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date = "2018-12-07T21:08:39-08:00"
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date = "2018-12-07T00:00:00-08:00"
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title = "Privacy is Consent (alt: On Optimal Privacy vs. System Efficiency)"
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@ -47,11 +47,11 @@ $$f(x,y) = \begin{pmatrix}0 & 1 & 0 & 1 & 0 & 1 & \dots \\\ 1 & 0 & 1 & 0 & 1 &
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Such a function can be partitioned based on a single bit of information e.g. whether $y$ is divisible by 2:
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$$f(x,y) = \begin{pmatrix}1 & 1 & 1 & \dots\\\ 0 & 0 & 0 & \dots\\\ 1 & 1 & 1 & \dots \\\ 0 & 0 & 0 & \dots \\\ \vdots & \vdots & \vdots & \vdots \ddots \end{pmatrix}$$
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$$f(x,y) = \begin{pmatrix}1 & 1 & 1 & \dots\\\ 0 & 0 & 0 & \dots\\\ 1 & 1 & 1 & \dots \\\ 0 & 0 & 0 & \dots \\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$$
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or not:
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$$f(x,y) = \begin{pmatrix}0 & 0 & 0 & \dots \\\ 1 & 1 & 1 & \dots\\\ 0 & 0 & 0 & \dots\\\ 1 & 1 & 1 & \dots \\\ \vdots & \vdots & \vdots & \vdots \ddots \end{pmatrix}$$
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$$f(x,y) = \begin{pmatrix}0 & 0 & 0 & \dots \\\ 1 & 1 & 1 & \dots\\\ 0 & 0 & 0 & \dots\\\ 1 & 1 & 1 & \dots \\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$$
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We can partition each matrix again, based on whether $x$ is divisible by 2 or not. Regardless of which of the above partitioned matrices we start with, the resulting matrices are the same:
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@ -0,0 +1,36 @@
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e 0 1 2 3
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-------
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0| 0 1 0 1
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1| 1 0 1 0
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2| 0 1 0 1
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3| 1 0 1 0
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f(2,1) = 1 (y is either 1 or 3 | x is either 0 or 2)
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p1 sends 0
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0 1 2 3
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0| 0 1 0 1
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2| 0 1 0 1
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p2 sends 1
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1 3
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---
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0| 1 1
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2| 1 1
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Why AND Fails:
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0 1 2 3
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--------
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0| 0 0 0 0
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1| 0 1 0 1
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2| 0 0 1 1
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3| 0 1 1 1
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f(1,2) = 1
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p1 sends....there is no way to send information without revealling out input...the matrix is not partitionable.
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