APARATO OSTEO-ARTICULAR

Algorithm 6.3 shows how to calculate the top-K POIs for a given query cell cref. Two

lists are maintained – (i) a list T representing the top-K nodes (POIs), ordered according to the rank of the nodes with respect to cref, and (ii) a list L of nodes to be explored,

sorted by a lower bound value. The lower bound value for a node represents the minimum possible rank achievable for the subtree rooted at that node.

To compute the lower bound value for a node we associate two distance estimates to every node c, dx(c) and dy(c), representing the minimum possible distance we expect to

find in the subtree rooted at node c, along x- and y- dimensions, respectively. Equation 6.2 gives the lower bound of the rank values in the subtree at node c.

Algorithm 6.3 Compute the top-K set for reference cell.

Input: Root node root of kd-tree; Reference cell cref; Parameter K

Output: Ordered list of top-K POIs with respect to the cell cref

1: function CellTopPOIs(root, cref, K)

2: T Ω empty list 3: LΩ empty list 4: L.Append(root)

5: while L is not empty do

6: ntestΩ L.Remove-And-Return-Head()

7: i_{Ω 1}

8: if T.size = K and @isuch thatntest.lboundÆ T [i].rank then exit while

9: ntest.CalcRank(Cref)

10: T.PriorityInsert(ntest) Û rank based.

11: if T.size > K then T .RemoveLast() 12: If there exists ntest.lef tnode then

13: L.PriorityInsert(ntest.lef t) Û lbound based.

14: If there exists ntest.rigthnode then

15: L.PriorityInsert(ntest.rigth) Û lbound based.

16: end while 17: return T 18: end function LB(c) = Ò dx(c)2+ dy(c)2 N + “min(c), (6.2)

where N is normalization factor for distance (see Section 4.6). Both distance estimates are initialized to zero for the root of the tree and then updated as we explore the tree. Let cref be (xref, yref) and we want to calculate the lower bound for the left and right child

nodes of the node ci= (xi, yi). Further, assume that x-axis is the cutting dimension at the

node ci.

• If xref < xi, then the minimum possible x-distance for the nodes in the “right”

subtree of ci with respect to cref is dx(right) = |xref ≠ xi|, because all nodes in the

right subtree of ci have x-coordinate values greater than xi. We cannot estimate the

minimum y-distance for the right subtree here because we split on x-dimension at this node. However, we can still retain the estimate from the node ci itself, effectively

• If xref Ø xi, then the minimum possible x-distance for the nodes in the “left” subtree

of ci with respect to cref is dx(left) = |xref≠ xi|, because all nodes in the left subtree

of ci have x-coordinate values less than xi. As in the previous case, dy(left) = dy(ci),

dx(right) = dx(ci) and dy(right) = dy(ci).

The same line of reasoning can can be applied if the splitting dimension on node ci was

y-axis instead of x-axis.

During the search, we explore the nodes in L in their order of appearance, and terminate when L becomes empty (line 5 in Algorithm 6.3), or it is determined that no node in the subtree can potentially change the existing T list (line 8 in Algorithm 6.3). The latter case can happen when the lower bound value of the first node in L is greater than the rank value of the last node in T . Exploring a node involves the steps of checking if the node can be inserted in the T list based on its rank (lines [9 ≠ 11] in Algorithm 6.3), computing the lower bounds for the left and right children, and then inserting them in L (lines [12 ≠ 15] in Algorithm 6.3).

Example. We present this example for further illustration of the process. In this example, a grid size is assumed to be 320 ◊ 320 cells, giving us a normalization factor N = 1

320Ô2.

We consider a top-3 search query for the cell cref. In state–1, the first node c1 in L is the

root of the subtree shown in Figure 6.2. The minimum distance estimates for the root node c1 are assumed as dx(c1) = 15 and dy(c1) = 5. The location and the “ values for cref and

for each node in our example subtree are listed below. Location (x, y) “ “min cref (204, 115) c1 (190, 72) 0.26 c2 (180, 125) 0.25 c3 0.47 c4 0.38 c5 0.35

Split on y-dimension 1 c min=0.25 1 c min=0.25 2 c min=0.25 5 c min=0.38 4 c min=0.35 3 c min=0.47 1 c : 0.285c : 0.425a

L

T

T

L

L

T

Figure 6.2: Example of kd-tree search

Following Algorithm 6.3, if L is not empty we remove the first node, c1, and check

whether is it possible to add this node to T . Since the rank of c1 is

Ô

(204≠190)2_+(115≠72)2

320Ô2 +

0.26 = 0.360 it will be inserted into T , after the node cb, as shown in state–2. Then we

look for the left and right children of c1. Since on c1 we split on x-dimension, all nodes

on the left subtree of c1 have x-coordinate values < 190. Since the x-coordinate of cref is

greater than the x-coordinate of c1, the minimum x-distance for the left subtree of c1 is

dx(c2) = |204 ≠ 190| = 14. The same argument is not possible for the minimum y-distance

of c2, so we retain this value from the parent node, dy(c2) = dy(c1) = 5. Thus, the lower

bound for c2 is LB(c2) = Ô14_320Ô22+52 + 0.25 = 0.283. For the right child, c3, we can only

retain the values from the parent, i.e., dx(c3) = dx(c1) = 15, dy(c3) = dy(c1) = 5. Thus,

the lower bound for c3 is 0.508. Now, nodes c2 and c3 are inserted into L based on their

lower bound as shown in state–2. In the next iteration, we remove the node c2 from L,

so the cutting dimension this time is the y-axis. Since y-coordinate of cref is less than

the y-coordinate of c2, all nodes on the right subtree of c2 will have y-coordinate > 125,

giving dy(c5) = |115 ≠ 125| = 10. With dx(c5) = dx(c2) = 14, the lower bound for c5 is Ô142₊₁₀2

320Ô2 + 0.38 = 0.418. For the left child, c4, dx(c4) = dx(c2) = 14, dy(c4) = dy(c2) = 5

and the lower bound is 0.383. The rank of c2 is 0.307. The node c2 is inserted into T based

on its rank, and the nodes c4 and c5 are inserted into L based on their lower bound. The

we observe that the lower bound value for c4 (the first node in L) is greater then the rank

of the last node in T , we can terminate the search and return the list T .