如何根据两个节点的位置确定第三个节点的位置？

Question 1

您需要选项on grid=true，它将使用节点中心（而不是其边界）进行定位。但随后您将需要增加node distance。

对于虚线框，我建议使用fit库。

\documentclass{article} 
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{shapes,snakes}
\usepackage{siunitx} %SI单位
%\usepackage{xparse}
\usetikzlibrary{positioning, fit}


\begin{document}
    \begin{tikzpicture}[on grid=true, node distance=3cm]
    
    \node[rectangle,draw, very thick, minimum size=10mm] (A) {
    A};
    \node[rectangle,draw, very thick, minimum size=10mm] (B) [below=of A]{B};
    \node[rectangle,draw=red, very thick, minimum size=10mm] (C) [right=of B]{C};
    \node[rectangle,draw, very thick, minimum size=10mm] (D) [right=of C]{D};
    \node[rectangle,draw, very thick, minimum size=10mm] (E) [above=of D] {E};
    \node[diamond,draw, very thick, minimum size=10mm] (F) [above=of C] {FFFFFfF};
    
        %Lines
    \draw[->] (A) -> (B);
    \draw[->] (B) -> (C);
    \draw[->] (C) -> (D);
    \draw[->] (D) -> (E);
    \draw[->] (E) -> (F);
    \draw[->] (F) -> (C);
    \node[draw, dashed, fit = {(B)(D)}, text height=2cm]{for $i=1,2,\dots,N$};
    \end{tikzpicture}
\end{document}

Answer

您需要选项on grid=true，它将使用节点中心（而不是其边界）进行定位。但随后您将需要增加node distance。

对于虚线框，我建议使用fit库。

\documentclass{article} 
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{shapes,snakes}
\usepackage{siunitx} %SI单位
%\usepackage{xparse}
\usetikzlibrary{positioning, fit}


\begin{document}
    \begin{tikzpicture}[on grid=true, node distance=3cm]
    
    \node[rectangle,draw, very thick, minimum size=10mm] (A) {
    A};
    \node[rectangle,draw, very thick, minimum size=10mm] (B) [below=of A]{B};
    \node[rectangle,draw=red, very thick, minimum size=10mm] (C) [right=of B]{C};
    \node[rectangle,draw, very thick, minimum size=10mm] (D) [right=of C]{D};
    \node[rectangle,draw, very thick, minimum size=10mm] (E) [above=of D] {E};
    \node[diamond,draw, very thick, minimum size=10mm] (F) [above=of C] {FFFFFfF};
    
        %Lines
    \draw[->] (A) -> (B);
    \draw[->] (B) -> (C);
    \draw[->] (C) -> (D);
    \draw[->] (D) -> (E);
    \draw[->] (E) -> (F);
    \draw[->] (F) -> (C);
    \node[draw, dashed, fit = {(B)(D)}, text height=2cm]{for $i=1,2,\dots,N$};
    \end{tikzpicture}
\end{document}

Question 2

您可以先在选项中定义节点和边的样式tikzpicture，然后编写图像的短代码。
虚线矩形绘制为节点，围绕节点 B ... D：

\documentclass[border=3.141592]{standalone}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{arrows.meta,  % new
                fit,          % new
                positioning,
                shapes}


\begin{document}
    \begin{tikzpicture}[% nodes, edges styles
node distance = 10mm and 10mm,
D/.style = {diamond, aspect=1.3, 
            inner xsep=0mm, draw, thick}, 
F/.style = {draw, dashed, inner ysep=3ex, yshift=-1ex,
            fit=#1},
N/.style = {draw=#1, thick, minimum size=10mm},
N/.default = black,
every edge/.append style = {draw, -Straight Barb}
                      ]
\node (A) [N]               {A};
\node (B) [N, below=of A]   {B};
\node (C) [N=red, 
              right=of B]   {C};
\node (D) [N, right=of C]   {D};
\node (E) [N, above=of D]   {E};
\path (E) -- node (F) [D] {FFFFFfF} (A);    % <---
% dashed rectangle
\node[F=(B) (D), 
      label={[anchor=south]below:{for $i=1,2,\dotsc,N$}}] {};
% arrows
\draw   (A) edge (B) 
        (B) edge (C)
        (C) edge (D) 
        (D) edge (E) 
        (E) edge (F) 
        (F) edge (C);
    \end{tikzpicture}
\end{document}

Answer

您可以先在选项中定义节点和边的样式tikzpicture，然后编写图像的短代码。
虚线矩形绘制为节点，围绕节点 B ... D：

\documentclass[border=3.141592]{standalone}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{arrows.meta,  % new
                fit,          % new
                positioning,
                shapes}


\begin{document}
    \begin{tikzpicture}[% nodes, edges styles
node distance = 10mm and 10mm,
D/.style = {diamond, aspect=1.3, 
            inner xsep=0mm, draw, thick}, 
F/.style = {draw, dashed, inner ysep=3ex, yshift=-1ex,
            fit=#1},
N/.style = {draw=#1, thick, minimum size=10mm},
N/.default = black,
every edge/.append style = {draw, -Straight Barb}
                      ]
\node (A) [N]               {A};
\node (B) [N, below=of A]   {B};
\node (C) [N=red, 
              right=of B]   {C};
\node (D) [N, right=of C]   {D};
\node (E) [N, above=of D]   {E};
\path (E) -- node (F) [D] {FFFFFfF} (A);    % <---
% dashed rectangle
\node[F=(B) (D), 
      label={[anchor=south]below:{for $i=1,2,\dotsc,N$}}] {};
% arrows
\draw   (A) edge (B) 
        (B) edge (C)
        (C) edge (D) 
        (D) edge (E) 
        (E) edge (F) 
        (F) edge (C);
    \end{tikzpicture}
\end{document}

Question 3

您可以将节点 F 放置在十字路口一条垂直线穿过C，一条水平线穿过A，或者E：

\node[diamond] (F) at (C|-E) {FFFFFfF};

使用calc库，您可以将其放置在（中心）A和中间E：

\node[diamnond] (F) at ($(A)!.5!(E)$) {FFFFFfF};

A或者在和的边界中间，E沿路径有一个节点：

\path (A) -- node[diamond] (F) {FFFFFfF} (E);

但是，在后者中您也可以使用A.center和，E.center而在前者中您可以使用($(A.east)!.5!(E.west)$)，但由于您的节点A和E具有相同的宽度，因此将节点放置在它们的中心中间和边界中间没有区别。

当然，在所有这些情况下，node distance将不再被观察到，但对于F节点来说已经太晚了。

您可以使用on grid选项库positioning，以便它测量节点中心之间的节点距离，这需要您仔细设置适当的节点距离。（→Sandy G 的回答）

您可以将节点放置在中，matrix of nodes这将使它们位于网格中，但将确保它们之间的最小节点距离由和表示row sep。column sep通过使用关键字between origins在row sep和/或column sep类似的定位算法作为on grid选项可以用于矩阵。

\matrix[
  matrix of nodes,
  row sep=1cm,
  column sep=1cm,
  cells={nodes={draw, very thick, minimum size=10mm}}]{
|(A)| A & |[diamond]  (F)| FFFFFfF & |(E)| E \\
|(B)| B & |[draw=red] (C)| C       & |(D)| D \\};

代码

\documentclass[tikz,border=2pt]{standalone}
\usetikzlibrary{fit, graphs, matrix, positioning, shapes.geometric}
\begin{document}
\begin{tikzpicture}
\begin{scope}[nodes={draw, very thick, minimum size=10mm}]
\node           (A)              {A};
\node           (B) [below=of A] {B};
\node[draw=red] (C) [right=of B] {C};
\node           (D) [right=of C] {D};
\node           (E) [above=of D] {E};
\node[diamond]  (F) at (C|-E)    {FFFFFfF};
% with calc
% \node[diamnond]  (F) at ($(A)!.5!(E)$) {FFFFFfF};
% or without
% \path (A) -- node[diamond] (F) {FFFFFfF} (E);
\end{scope}
\graph[path, ->, use existing nodes] {A, B, C, D, E, F -> C};

\node[
  fit={(B)(D)([yshift=+-1em]C.south)}, dashed, inner sep=+.5em, thick, draw,
  label={[above, inner sep=+.2em]south:for $i = 1, 2, \dots, N$}]{};
\end{tikzpicture}

\begin{tikzpicture}
\matrix[
  matrix of nodes,
  row sep=1cm,
  column sep=1cm,
  cells={nodes={draw, very thick, minimum size=10mm}}]{
|(A)| A & |[diamond]  (F)| FFFFFfF & |(E)| E \\
|(B)| B & |[draw=red] (C)| C       & |(D)| D \\};
\graph[path, ->, use existing nodes] {A, B, C, D, E, F -> C};

\node[
  fit={(B)(D)([yshift=+-1em]C.south)}, dashed, inner sep=+.5em, thick, draw,
  label={[above, inner sep=+.2em]south:for $i = 1, 2, \dots, N$}]{};
\end{tikzpicture}
\end{document}

输出

Answer