alternative rank function RBTree (red black tree)

I have an order-statistic augmented red black tree.

it works for the most part. but i need to implement a fast function (O(lg n)) that mostly returns the place of a node in sorted order. like the OS-rank function from my textbook. but with one twist: the return value if two nodes have the same score, should be the same. here is the os-rank function (in pseudocode, for a given node x, where root is the root of the tree).

OS-Rank(x)
r=x.left.size+1
y=x
while y!=root
  if y==y.p.right
    r+=y.p.left.size+1
y=y.p
return r

But: what i need is something where if A has key 1 and Node B has key 1, the function returns 1 for both. and so on. I tried myself with something like this.

rank(x)
start with value r=1
check that x.right is not Nil
  case x.right has the same key as x
    add x.right.#nodeswithkeyhigher(x.key) to r
  other cases: add x.right.size to r
y=x
while y != root
  if y.parent.left == y
    case y.parent.right.key>x.key
      add y.parent.right to r
    other cases
      add y.parent.right.#nodeswithkeyhigher(x.key) to r
  y=y.parent
return r

Guess what: a testcase failed. I'd like to know if this is a correct way of doing things, or if perhaps i made some mistake i am not seeing (else the mistake is in the Node.#nodeswithkeyhigher(key) function).

Solution

edit: final paragraph for answer, thanks to Sticky.

tl;dr: skip to last paragraphs

This is the same issue I'm having trouble with. (Yes DS aswell). So far all runs except 5 are correct. I've tested several things, one being a very simple one: Just exchange left and right in OSRank. In some cases it gave a correct answer but in the harder cases it was quite a bit off. Oh I also added that if y.score == y.parent.score I only add the right size of y.parent, if not I add the right size + 1.

public int OSRank(Node x)
    {
        int r = x.Right.Size + 1;
        Node y = x;
        while (y != root)
        {
            if (y == y.Parent.Left)
            {
                if (y.Score == y.Parent.Score)
                    r = r + y.Parent.Right.Size;
                else
                    r = r + y.Parent.Right.Size + 1;
            }
            y = y.Parent;
        }
        return r;
    }

Let's first test this method on the tree on page 340 (figure 14.1). We'll search for the rank of 38 (which should return 4 because 39, 47 and 41 are higher):

r = 1 + 1 = 2 //Right side + 1
r = 2 //nothing happens because we're a right child
r = r + 1 + 1 = 4 //we're a left child, the key of our parent is larger and parent.Right.size = 1
r = 4 //nothing happens because we're a right child

So in this case the result is correct. But what if we add another node with key 38 to our tree. That reshapes our tree a bit, the right part of node 26 now looks like:

(I'm not allowed to add images yet so look here:http://i47.tinypic.com/358ynhh.png)

If we would use the same algorithm we'd get the following result (picking the red one):

r = 0 + 1 = 1 //no right side
r = 1 //we're a right child
r = 1 //we're a right child
r = 1 + 3 + 1 = 5 //The 3 comes from the size of node 41.
r = 5 //we're a right child

Though we expect rank 4 here. While I was typing this out I noticed that we check if y.Score == y.Parent.Score, but I completely forgot y changes. So in line 4 the clause "y.Score == y.Parent.Score" was false because we compared node 30 with 38. So if we change that line to:

if (x.Score == y.Parent.Score)

The algorithm outputs rank 4, which is correct. This means we eliminated another issue. But there are more, which I didn't figure out either:

The case in which Y.Parent.Right contains duplicate keys. Technically if we have 3 nodes with the same key, they should count as 1.
The case in which Y.Parent.Right contains keys that are equal to x.Key (the node you want the rank of). That would put us a few ranks back, incorrectly.

I suppose you could keep another integer which holds the amount of nodes with a higher score. Upon insertion you could climb the tree and adjust values if the subtree of that node doesn't contain a node with the same score. But how this is done (and efficiently) is unknown to me right now.

edit: First find the final successor of x with the same score x. Then calculate the rank the normal way. The code above works.