Editing Open Problems:78
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|who=Grigory Yaroslavtsev | |who=Grigory Yaroslavtsev | ||
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| β | For a function $f:\{0,1\}^n\rightarrow\{0,1\}$, we define its deterministic linear sketch complexity $D^\text{lin}(f)$ as a the smallest number $k$ such that there exist $k$ sets $S_1,\ldots,S_k \subseteq [n]$ such that for any $x\in \{0,1\}^n$, we can compute $f(x)$ using $\sum_{i\in S_1} x_i, \ldots, \sum_{i\in S_k} x_i$, where the sum is mod $2$. For randomized linear sketch complexity, which is denoted by $R^\text{lin}(f)$, the $k$ sets are chosen in advance from a joint distribution and are available for recovering $f(x)$. Please see the paper by Kannan, Mossel, | + | For a function $f:\{0,1\}^n\rightarrow\{0,1\}$, we define its deterministic linear sketch complexity $D^\text{lin}(f)$ as a the smallest number $k$ such that there exist $k$ sets $S_1,\ldots,S_k \subseteq [n]$ such that for any $x\in \{0,1\}^n$, we can compute $f(x)$ using $\sum_{i\in S_1} x_i, \ldots, \sum_{i\in S_k} x_i$, where the sum is mod $2$. For randomized linear sketch complexity, which is denoted by $R^\text{lin}(f)$, the $k$ sets are chosen in advance from a joint distribution and are available for recovering $f(x)$. Please see the paper by Kannan, Mossel, and Yaroslavtsev {{cite|KannanMSY-18}} for more details. |
Given $f$, we also define $f^+:\{0,1\}^n\times \{0,1\}^n \rightarrow\{0,1\}$ as $f^+ (x,y) = f(x\oplus y)$ for all $x,y\in\{0,1\}^n$, where $\oplus$ denotes bitwise XOR. It is known that $D^\text{lin}(f) = D^\rightarrow(f^+)$ {{cite|MontanaroO-09}}, where $D^\rightarrow$ denotes one-way communication complexity (Alice sends one message to Bob). | Given $f$, we also define $f^+:\{0,1\}^n\times \{0,1\}^n \rightarrow\{0,1\}$ as $f^+ (x,y) = f(x\oplus y)$ for all $x,y\in\{0,1\}^n$, where $\oplus$ denotes bitwise XOR. It is known that $D^\text{lin}(f) = D^\rightarrow(f^+)$ {{cite|MontanaroO-09}}, where $D^\rightarrow$ denotes one-way communication complexity (Alice sends one message to Bob). | ||
Prove (or disprove) the following conjecture: $R^\text{lin}(f) = \tilde{\Theta}(R^\rightarrow(f^+))$. | Prove (or disprove) the following conjecture: $R^\text{lin}(f) = \tilde{\Theta}(R^\rightarrow(f^+))$. | ||
