Abstract: We study the problem of hardness amplification in NP<math><mstyle mathvariant="sans-serif" is="true"><mi is="true">NP</mi></mstyle></math>. We prove that if there is a balanced function in NP<math><mstyle mathvariant="sans-serif" is="true"><mi is="true">NP</mi></mstyle></math> such that any circuit of size s(n)=2Ω(n)<math><mi is="true">s</mi><mrow is="true"><mo is="true">(</mo><mi is="true">n</mi><mo is="true">)</mo></mrow><mo is="true">=</mo><msup is="true"><mrow is="true"><mn is="true">2</mn></mrow><mrow is="true"><mi is="true">Ω</mi><mrow is="true"><mo is="true">(</mo><mi is="true">n</mi><mo is="true">)</mo></mrow></mrow></msup></math> fails to compute it on a 1/poly(n)<math><mn is="true">1</mn><mo is="true">/</mo><mstyle mathvariant="normal" is="true"><mi is="true">poly</mi></mstyle><mrow is="true"><mo is="true">(</mo><mi is="true">n</mi><mo is="true">)</mo></mrow></math> fraction of inputs, then there is a function in NP<math><mstyle mathvariant="sans-serif" is="true"><mi is="true">NP</mi></mstyle></math> such that any circuit of size s′(n)<math><msup is="true"><mrow is="true"><mi is="true">s</mi></mrow><mrow is="true"><mo is="true">′</mo></mrow></msup><mrow is="true"><mo is="true">(</mo><mi is="true">n</mi><mo is="true">)</mo></mrow></math> fails to compute it on a 1/2−1/s′(n)<math><mn is="true">1</mn><mo is="true">/</mo><mn is="true">2</mn><mo is="true">−</mo><mn is="true">1</mn><mo is="true">/</mo><msup is="true"><mrow is="true"><mi is="true">s</mi></mrow><mrow is="true"><mo is="true">′</mo></mrow></msup><mrow is="true"><mo is="true">(</mo><mi is="true">n</mi><mo is="true">)</mo></mrow></math> fraction of inputs, with s′(n)=2Ω(n2/3)<math><msup is="true"><mrow is="true"><mi is="true">s</mi></mrow><mrow is="true"><mo is="true">′</mo></mrow></msup><mrow is="true"><mo is="true">(</mo><mi is="true">n</mi><mo is="true">)</mo></mrow><mo is="true">=</mo><msup is="true"><mrow is="true"><mn is="true">2</mn></mrow><mrow is="true"><mi is="true">Ω</mi><mrow is="true"><mo is="true">(</mo><msup is="true"><mrow is="true"><mi is="true">n</mi></mrow><mrow is="true"><mn is="true">2</mn><mo is="true">/</mo><mn is="true">3</mn></mrow></msup><mo is="true">)</mo></mrow></mrow></msup></math>. This improves the result of Healy et al. (STOC’04), which only achieves s′(n)=2Ω(n1/2)<math><msup is="true"><mrow is="true"><mi is="true">s</mi></mrow><mrow is="true"><mo is="true">′</mo></mrow></msup><mrow is="true"><mo is="true">(</mo><mi is="true">n</mi><mo is="true">)</mo></mrow><mo is="true">=</mo><msup is="true"><mrow is="true"><mn is="true">2</mn></mrow><mrow is="true"><mi is="true">Ω</mi><mrow is="true"><mo is="true">(</mo><msup is="true"><mrow is="true"><mi is="true">n</mi></mrow><mrow is="true"><mn is="true">1</mn><mo is="true">/</mo><mn is="true">2</mn></mrow></msup><mo is="true">)</mo></mrow></mrow></msup></math> for the case with s(n)=2Ω(n)<math><mi is="true">s</mi><mrow is="true"><mo is="true">(</mo><mi is="true">n</mi><mo is="true">)</mo></mrow><mo is="true">=</mo><msup is="true"><mrow is="true"><mn is="true">2</mn></mrow><mrow is="true"><mi is="true">Ω</mi><mrow is="true"><mo is="true">(</mo><mi is="true">n</mi><mo is="true">)</mo></mrow></mrow></msup></math>.
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