-1

Could anyone please help me draw this with TikZ ? please notice that there is curved lines I tried to made by hand in office enter image description here

  • 6
    This is not super hard, but this is a question and answers site, not a "do this for me" service. What do you have so far? – daleif Oct 4 '18 at 5:44
6

Could you please provide some MWE when you ask the next question? You will see that it is much more fun to try out the things yourself (rather than depend on others).

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{positioning,calc,arrows.meta}
\begin{document}
\begin{tikzpicture}[bullet/.style={draw,circle,minimum width=3mm,inner sep=0pt,
fill=cyan!50}]
 \node[bullet,label={[yshift=-2mm]below:$R_1$}] (R1){};
 \node[right=2cm of R1,bullet,label={[yshift=-2mm]below:$R_2$}] (R2){};
 \node[right=2cm of R2,yshift=6mm,bullet,label={[yshift=2mm]above:$R_3$}] (R3){};
 \node[right=2cm of R3,yshift=-6mm,bullet,label={[yshift=-2mm]below:$R_4$}] (R4){};
 \node[right=2cm of R4,bullet,label={[yshift=-2mm]below:$R_5$}] (R5){};
 \foreach \X [evaluate=\X as \Y using {int(\X+1)}] in {1,...,4}
 {\draw[thick,-latex,cyan!50] (R\X) -- (R\Y);}
 %
 \draw[thick,red,-latex] let \p1=($(R3)-(R2)$), \p2=($(R4)-(R3)$),
 \n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)} in 
 ([yshift=-1mm]R2.south) to[out=\n1,in=180,looseness=0.5] ([yshift=-1mm]R3.south) 
 to[out=0,in=\n2-180,looseness=0.5] ([yshift=-1mm]R4.south);
 %
 \draw[thick,red,{Circle}-latex,dashed] let \p1=($(R2)-(R1)$), \p2=($(R3)-(R2)$),
 \n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)} in 
 ([yshift=1mm]R1.north) to[out=\n1,in=180,looseness=0.5] ([yshift=1mm]R2.north) 
 to[out=0,in=\n2-180,looseness=0.5] ([yshift=1mm,xshift=-2mm]R3.north);
 %
 \draw[thick,{Circle}-latex,dashed] let \p1=($(R4)-(R3)$), \p2=($(R5)-(R4)$),
 \n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)} in 
 ([yshift=1mm,xshift=2mm]R3.north) to[out=\n1,in=180,looseness=0.5] 
 ([yshift=1mm]R4.north) 
 to[out=0,in=\n2-180,looseness=0.5] ([yshift=1mm]R5.north);
\end{tikzpicture}
\end{document}

enter image description here

Not the answer you're looking for? Browse other questions tagged or ask your own question.