Question

I am currently doing some calculations with trees. Each node has 5 values I am trying to calculate and a type deciding how these values are calculated. Some calculations can be pretty complicated algorithms. All calculations within a node depend solely on the values of its child nodes, so I am doing calculations from down to top. For each node type, a value depends on different values of the childnodes. I am interested mainly in the 5 values in the root node, which depend on all values in all other nodes ofc. All this is working just fine. A node can only have 1 or 2 childnodes, and the tree usually is no deeper than 5 levels.

For some node-types, there is a tolerance; meaning some values there would not matter, see this picture, I marked those with XX. Sometimes even, some values would be in relation, like C = XX * A. Currently, these values are just set to some default values. Sometimes there would be a complicated relationship even, like multiple possible solutions of an algorithm like Newton’s Method, depending on starting values.

Now there is a rating I can apply on the values of the root node. What I would like is to optimize this rating by adjusting the XX-values deep within the tree. The calculations within each node can be a range of many possible formulas and the tolerance can be one of many possible patterns, so I cannot just figure out some formula but I would need some algorithm which is very flexible. I do not know of such an algorithm. Does anyone have an idea?

/Edit: To clarify, it is unclear how many values in the tree will be free. There is not just one XX, but there may be any number of them (I guess max. 10), so my first step would be identifying these values. Also, I will be doing this on many generated trees within a time window, so speed is not unimportant as well. Thanks:)

Answer 1

If you have 3 input values XX, YY, and ZZ, you are searching a 3 dimension space. What you are looking to do is to apply an optimisation algorithm, or Heuristic algorithm. Your choice of algorithm is key, a cost benefit between your time and the computer’s time. I am guessing that you just want to do this once.

What ever method you use, you need to understand the problem, which means to understand how your algorithm changes with different input values. Some solutions have a very nice minimum that is easy to find (e.g. using Newton’s Method), some don’t.

I suggest starting simple. One of the most basic is just to do an iterative search. It’s slow, but it works. You need to make sure that your iteration step is not too large, such that you don’t miss some sweet spots.

For XX = XXmin to XXmax
  For YY = YYmin to YYmax
    For ZZ = ZZmin to ZZmax

      result = GetRootNodeValue(XX, YY, ZZ)

      If result < best_result then
        print result, XX, YY, ZZ
        best_result = result
      End if

    End For
  End For
End For

Below is another method, it’s a stochastic optimisation method (uses random points to converge on the best solution), it’s results are reasonable for most conditions. I have used this successfully and it’s good at converging to the minimum value. This is good to use if there is no clear global minimum. You will have to configure the parameters for your problem.

' How long to search for, a larger value will result in long search time
max_depth = 20000

' Initial values
x0 = initial XX value
y0 = initial YY value
z0 = initial ZZ value

' These are the delta values, how far should the default values range
dx = 5
dy = 5
dz = 5

' Set this at a large value (assuming the best result is a small number)
best_result = inf

' Loop for a long time
For i = 1 To max_depth

    ' New random values near the best result
    xx = x0 + dx * (Rnd() - 0.5) * (Rnd() - 0.5)
    yy = y0 + dy * (Rnd() - 0.5) * (Rnd() - 0.5)
    zz = y0 + dy * (Rnd() - 0.5) * (Rnd() - 0.5)

    ' Do the test
    result = GetRootNodeValue(xx, yy, zz)

    ' We have found the best solution so far
    If result < best_result Then
        x0 = xx
        y0 = yy
        z0 = zz
        best_result = result
    End If

    Print progress
Next i

There are many optimisation algorithms to choose from. Above are some very simple ones, but they may not be the best for your problem.