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UDC 51.01:518.5
MATHEMATICS
Ya. M. BARZDIN’
ON A FREQUENCY SOLUTION OF ALGORITHMICALLY UNSOLVABLE MASS PROBLEMS
(Presented by Academician P. S. Novikov on 4 IX 1969)
1°. Suppose a certain mass problem \(\mathcal M\) with a fixed numbering of individual problems is given. If \(\mathcal M\) is an algorithmically unsolvable problem, then it is impossible to solve all these problems by a single method. But the question arises: perhaps the “majority” of the individual problems can nevertheless be solved by a single method?* We shall say that the mass problem \(\mathcal M\) is algorithmically solvable with frequency \(1-\varepsilon\) on infinitely many initial segments if there exists an algorithm \(\Omega\) such that, for infinitely many natural \(n\),
\[ [\mathcal M,\Omega,n]/n \geqslant 1-\varepsilon, \]
where \([\mathcal M,\Omega,n]\) is the number of those individual problems among the first \(n\) which the algorithm \(\Omega\) solves correctly; here it is required that the algorithm \(\Omega\) be applicable to every individual problem from \(\mathcal M\) (i.e., halt when solving it). The main purpose of this article is to show that a number of typical unsolvable mass problems can be algorithmically solved with frequency \(1-\varepsilon\) on infinitely many initial segments (\(\varepsilon\) is an arbitrary preassigned number greater than 0).
2°. As a typical example of mass problems let us consider the problem of membership in a set \(M\), where \(M\) is an arbitrary recursively enumerable set. In this case the notion of solvability with frequency \(1-\varepsilon\) can be made precise as follows: the problem of membership in the set \(M\) is algorithmically solvable with frequency \(1-\varepsilon\) on infinitely many initial segments if there exists a general recursive predicate \(P(x)\) such that, for infinitely many natural \(n\),
\[ [M,P,n]/n \geqslant 1-\varepsilon, \]
where \([M,P,n]\) is the number of those natural \(x\), not exceeding \(n\), for which \(P(x)=\chi_M(x)\) (where \(\chi_M(x)\) is the characteristic function of the set \(M\)). In this case we shall say of the predicate \(P(x)\) that it solves the problem of membership in the set \(M\) with frequency \(1-\varepsilon\) on infinitely many initial segments.
Theorem 1. Whatever recursively enumerable set \(M\) and \(\varepsilon>0\) we take, the problem of membership in the set \(M\) is algorithmically solvable with frequency \(1-\varepsilon\) on infinitely many initial segments.
Proof. Let \(M\) be an arbitrary recursively enumerable set and \(\varepsilon>0\). Put \(M^{(n)}=\{x\mid x\in M\ \&\ x\leqslant n\}\). Denote by \(i_0\) the greatest natural number for which, for infinitely many natural \(n\), the inequality
\[ (i_0-1)\cdot{}^{1}/_{2}\varepsilon n \leqslant |M^{(n)}| < i_0\cdot{}^{1}/_{2}\varepsilon n \]
still holds (where \(|M^{(n)}|\) is the cardinality of the set \(M^{(n)}\)). Taking into account
* An analogous question also arises in the decoding of automata (without the use of a priori information), and in \((^1)\) it is, in a certain sense, answered positively.
** Here use will be made of ideas similar to those used in the proof of Theorem 4 from \((^2)\) and, apparently, also of Theorem 4.4 from \((^3)\). In particular, Theorem 4 of the work \((^2)\) follows easily from Theorem 1.
that the set \(\{n\mid |M^{(n)}|\ge i_0\}\) is finite, we easily obtain that the set \(\{n\mid (i_0-1)\cdot {}^{1}/_{2}\varepsilon n \le |M^{(n)}|< i_0\cdot {}^{1}/_{2}\varepsilon n\}\) is recursively enumerable. Moreover, this set is infinite. Therefore there exists a general recursive function \(\eta(s)\) such that, for all natural \(s\): a) \((i_0-1)\cdot {}^{1}/_{2}\varepsilon\eta(s)\le |M^{(\eta(s))}|< i_0\cdot {}^{1}/_{2}\varepsilon\eta(s)\); b) \(\eta(s+1)\ge 2\eta(s)\).
Using the function \(\eta(s)\), we shall now construct the desired predicate \(P(x)\) (more precisely, a predicate \(P(x)\) for which, for every \(s\), the inequality \([M,P,\eta(s)]/\eta(s)\ge 1-\varepsilon\) holds). First note that, using the function \(\eta(s)\), for every \(s\) one can effectively construct a set \(R_s\) such that: a) \(R_s\subset M^{(\eta(s))}\); b) \(|M^{(\eta(s))}\setminus R_s|<{}^{1}/_{2}\varepsilon\eta(s)\).
This can be done, for example, in the following way: take \(\eta(s)\) copies of Turing machines computing the partial characteristic function of the set \(M\), write on their tapes respectively the numbers \(1,2,\ldots,\eta(s)\), start them running, and wait until \(\lfloor(i_0-1)\varepsilon/2\,\eta(s)\rfloor\) machines halt\(^*\) (the fact that this will happen eventually follows from property a) of the function \(\eta(s)\)); the numbers that were written on the tapes of these \(\lfloor(i_0-1)\cdot {}^{1}/_{2}\varepsilon\eta(s)\rfloor\) machines (for them the partial characteristic function is defined, hence they belong to \(M\)) will form the desired set \(R_s\). Put \(R_1^*=R_1\), \(R_s^*=\{x\mid x\in R_s\ \&\ x\ge \eta(s-1)\}\), \(s=2,3,\ldots\), and
\[ R=\bigcup_{s=1}^{\infty} R_s^* . \]
Let \(\alpha(s)\) be the number of elements of the set \((M\setminus R)\cup(R\setminus M)\) not exceeding \(\eta(s)\). Since \(R\subseteq M\) and
\[ \{x\mid x\in R\ \&\ x\le \eta(s)\}=\bigcup_{i=1}^{s} R_i^*, \]
we have
\[ \alpha(s)=\left|M^{(\eta(s))}\setminus \bigcup_{i=1}^{s} R_i^*\right|. \]
Using \(\eta(i+1)\ge 2\eta(i)\) and \(|M^{(\eta(i))}\setminus R_i|<{}^{1}/_{2}\varepsilon\eta(i)\), it is not hard to show (by induction on \(s\)) that
\[ \left|M^{(\eta(s))}\setminus \bigcup_{i=1}^{s} R_i^*\right|<\varepsilon\eta(s). \]
Thus, \(\alpha(s)<\varepsilon\eta(s)\). Now choose as the desired predicate \(P(x)\) the characteristic function of the set \(R\). We obtain that \([M,P,\eta(s)]/\eta(s)=1-\alpha(s)/\eta(s)>1-\varepsilon\). Since \(R\) is a recursive set and, consequently, \(P(x)\) is a general recursive predicate, the validity of Theorem 1 follows.
Above, in defining the notion of solvability of the membership problem with frequency \(1-\varepsilon\), we proceeded from the idea that the natural numbers are taken in their natural order. However, of interest is also the more general case in which a certain general recursive function \(\pi(s)\) is given and we first take \(\pi(1)\), then \(\pi(2),\pi(s),\ldots\), and try to solve the membership problem for these numbers. We shall say that the membership problem in a set \(M\), under the numbering \(\pi(s)\), is algorithmically solvable with frequency \(1-\varepsilon\) on infinitely many initial segments if there exists a general recursive predicate \(P(x)\) such that, for infinitely many natural \(n\), the inequality \([M,P,\pi,n]/n\ge 1-\varepsilon\) holds, where \([M,P,\pi,n]\) is the number of those \(x\), not exceeding \(n\), for which \(P(\pi(x))=\chi_M(\pi(x))\).
Since the preimage of a recursively enumerable set under a general recursive mapping \(\pi(s)\) is also a recursively enumerable set, it follows from Theorem 1 that
Corollary. Whatever the general recursive function \(\pi(s)\), the recursively enumerable set \(M\), and \(\varepsilon>0\) may be, the membership problem in the set \(M\), under the numbering \(\pi(s)\), is algorithmically solvable with frequency \(1-\varepsilon\) on infinitely many initial segments.
The following question also arises: can Theorem 1 be strengthened so that, in its formulation, the phrase “solvable with frequency \(1-\varepsilon\) on infinitely many initial segments” is replaced by the phrase “solvable with frequency \(1-\varepsilon\) on all initial segments”? A negative answer to this question follows from Theorem 4 of paper \((^4)\).
\[ {}^* \lfloor l\rfloor \text{ is the smallest natural number } \ge l. \]
3°. In Theorem 1 nothing is said about whether, from the number of a recursively enumerable set, one can effectively find the corresponding predicate \(P(x)\). Therefore the following question arises. Suppose that some \(\varepsilon_0\), \(0<\varepsilon_0<1\), and some numbering \(\tau\) of recursively enumerable sets are fixed (\(\tau n\) is the set with number \(n\)). Is the following mass problem solvable: for any natural \(n\), find the Kleene number of a general recursive predicate that solves the membership problem for the set \(\tau n\) with frequency \(1-\varepsilon\) on infinitely many initial segments? In what follows we shall call such a mass problem the metaproblem \(\mathcal M_{\varepsilon_0,\tau}\).
Theorem 2. For any \(\varepsilon_0\), \(0<\varepsilon_0<1\), and any principal numbering \(\tau\) of recursively enumerable sets \({}^{(5)}\), the metaproblem \(\mathcal M_{\varepsilon_0,\tau}\) is algorithmically undecidable.
Proof. Suppose, to the contrary, that \(\mathcal M_{\varepsilon_0,\tau}\) is algorithmically decidable. Consider a universal recursive predicate \(F(s,x)\) corresponding to some principal numbering of one-place recursive predicates (the principal numbering is needed here in order that the fixed-point theorem apply). By an \(F\)-number of a recursively enumerable set \(M\) we shall mean any \(s\) for which \(M=\{x\mid F(s,x)=1\}\). It is obvious that there exists a general recursive function that reduces the \(F\)-numbering to the principal numbering \(\tau\). Therefore, from the assumption that \(\mathcal M_{\varepsilon_0,\tau}\) is algorithmically decidable it follows that there must exist a two-place general recursive predicate \(P(s,x)\) which assigns to any recursively enumerable set \(M_s\) with \(F\)-number \(s\) the one-place predicate \(P_s(x)=P(s,x)\) such that for infinitely many \(n\) one has \([M_s,P_s,n]/n\ge 1-\varepsilon\). Consider, instead of the predicate \(P(s,x)\), the opposite predicate \(\overline P(s,x)\). By the fixed-point theorem we obtain that there exists a number \(a\) for which \(F(a,x)=\overline P(a,x)\). Consider the set \(M_a\) with \(F\)-number \(a\). We obtain
\[
M_a=\{x\mid F(a,x)=1\}=\{x\mid \overline P(a,x)=1\}=\{x\mid P(a,x)=0\}.
\]
Thus, for every \(x\), the value of the predicate \(P_a(x)=P(a,x)\) is opposite to the value of the characteristic function \(\chi_{M_a}(x)\) of the set \(M_a\). Therefore, for every \(n\), \([M_a,P_a,n]=0\) and, consequently, \([M_a,P_a,n]/n<1-\varepsilon\). The contradiction obtained proves the theorem.
In view of the negative result of Theorem 2, the following question is of interest: perhaps with frequency \(1-\varepsilon\) we nevertheless can solve the problem \(\mathcal M_{\varepsilon_0,\tau}\)? We shall say that the metaproblem \(\mathcal M_{\varepsilon_0,\tau}\) is algorithmically decidable with frequency \(1-\varepsilon\) on infinitely many initial segments if there exists a general recursive function \(F(x)\) such that for infinitely many natural \(n\) the inequality \([\mathcal M_{\varepsilon_0,\tau},F,n]/n\ge 1-\varepsilon\) holds, where \([\mathcal M_{\varepsilon_0,\tau},F,n]\) is the number of those natural \(x\), not exceeding \(n\), for which \(F(x)\) is the Kleene number of a general recursive predicate solving the membership problem for the set \(\tau x\) with frequency \(1-\varepsilon_0\) on infinitely many initial segments. The following theorem gives a positive answer to the question posed.
Theorem 3. Whatever \(\varepsilon_0\), \(0<\varepsilon_0<1\), a computable numbering \(\tau\) of recursively enumerable sets, and \(\varepsilon>0\) may be, the metaproblem \(\mathcal M_{\varepsilon_0,\tau}\) is algorithmically decidable with frequency \(1-\varepsilon\) on infinitely many initial segments.
The proof of Theorem 3 is based on the same idea as was used in the proof of Theorem 1. However, technically the proof of Theorem 3 is considerably more complicated.
4°. In connection with Theorem 1 there arises still the following question (in a certain sense the opposite of the question of the preceding item): is it true that for any fixed recursively enumerable set \(M\) there exists an algorithm which, for any rational \(\varepsilon>0\), constructs the Kleene number of a predicate solving the membership problem in the set \(M\) with frequency \(1-\varepsilon\) on infinitely many initial segments?
It can be shown, by reducing to a contradiction with Theorem 3 of \((^{2})\), that the answer to this question is negative. However, the question of whether the given mass problem is decidable with frequency \(1-\varepsilon\) remains open.
5°. Alongside the complexity of programs recognizing, among the natural numbers not exceeding \(n\), membership in the set \(M\) \((^{2})\), one may also consider the complexity of programs that do the same only with frequency \(1-\varepsilon\). This complexity (denote it by \(K(M,n,\varepsilon)\)) may be defined as follows: let \(A(p,x)\) be a partial recursive function (a “programming method”) for which the minimal complexity is obtained, up to an additive constant (see \((^{2})\)), and let \(A_p(x)=A(p,x)\); then \(K(M,n,\varepsilon)\) is the minimal length of numbers \(p\) such that \([M,A_p,n]/n \ge 1-\varepsilon\), where \([M,A_p,n]\) is the same as in 2°. It is not hard to see that for any recursively enumerable set \(M\) and any natural \(n\),
\(K(M,n,\varepsilon) \le \log_2 1/\varepsilon + C\), where \(C\) is a constant independent of \(n\) and \(\varepsilon\). The question of whether this upper bound (as a function of \(\varepsilon\)) is final remains open.
By analogy with \(K_{\mathfrak A}^{t}(M;n)\) from \((^{2})\), one may also introduce \(K_{\mathfrak A}^{t}(M,n,\varepsilon)\), characterizing the complexity of programs recognizing membership, among the natural numbers not exceeding \(n\), in the set \(M\) with frequency \(1-\varepsilon\) under the restriction \(t(x)\) on the admissible difficulty of processing programs. In that case, using Theorem 3 of \((^{2})\), one can show that there exists a recursively enumerable set \(M\) with the following property: for every general recursive function \(t(x)\) there exists \(\varepsilon_0>0\) such that, for every \(\varepsilon \le \varepsilon_0\) and every natural \(n\), one has
\(K_{\mathfrak A}^{t}(M,n,\varepsilon) \ge C_{t,\varepsilon} n\), where \(C_{t,\varepsilon}\) is a positive constant independent of \(n\) (but dependent on \(t\) and \(\varepsilon\)). This shows that, for sufficiently small \(\varepsilon\), the estimate from Theorem 3 \((^{2})\) cannot be improved in order of magnitude by replacing ordinary decidability with decidability with frequency \(1-\varepsilon\).
Computing Center
of the P. Stučka Latvian State University
Riga
Received
4 IX 1969
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