在 Tikz 中定義變數向量

Question

我在投影機中有以下動畫：

使用以下程式碼（可能不是最佳的）：

\documentclass{beamer}
\usepackage{tikz}
\usepackage{multimedia}
\begin{document}
\begin{frame}[label=persistence]
\animate<1-10>
\begin{columns}
    \begin{column}{5cm}
    \foreach \n in {1,...,10} {
    \begin{tikzpicture}[radius=2pt]
    \only<\n>{
        \node  at (-1,6){};
        \node  at (6.5,0){};
        \begin{scope}[fill opacity=0.2]
        \filldraw[fill=yellow,draw=black] (1,1) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (2.3,1.1) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (4.5,0.8) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (5.1,1.8) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (0.4,3.3) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (2.1,2.8) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (3.8,3.5) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (4.8,4.2) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (0.8,4.9) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (2.1,4.1) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (3.8,2.0) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (3.5,0.6) circle (2+3*\n pt);
        \filldraw[fill=yellow,draw=black] (3.0,5.0) circle (2+3*\n pt);   
        \filldraw[fill=yellow,draw=black] (4.1,5.1) circle (2+3*\n pt);          
        \filldraw[fill=yellow,draw=black] (0.9,2.1) circle (2+3*\n pt);
        \end{scope}
        \filldraw[red] (1,1) circle ;
        \filldraw[red] (2.3,1.1) circle ;
        \filldraw[red] (4.5,0.8) circle ;
        \filldraw[red] (5.1,1.8) circle ;
        \filldraw[red] (0.4,3.3) circle ;
        \filldraw[red] (2.1,2.8) circle ;
        \filldraw[red] (3.8,3.5) circle ;
        \filldraw[red] (4.8,4.2) circle ;
        \filldraw[red] (0.8,4.9) circle ;
        \filldraw[red] (2.1,4.1) circle ;
        \filldraw[red] (3.8,2.0) circle ;
        \filldraw[red] (3.5,0.6) circle ;
        \filldraw[red] (3.0,5.0) circle ;   
        \filldraw[red] (4.1,5.1) circle ;          
        \filldraw[red] (0.9,2.1) circle ;
        }   
    \end{tikzpicture}
    }
    \end{column}
    \begin{column}{.4\textwidth}
    \begin{tikzpicture}     
    \filldraw[red] (1,1) circle ;
        \filldraw[red] (2.3,1.1) circle ;
        \filldraw[red] (4.5,0.8) circle ;
        \filldraw[red] (5.1,1.8) circle ;
        \filldraw[red] (0.4,3.3) circle ;
        \filldraw[red] (2.1,2.8) circle ;
        \filldraw[red] (3.8,3.5) circle ;
        \filldraw[red] (4.8,4.2) circle ;
        \filldraw[red] (0.8,4.9) circle ;
        \filldraw[red] (2.1,4.1) circle ;
        \filldraw[red] (3.8,2.0) circle ;
        \filldraw[red] (3.5,0.6) circle ;
        \filldraw[red] (3.0,5.0) circle ;   
        \filldraw[red] (4.1,5.1) circle ;          
        \filldraw[red] (0.9,2.1) circle ;
    \end{tikzpicture}
    \end{column}

\end{columns}
\end{frame}
\end{document}

現在，使用相同的點集（紅色圓圈），我需要在幻燈片的右側創建一個動畫圖形，如果鄰域之間的交集不存在，則點 $P$ 和 $Q$ 之間的邊緣將出現空的。 Tikz 可以嗎？我想到的解決方案是為點定義一個向量，然後使用兩個巢狀向量。但我不知道該怎麼做。

Answer 1

對於需求的第一部分，TikZ\foreach指令可以解析可以儲存在巨集中的座標列表。下面說明如何做到這一點。根據所需的用例調整程式碼應該很簡單：

\documentclass[tikz,border=5]{standalone}
\begin{document}
\def\pointlist{
  (1.0,1.0), (2.3,1.1), (4.5,0.8), 
  (5.1,1.8), (0.4,3.3), (2.1,2.8),
  (3.8,3.5), (4.8,4.2), (0.8,4.9), 
  (2.1,4.1), (3.8,2.0), (3.5,0.6),
  (3.0,5.0), (4.1,5.1), (0.9,2.1) 
}
\begin{tikzpicture}[radius=2pt]
  \begin{scope}[fill opacity=0.2]
    \foreach \point in \pointlist
      \filldraw[fill=yellow,draw=black] \point circle [radius=5pt];
   \end{scope}
   \foreach \point in \pointlist
     \filldraw[red] \point circle;
\end{tikzpicture}
\end{document}

假設我理解第二部分：

\documentclass[tikz,border=5]{standalone}
\begin{document}
\def\pointlist{
  (1.0,1.0), (2.3,1.1), (4.5,0.8), 
  (5.1,1.8), (0.4,3.3), (2.1,2.8),
  (3.8,3.5), (4.8,4.2), (0.8,4.9), 
  (2.1,4.1), (3.8,2.0), (3.5,0.6),
  (3.0,5.0), (4.1,5.1), (0.9,2.1) 
}
\foreach \N in {1,...,10}{
\begin{tikzpicture}[radius=2pt]
  \useasboundingbox (-1,-.5) rectangle (6.25,6.25);
  \begin{scope}[fill opacity=0.2]
    \foreach \point in \pointlist
      \filldraw[fill=yellow,draw=black] \point circle [radius=2pt+3*\N];
   \end{scope}
   \foreach \point in \pointlist
     \filldraw[red] \point circle;
   \foreach \P [count=\i] in \pointlist
     \foreach \Q [count=\j]in \pointlist {
       \ifnum\j>\i
       \else
         \path \P coordinate (P) \Q coordinate (Q);
         \pgfpointdiff{\pgfpointanchor{P}{center}}{\pgfpointanchor{Q}{center}}
         \pgfgetlastxy\x\y
         \pgfmathparse{int(veclen(\x,\y)/2 < 2+3*\N)}
         \ifnum\pgfmathresult=1
           \draw [thick] (P) -- (Q);
         \fi
       \fi
     }
\end{tikzpicture}
}
\end{document}

根據 amorvincomni 的回答，這是使用該程式庫執行操作的另一種方法math：

\documentclass[tikz,border=5]{standalone}
\usetikzlibrary{math}
\begin{document}
\def\pointlist{
  (1.0,1.0), (2.3,1.1), (4.5,0.8), 
  (5.1,1.8), (0.4,3.3), (2.1,2.8),
  (3.8,3.5), (4.8,4.2), (0.8,4.9), 
  (2.1,4.1), (3.8,2.0), (3.5,0.6),
  (3.0,5.0), (4.1,5.1), (0.9,2.1) 
}
\foreach \N in {1,...,10}{
\begin{tikzpicture}[radius=2pt]
  \useasboundingbox (-1,-.5) rectangle (14.25,6.25);
  \begin{scope}[fill opacity=0.2]
    \foreach \point in \pointlist
      \filldraw[fill=yellow,draw=black] \point circle [radius=2pt+3*\N];
   \end{scope}
   \foreach \point in \pointlist
     \filldraw[red] \point circle;
   \begin{scope}[xshift=8cm]
     \foreach \point in \pointlist
       \filldraw[red] \point circle;
     \foreach \P [count=\i] in \pointlist
       \foreach \Q [count=\j]in \pointlist {
         \ifnum\j>\i
           \tikzmath{%
             coordinate \p, \q, \r;
             \p = \P; \q = \Q; 
             \pq = veclen(\px-\qx, \py-\qy)/2;
             \d = 2pt+3*\N;
             if (\pq < \d) then {
               { 
                 \draw \P -- \Q;
                 \foreach \R [count=\k] in \pointlist {
                   \ifnum\k>\j
                     \tikzmath{%
                       \r = \R;
                       \pr = veclen(\px-\rx, \py-\ry)/2;
                       \qr = veclen(\qx-\rx, \qy-\ry)/2;             
                       if (\pr < \d) && (\qr < \d) then {
                         { 
                           \fill [fill=yellow, fill opacity=.2] \P -- \Q -- \R;                        
                         };
                       };
                     }
                   \fi
                 }
               };
             };
           }
        \fi
      }
  \end{scope}
\end{tikzpicture}
}
\end{document}

Answer 2

只是為了好玩：根據 Mark Wibrow 的好答案，我編寫了以下程式碼。它不是最優的，但當與三角形頂點對應的三個鄰域的交集不為空時，它會創建填充三角形：

\begin{frame}[label=persistence]
\animate<1-27>
\scalebox{.8}{%
\begin{columns}
    \begin{column}{5cm}
    \begin{tikzpicture}[radius=2pt]%        
    \foreach \n in {1,...,27}% 
    {%      
    \only<\n>{%
        \useasboundingbox (-1,-1.5) rectangle (6.25,7.25);      
        \begin{scope}[fill opacity=0.2]
            \foreach \point in \pointlist {
        \filldraw[fill=yellow,draw=black] \point circle (5+\n pt);
        }
        \end{scope}
        \foreach \point in \pointlist {
        \filldraw[red] \point circle;}
        }
    }
    \end{tikzpicture}

    \end{column}
    \begin{column}{5cm}
        \begin{tikzpicture}[radius=2pt]
            \useasboundingbox (-1,-1.5) rectangle (6.25,7.25);              
            \foreach \n in {1,...,27}{%
                \only<\n>{%
                    \foreach \point in \pointlist
                        \filldraw[red] \point circle;
                    \foreach \P [count=\i] in \pointlist
                        \foreach \Q [count=\j]in \pointlist {
                        \ifnum\j>\i
                            \path \P coordinate (P) \Q coordinate (Q);
                            \pgfpointdiff{\pgfpointanchor{P}{center}}{\pgfpointanchor{Q}{center}}
                            \pgfgetlastxy\x\y
                            \pgfmathparse{int(veclen(\x,\y)/2 < 5+\n)}
                            \ifnum\pgfmathresult=1
                                \draw [thick] (P) -- (Q);
                                \foreach \T [count=\k] in \pointlist {  
                                \ifnum\k>\j
                                \path \P coordinate (P) \T coordinate (T);
                                \pgfpointdiff{\pgfpointanchor{P}{center}}{\pgfpointanchor{T}{center}}
                                \pgfgetlastxy\x\y
                                \pgfmathparse{int(veclen(\x,\y)/2 < 5+\n)}
                                \ifnum\pgfmathresult=1
                                \coordinate  (A) at ($(P)!0.5!(Q)$);
                                \coordinate  (C) at ($(P)!0.5!(T)$);
                                \coordinate  (A') at ($(A)!2cm!90:(P)$);
                                \coordinate  (C') at ($(C)!2cm!90:(P)$);
                                \coordinate  (O) at (intersection of A--A' and C--C');
                                \pgfpointdiff{\pgfpointanchor{O}{center}}{\pgfpointanchor{T}{center}}
                                \pgfgetlastxy\x\y
                                \pgfmathparse{int(veclen(\x,\y) < 5+\n)}
                                \ifnum\pgfmathresult=1
                                \pgfpointdiff{\pgfpointanchor{O}{center}}{\pgfpointanchor{Q}{center}}
                                \pgfgetlastxy\x\y
                                \pgfmathparse{int(veclen(\x,\y) < 5+\n)}
                                \ifnum\pgfmathresult=1
                                \pgfpointdiff{\pgfpointanchor{O}{center}}{\pgfpointanchor{P}{center}}
                                \pgfgetlastxy\x\y
                                \pgfmathparse{int(veclen(\x,\y) < 5+\n)}
                                \ifnum\pgfmathresult=1
                                \begin{scope}[fill opacity=0.3]
                                    \draw[fill=yellow] (P) --  (Q) --  (T) -- cycle;
                                    \end{scope}
                                \fi
                                \fi
                                \fi
                                \fi
                                \fi
                                }%
                            \fi
                        \fi
                        }%
                }
            }
        \end{tikzpicture}

    \end{column}
\end{columns}
}
\end{frame}

\end{document}

這是結果：