Set node direction on graphviz - graphviz

Question

Suppose this code using neato:
graph sample {
layout=neato
overlap=false
splines=true
tailclip=false
headclip=false
A -- I
A -- J
A -- B
A -- H
A -- E
A -- K
B -- D
B -- C
B -- L
C -- M
C -- N
C -- O
D -- P
D -- Q
E -- R
F -- A
G -- F
H -- J
}
This gives us this diagram:
What I need is to place a node X, always fixed in a position south from his parent node. i.e. If I put another relation A -- X, X should be always placed south from A. And I don't really care where everything else ends up.
I've looked into the pos attribute, but it doesn't seems to be the solution since X is not really in a fixed position, but on a position relative to his relation.
Also tailport and headport, but they only define from where the edge will come out/in, but don't really affect the direction of the node.
Update
An additional image to make things clearer:
I don't require neato, but I don't want the graph to look like a UD or LR dot tree, I don't want it to be linearly ordered. circo, fdp, sfdp, twopi are alright too.

Answer 1

The neato program supports multiple modes, one of which can probably give you what you want. In particular, if you set mode=ipsep, you can specify dot-like constraints that are honored during the layout. For example, I take your graph and use the graph attributes
mode=ipsep
diredgeconstraints=true
levelsgap=0.5
The first turns on ipsep mode, the second tells the model to support directed edges as in dot,
and the last specifies how strong the separation should be. I then set the edge dir attribute to none
edge[dir=none]
and add an edge A -- X [dir=1]
The dir=1 indicates this edge should induce a directional constraint. If I then run neato, I get the appended picture.
The Graphviz attribute documentation http://www.graphviz.org/content/attrs provides more information about these attributes.

Answer 2

In response to the updated constraints, one solution is to pin A and X, and then lay out the graph around them:
graph sample {
overlap=false;
splines=true;
tailclip=false;
headclip=false;
A [pin=true,pos="0,.2"]
X [pin=true,pos="0,.1"]
A -- I
A -- J
A -- B
A -- H
A -- E
A -- K
B -- D
B -- C
B -- L
C -- M
C -- N
C -- O
D -- P
D -- Q
E -- R
F -- A
G -- F
H -- J
A -- X
I tried layout with both neato and fdp, and it seems to produce a graph like what you want. Naturally, if you want to impose such a constraint on arbitrary pairs of nodes in the same graph, this solution may not work.
--- Earlier answer ---
If you're committed to using neato, I'm not certain there is a way to solve the problem without modifying the graph in a post-processing step. If neato is just a convenient default, then you should be able to solve your problem by using dot as your layout engine instead, and using "rankdir=UD", plus a couple of additional kludges if X needs to be due south.
In the event that you only need the constraint to apply for a single node X, then putting X and A together in a cluster should do the job:
graph sample {
rankdir=UD
layout=dot
overlap=false
// .. as before
A -- X
subgraph clusterone {
style=invisible
A
X
}
}
If you need a strictly-south constraint to apply to arbitrary children of A, then that kind of clustering followed by the trick described in:
How to force all nodes in the same column in graphviz?
might do the trick. The attribute clusterrank=local might also be useful in that case, but I'm not certain. Hope this helps.