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第629行：
第629行： − Turing(1936)除了在脚注中描述了如何使用自动机来“找到所有希尔伯特微积分的可证明公式”，而不是使用选择机之外，没有做进一步的说明。他"假设选择总是在0和1之间。那么每个证明将由i<sub>1</sub>, i<sub>2</sub>, ..., i<sub>n</sub> (i<sub>1</sub> = 0 或 1, i<sub>2</sub> = 0 或 1, ..., i<sub>n</sub> = 0 或 1) 的选择序列决定，因此数字 2<sup>n</sup> + i<sub>1</sub>2<sup>n-1</sup> + i<sub>2</sub>2<sup>n-2</sup> + ... +i<sub>n</sub> 完全决定了证明。自动机依次执行验证1、验证2、验证3，……”+
→选择C型机器、Oracle的O型机器
Turing (1936) does not elaborate further except in a footnote in which he describes how to use an a-machine to "find all the provable formulae of the [Hilbert] calculus" rather than use a choice machine. He "suppose[s] that the choices are always between two possibilities 0 and 1. Each proof will then be determined by a sequence of choices i<sub>1</sub>, i<sub>2</sub>, ..., i<sub>n</sub> (i<sub>1</sub> = 0 or 1, i<sub>2</sub> = 0 or 1, ..., i<sub>n</sub> = 0 or 1), and hence the number 2<sup>n</sup> + i<sub>1</sub>2<sup>n-1</sup> + i<sub>2</sub>2<sup>n-2</sup> + ... +i<sub>n</sub> completely determines the proof. The automatic machine carries out successively proof 1, proof 2, proof 3, ..." (Footnote ‡, The Undecidable, p. 138)
Turing (1936) does not elaborate further except in a footnote in which he describes how to use an a-machine to "find all the provable formulae of the [Hilbert] calculus" rather than use a choice machine. He "suppose[s] that the choices are always between two possibilities 0 and 1. Each proof will then be determined by a sequence of choices i<sub>1</sub>, i<sub>2</sub>, ..., i<sub>n</sub> (i<sub>1</sub> = 0 or 1, i<sub>2</sub> = 0 or 1, ..., i<sub>n</sub> = 0 or 1), and hence the number 2<sup>n</sup> + i<sub>1</sub>2<sup>n-1</sup> + i<sub>2</sub>2<sup>n-2</sup> + ... +i<sub>n</sub> completely determines the proof. The automatic machine carries out successively proof 1, proof 2, proof 3, ..." (Footnote ‡, The Undecidable, p. 138)
图灵(1936)除了在脚注中描述了如何使用自动机来“找到所有希尔伯特微积分的可证明公式”，而不是使用选择机之外，没有做进一步的说明。他"假设选择总是在0和1之间。那么每个证明将由i<sub>1</sub>, i<sub>2</sub>, ..., i<sub>n</sub> (i<sub>1</sub> = 0 或 1, i<sub>2</sub> = 0 或 1, ..., i<sub>n</sub> = 0 或 1) 的选择序列决定，因此数字 2<sup>n</sup> + i<sub>1</sub>2<sup>n-1</sup> + i<sub>2</sub>2<sup>n-2</sup> + ... +i<sub>n</sub> 完全决定了证明。自动机依次执行验证1、验证2、验证3，……”
This is indeed the technique by which a deterministic (i.e., a-) Turing machine can be used to mimic the action of a nondeterministic Turing machine; Turing solved the matter in a footnote and appears to dismiss it from further consideration.
This is indeed the technique by which a deterministic (i.e., a-) Turing machine can be used to mimic the action of a nondeterministic Turing machine; Turing solved the matter in a footnote and appears to dismiss it from further consideration.