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@ -47,11 +47,11 @@ $$f(x,y) = \begin{pmatrix}0 & 1 & 0 & 1 & 0 & 1 & \dots \\\ 1 & 0 & 1 & 0 & 1 &
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Such a function can be partitioned based on a single bit of information e.g. whether $y$ is divisible by 2:
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Such a function can be partitioned based on a single bit of information e.g. whether $y$ is divisible by 2:
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$$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 & \ddots \end{pmatrix}$$
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or not:
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or not:
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$$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 & \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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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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