A simple algebraic expression for Kronecker delta when both arguments take values 0 or 1 I am not sure whether this question makes sense but if it does and if it has some answer, then that would hugely simplify my task. I am looking for an algebraic expression for Kronecker delta $\delta_{ij}$ when $i,j \in \{0, 1\}$. In essence, I want an expression for map $f$ that involves only addition, multiplication and powers such that:
$$ f(0, 0) = 1$$
$$ f(1, 1) = 1$$
$$ f(0, 1) = 0$$
$$ f(1, 0) = 0$$
I hope that such expression exists because for a similar case where we need:
$$g(0) = 1$$
$$g(1) = 0$$
we have : $$g(i) = 1 - i$$
Does somebody have any idea about this? Thanks in advance. 

$$g(x,y)=1-(x-y)^2$$
$$g(0,0)=1-(0-0)^2=1-0=1$$
$$g(1,1)=1-(1-1)^2=1-0=1$$
$$g(0,1)=1-(0-1)^2=1-1=0$$
$$g(1,1)=1-(1-0)^2=1-1=0$$
