Relative positioning of node with TikZ Positioning

Question

I am trying to replicate an automata with tikz and was wondering if there is a keyword that would allow be to mimic the position of the states with relative positioning without a "dummy" node. Below is my current solution with a blank node and an image of what I am trying to replicate the state positions of:

\documentclass[12pt,letterpaper]{article}
\usepackage[margin=1in]{geometry}
\usepackage{fancyhdr, float}

\usepackage{amsmath, amssymb, amsthm}
\usepackage{cleveref}
\usepackage{algorithm2e}
\usepackage{enumerate}
\usepackage{graphicx}

\usepackage{tikz}
\usetikzlibrary{automata,positioning}

\renewcommand{\a}{\mathtt{a}}
\renewcommand{\b}{\mathtt{b}}

\begin{document}
\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto] 
    \node[state,initial] (A_2) {$A_2$};
    \node (Z) [right=of A_2] {};
    \node[state,accepting] (B_2) [right=of Z] {$B_2$};
    \node[state] (C_2) [below=of Z] {$C_2$};
    \path[->]
        (A_2) edge [bend left=15] node {$\varepsilon$} (B_2)
    edge [swap] node {$\a$} (C_2)
        (B_2) edge [bend left=15] node {$\a$} (A_2)
        (C_2) edge [loop right] node {$\b$} ()
    edge [swap] node {$\a,\b$} (B_2);
\end{tikzpicture}
\end{document}

Answer 1

I changed \a,\b to \tta,\ttb, because it's not a good idea to overwrite existing macros. Their original definition (in article) is

> \a=macro:
#1->\expandafter \@changed@cmd \csname \string #1\endcsname \relax .

> \b=macro:
->\OT1-cmd \b \OT1\b .

To get the gray background use state/.append style={fill=black!10}. The arrow can be mimicked by >=stealth.

\documentclass[tikz]{standalone}
\usetikzlibrary{automata,positioning}
\newcommand{\tta}{\mathtt{a}}
\newcommand{\ttb}{\mathtt{b}}
\begin{document}
\begin{tikzpicture}[>=stealth,auto,state/.append style={fill=black!10}] 
    \node[state,initial] (A2) {$A_2$};
    \node[state,below right=of A2] (C2) {$C_2$};
    \node[state,accepting,above right=of C2] (B2) {$B_2$};
    \draw (A2) edge[->,bend left=15] node {$\varepsilon$} (B2);
    \draw (B2) edge[->,bend left=15] node {$\tta$} (A2);
    \draw (C2) edge[<-] node {$\tta$} (A2);
    \draw (B2) edge[<-] node {$\tta,\ttb$} (C2);
    \draw (C2) edge [loop right] node {$\ttb$} (C2);
\end{tikzpicture}
\end{document}

Answer 2

Place node C_2 first then use right/left of=C_2,above of=C_2 together to place A_2 and B_2 diagonally. The result of following code looks exactly the same as in OP.

\documentclass[12pt,letterpaper]{article}
\usepackage[margin=1in]{geometry}
\usepackage{fancyhdr, float}

\usepackage{amsmath, amssymb, amsthm}
\usepackage{cleveref}
\usepackage{algorithm2e}
\usepackage{enumerate}
\usepackage{graphicx}

\usepackage{tikz}
\usetikzlibrary{automata,positioning}

\renewcommand{\a}{\mathtt{a}}
\renewcommand{\b}{\mathtt{b}}

\begin{document}
\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto] 
    \node[state] (C_2) {$C_2$};
    \node[state,initial] (A_2) [left of=C_2,above of=C_2] {$A_2$};
    \node[state,accepting] (B_2) [right of=C_2,above of=C_2] {$B_2$};
    \path[->]
        (A_2) edge [bend left=15] node {$\varepsilon$} (B_2)
    edge [swap] node {$\a$} (C_2)
        (B_2) edge [bend left=15] node {$\a$} (A_2)
        (C_2) edge [loop right] node {$\b$} ()
    edge [swap] node {$\a,\b$} (B_2);
\end{tikzpicture}
\end{document}