GraphViz - How to connect subgraphs?

Viewed 119344

In the DOT language for GraphViz, I'm trying to represent a dependency diagram. I need to be able to have nodes inside a container and to be able to make nodes and/or containers dependent on other nodes and/or containers.

I'm using subgraph to represent my containers. Node linking works just fine, but I can't figure out how to connect subgraphs.

Given the program below, I need to be able to connect cluster_1 and cluster_2 with an arrow, but anything I've tried creates new nodes instead of connecting the clusters:

digraph G {

    graph [fontsize=10 fontname="Verdana"];
    node [shape=record fontsize=10 fontname="Verdana"];

    subgraph cluster_0 {
        node [style=filled];
        "Item 1" "Item 2";
        label = "Container A";
        color=blue;
    }

    subgraph cluster_1 {
        node [style=filled];
        "Item 3" "Item 4";
        label = "Container B";
        color=blue;
    }

    subgraph cluster_2 {
        node [style=filled];
        "Item 5" "Item 6";
        label = "Container C";
        color=blue;
    }

    // Renders fine
    "Item 1" -> "Item 2";
    "Item 2" -> "Item 3";

    // Both of these create new nodes
    cluster_1 -> cluster_2;
    "Container A" -> "Container C";
}

enter image description here

4 Answers

The DOT user manual gives the following example of a graph with clusters with edges between clusters:

IMPORTANT: The initial compound=true statement is required.

digraph G {
  compound=true;
  subgraph cluster0 {
    a -> b;
    a -> c;
    b -> d;
    c -> d;
  }
  subgraph cluster1 {
    e -> g;
    e -> f;
  }
  b -> f [lhead=cluster1];
  d -> e;
  c -> g [ltail=cluster0,lhead=cluster1];
  c -> e [ltail=cluster0];
  d -> h;
}

... and edges between nodes and clusters:

enter image description here

Make sure that compound=true in the digraph options (reference):

digraph {
  compound=true;

  subgraph cluster_a {
    label="Cluster A";
    node1; node3; node5; node7;
  }
  subgraph cluster_b {
    label="Cluster B";
    node2; node4; node6; node8;
  }

  node1 -> node2 [label="1"];
  node3 -> node4 [label="2" ltail="cluster_a"];
  
  node5 -> node6 [label="3" lhead="cluster_b"];
  node7 -> node8 [label="4" ltail="cluster_a" lhead="cluster_b"];
}

enter image description here

Related