当前位置: X-MOL 学术Comput. Complex. › 论文详情
Our official English website, www.x-mol.net, welcomes your feedback! (Note: you will need to create a separate account there.)
An adaptivity hierarchy theorem for property testing
computational complexity ( IF 0.7 ) Pub Date : 2018-05-24 , DOI: 10.1007/s00037-018-0168-4
Clément L. Canonne , Tom Gur

AbstractAdaptivity is known to play a crucial role in property testing. In particular, there exist properties for which there is an exponential gap between the power of adaptive testing algorithms, wherein each query may be determined by the answers received to prior queries, and their non-adaptive counterparts, in which all queries are independent of answers obtained from previous queries. In this work, we investigate the role of adaptivity in property testing at a finer level. We first quantify the degree of adaptivity of a testing algorithm by considering the number of “rounds of adaptivity” it uses. More accurately, we say that a tester is k-(round) adaptive if it makes queries in $${k+1}$$k+1 rounds, where the queries in the $${i}$$i ’th round may depend on the answers obtained in the previous $${i-1}$$i-1 rounds. Then, we ask the following question: Does the power of testing algorithms smoothly grow with the number of rounds of adaptivity? We provide a positive answer to the foregoing question by proving an adaptivity hierarchy theorem for property testing. Specifically, our main result shows that for every $${n \in \mathbb{N}}$$n∈N and $${0 \le k \le n^{0.33}}$$0≤k≤n0.33 there exists a property $${\mathcal{P}_{n,k}}$$Pn,k of functions for which (1) there exists a $${k}$$k-adaptive tester for $${\mathcal{P}_{n,k}}$$Pn,k with query complexity $${{\tilde O}{(k)}}$$O~(k), yet (2) any $${(k-1)}$$(k-1)-adaptive tester for $${\mathcal{P}_{n,k}}$$Pn,k must make $${{\tilde \Omega}{(n/k^2)}}$$Ω~(n/k2) queries. In addition, we show that such a qualitative adaptivity hierarchy can be witnessed for testing natural properties of graphs.

中文翻译:

属性测试的适应性层次定理

摘要众所周知,适应性在属性测试中起着至关重要的作用。特别是,存在自适应测试算法的能力之间存在指数差距的属性,其中每个查询都可以由先前查询收到的答案及其非自适应对应项确定,其中所有查询都独立于答案从以前的查询中获得。在这项工作中,我们在更精细的水平上研究了适应性在属性测试中的作用。我们首先通过考虑它使用的“自适应轮数”的数量来量化测试算法的自适应程度。更准确地说,如果测试者在 $${k+1}$$k+1 轮中进行查询,则我们说测试器是 k-(round) 自适应的,其中 $${i}$$i 'th 轮中的查询可能取决于在之前的 $${i-1}$$i-1 轮中获得的答案。然后,我们提出以下问题:测试算法的能力是否会随着自适应轮数的增加而平稳增长?我们通过证明属性测试的适应性层次定理为上述问题提供了肯定的答案。具体来说,我们的主要结果表明,对于每一个 $${n \in \mathbb{N}}$$n∈N 和 $${0 \le k \le n^{0.33}}$$0≤k≤n0.33存在属性 $${\mathcal{P}_{n,k}}$$Pn,k 的函数,其中 (1) 存在 $${\ mathcal{P}_{n,k}}$$Pn,k 查询复杂度为 $${{\tilde O}{(k)}}$$O~(k),但 (2) 任何 $${( k-1)}$$(k-1)-$${\mathcal{P}_{n,k}}$$Pn,k 的自适应测试器必须使 $${{\tilde \Omega}{(n /k^2)}}$$Ω~(n/k2) 查询。此外,我们表明可以见证这种定性的自适应层次结构来测试图的自然属性。测试算法的威力是否会随着自适应轮数的增加而平稳增长?我们通过证明属性测试的适应性层次定理为上述问题提供了肯定的答案。具体来说,我们的主要结果表明,对于每一个 $${n \in \mathbb{N}}$$n∈N 和 $${0 \le k \le n^{0.33}}$$0≤k≤n0.33存在属性 $${\mathcal{P}_{n,k}}$$Pn,k 的函数,其中 (1) 存在 $${\ mathcal{P}_{n,k}}$$Pn,k 查询复杂度为 $${{\tilde O}{(k)}}$$O~(k),但 (2) 任何 $${( k-1)}$$(k-1)-$${\mathcal{P}_{n,k}}$$Pn,k 的自适应测试器必须使 $${{\tilde \Omega}{(n /k^2)}}$$Ω~(n/k2) 查询。此外,我们表明可以见证这种定性的自适应层次结构来测试图的自然属性。测试算法的威力是否会随着自适应轮数的增加而平稳增长?我们通过证明属性测试的适应性层次定理为上述问题提供了肯定的答案。具体来说,我们的主要结果表明,对于每一个 $${n \in \mathbb{N}}$$n∈N 和 $${0 \le k \le n^{0.33}}$$0≤k≤n0.33存在属性 $${\mathcal{P}_{n,k}}$$Pn,k 的函数,其中 (1) 存在 $${\ mathcal{P}_{n,k}}$$Pn,k 查询复杂度为 $${{\tilde O}{(k)}}$$O~(k),但 (2) 任何 $${( k-1)}$$(k-1)-$${\mathcal{P}_{n,k}}$$Pn,k 的自适应测试器必须使 $${{\tilde \Omega}{(n /k^2)}}$$Ω~(n/k2) 查询。此外,我们表明可以见证这种定性的自适应层次结构来测试图的自然属性。
更新日期:2018-05-24
down
wechat
bug