Given an $n$-bit Boolean function with a complexity measure (such as block sensitivity, query complexity, etc.) $M(f) = k$, the hardness condensation question asks whether $f$ can be restricted to $O(k)$ variables such that the complexity measure is $Ω(k)$? In this work, we study the condensability of block sensitivity, certificate complexity, AND (and OR) query complexity and Fourier sparsity. We show that block sensitivity does not condense under restrictions, unlike sensitivity: there exists a Boolean function $f$ with query complexity $k$ such that any restriction of $f$ to $O(k)$ variables has block sensitivity $O(k^{\frac{2}{3}})$. This answers an open question in Göös, Newman, Riazanov, and Sokolov (2024) in the negative. The same function yields an analogous incondensable result for certificate complexity. We further show that $\mathsf{AND}$(and $\mathsf{OR}$) decision trees are also incondensable. In contrast, we prove that Fourier sparsity admits a weak form of condensation.

翻译：给定一个$n$位布尔函数，其复杂度度量（如块灵敏度、查询复杂度等）为$M(f) = k$，硬度凝聚问题询问：是否可以将$f$限制在$O(k)$个变量上，使得复杂度度量仍为$Ω(k)$？本文研究了块灵敏度、证书复杂度、$\mathsf{AND}$（及$\mathsf{OR}$）查询复杂度与傅里叶稀疏性的凝聚性。我们证明，与灵敏度不同，块灵敏度在限制下不具备凝聚性：存在一个查询复杂度为$k$的布尔函数$f$，使得$f$在任意$O(k)$个变量上的限制的块灵敏度仅为$O(k^{\frac{2}{3}})$。这否定了Göös、Newman、Riazanov与Sokolov（2024）中提出的一个开放性问题。同一函数对证书复杂度也产生了类似的不可凝聚结果。我们进一步证明$\mathsf{AND}$（及$\mathsf{OR}$）决策树同样不具备凝聚性。相比之下，我们证明了傅里叶稀疏性允许一种弱形式的凝聚。