Javier Cordova. Abstract. In this paper the following problem is considered: given a root node R in a

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1 TR94025 A Heuristic Algorithm for the Rectilinear Steiner Arborescence Problem Javier Cordova Computer Science Department University of Puerto Rico Arecibo, PR 0013 YannHang Lee Computer and Information Sciences Department University of Florida Gainesville, FL 3211 August 17, 1994 Abstract In this paper the following problem is considered: given a root node R in a mesh and a set D of nodes from the mesh, construct a shortestpath tree rooted at R that spans the set D and minimizes the number of links used. The problem is equivalent as nding a Steiner tree in a directed mesh in which all the links point away from the root node R. The problem of nding a Steiner tree in such grid has been known in the literature as the Rectilinear Steiner Arborescence (RSA) problem. Rao et. al [9] have proposed an ecient heuristic algorithm for a special case of this problem, in which all the nodes in D lie in the rst quadrant of E 2 (assuming the tree is rooted at the origin). The algorithm has time complexity of O(j D j log j D j). They proved that the trac generated by the heuristic has an upper bound of twice of that generated by any optimal algorithm. They have proposed an extension of this algorithm that solves the general case in which the nodes in D lie anywhere in the plane that runs with O(j D j 3 log j D j) time complexity. We propose in this work a similar heuristic with same upper bound that also has time complexity O(j D j log j D j) but solves this general case in which the nodes are allowed to be anywhere in E 2. 1

2 1 Introduction Consider a mesh network in which a root node R is given. Let D be a set of nodes from the mesh. We want to build a shortestpath tree rooted at R that spans the set D and minimizes the number of links used. In a shortestpath tree the distance between the root and every node in the tree is minimized. Constructing a shortestpath tree with lowest trac rooted at node R (grid M in the gure below) is equivalent to nding a rectilinear directed Steiner tree rooted at R in a directed mesh in which all the links are directed away from R (M 0 in the gure). M M 0 R R s s The problem of nding a Steiner tree given a set of points in a plane using rectilinear distances has been proved to be an NPComplete problem [4]. On the other hand, the problem of nding a rectilinear directed Steiner tree has been known in the literature as the Rectilinear Steiner Arborescence (RSA) problem. The problem here is to nd a Steiner directed tree for a set D of nodes lying in the grid, rooted at point R, using only horizontal and vertical arcs oriented away from R. Rao et. al [9] have studied the Rectilinear Steiner Arborescence problem when the set D lies in the rst quadrant ofe 2, and the directed tree is rooted at the origin. They proposed a heuristic algorithm in which a Steiner tree H for D is generated by iterative substitution. The algorithm has time complexity O(j D j log j D j). They have shown that l(h) 2l(RSMA), where l(a) isthenumber of links in the tree A and an RSM A (Rectilinear Steiner Minimum Arborescence) is an optimal RSA. They proposed a solution to the general RSA problem in which D is allowed to lie anywhere in E 2. They reduced the problem to the rst quadrant version by restricting the arborescence to contain the xaxis only between a and b (a 0 b), and the yaxis only between c and d (c 0 d). Using the previous algorithm the best RSA in each quadrant satisfying these conditions can be determined (for the best restricted RSA in the rst quadrant, for example, all j D j 2 possible choices of b and d have to be examined, taking 2

3 O(j D j 3 log j D j) time). The best RSA in E 2 is obtained by taking the best among all restricted RSAs. We propose a similar heuristic, on the other hand, which is a simple extension of their algorithm and generates a suboptimal RSMA when D lies anywhere in the grid, maintains the same upper bound for the number of links in the tree, and can be implemented in a similar way with time complexity being also O(j D j log j D j). The heuristic is summarized in the following. 2 Heuristic Algorithm For two grid points p and q, let max o (p q) be the grid point that maximizes the overlapping in the shortest paths from the origin to the points p and q. The point max o (p q) is dened in the following way, given that p =(x p y p ), q =(x q y q ), for z = x or y coordinates: if z p z q 0 then the zcoordinate of max o is given by min(z p z q ). Else if z p z q 0 then the zcoordinate is max(z p z q ). Otherwise it is 0. We assume that the root of the tree is the origin. This denition of max o (p q) allows us to combine points from adjacent quadrants. This was not handled by the algorithm proposed in [9]. Let L z denote the four lines lying in the grid given by j x j + j y j= z. A set of points P is a cover for a set of points Q if 8q 2 Q there exists a point p 2 P with a path from p to q. Let Q z be the subset of Q lying above L z. Then let MC(Q z) denote the set of points on L z with minimum cardinality thatcovers Q z. It follows directly from this Denition that any RSA for D intersects L z at least MC(D z) times for every z. A heuristic algorithm generates a suboptimal RSA (SRSA)by iteratively substituting two points in D by the point max o (p q)until the origin remains. Again, the pair of points p q are chosen to maximize jj max o (p q) jj. Let D z denote the set D at the point in the construction of the SRSA in which the last possible pair of points p q having jj max o (p q) jj z have been joined, and let W z = fp 2 D z :jj p jj zg. The nodes in the initial set D are called sinks. Lemma 1 During the construction of the SRSA, for any z 0, there can be at most one point in W z on or above the horizontal line y =(0 z) (also, there is at most one point in W z on or below the horizontal line y =(0 ;z), at most one point on or to the right of the vertical line x =(z 0) and at most one on or to the left of the vertical line x =(;z 0)). 3

4 Proof: Let p q 2 W z with y p y q z. Let y p y q. Then max o (p q) =(x y q ) for some value x. Hence jj (x y q ) jjz, contradicting the fact that p q 2 W z. A similar proof works for the other 3 lines. 2 The following lemma is an adapted version of lemma 4 in [9]. Lemma 2 Let p i =(x i y i ) be apoint in a SRSA generated by the heuristic. If x i =0 and y i =0, then there must exist a sink on the vertical path x = x i (horizontal path y = y i ) starting at p i. Proof: Assume, w:l:g:, that p i lies in the rst quadrant. The proof is by induction on the number of descendents in the tree of p i.ifp i does not have descendents then it is a sink, and the result immediately follows. Suppose, on the contrary, that p i = max o (p j p k ). Observe that these two points belong also to the rst quadrant. Thus x i = min(x j x k ), say x k, and y k y i. By induction there is a sink on the vertical path x = x k starting at p j.the fact that y k y i implies that this sink is also on the vertical path x = x i = x k staring at p i. A similar argument can be made for the ycoordinate. 2 Now assume that the points p and q lie in the rst quadrant. lemmas are taken directly from [9]. The proofs can be found in their paper. The following two Lemma 3 During the construction of the SRSA, for any z 0 and p q 2 W z, 1. x p = x q and y p = y q, 2. x p <x q if and only if y p >y q, 3. if x p <x q, then x p + y q <z. Lemma 4 Let p q 2 W z with x p <x q. Then there exists a sink either on the horizontal path between p and (x p y p ) or on the vertical path between (x q y p ) and q. It is straight forward to prove similar results when both of the points p and q lie in any of the other three quadrants. The following theorem provides an upper bound for the trac generated for the multicast tree by the heuristic. Its proof is similar to the proof of the upper bound provided in [9]. 4

5 Theorem 1 l(srsa) 2l(RSM A). Proof: First, note that for any RSA A, l(a) = P z(a\l z ). Weknowthatj L z \A jj MC(D z) j for any RSA A. Thus it is sucient toshowthatj SRSA \ L z j 2 j MC(D z) j for any z. For any given z, order the points in W z into p1 p2 ::: p jwzj circularly as follows: rst take the nodes in the rst quadrant and order them by increasing values of the xcoordinates (with x1 > 0), then the nodes in the second quadrant by decreasing order of the xcoordinate, and so on. It can be shown that MC(D z) must contain distinct points for each pair of points p2i;1 p2i, fori =1 2 ::: bj W z j =2c. Consider the two pairs of points (p2i;1 p2i), and (p2i+1 p2i+2), for any given i in the above range. Assume, for simplicity, that these points lie in the rst quadrant. By Lemma 4 there must exist a sink on the horizontal path y = y2i;1 between p2i;1 and (x2i y2i;1) oronthevertical path x = x2i between (x2i y2i;1) and p2i. Hence MC(D z) must contain a point whose xcoordinate lies between z ;y2i;1 and x2i. Also, there is a sink on the horizontal path y = y2i+1 between p2i+1 and (x2i+2 y2i+1) or on the vertical path x = x2i+2 between (x2i+2 y2i+1) andp2i+2. Hence MC(D z) must contain a point whose xcoordinate lies between z ; y2i+1 and x2i+2. Since x2i + y2i+1 <zby Lemma 3, then the above two points are dierent. A similar result holds no matter where the points lie. Thus MC(D,z) contains at least bj W z j =2c points. This implies that if j W z j is even, then the SRSA generated by the heuristic intersects L z If j W z j is odd, then consider p1. Case 1: x p1 =0. By Lemma 2, there is a sink s on the vertical path x = x p1 at most 2 j MC(D z) j times. starting at p1. Thus MC(D z) must contain a point whose xcoordinate is between 0 and x p1. For the rest of the j W z j;1 points in W z there correspond at least (j W z j ;1)=2 distinct points in MC(D z), and it is easy to see that these points mustbedierent from the point in MC(D z) corresponding to p1. 1 Thus MC(D z) contains at least j Wz points. j =2 distinct 1 We can assume here that jj pw z jj< z. Otherwise, we can order the nodes circularly the other way around, starting with pw z and nishing with p 1. 5

6 Case 2: x p1 =0. We can assume now thatw z contains points on or above the points (0 z) (z 0) (0 ;z) (;z 0) in the same vertical or horizontal lines. Otherwise, we could order the points in W z starting with the points in some other quadrant and reduce the problem to Case 1. Assume, w:l:g:, that there are 2c points in W z in the rst quadrant, for some positive integer c. Note that since now we are assuming that j W z j is odd, at least one quadrant must have aneven number of points in W z. Thus MC(D z) must contain at least c points in the rst quadrant. Case 2a: MC(D z) contains at least c +1points in the rst quadrant. MC(D z) must contain at least b(j W z j;2c)=2c distinct points in the other three quadrants. Thus MC(D z) contains at least bj W z j =2c ;c + c +1>j W z j =2 distinct points. Case 2b: MC(D z) contains exactly c points in the rst quadrant. We will show that at least one of the points (0 z)or(z 0) cannot be one of these c points in MC(D z). Consider the rst 2c ; 2points in W z in the rst quadrant, not including p1. There must be c ;1 distinct points in MC(D z) tocover them, and none of them is (0 z)or(z 0). Thus one of these two points is not in MC(D z). Assume, w:l:g:, that p1 2 MC(D z). Thus between x p Wz and 0 there is a dierent point inmc(d z) required to cover the sink on or above p Wz. For the rest of the j W z j;2c ; 1 points in W z there must be (j W z j;2c ; 1)=2 distinct points in MC(D z). Thus MC(D z) contains at least bj W z j =2c ;c + c +1>j W z j =2 distinct points. Thus W z contains at most 2 j MC(D z) j points, and therefore the SRSA generated intersects L z at most 2 j MC(D z) j times. Since z is arbitrary this holds for any z, and the theorem is proved. 2 The implementation presented in [9] for their algorithm can be adapted for our algorithm. The key idea is that Steiner points can only be introduced by the heuristic from adjacent points in W z. To implement the algorithm a priority queue is used containing all the sinks below the lines in L z and all the j W z j;1potential new Steiner points max o (p q) for each pair of adjacent points p q 2 W z. See [9] for details of the implementation. To evaluate the performance of the algorithm we ran simulation programs for an 8*8 mesh both for the optimal and the heuristic algorithm. The optimal solution for 100

7 randomly generated groups was found by exhaustive search of all shortestpath trees, and compared to the suboptimal solution found by the heuristic algorithm for the same multicast groups. The trac is the number of links in the multicast tree. The results are shown in the graph above. To get the worst case line we added to the optimal trac for each number of destinations the highest dierence found between the optimal trac and the trac computed by the heuristic. It can be seen that even in the worst case the values computed are much closer to the optimal values than to the upper bound Heuristic Optimal Worst Case Upper Bound 80.0 Traffic Number of Destinations Figure 1: Optimal Algorithm vs Heuristic References [1] Foulds L.R., Graham R.L., "The Steiner Problem in Phylogeny is NPComplete", Advances in Applied Mathematics, Vol. 3, 1982, pp

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