Latex tabularx issue with spacing

Question

So just trying to edit a table someone had done in tabularx as I liked the style however I have no idea what I'm doing really and as such I've come up against a problem. There is a big space after the entry 'setup', in the below code and I would like it all to be nice and symmetrical the space makes it very asymmetrical. Is there a way to remove this space?

\documentclass{article}
\usepackage{array}
\usepackage{booktabs}
\usepackage{ragged2e}
\usepackage{tabularx}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\newcolumntype{Y}{>{\RaggedRight\arraybackslash}X}


\begin{document}
    \noindent
    \edef\TabularRowHeight{\the\dimexpr-\arraystretch\baselineskip}

    \begin{tabularx}{\textwidth}{YY}
    \toprule
    Alice  &  Bob \\
    \midrule
    \emph{Setup}\\
    \cmidrule(lr){1-1}
    Alice \& Bob select a prime $p$ and a generator $g$ for the finite field $\mathbb{F}_p$
    &\\

    &\emph{Private Computation}\\
    \cmidrule(lr){2-2}
    &  Bob randomly selects $b \in \mathbb{F}_p$ and then computes the following:

     $B \equiv g^b \mod p$ once computed Bob sends Alice $B$

     \\[\TabularRowHeight] & \\

    & \emph{Key Computation}\\
    \cmidrule(lr){2-2}
    & Bob who now has $A$, calculates the following:$A^b \equiv (g^a)^b \equiv g^{ab} \mod p$

     \\


    \emph{Private Computation}\\ 
    \cmidrule(lr){1-1}
    Alice randomly selects $a \in \mathbb{F}_p$ and then computes the following:

    $A \equiv g^a \mod p$ once computed Alice sends Bob $A$ & \\ \pagebreak

    \emph{Key Computation}\\ 
    \cmidrule(lr){1-1}
    Alice who now has $B$, calculates the  following:$B^a \equiv (g^b)^a \equiv g^{ab} \mod p$ & \\


    \bottomrule


    \end{tabularx}
\end{document}

Picture of current output

Output I'd like to achieve

Answer 1

I don't think removing the alternating layout from the table you showed makes a lot of sense, but here is a try

\documentclass{article}
\usepackage{array}
\usepackage{booktabs}
\usepackage{ragged2e}
\usepackage{tabularx}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\newcolumntype{Y}{>{\RaggedRight\arraybackslash}X}


\begin{document}
        \emph{Setup}: Alice \& Bob select a prime $p$ and a generator $g$ for the finite field $\mathbb{F}_p$
    \begin{tabularx}{\textwidth}{@{}YY@{}}
    \toprule
    Alice  &  Bob \\
    \midrule
    \addlinespace[1.2em]
    \emph{Private Computation}:&\\ \addlinespace
%    \cmidrule(r){1-1}\cmidrule(l){2-2}
    Alice randomly selects $a \in \mathbb{F}_p$ and then computes the following:

    $A \equiv g^a \mod p$ once computed Alice sends Bob $A$& Bob randomly selects $b \in \mathbb{F}_p$ and then computes the following:

     $B \equiv g^b \mod p$ once computed Bob sends Alice $B$\\\addlinespace[1.2em]

     \emph{Key Computation}: &\\ \addlinespace
%    \cmidrule(r){1-1}\cmidrule(l){2-2}
    Alice who now has $B$, calculates the  following:$B^a \equiv (g^b)^a \equiv g^{ab} \mod p$  & Bob who now has $A$, calculates the following:$A^b \equiv (g^a)^b \equiv g^{ab} \mod p$
     \\
    \bottomrule


    \end{tabularx}
\end{document}

Instead I would prefer a 3-column layout:

\documentclass{article}
\usepackage{array}
\usepackage{booktabs}
\usepackage{ragged2e}
\usepackage{tabularx}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\newcolumntype{Y}{>{\RaggedRight\arraybackslash}X}


\begin{document}
        \emph{Setup}: Alice \& Bob select a prime $p$ and a generator $g$ for the finite field $\mathbb{F}_p$
    \begin{tabularx}{\textwidth}{@{}>{\RaggedRight\arraybackslash}p{2.2cm}YY@{}}
    \toprule
    & Alice  &  Bob \\
    \midrule
    \emph{Private \linebreak Computation} & %    
    Alice randomly selects $a \in \mathbb{F}_p$ and then computes the following:

    $A \equiv g^a \mod p$ once computed Alice sends Bob $A$& Bob randomly selects $b \in \mathbb{F}_p$ and then computes the following:

     $B \equiv g^b \mod p$ once computed Bob sends Alice $B$\\\addlinespace
     \emph{Key \linebreak Computation} & Alice who now has $B$, calculates the  following:$B^a \equiv (g^b)^a \equiv g^{ab} \mod p$  & Bob who now has $A$, calculates the following:$A^b \equiv (g^a)^b \equiv g^{ab} \mod p$
     \\
    \bottomrule


    \end{tabularx}
\end{document}