Dear Users, hello Stoyo (are you out there?)

below you can find two pictures of a tree I wish to generate, the pictures are showing different situations in the same tree.

The nodes should be positioned in a column/row manner. Important for this layout is the position of the nodes: all items must be positioned in virtual layers (horizontal and vertical).
In my application I use own nodes derived from ContainerNode. Currently I use a combination of LayeredLayout and OrthogonalRouter with some options set.
The first layer is sorted manually using LayoutTraits[LayeredLayoutTraits.Layer] == 0
The problem with the combination is, that the resulting graph is not that strictly displayed like in the pictures below.

The idea behind is something like that in Picture 1:
* AV 100 is at position 0,0 (x,y)
* AV 200 is at position 1,0 (x,y)
* ...
* AV 330 is at position 2,3 (x,y)

With such a logic it is easy to generate something like Picture 2:
* AV 100 is at position 0,0 (x,y)
* (1,0 is empty: tree logic cares about positioning)
* AV 111 is at position 1,1 (x,y)
* (2,0 is empty: tree logic cares about positioning)
* (2,1 is empty: tree logic cares about positioning)
* (2,2 is empty: tree logic cares about positioning)
* AV 114 is at position 2,3 (x,y)

With the parameters Nodedistance and behaviour control settings for the arrows it woluld be very easy to get a graphically strict positioned diagram.
I think something like this is already done with lane diagrams.

Can you give be advices or hints or do you have ideas how to construct such a tree?

Hello,

yes I have constrains: the numbers of the nodes.

On their basis the tree must be build and aligned.

So the layout function can not choose the column.

AV 100 is first node in 1. column,
  AV 110 is second node in 1. column,
  AV 120 is third node in 1. column,
  ...
AV 200 is first node 2. column
   ...

Due to the "Missing" nodes in 2. picture (empty space above AV111) I think this is difficult for current layout mechanisms regarding strict layout as shown in the pictures. Maybe you have a solution?

Try this code:

LayeredLayout ll = new LayeredLayout();
ll.Orientation = MindFusion.Diagramming.Layout.Orientation.Vertical;
ll.EnforceLinkFlow = true;
ll.StraightenLongLinks = true;
ll.NodeDistance /= 2;
ll.Arrange(diagram);

foreach (ShapeNode node in diagram.Nodes)
{
	int column = int.Parse(node.Text);
	node.Move(20 + column * 50, node.Bounds.Y);
}

foreach (DiagramLink link in diagram.Links)
{
	link.ControlPoints[0] = GetBottomCenter(link.Origin);
	link.ControlPoints[link.ControlPoints.Count - 1] = GetTopCenter(link.Destination);
}

diagram.RouteAllLinks();
 

Replace the int.Parse line with one that gets the column value from where you have it stored.

I hope that helps,
Stoyan

Hi,

Here's our version:

private PointF GetBottomCenter(DiagramNode node)
{
	RectangleF r = node.Bounds;
	return new PointF(r.Left + r.Width / 2, r.Bottom);
}

private PointF GetTopCenter(DiagramNode node)
{
	RectangleF r = node.Bounds;
	return new PointF(r.Left + r.Width / 2, r.Top);
}
 

I hope that helps,
Stoyan

Hi Dirk,

There shouldn't be any problems if you save the column number in LayoutTraits, as long as you don't use one of the keys used by other layout classes that might try to typecast the value to a type different than int.

I hope that helps,
Stoyan