次の回路を描きましたが、ロジック ゲートの入力を反転するにはどうすればよいでしょうか? (スキャンした画像のように、どのゲートかは関係なく、一般的なものです)
\documentclass[a4paper,12pt]{article}
\usepackage{tikz} %vectorgraphics
\usetikzlibrary{shapes,arrows,shadows,
decorations.pathreplacing,backgrounds,calc}
%graph/flowchart
\usepackage[siunitx]{circuitikz}
\begin{document}
\begin{circuitikz} \draw
(0,0) node[nand port] (nand1) {}
(3,0) node[nand port] (nand2) {}
(5,0) node[nand port] (nand3) {}
(3,2) node[nand port] (nand4) {}
(5,2) node[nand port] (nand5) {}
(3,4) node[nand port] (nand6) {}
(5,4) node[nand port] (nand7) {}
(0,4) node[nand port] (nand8) {}
(nand1.out) |- (nand2.in 2)
(nand2.out) -| node[circ,midway]{} (nand3.in 1)
(nand2.out) -| (nand3.in 2)
(nand8.out) |- (nand6.in 1)
(nand6.out) -| node[circ,midway]{} (nand7.in 2)
(nand6.out) -| (nand7.in 1)
(nand4.out) -| node[circ,midway]{} (nand5.in 2)
(nand4.out) -| (nand5.in 1)
(-2,1) node(B1)[anchor=east] {B}
(-2,1) -| node[circ,midway]{} (nand1.in 1)
to[short,*-] (nand1.in 2)
(-2,1) -| (nand4.in 2)
(1,1) node[circ]{} |- (nand6.in 2)
(-2,3) node[anchor=east] {A}
(-2,3) -| node[circ,midway]{} (nand8.in 2)
to[short,*-] (nand8.in 1)
(-2,3) -| (nand4.in 1)
(0.5,3) node[circ]{} |- (nand2.in 1)
(6,1.97) node[anchor=east] {O2}
(6,0.0) node[anchor=east] {O3}
(6,4) node[anchor=east] {O1}
;\end{circuitikz}
\end{document}
答え1
新鮮な新しいcircuitikzgit 数分前にアップロードされました、次の操作を実行できるようになりました。
\documentclass[border=10pt]{standalone}
\usepackage[siunitx, RPvoltages]{circuitikzgit}
\begin{document}
\begin{circuitikz}
\draw (0,3) node[american and port] (A) {P1};
\node at (A.bin 1) [ocirc, left]{} ;
\begin{scope}
\ctikzset{tripoles/american or port/height=1.6}
\draw (A.out) -- ++(0.5,0) node[american or port,
number inputs=5, anchor=in 1] (B) {P2};
\node at (B.bin 3) [ocirc, left]{} ;
\end{scope}
\draw (0,1.5) node[american or port] (C) {P3};
\node at (C.bin 2) [ocirc, left]{} ;
\draw (C.out) |- (B.in 2);
\end{circuitikz}
\end{document}
