PGF/TikZ Manual

The TikZ and PGF Packages
Manual for version 3.1.9a

Part II Installation and Configuration

by Till Tantau

This part explains how the system is installed. Typically, someone has already done so for your system, so this part can be skipped; but if this is not the case and you are the poor fellow who has to do the installation, read the present part.

(-tikz- diagram)

\usetikzlibrary {arrows.meta,automata,positioning,shadows}
\begin{tikzpicture}[->,>={Stealth[round]},shorten >=1pt,auto,node distance=2.8cm,on grid,semithick,
every state/.style={fill=red,draw=none,circular drop shadow,text=white}]

\node[initial,state] (A) {$q_a$};
\node[state] (B) [above right=of A] {$q_b$};
\node[state] (D) [below right=of A] {$q_d$};
\node[state] (C) [below right=of B] {$q_c$};
\node[state] (E) [below=of D] {$q_e$};

\path (A) edge node {0,1,L} (B)
edge node {1,1,R} (C)
(B) edge [loop above] node {1,1,L} (B)
edge node {0,1,L} (C)
(C) edge node {0,1,L} (D)
edge [bend left] node {1,0,R} (E)
(D) edge [loop below] node {1,1,R} (D)
edge node {0,1,R} (A)
(E) edge [bend left] node {1,0,R} (A);

\node [right=1cm,text width=8cm] at (C)
The current candidate for the busy beaver for five states. It is
presumed that this Turing machine writes a maximum number of
$1$'s before halting among all Turing machines with five states
and the tape alphabet $\{0, 1\}$. Proving this conjecture is an
open research problem.