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首页> 外文期刊>Journal of Symbolic Logic >The complexity of ODD_n~A
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The complexity of ODD_n~A

机译:ODD_n〜A的复杂度

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摘要

For a fixed set A, the number of queries to A needed in order to decide a set S is a measure of S's complexity. We consider the complexity of certain sets defined in terms of A: ODD_n~A = {(x_1, …, x_n) : #_n~4(x_1, …, x_n) is odd} and, for m ≥ 2. MODm_n~A = {(x_1, …, x_n):#_n~A(x_1, …, x_n) not ≡ 0 (mod m)}. where #_n~A(x_1, …, x_n) = A(x_1) + … + A(x_n). (We identify A(x) with χ_A(x), where χ_A is the characteristic function of A.) If A is a nonrecursive semirecursive set or if A is a jump, we give tight bounds on the number of queries needed in order to decide ODD_n~A and MODm_n~A: ·ODD_n~A can be decided with n parallel queries to A. but not with n - 1. ·ODD_n~A can be decided with [log(n + 1)] sequential queries to A but not with [log(n + 1)] - 1. ·MODm_n~A can be decided with[n/m] + [n/m] parallel queries to A but not with [n/m] + [n/m] - 1. ·MODm_n~A can be decided with [log([n/m] + [n/m] + 1] sequential queries to A but not with [log([n/m] + [n/m] + 1)] - 1. The lower bounds above hold for nonrecursive recursively enumerable sets A as well. (Interestingly, the lower bonds for recursively enumerable sets follow by a general result from the lower bounds for semirecursive sets.) In particular, every nonzero truth-table degree contains a set A such that ODD_n~A cannot be decided with n - 1 parallel queries to A. Since every truth-table degree also contains a set B such that ODD_n~B)n can be decided with one query to B, a set's query complexity depends more on its structure than on its degree. For a fixed set A, Q(n, A) = {S: S can be decided with n sequential queries to A}. Q_‖(n, A) = {S: S can be decided with n parallel queries to A}. We show that if A is semirecursive or recursively enumerable, but is not recursive, then these classes form non-collapsing hierarchies: ·Q(0, A) is contained in Q(1, A) is contained in Q(2, A) is contained in … ·Q_‖(0, A) is contained in Q_‖(1, A) is contained in Q_‖(2, A) is contained in … The same is true if A is a jump.
机译:对于固定的集合A,确定集合S所需的对A的查询次数是S复杂度的度量。我们考虑用A定义的某些集合的复杂度:ODD_n〜A = {(x_1,…,x_n):#_n〜4(x_1,…,x_n)是奇数},并且对于m≥2。MODm_n〜A = {((x_1,…,x_n):#_ n〜A(x_1,…,x_n)不≡0(mod m)}。其中#_n〜A(x_1,...,x_n)= A(x_1)+…+ A(x_n)。 (我们将A(x)标识为χ_A(x),其中χ_A是A的特征函数。)如果A是非递归半递归集,或者如果A是跳转,则我们对所需的查询数量给出严格的限制,以便确定ODD_n〜A和MODm_n〜A:·可以通过对A的n个并行查询来确定ODD_n〜A,但是不能通过n-1来确定。·可以通过对[A]的[log(n + 1)]顺序查询来确定ODD_n〜A。但不能使用[log(n + 1)]-1。·可以通过对[A]的[n / m] + [n / m]并行查询来确定MODm_n〜A,但不能使用[n / m] + [n / m] ]-1.·MODm_n〜A可以通过对A的[log([n / m] + [n / m] + 1]顺序查询来确定,但不能通过[log([n / m] + [n / m] + 1)]-1.非递归可枚举集合A的上述下界也成立(有趣的是,递归可枚举集合的下键后面是半递归集合的下界的一般结果。)尤其是每个非零真值表度包含一个集合A,使得无法通过对A的n-1个并行查询来确定ODD_n〜A。 le程度还包含一个集合B,这样就可以通过对B的一个查询来确定ODD_n〜B)n,一个集合的查询复杂度更多地取决于其结构而不是程度。对于固定集A,Q(n,A)= {S:S可以通过对A的n个顺序查询来确定。 Q_‖(n,A)= {S:S可以通过对A的n个并行查询来确定。我们证明如果A是半递归的或递归可枚举的,但不是递归的,则这些类形成非折叠层次结构:·Q(0,A)包含在Q(1,A)包含在Q(2,A)包含在...中。·Q_‖(0,A)包含在Q _′(1,A)包含在Q_‖(2,A)包含在…中。如果A是跳跃,则同样如此。

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