Difference between revisions of "Iterated majority"
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Let <math>n = 3^k</math>. The '''iterated majority function''' <math>f:\{-1,1\}^n \to \{-1,1\}</math> is a recursively-defined variation on the [[majority function]]: | Let <math>n = 3^k</math>. The '''iterated majority function''' <math>f:\{-1,1\}^n \to \{-1,1\}</math> is a recursively-defined variation on the [[majority function]]: | ||
| − | <math> f(x) = \begin{cases} \textrm{maj}(x), & \textrm{if}~ n = 3 \\ | + | :::{| class="wikitable" |
| + | |- | ||
| + | |<math> f(x) = \begin{cases} \textrm{maj}(x), & \textrm{if}~ n = 3 \\ | ||
\textrm{maj}(f(x^{(1)}),f(x^{(2)}),f(x^{(3)}) & \textrm{otherwise}, \end{cases}</math> | \textrm{maj}(f(x^{(1)}),f(x^{(2)}),f(x^{(3)}) & \textrm{otherwise}, \end{cases}</math> | ||
| + | |} | ||
where <math>x^{(1)} = (x_1,x_2\ldots x_{n/3})</math>, <math>x^{(2)} = (x_{n/3+1},\ldots x_{2n/3})</math> and <math>x^{(3)} = (x_{2n/3+1},\ldots x_{n})</math>. | where <math>x^{(1)} = (x_1,x_2\ldots x_{n/3})</math>, <math>x^{(2)} = (x_{n/3+1},\ldots x_{2n/3})</math> and <math>x^{(3)} = (x_{2n/3+1},\ldots x_{n})</math>. | ||
Revision as of 13:38, 20 November 2019
Definition
Let [math]n = 3^k[/math]. The iterated majority function [math]f:\{-1,1\}^n \to \{-1,1\}[/math] is a recursively-defined variation on the majority function:
[math] f(x) = \begin{cases} \textrm{maj}(x), & \textrm{if}~ n = 3 \\ \textrm{maj}(f(x^{(1)}),f(x^{(2)}),f(x^{(3)}) & \textrm{otherwise}, \end{cases}[/math]
where [math]x^{(1)} = (x_1,x_2\ldots x_{n/3})[/math], [math]x^{(2)} = (x_{n/3+1},\ldots x_{2n/3})[/math] and [math]x^{(3)} = (x_{2n/3+1},\ldots x_{n})[/math].
Properties
- TODO
- TODO: add influence.
