Sorry if duplicate (I'm sure duplicate) but I really can't find the way to do it in manual. I have created next block chain.

enter image description here

Here you can see curved line between last and third blocks. But I want something like this:enter image description here

And the question: how can I сonnect them in this way? I'm sure that it is pretty simple. MWE:



\node[draw, thick, rectangle] (0) {Data arrival};
\node[draw, thick, rectangle, below of=0] (1) {Coks backoff procedure};
\node[draw, thick, rectangle, below of= 1] (2) {Get trigger-frame?};
\node[draw, thick, rectangle, below of= 2] (3) {Performs backoff procedure};
\node[draw, thick, shape aspect=2.7, diamond, below =0.5cm of 3] (4) {$b<0$?};

 \path[>=latex, auto = right,every loop]

(0) edge[] node {} (1)
(1) edge node {} (2)
(2) edge node {} (3)
(3) edge node {} (4)
(4.east) edge[in=0, out=0, looseness=3] node[right] {N} (2.east)

The code below uses the following path constructions. There are many more, see the tikz manual or any of the many examples on the internet).

  • (a) -- (b) describes a straight path from a to b.

  • (a) -| (b) describes a path from a to b, first horizontally until it is below or above b and then vertically.

  • (a) |- (b) is the same but starting vertically and then continuing horizontally.

  • ++(1,-2) denotes a position relative to the preceding one, 1 unit to the right and 2 units down. So (a) -| ++(1,-2) is a path starting at a, going 1 unit to the right and 2 units down. As a side effect, the new starting position has moved to the end of the path. So

    \draw (a) -| ++(1,-2) -| ++(-1,2);

    draws a rectangle with the final position being again at a.

  • +(1,-2) is basically the same as ++(1,-2), except that the position does not move.

    \draw (a) -| +(1,-2) -| +(-1,2);

    draws two lines with an angle each, one to the right of a and one to the left of a.

enter image description here



   test/.style={draw, thick, shape aspect=2.7, diamond}
\node[action] (0) {Data arrival};
\node[action, below=of 0] (1) {Coks backoff procedure};
\node[action, below=of 1] (2) {Get trigger-frame?};
\node[action, below=of 2] (3) {Performs backoff procedure};
\node[test, below= 0.5cm of 3] (4) {$b<0$?};
\node[action, left=of 3] (5) {Do something};
  (0) edge node {} (1)
  (1) edge node {} (2)
  (2) edge node {} (3)
  (3) edge node {} (4);
\draw[->] (4) -- node[below right,pos=0.2]{N} ++(3,0) |- (2);     
\draw[->] (4) -| node[below left,pos=0.1]{Y} (5);
\draw[->] (5) |- (2); 
\draw[->] (4) --node[right] {maybe} +(0,-1.5);     
  • Thanks! appreciate that. Gods, it is called flowchart in English (I've searched examples on that site)... btw I was so close to a solution actually with my own trying but no matter. Thanks! now I understand that. – Lust_For_Love Apr 12 '17 at 17:57

Slightly modified gernot answer (for joy and exercise):

\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{arrows, chains, positioning, shapes}% added chains
\tikzset{supress chain/.code={\def\tikz@after@path{}}}% added for suppress joining of nodes

            > = latex,
node distance = 5mm and 7mm,% added (not used default value)
  start chain = going below,% activation of chains
action/.style = {draw, thick, on chain, join= by ->},% nodes are in chain and connected by ->
  test/.style = {diamond, draw, thick, shape aspect=2.4, on chain, join= by ->}% node is in the chain and connected by -> with previous node
\node[action]   (n0) {Data arrival};
\node[action]   (n1) {Coks backoff procedure};
\node[action]   (n2) {Get trigger-frame?};
\node[action]   (n3) {Performs backoff procedure};
\node[test]     (n4) {$b<0$?};
      supress chain, % this node is not connected with join
      left=of n3]   (n5) {Do something};
\draw[->] (n4) -| node[below,pos=0.25] {Y}  (n5); % left feedback loop
\draw[->] (n5) |- (n0); % left feedback loop
\draw[->] (n4) -|  ([xshift=5mm] n3.east) node[below,pos=0.25] {N} |- (n2); % right feedback loop
\draw[->] (n4.south) -- node[right] {maybe} ++ (0,-1.1);

The results is almost the same:

enter image description here

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