Past exam of the computer science course of the University of Oxford / 2015 / Computational complexity / 1 / d

${\mathcal{L}}_{\ell 3}$ is undecidable. The condition depends only on the accepted formal language. It holds for a Turing machine accepting the empty language, and fails for a Turing machine accepting the singleton $\{aaaa\}$, where $a$ is an alphabet symbol. It is therefore a nontrivial extensional property of programs, and Rice's theorem applies. In particular, the condition does not assert that the machine halts on every input; nonacceptance of long words over an alphabet may result from rejection or from an infinite computation.

${\mathcal{L}}_{1\mathrm{T}}$ is a decidable language: it consists of every valid Turing machine encoding. Every multitape Turing machine has an equivalent single-tape Turing machine. To see this, represent the finitely used portions of its tapes on one tape, separated by delimiters, and mark each simulated head position. To simulate one step, scan these portions to read the marked symbols, retain this fixed number of symbols and the control state in finite control, and scan again to update the symbols and head marks. Extend a represented portion with a blank cell when necessary. Each simulated step takes finitely many real steps.

The single-tape Turing machine accepts exactly when the simulated multitape Turing machine accepts; infinite computations remain infinite. The simulation may increase time complexity, but the question only asks about accepted formal languages. Hence a decider simply checks the syntax of the Turing machine encoding and accepts every valid description. Rice's theorem does not make this property undecidable, because the property is true of every computably enumerable language and is therefore trivial in the theorem's sense.