MathDB

ICMC 2

Part of ICMC

Subcontests

(6)
4
2

ICMC 2018/19 Round 2, Problem 4

Let f:{0,1}n{0,1}Rf:\{0, 1\}^n \to \{0, 1\} \subseteq \mathbb{R} be a function. Call such a function a Boolean function. Let \wedge denote the component-wise multiplication in {0,1}n\{0,1\}^n. For example, for n=4n = 4, (0,0,1,1)(0,1,0,1)=(0,0,0,1).(0,0,1,1) \wedge (0,1,0,1) = (0,0,0,1).
Let S={i1,i2,,ik}{1,2,,n}S = \left\{i_1,i_2,\ldots ,i_k\right\} \subseteq \left\{1,2,\ldots ,n\right\}. ff is called the oligarchy function over SS if f(x)=xi1,xi2,,xik  (with the usual multiplication),f (x) = x_{i_1},x_{i_2},\ldots,x_{i_k}\ \text{ (with the usual multiplication),}
where xix_i denotes the ii-th component of xx. By convention, ff is called the oligarchy function over \emptyset if ff is constantly 1.
(i) Suppose ff is not constantly zero. Show that ff is an oligarchy function if and only if ff satisfies f(xy)=f(x)f(y), x,y{0,1}n.f(x\wedge y)=f(x)f(y),\ \forall x,y\in\left\{0,1\right\}^n. Let YY be a uniformly distributed random variable over {0,1}n\left\{0, 1\right\}^n. Let TT be an operator that maps Boolean functions to functions {0,1}nR\left\{0, 1\right\}^n\to\mathbb{R}, such that
(Tf)(x)=EY(f(xY)), x{0,1}n(Tf)(x)=E_Y(f(x\wedge Y)),\ \forall x\in\left\{0,1\right\}^n
where EY()E_Y() denotes the expectation over YY. ff is called an eigenfunction of TT if λR\{0}\exists\lambda\in\mathbb{R}\backslash\left\{0\right\} such that (Tf)(x)=λf(x), x{0,1}n(Tf)(x)=\lambda f(x),\ \forall x\in\left\{0,1\right\}^n
(ii) Prove that ff is an eigenfunction of TT if and only if ff is an oligarchy function.