On Jul 9, 2013, at 4:14 AM, Bojan Janisch wrote:

> what do I have to do, to build a complete GPTree and use it for evaluation?
>
> Background is that I have now successfully created my first rules from a GPTree.
> But I want to know if the results would be better if I use a rule that I manually
> wrote, as a GPIndividuum in my population.


If you just want to build a GPTree manually, you just need to allocate various GPNodes and hang them together. I'll presume you don't care about making a new GPIndividual, just constructing a GPTree and evaluating it.

To allocate GPNodes you might need your function set. Let's say that your GPTree was intended to fit into tree slot s of your GPIndividual (in most cases you just have one tree per individual, so s=0). You might do:

final EvolutionState state = ...
final Subpopulation subpop = ...
final GPInitializer init = (GPInitializer)(state.initializer);
final GPSpecies gps = (GPSpecies)(subpop.species);
final GPIndividual indproto = ((GPIndividual)(gps.i_prototype));
final GPTree gptproto = indproto.trees[0];
final GPTreeConstraints gptc = gptproto.constraints(init);
final GPFunctionSet gpfs = gptc.functionset;

Okay, now you have your function set. See GPFunctionSet to extract specific GPNodes. For example, you can make a node whose name is "+" like this:

GPNode plus = (GPNode)(gpfs.nodesByName.get("+").clone()); // ALWAYS clone

A GPNode has three critical variables that you will need to set.

    public GPNodeParent parent;
    public GPNode children[];
    public byte argposition;

Parent points to the parent GPNode of the node, or to a GPTree if the node is the root. You probably don't need to attach to a tree if you're just evaluating the tree, just let that value be null.

children[] is an array of the children nodes to the node. children[] can be null, or can be zero length, if the node is a leaf.

argposition is the index in the parent's children array in which this node is found. It says "I'm child number 4 of my parent", for example.

Let's say we're trying to create the tree: (+ (cos x) x)

I'd do this:

GPNode plus = (GPNode)(gpfs.nodesByName.get("+").clone());
GPNode cos = (GPNode)(gpfs.nodesByName.get("cos").clone());
GPNode x_one = (GPNode)(gpfs.nodesByName.get("+").clone());
GPNode x_two = (GPNode)(gpfs.nodesByName.get("+").clone());

plus.parent = null; // no GPTree :-(
plus.argposition = 0;
plus.children[0] = cos;
plus.children[1] = x_one;

cos.parent = plus;
cos.argposition = 0;
cos.children[0] = x_two;

x_one.parent = plus;
x_one.argposition = 1;
x_one.children = null; // it's probably already null anyway

x_two.parent = cos;
x_two.argposition = 0;
x_two.children = null; // it's probably null already

Now you're ready to evaluate the node.

GPProblem prob = ...
GPData gpd = ...
int threadnum = ...
GPIndividual ind = null; // our GPNodes don't need the individual to evaluate
plus.eval(state, threadnum, gpd, new ADFStack(), ind, prob);

If you need the GPTree or the GPIndividual constructed as well, then there's a bit more to that.

Sean