Graphviz .dot节点排序

Question

我正在构建一个epsilon NFA,以使用规范构造识别正则表达式.我正在使用子图来对正则表达式的各个部分进行分组.由于dot决定移动节点的顺序,*运算符给了我特别的麻烦.我已经尝试添加边缘权重以强制特定边缘变短以保持边缘的顺序,但这似乎不起作用.

我想要做的是强制子图中的节点按特定顺序放置,以便输出图可以识别为特定类型的(众所周知的)构造.在下面的示例中,我希望边缘3,4,5和6按此顺序放置,但是点将它们按6,3,4,5的顺序放置.任何指针都会被欣赏.

请注意,当前权重参数与完全没有权重参数没有任何区别.

我有以下内容

digraph G {
    rankdir = LR;
    node [shape = none];
            0 [label = "start"];
    node [shape = circle];
            1 [label = "q1"];
            2 [label = "q2"];
            3 [label = "q3"];
            4 [label = "q4"];
            5 [label = "q5"];
    node [shape = doublecircle];
            6 [label = "q6"];
    subgraph re1 {
            rank = same;
            edge[label = "0"];
            1 -> 2;
    };
    subgraph re2 {
            rank = same;
            edge[label = "ε"];
                    3 -> 4 [weight = 10];
            edge[label = "1"];
                    4 -> 5 [weight = 10];
            edge[label = "ε"];
                    5 -> 6 [weight = 10];
                    5 -> 4 [weight = 1];
                    6 -> 3 [weight = 1];
    };
    edge[color=black];
            0 -> 1
    edge[label = "ε"];
            2 -> 3;
}

Answer 1

这是我写这个图的方式:

• 首先,对我来说,这是一个从上到下,而不是从左到右的图形,因此我删除了仅为节点0/1和节点2/3 rankdir=LR添加的图形rank=same.
• 我删除了所有的重量
• 最重要的是,我添加constraint=false到了与图形方向相反的边缘 - 从节点4到节点5的边缘,以及从节点6到节点3的边缘.

这里的来源:

digraph G {
    0 [label = "start", shape = none]; 
    node [shape = circle];
    1 [label = "q1"];
    2 [label = "q2"];
    3 [label = "q3"];
    4 [label = "q4"];
    5 [label = "q5"];
    6 [label = "q6", shape = doublecircle];

    {rank = same; 0 -> 1; }
    1 -> 2 [label = "0"];
    {rank = same; 2 -> 3 [label = "ε"]; }
    4 -> 5 [label = "1"];
    edge [label = "ε"];
    3 -> 4;
    5 -> 6;
    5 -> 4 [constraint = false];
    6 -> 3 [constraint = false];
}

这是结果:

现在,如果你愿意,你可以保留rankdir=LR,只需取你发布的标记,删除权重并添加constraint=false到我所做的相同边缘,它也可以.