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There's an exercise in Arora-Barak on sensitivity and block sensitivity. The statement of the problem is:

Let $f$ be a function on $n = k^2$ variables that is an OR of $k$ applications of $g: \{0, 1\}^n \rightarrow \{0,1\}$, all $k$ blocks of variables are disjoint. $g(x_1, \ldots, x_k) = 1$ if there exists $i \in [k-1]$ s.t. $x_i = x_{i=1} = 1$ and $x_j = 0$ for all $j \neq i$. Prove that $s(f) = \sqrt{n}, bs(f) = n/2$.

I am confused about $x_{i=1} = 1$. Is there a typo and should it be $x_{i-1} = 1?$ If it is the case, what about $i=1$? Or is it simply $x_1 = 1?$

Another question is whether $g = 1$ iff the condition holds, or if it is only a sufficient condition.

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    $\begingroup$ The Arora-Barak book has many typos, esp. in the exercises. To the best of my knowledge, unfortunately they don't have an errata page (like, e.g., Goldreich has). $\endgroup$ – dkaeae Jul 8 at 9:35
  • $\begingroup$ Here, possibly $x_{i=1}$ should be simply ignored, that is, it should simply read $x_i = 1$. $\endgroup$ – dkaeae Jul 8 at 9:36
  • $\begingroup$ The statement says that OR of $k$ clauses has sensitivity $k$, which is problematic: consider some assignment $y$ s.t. $g(y) = 0$. Let $x$ be $k$ copies of $y$. $s(f) = k$ implies that $s(g) = 1$ which is hardly believable for any interpretation of the definition of $g$. $\endgroup$ – diplodoc Jul 8 at 16:43
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The definition of $g$ should be: $g(x_1,\ldots,x_k) = 1$ if there exists $i \in [k-1]$ such that $x_i=x_{i+1}=1$ and $x_j = 0$ for $j \neq i,i+1$.

As for your question, whether $g=1$ iff the condition holds, or whether it is just a sufficient condition: in mathematics, when we define something, we use the word "if" to mean "iff". Check your favorite textbook, monograph or paper for many examples.

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  • $\begingroup$ Thanks, it makes sense now. Also, there should be $s(f) = \Theta (\sqrt{n})$, since $s(f ) = \sqrt{n}$ is clearly incorrect. $\endgroup$ – diplodoc Jul 9 at 6:40
  • $\begingroup$ I’m not so sure it’s incorrect. $\endgroup$ – Yuval Filmus Jul 9 at 6:42
  • $\begingroup$ Suppose that $y = [0, 1, 0, 0, \ldots, 0], y \in \{0, 1\}^k$. Suppose that $x$ is $k$ copies of $y$. Then, $f(x) = 0$. However, by switching first or third bit in any block, we change the value of the function. It means that $s_f(x) \ge 2\sqrt{n}$. I've checked, and it seems that $s(f) = 2\sqrt{n}$ $\endgroup$ – diplodoc Jul 9 at 6:52
  • $\begingroup$ You can fix that by requiring $i$ to be even. At any rate, looking at the literature, people care about the constant. $\endgroup$ – Yuval Filmus Jul 9 at 7:00

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