M\mathfrak {M} on input xx.

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Problem 613

Prove the algorithmic undecidability of the zero problem ZERO\mathrm{ZERO}. The problem ZERO\mathrm{ZERO} consists of codes π(M)\pi (\mathfrak {M}) of Turing machines M\mathfrak {M} such that M(x)=ε\mathfrak {M}(x)=\varepsilon for all xx. Hint. Reduce TOTAL\mathrm{TOTAL} to ZERO\mathrm{ZERO}.

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Problem 614

Prove the algorithmic undecidability of the following equivalence problem EQU\mathrm{EQU}. The problem EQU\mathrm{EQU} consists of pairs (π(M),π(N))(\pi (\mathfrak {M}), \pi (\mathfrak {N})) such that the machines M\mathfrak {M} and N\mathfrak {N} are equivalent. Hint. Reduce ZERO\mathrm{ZERO} to EQU\mathrm{EQU}.

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Problem 615

Prove the algorithmic undecidability of the constant problem CONST\mathrm{CONST}. The problem CONST\mathrm{CONST} consists of those π(M)\pi (\mathfrak {M}) such that the Turing machine M\mathfrak {M} returns the same result on any input, whenever it halts at all. Hint. Reduce the complement of SELF\mathrm{SELF} to CONST\mathrm{CONST}.

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Problem 616

Prove the algorithmic undecidability of the growth problem GROW\mathrm{GROW}. The problem GROW\mathrm{GROW} consists of those π(M)\pi (\mathfrak {M}) such that the Turing machine M\mathfrak {M} computes an arithmetic function and M(x)<M(y)\mathfrak {M}(x)<\mathfrak {M}(y) holds for all x<yx<y, whenever both values are defined. Hint. Reduce the complement of SELF\mathrm{SELF} to GROW\mathrm{GROW}.

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Problem 617

Prove that

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(a)

the intersection of two decidable sets is a decidable set;

(b)

the union of two decidable sets is a decidable set;

(c)

the Cartesian product of two decidable sets is a decidable set.

Problem 618

Prove that for two decidable sets AA and BB of natural numbers, their "sum" A+B={x+y:x∈A,y∈B}A+B=\left\{ x+y: x \in A, y \in B\right\} and "product" (not Cartesian!) A⋅B={x⋅y:x∈A,y∈B}A \cdot B=\left\{ x \cdot y: x \in A, y \in B\right\} are also decidable sets.

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Problem 619

Prove that for two decidable languages LL and KK over the alphabet Σ\Sigma, their concatenation LKL K and iteration L∗L^{*} are also decidable languages.

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Problem 620

Let AA be a decidable set, and let g(x)g(x) and h(x)h(x) be g.r.f.'s. Prove that the function

F(x)={g(x), if x∈A,h(x), otherwise  F(x)= \begin{cases} g(x), & \text{ if } x \in A, \\ h(x), & \text{ otherwise }\end{cases}

is also general recursive.

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Problem 621

Prove that the bounded halting problem is decidable. The problem consists of triples of the form (π(M),x,t)(\pi (\mathfrak {M}), x, t) such that the computation of the Turing machine M\mathfrak {M} on input xx halts in at most tt steps.

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Problem 622

Show that under projection, a decidable language can become undecidable.

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Problem 623

Show that the function arc grows unboundedly but is not monotonic: it is possible that arc⁡(u)<arc⁡(v)\operatorname {arc}(u)<\operatorname {arc}(v), even though the word vv is shorter than the word uu.

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Problem 624

Show that the function arc grows more slowly than every computable function: if a g.r.f. ff satisfies f(x)⩽arc⁡(x)f(x) \leqslant \operatorname {arc}(x) for all xx, then ff is bounded.

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Problem 625

Let T(n)T(n) be the longest running time of a Turing machine with nn states on a blank tape. Prove that the function T(n)T(n) is not computable.

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Problem 626

Call the working region of a Turing machine M\mathfrak {M} on input xx the set of cells visited by the machine's head before it halts. Denote by S(M,x)S(\mathfrak {M}, x) the size of the working region of the machine M\mathfrak {M} on input xx. If the machine M\mathfrak {M} does not halt on xx, then the value S(M,x)S(\mathfrak {M}, x) may be arbitrary. Suppose that for the machine M\mathfrak {M} the function S(M,x)S(\mathfrak {M}, x) is majorized by a general recursive function S∗(x)S^{*}(x): S(M,x)⩽S∗(x)S(\mathfrak {M}, x) \leqslant S^{*}(x) for all xx. Prove that the halting problem HALTM\mathrm{HALT}_{\mathfrak {M}} for the machine M\mathfrak {M} is algorithmically decidable.

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Problem 627

Prove that the function S(π(M),x)S(\pi (\mathfrak {M}), x) from the previous problem is not computable.

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Problem 628

For any pair of sets AA and BB define

A∨B={2x:x∈A}∪{2x+1:x∈B}. A \vee B=\left\{ 2 x: x \in A\right\} \cup \left\{ 2 x+1: x \in B\right\} .

Prove that for any set CC the condition A∨B⩽mCA \vee B \leqslant_{m} C holds if and only if A⩽mCA \leqslant_{m} C and B⩽mCB \leqslant_{m} C.

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Problem 629

A nonempty set AA is called recursively enumerable if it is the range of some g.r.f. ff: A={f(0),f(1),f(2),…}A=\left\{ f(0), f(1), f(2), \ldots \right\}. Prove that

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(a)

every nonempty decidable set is recursively enumerable;

(b)

the nonempty intersection of a recursively enumerable set with a decidable set is also a recursively enumerable set;

(c)

the union of two recursively enumerable sets is also a recursively enumerable set;

(d)

for any recursively enumerable set AA there exists a p.r.f. gg such that dom⁡g=A\operatorname {dom} g=A.

Problem 630

Prove that the set TOTAL\mathrm{TOTAL} is not recursively enumerable. Hint. For the universal Turing machine U\mathfrak {U}, consider the function U(f(x),x)&a\mathfrak {U}(f(x), x) \& a (concatenation with a nonempty symbol).

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Problem 631

Prove that there exists a p.r.f. ff that cannot be extended to any g.r.f. gg, that is, such that f(x)=g(x)f(x)=g(x) for all x∈dom⁡fx \in \operatorname {dom} f. Hint. For the universal Turing machine U\mathfrak {U}, consider the arithmetic function f(x)=U(x,x)+1f(x)=\mathfrak {U}(x, x)+1.

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