Answer:
See below
Step-by-step explanation:
We are looking for a function f such that
f(0) = -1
f(1) = 1
One of the simplest ones is the line that joins (0, -1) and (1,1)
y = 2x -1
So, if we have a Boolean n-vector in the first codification (0,1)
[tex]\large X=(x_1,x_2,...x_n)[/tex]
then
[tex]\large (2x_1-1,2x_2-1,...2x_n-1)=(y_1,y_2,...,y_n) =Y[/tex]
is the expression of Y in terms of X.
Similarly, the inverse function of the line, x = (y+1)/2, does the inverse transformation.
That is to say, if we have a vector
[tex]\large Y=(y_1,y_2,...,y_n)[/tex]
in the second codification -1 and +1
[tex]\large (\frac{y_1+1}{2},\frac{y_2+1}{2},...\frac{y_n+1}{2})=(x_1,x_2,...x_n)=X[/tex]
is the expression of X in terms of Y