Szeszlér, Dávid

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Additive Approximation for LayerMinimization of Manhattan Switchbox RoutingDa´vid Szeszle´r 1,2Department of Computer Science and Information TheoryBudapest University of Technology and EconomicsBudapest, HungaryAbstractSwitchbox routing is one of the many problems arising in the field of VLSI routing.It requires interconnecting given sets of terminals that are placed on the boundariesof a rectangular circuit board using a 3-dimensional grid in a vertex-disjoint way.An important special case is the Manhattan Switchbox Routing problem. Hereminimizing the number of layers of a routing (that is, the height of the grid) isknown to be NP-hard. In this paper we provide a linear time algorithm that solvesany such problem on a number of layers that is greater by at most 5 than theoptimum.Keywords: VLSI, Switchbox Routing, Additive Approximation1 IntroductionIn the detailed routing phase of the design process of very large scale integrated(or VLSI) circuits the task is to interconnect certain given subsets (or nets) of1 Research partially supported by the Hungarian National Research Fund and by the Na-tional Office for Research and Technology (Grant Number OTKA 67651)2 Email: szeszler@cs.bme.hupins (or terminals) of the circuit elements that have already found their finalpositions on the circuit board. Since a minimum distance is to be kept betweenany two wires (corresponding to different nets), wires must go along the edgesof a rectangular grid. However, this grid is not planar, it consists of a fewplanar layers (each parallel to the circuit board) and a wire can leave a layerfor an adjacent one at any grid point. In some cases, certain technologicalconstraints require that consecutive layers contain non-parallel wire segmentsonly; if such a restriction is imposed, we speak of a Manhattan routing.In case of the switchbox routing problem, terminals are situated on theboundaries of a rectangular circuit board. Minimizing the number of layersneeded for solving a switchbox routing problem in the Manhattan model isknown to be NP-hard; even very special cases of this problem are NP-hard.(For example: deciding solvability on two layers is NP-complete even if ter-minals are situated on two opposite boundaries only [5].) Consequently, anddue to the importance of the problem, many heuristics have been proposed.In this paper we present a linear time algorithm with a provably goodperformance: the number of layers used will be at most 5 greater than theoptimum. Considering that the number of layers allowed by the present stateof technology is typically not more than 10, this additive constant of 5 sug-gests that the presented construction is not intended for designing ready-to-userouting patterns for given specifications. Contrarily, we rather aim at under-standing the nature of the problem and thus helping heuristics to be designedand tested. We emphasize however, that the number of layers used will staywithin the realizable range of at most 10 for many realistic specifications.2 Basic Definitions and Preliminary ResultsThe switchbox routing problem (or SRP for short) is defined as follows. Assumethat a (w+2)×(n+2) rectangular grid formed by w+2 horizontal line segmentsnumbered from 0 to w + 1 (called tracks) and n + 2 vertical line segmentsnumbered from 0 to n + 1 (called columns) is given. Non-corner boundarypoints of the grid are called terminals. A net is a subset of terminals. Aninstance of the SRP is a set N = {N1, . . . , Nt} of pairwise disjoint nets. Theleft hand side part of Figure 1 shows an SRP instance with w = n = 5; setsof terminals marked with a common number form the nets.Terminals with a column number of 0 or n+1 are called western or eastern,respectively; terminals with a track number of 0 or w + 1 are called southernor northern, respectively. w and n are called the width and the length of theSRP, respectively. The w × n grid obtained by deleting all terminals andcorners is called the routing area. Of course, w ≤ n can be assumed withoutloss of generality; we will assume this throughout this paper. We will alsoassume that every net has at least two terminals. A net is called trivial if itconsists of two terminals situated in a common column.In order to define a k-layer solution of a SRP we first replace the w × nrouting area with a 3-dimensional w×n×k grid consisting of k parallel copies(called layers) of the original routing area. In the resulting routing graph eachterminal is adjacent to all the k vertices of the k layers corresponding to itssingle neighbour in the original (planar) routing area. (This is to ensure thatterminals are accessible from any layer.)A k-layer solution (or routing) of a SRP instance N is a set H ={H1, . . . , Ht} of pairwise vertex disjoint, connected subgraphs (called wires)of the routing graph such that Hi contains all terminals of Ni for i = 1, . . . , t.Edges of the wires that join adjacent vertices of two consecutive layers arecalled vias. Since wires can be chosen to be trees (which usually happens inevery application), the SRP can be regarded as a special Steiner-tree packingproblem.A solution of a SRP belongs to the Manhattan model (or is simply calleda Manhattan solution) if consecutive layers contain wire segments of differentdirections only. Thus layers with horizontal (east-west) and with vertical(north-south) wire segments alternate. In this paper we will restrict ourselvesto the Manhattan model.Denote the ith column of the grid by e (for some 1 ≤ i ≤ n). A nontrivialnet N is separated by e if N has a terminal with a column number less than orequal to i and another terminal with a column number greater than or equalto i. The congestion of e, denoted by c(e), is the number of nontrivial netsseparated by e. The density of a SRP instance, denoted by d, is the maximumcongestion of all columns (of the routing area).Assume that a SRP instance with density d is given that is solvable onk layers in the Manhattan model with h layers reserved for horizontal wiresegments. Naturally, wires corresponding to all the d nontrivial nets separatedby a column e with congestion c(e) = d must intersect e on one of these hlayers. At most w wires can intersect e on a single layer, hence h · w ≥ dmust hold. This shows (by k ≥ 2h − 1) that the minimum number of layersrequired for solving a SRP in the Manhattan model is at least 2⌈ dw⌉ − 1 (andthe corresponding lower bound is ⌈ dw⌉ if the Manhattan model is not assumed).Hambrusch [2] observed that a SRP with d = n+w can be constructed byletting the nets be terminal pairs, all separated by column ⌊n2⌋. This showsthat the number of layers required for a SRP can be as large as 2⌈ nw⌉ + 1 inthe Manhattan model (and ⌈ nw⌉+1 otherwise) in the worst case. In particular,there is no fixed number of layers that would suffice for every SRP instance.However, there is a general upper bound for the number of layers if theratio nwis fixed. This was first shown for the non-Manhattan case by Boros,Recski and Wettl [1]: they provided an upper bound of max{18, 2 nw+ 14}.This was later improved in [4] by giving an upper bound of 2⌈ nw⌉ + 4 in theManhattan model. (Both results involve an algorithm with a linear runningtime.) Naturally, although the latter upper bound is only larger by 3 thanthe lower bound of 2⌈ nw⌉ + 1 shown in the previous paragraph, the resultof [4] does not give an additive approximation for the optimum since thelower bound of 2⌈ nw⌉+ 1 is for the worst case only. In this paper we improvethe algorithm presented in [4] to achieve an additive approximation for theoptimum. (However, we do not rely on the results of [4].)3 Main ResultIn the previous section we have seen that the number of layers required for aSRP is at least 2⌈ dw⌉−1 in the Manhattan model, where d denotes the densityof the problem. This shows that the following theorem indeed guarantees asolution within an additive constant of 5 compared to the optimum.Theorem 3.1 There is a linear time algorithm that solves every switchboxrouting problem on at most 2⌈ dw⌉+ 4 layers in the Manhattan model.Proof. Assume that a SRP instance N with density d is given. We start byclassifying its nets for easier reference: we say, for example, that a net is NEtype if it has terminals on the northern and eastern boundaries, but none onthe western and southern ones; we say that a net is N1W type if it has oneterminal on the north, some (maybe one) on the west and none elsewhere,etc. Furthermore, denote by TW, TE, TN and TS the set of all western, eastern,northern and southern terminals, respectively.We can assume without loss of generality that out of NE, NW, SE and SWtype nets NE nets are (one of) the greatest in number.Denote the set of S1W1 type nets by N0. Partition N0 into two subsetsN 10and N 20such that |N 10| ≤ |N 20| ≤ |N 10|+ 1. We define two further subsetsof N , N1 and N2. Let a nontrivial net N ∈ N belong to N1 if• |N ∩ (TN ∪ TS) | ≥ 2, but N is not NE type, or• N is N1W type, or• N ∈ N 10.Similarly, let N ∈ N belong to N2 if• |N ∩ (TW ∪ TE) | ≥ 2, but N is not NE type, or• N is SE1 type, or• N ∈ N 20.We assign an interval to each net of N1 and N2: if N ∈ N1 then thehorizontal interval IH(N) stretches from the column of the westernmost tothe column of the easternmost terminal of N \ TE. Similarly, if N ∈ N2 thenthe vertical interval IV (N) stretches from the track of the northernmost to thetrack of the southernmost terminal of N \ TN. (In both cases, each boundaryis considered when determining the endpoints of the intervals, except for theeastern boundary in case of N1 and the northern one in case of N2.) Forthe SRP instance of Figure 1 nets of N1 together with the assigned intervalsIH(N) are shown in the middle part of the figure, while nets of N2 with theintervals IV (N) are shown in the right hand side part.12 7 5 8476428 745 33866112 7 5 8476428 745 33866112 7 5 8476428 745 338661851674 3Figure 1We claim that |N2| ≤ w. Denote by N 12 and N22the first two of the three(pairwise disjoint) subsets of N that form N2 as listed on the top of this page(in the same order). Let |N 12| = x1, |N 22 | = x2, |N0| = y and let z be thenumber of NE type nets. Since |TW∪TE| = 2w, obviously 2x1+x2+y+z ≤ 2w.Because of the assumption made at the beginning of the proof, x2 ≤ z holds,thus 2x1 + 2x2 + y ≤ 2w is also true. Dividing by 2 we get x1 + x2 +12y ≤ w,which establishes the claim by |N2| = x1+x2+⌈12y⌉. (We remark that |N1| ≤ nis obviously also true with an analogous proof.)After all this preparation we start describing the required routing. We saythat a layer is a W-comb if on that layer a horizontal wire segment leads fromeach western terminal to the easternmost column of the routing area. Theterms E-comb, N-comb and S-comb are defined analogously; however, in caseof (say) N-combs we declare that if a column contains the two terminals ofa trivial net then the wire segment of that column is extended by one unitto the south to connect these two terminals. (We remark that the usage ofcomb layers simplifies the description of the construction, but obviously inmost cases wire segments on the comb layers need not stretch to the oppositeend of the routing area. Of course, unnecessary wire ends can be removed butwe will disregard this in the sequel.)Number the layers from bottom upwards. We declare that layer 1 is aW-comb, layer 3 is an E-comb, layers 4, 8, 12, . . . are N-combs and layers 6,10, 14, . . . are S-combs. (Since the number of layers 2⌈ dw⌉ + 4 is even and atleast 6, the topmost layer is either a N-comb or a S-comb.)Columns of layer 2 are used for holding vertical wire segments correspond-ing to the intervals IV (N) assigned to the nets of N2. (We will also use thenotation IV (N) to refer to the wire segment corresponding to the intervalIV (N), this will cause no ambiguity.) We have seen that |N2| ≤ w and recallthat w ≤ n was assumed. Consequently, each net N ∈ N2 can be assigned aseparate column on layer 2 to hold IV (N). However, if N ∈ N2 and N∩TS 6= ∅then IV (N) should be placed in the column of an (arbitrary) element of N∩TS;this way the wire segment also reaches this southern terminal of N , see nets3 and 4 in Figure 1.Consider the interval graph G defined by the intervals IH(N) assigned tothe nets of N1. (That is, V (G) is the set of intervals IH(N) and edges of Gcorrespond to intersecting pairs of intervals. The intervals are assumed to beclosed, thus two intervals are adjacent in G even if they only have a commonendpoint.) By definition, the clique number ω(G) of G is the density d1 of theSRP defined by N1. Since d1 ≤ d is obvious from N1 ⊆ N , vertices of G canbe coloured with d colours. (Here we use the well-known fact that intervalgraphs are perfect; this result is usually dedicated to Tibor Gallai.) Denotethe colour classes by I1, . . . , Id.Tracks of layers 5, 7, 9, 11, . . . are used for holding horizontal wire segmentscorresponding to the intervals IH(N) assigned to the nets of N1. (Again, thesewire segments will also be denoted by IH(N).) However, as opposed to thecase of layer 2, we do not assign a separate track to each interval. Instead,each track is assigned to one of the colour classes Ik and the track holdswire segments corresponding to all the intervals of Ik. (This is possible sinceintervals belonging to Ik are pairwise disjoint by the construction.) Again,if N ∈ N1 and N ∩ TW 6= ∅ then the wire segments corresponding to thecolour class of IH(N) should be placed in the track of an (arbitrary) elementof N ∩ TW. (This is obviously possible since no colour class can contain morethan one intervals corresponding to nets having western terminals.) Since thetotal number of layers is 2⌈ dw⌉+4, the number of layers 5, 7, 9, 11, . . . is ⌈ dw⌉;hence the total number of tracks on these layers is at least d which shows thateach colour class Ik can be assigned an appropriate track.The routing is now completed by introducing vias between consecutivelayers appropriately. For every N ∈ N2 the vertical wire segment IV (N) onlayer 2 can be connected to the horizontal wire segments of layers 1 and 3coming from the terminals of N ∩ (TW ∪ TE). Similarly, for every N ∈ N1 thehorizontal wire segment IH(N) can be connected to the vertical wire segmentscoming from the terminals of N ∩ (TN ∪ TS) in the layers below and above thelayer of IH(N). Finally, we introduce vias between layers 3 and 4 suitably:for every s ∈ TN and t ∈ TE if s and t belong to the same net then the verticalwire segment coming from s on layer 4 and the horizontal wire segment comingfrom t on layer 3 can be connected through an appropriate via. (Evidently, itis not necessary to take advantage of all the possibilities to introduce the abovementioned vias between layers 3 and 4. Moreover, it is not only unnecessary,but for certain SRP instances the number of – mainly needless – vias couldbecome so high that the algorithm would not be linear any more. We disregardthis detail in the sequel with mentioning that for each net N the number ofvias needed between layers 3 and 4 is at most |N ∩ (TN ∪ TE)|.)We claim that the above described routing is good. Choose a net N ∈ Nand two terminals s, t ∈ N arbitrarily. It is sufficient to show that s and tare connected by the routing. If N is trivial then s and t are joined in thecorresponding column of a N-comb. If N is NE type then the routing of Nis completely solved by the vias between layers 3 and 4. So we can assumethat N is neither trivial nor NE type; it is easy to check that this impliesN ∈ N1 ∪ N2. If {s, t} ⊆ TW ∪ TE then N ∈ N2 and hence s and t areconnected through IV (N). Similarly, if {s, t} ⊆ TN ∪ TS then N ∈ N1 and sand t are connected through IH(N). So we can assume that s and t belongto two adjacent boundaries.Assume first that s ∈ TW and t ∈ TN. If N ∩ TE 6= ∅ and u ∈ N ∩ TEthen N ∈ N2, u and s are connected through IV (N), u and t are connectedthrough the vias between layers 3 and 4 and hence s and t are also connectedvia u. If, on the other hand, N ∩ TE = ∅ then N ∈ N1 by definition andhence t is connected to a terminal s′ ∈ N ∩ TW through IH(N). If s′ 6= s thenN ∈ N2 is also true, s and s′ are connected through IV (N) so s and t areconnected via s′. The case s ∈ TE, t ∈ TS is analogous. If s ∈ TW and t ∈ TSthen either N ∈ N1 and t is connected to a terminal s′ ∈ N ∩ TW throughIH(N) or N ∈ N2 and s is connected to a terminal t′ ∈ N ∩ TS throughIV (N); in the first case s 6= s′ implies N ∈ N2 and s and s′ are connectedthrough IV (N) while in the second case t 6= t′ implies N ∈ N1 and t and t′are connected through IH(N). Finally, if s ∈ TE and t ∈ TN then s and t aresimply connected by the vias between layers 3 and 4.We omit the proof that the above construction can be realized by a lineartime algorithm for space reasons. We restrict ourselves to mentioning that theonly nontrivial step is the colouring of the interval graph G defined above; thereader is referred to [3] for the details. 24 Concluding RemarksWe remark that the above proof gives a slightly better upper bound on thenumber of layers than the one claimed by Theorem 3.1: 2⌈d1w⌉+4, where d1 isthe density of the SRP N1 (defined in the proof). Observe that the maximumvalue of d1 is n (since |N1| ≤ n was mentioned in the proof) while the valueof d can be as large as n+w (as mentioned in the Introduction). This impliesthat the above upper bound of 2⌈d1w⌉ + 4 can indeed be less than the oneclaimed in Theorem 3.1. Furthermore, it also shows that the construction ofthe above proof generalizes the result of [4] mentioned in the Introduction by2⌈d1w⌉+ 4 ≤ 2⌈ nw⌉+ 4.It is also worth mentioning that although the additive constant guaranteedby Theorem 3.1 is 5 in general, however, in certain special cases this can beimproved. For example, if d1 ≤ w holds (which is definitely true in the n = wcase) then the construction guarantees a routing on 6 layers; since a 1-layersolution is clearly impossible (except for trivial cases), the additive constantof approximation is 4 in this case.References[1] Boros, E., A. Recski and F. Wettl, Unconstrained multilayer switchbox routing,Annals of Operations Research 58 (1995), 481–491.[2] Hambrusch, S. E., Channel routing in overlap models, IEEE Trans. Computer-Aided Design of Integrated Circ. Syst. CAD-4 (1985), 23–30.[3] Szeszle´r, D., Combinatorial Algorithms in VLSI Routing, PhD Dissertation,2005. http://www.cs.bme.hu/~szeszler/szeszler_dissertation.pdf[4] Szeszle´r, D., Switchbox routing in the multilayer Manhattan model, AnnalesUniv. Sci. Budapest. Eo¨tvo¨s Sect. Math. 40, (1997), 155–164.[5] Szymanski, T. G., Dogleg channel routing is NP-complete, IEEE Trans.Computer-Aided Design of Integrated Circ. Syst. CAD-4 (1985), 31–41.

Additive Approximation for Layer Minimization of Manhattan Switchbox Routing

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