Group Title: Department of Computer and Information Science and Engineering Technical Reports
Title: Minimum area joining of compacted cells
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Title: Minimum area joining of compacted cells
Alternate Title: Department of Computer and Information Science and Engineering Technical Report ; TR-94-013
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Language: English
Creator: Cho, Seonghun
Sahni, Sartaj
Publisher: Department of Computer and Information Sciences, University of Florida
Place of Publication: Gainesville, Fla.
Copyright Date: 1994
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Minimum Area Joining of Compacted Cells1


Seonghun Cho and Sartaj Sahni
Department of Computer and Information Science, U,.i' -.' of Florida
Gainesville, FL 32611, USA


Technical Report 94-013


Abstract. We consider the problem of joining a row of compacted cells so as to minimize
the area occupied by the cells and the interconnects. The cell joining process includes cell
stretching and river routing. Since we consider multilayer routing, we first develop necessary
and sufficient conditions for feasible multilayer river routing. This is followed by a develop-
ment of fast algorithms for the multilayer river routing problem. We then propose several
heuristics to join a row of cells in such a way that area is minimized. The proposed heuristics
are compared, experimentally with that proposed by Cheng and Despain [CHEN89].


Keywords and Phrases. Cell joining, cell stretching, compacted cells, multilayer river
routing, symbolic sticks designs, VLSI design.



1 Introduction

When designing circuits with compacted symbolic sticks basic cells, the circuit is realized by
a collection of compacted cells that tile a two-dimensional area. The intercell interconnects
are such that each interconnect connects two terminals that are on adjacent boundaries of
neighboring cells. So, for example, if cells A and B (Figure l(a)) are neighboring cells of
the circuit, then the right boundary of A is adjacent to the left boundary of B. The number
of terminals on each of these boundaries will be the same and the i'th terminal (from the
bottom) on the right boundary of A is to be connected to the i'th terminal (from the bottom)
on the left boundary of B.
Since the cells are available in compacted form, it is not possible to reduce the distance
'This research was supported, in part, by the National Science Foundation under grant MIP-9103379.














4 15
11 4 3 B 10
--- 3 2
1-...2 1..









12
(a) Horizontal adjacent cells (b) Joining by stretching












(c) Joining by river routing (d) Combination cell joining

Figure 1: Cell Joining


between any pair of terminals on any side a cell. However, this distance can be increased
by stretching the cell. In the example of Figure l(a), we can stretch either cell vertically by
defining a horizontal cut line at any position and pulling the two cell pieces apart by any
desired amount (the cell can also be stretched horizontally by using a vertical cut line).
The required interconnects between cells A and B of Figure l(a) can be accomplished by
stretching cells A and B so that the terminals of A and B line up as in Figure l(b). The
broken lines in Figure l(a) indicate the cut lines used for stretching. The stretching enables
us to join cells A and B using no routing tracks (by "join" we mean make the interconnects
between cells A and B). This method of joining cells is also called pitch matching.
Another way to join cells A and B is to river route the interconnects as in Figure l(c).
This uses routing tracks in a channel between cells A and B but does not increase cell
height. The pitch matching and river routing approaches to cell joining have been studied
in [BOYE88] and [WEST81]. Algorithms for single-layer river routing can be found in









[LEIS83, MIRZ87, PINT82, PINT83].
Cell stretching (or pitch matching) increases the height of the layout while river routing
increases its width. Both affect the layout area. The layout of Figure 1(b) has area 150.
To compute the area of the layout of Figure l(c), we assume tracks have unit separation.
So, the layout width is 14 and height is 11. The layout has area 154. Cheng and Despain
[CHEN89] have proposed using a combination of cell stretching and river routing so as to
obtain layouts with smaller area than possible when only one of these joining methods is
used. Figure 1(d) shows the result of joining cells A and B using both stretching and river
routing. The area of this layout is 144. This is minimum for the instance of Figure l(a).
Cheng and Despain [CHEN89] have proposed a heuristic for single layer joining of com-
pacted cells. At each step of their heuristic either a row or column of compacted cells is
joined. Following this, the row or column of joined cells is replaced by a composite cell that
represents the result of joining. Notice that when a row (column) of cells is joined, cells
may be stretched vertically (horizontally) and river routing is done in a vertical (horizontal)
channel. To join a row of cells, Cheng and Despain [CHEN89] bound the maximum height
to which a cell may be stretched. This bound is

hmax + h av/(4 hrax)

where hax is the height of the tallest compacted cell being joined and h,,, is the average
height of the cells being joined.
Using this bound, cells are joined one-at-a-time using a penalty/reward scheme to deter-
mine if a pair of terminals is to be joined by stretching or by river routing.
Lim, Cheng, and Sahni [LIM93] have considered the case when only two cells are to be
joined. They develop fast polynomial time algorithms to obtain the minimum area join of
two cells. In addition, they are able to obtain, in low order polynomial time, minimum area
joins that minimize the length of the longest wire or the total wire length. Lim [LIM92]
has proposed an O(n(n/c)c-1) algorithm to find the minimum area join of c cells having a
total of n terminals. This algorithm does an exhaustive search over all possible numbers of
tracks in the c 1 routing channels between adjacent cells. A constraint graph is used to
determine the minimum height layout for each assignment of number of tracks to routing
channels. The time required per track assignment is O(n) and the worst case number of track



























(a) 1-layer (b) 2-layer

Figure 2: /-layer river routing



assignments is O((n/c)'-1). The algorithm of [LIM92] is flawed as it handles channels with
zero routing tracks by joining the adjacent cells using minimum height cell stretching and
then considers the joined cells as one. This problem is easily fixed, however, by combining,
in the constraint graph, pairs of vertices that represent corresponding terminals of the two
cells (i.e., i'th terminals of each cell) with zero routing tracks in between.
In this paper, we consider the case when 1 > 1 routing layers are available to river route
the inter cell connections. Note that while multiple layers do not affect layout area when
cell stretching alone is used, a reduction in area is possible when cell stretching is combined
with river routing or when river routing alone is used. We assume that in each layer of each
routing channel, the interconnects are to be accomplished using river routing. When the
number of layers available for river routing is increased, one may see a dramatic reduction
in the number of routing tracks needed per layer. Figure 2 shows an instance that needs n
tracks when routed in one layer but only one track/layer when routed in two layers.
We begin, in section 2, by developing a necessary and sufficient condition for a river
routing instance to be routable in 1 layers using at most t tracks per layer. This development
includes the formulation of two algorithms to perform /-layer river routing when such a
routing is possible. In Section 3, we describe the constraint graph used to determine minimum









height stretching of c cells. Heuristics for the minimum area joining of c cells are proposed in
Section 4 and the results of experiments with these are provided in Section 5. Our conclusions
appear in Section 6.



2 1-layer River Routing

Let (Ai, Bi), 1 < i < m, be a set of terminal pairs such that the Ai's are on one side (say left
or top) of a routing channel and the Bi's are on the other (right or bottom) side. Terminal
A, is to be connected to terminal Bi, 1 < i < m. For this channel routing instance to be an
instance of river routing, it must be the case that al < a2 < ... < am and bl < b2 < ... < bm
where ai and bi, respectively, give the positions of terminals Ai and Bi, 1 < i < m. We
may assume an underlying grid with each terminal being at a grid position. In the case of a
horizontal (vertical) channel the ais and bis are grid column (row) numbers. Leiserson and
Pinter [LEIS83] have obtained the following necessary and sufficient condition for a river
routing instance to be routable in a single layer using at most t > 0 tracks.

Theorem 1 [LEIS83] The river routing instance defined above is routable in a single i., I
using at most t > 0 tracks if and o,,li if
(a) ai+t bi > t
(b) bi+~ ai > t
for every i < m t.

For the general case of 1 > 1 layers, we obtain the necessary and sufficient condition of
Theorem 2.

Theorem 2 The river routing instance defined above is routable in 1 > 1 ,,, i (each 1, i;,
routing whole nets) using at most t > 0 tracks per ,,u, if and c,,lii if
(a) ai+lt bi > t
(b) bi+lt ai > t
for every i < m t.

Proof: First, we establish that (a) and (b) are necessary for 1 layer routing. Since the proofs
for (a) and (b) are similar, we provide that for (a) only. Suppose that ai+it b < t for some








procedure RoundRobin ;
{ Assign the m nets to 1 layers. }
begin
for i := 1 to m do
assign net (Ai, Bi) to layer (i mod 1) + 1 ;
end;

Figure 3: Round robin layer assignment

i. Consider the It + 1 terminal pairs (Aj, Bj), i at least one layer has to be assigned > t + 1 terminal pairs. So, suppose that terminal pairs
(A, B'), (A, B), ..., (A+, B'+), ... are assigned to the same layer for river routing. We
may assume that a' < a' < ... < a+1 and b < b' < ... < b+1. Since a'+ < a,+it and
bi > bi, a' b' < ai+lt b < t. From Theorem 1, it follows that the terminal pairs (A/, B),
1 be river routed on 1 layers. So, (Aj, Bj), 1 < j < m, cannot be river routed on 1 layers. As
a result, (a) is a necessary condition.
To show that (a) and (b) are sufficient conditions for routability, we present two algo-
rithms (RoundRobin and Greedy) that assign the nets to layers in such a way that each layer
is river routable when both (a) and (b) are satisfied. The correctness of these algorithms is
established in Theorems 3 and 4, respectively. O

Theorem 3 The 1,i', assignment produced by the RoundRobin procedure is river routable
if
(a) ai+it b > t
(b) bi+l ai > t
for all i < m It.

Proof: Let (A/,B.), (A/, B),... (Aj, B), (Aj+i, Bj+), (Aj+21, B+21), ... be the nets
assigned to layer (j mod 1) + 1, j < 1. So, a' = aj+(~ 1), and b' = bj+(~ _i). Hence,

a'+ bi = aj+(i+t-1), bj+(_i-)l

= aj+(i_l)l+tl bj+(i_l)l
> t (from (a))

Similarly, b'+t a' > t. So, the layer assignment satisfies the conditions of Theorem 1 and
is river routable using t tracks. o








procedure Greedy ;
{ Assign the m nets to 1 layers. }
begin
for i:= 1 to m do
assign net (Ai, Bi) to layer q such that q is the smallest integer < 1
for which the conditions of Theorem 1 are not violated on layer q
(if there is no such q, then fail) ;
end ;

Figure 4: Greedy layer assignment

Theorem 4 If (a) ai+lt b, > t and (b) bi+lt a, > t for all i < m It, then procedure
Greedy assigns nets to 1,11i, such that the assignment to each 1,.i, is routable using t tracks.

Proof: If procedure Greedy is able to assign each of the m nets to a layer, then the layer
assignments satisfy the conditions of Theorem 1 and so are routable using t tracks. Suppose
the algorithm fails while trying to assign net (A,, B,) to a layer. At this time nets (A,, Bi),
1 < i < r, have been assigned to layers so as to satisfy the conditions of Theorem 1 and the
assignment of net (A,, B,) to each of these layers violates these conditions. Consider first
those layers, L,, on which condition (a) is violated. For a layer s E L,, suppose that the
assigned nets are ..., (A _, BJ_), ..., (A_ 1, BJ_). Let (A, BJ) = (A,,B,). Since s E L,,
we have a b'_, < t. Now, if br-lt > b _, then a, b,_-I < t which violates condition (a)
of this theorem. So, b,_-t < b _,. Since, b,_-t < b' < ... < b -2 < b' _, t of the It 1 nets
(A.r-t+,, B,_-t+,), ..., (A,._, B,_1) have been assigned to layer s e La. Consequently, the
layers in L, account for tlL~, of these It 1 nets.
In a similar way, we can show that the remaining 1 IL, layers account for another
t(1 IL, ) of these nets. This gives us a total of tl nets, whereas we had only tl 1. This
contradiction implies that procedure Greedy cannot fail unless conditions (a) and (b) are
not satisfied. o


Procedure RoundRobin is easily seen to have complexity of O(m). A straightforward
implementation of procedure Greedy will have complexity of O(ml). However, by using
priority search trees [MCCR "] the complexity can be reduced to O(mlog ). In practice,
since 1 is quite small, it is unlikely that the priority search tree implementation will run faster
than the straightforward implementation in which the 1 layers are checked in sequence. The








procedure MinimizeTracks ; {or MinimizeLayers}
{ Determine the minimum number of tracks per layer (or minimum number
of layers) needed for multilayer river routing }
begin
t := 0 ; {or / := 1}
i:= 1 ;
while (i < m It) do
if (ai+l bi < t) or (bi+l ai < t)
then :=t + 1 {orl:= + 1}
else i := i+ 1 ;
end ;

Figure 5: Minimizing the number of tracks or layers

actual routing for all 1 layers can be done in O(mt) time using the computed layer assignment
and the single layer routing algorithm of [LEIS83].
Using Theorem 2, we can develop a linear time algorithm to determine the minimum
number of tracks needed to route an instance in 1 layers as well as to find the minimum
number of layers needed for a t track routing. The algorithm for the former is given in
Figure 5. This figure also shows the changes needed in case t is given and we wish to
determine the minimum number of layers. The correctness of the algorithm follows from
that of Theorem 2 and the fact that t (or 1) is increased only if the current t (1) is found to
be infeasible. The complexity is O(m) as neither i nor t (1) can exceed m. So, neither clause
of the if statement can be executed more than m 1 times.
The developments of this section allow us to trivially extend all the results of [LIM93] to
the case of multilayer joining of compacted cells. So, the multilayer minimum area join of
two compacted cells with m nets can be obtained in O(m2) time. If we wish to minimize the
maximum wirelength while keeping area minimum, the asymptotic time complexity is still
O(m2). The total wire length can be minimized while keeping area minimum in O(m2 log m)
time.



3 Constraint Graph Representation

Lim [LIM92] has proposed the use of a constraint graph to determine the terminal positions
in a row of compacted cells. This is for the case when the number of tracks in each routing










sink


(a) Cell joining with 2 layers


source


(b) Constraint graph representation

Figure 6: Constraint graph representation



channel is given and we wish to minimize the layout height. In the constraint graph, each
cell is represented by a directed chain of vertices. Each cell terminal is represented by a
vertex. The exception is when a compacted cell has terminals at the same y-position on
both sides of the cell. In this case, the two terminals at the same y-position are represented

by a single vertex. The vertex chain is linked in the direction of increasing y-position. The
chain edges are labeled by the minimum allowable terminal separation. In addition, the
constraint graph contains a source vertex that represents the bottom of the layout and a
sink vertex that represents the layout top. The source vertex connects to the bottom of each
chain and the top of each chain is connected to the sink vertex.
Figure 6(b) shows the chains (solid edges) for the four cell row of Figure 6(a). To complete

the constraint graph directed edges are added to introduce the channel routing constraints
of Theorem 2. These are represented by the broken edges of Figure 6(b). Figure 6(b) is for
the two layer case.










sink








1 2

S'' 1 1 '
b 14ej


3 di 2
,' 1 1 1 1'
a ) k() k






source

Figure 7: Merge in constraint graph



Lim [LIM92] has shown that the constraint graph is acyclic provided the number of tracks

in each routing channel is > 0. He has proposed handling channels with zero tracks by finding

first the minimum area joining of the adjacent cells (only cell stretching is permitted now)

and then combining these two cells into one. I.e., the two cells are replaced by their minimum

area join. This strategy can be shown to result in non optimality of the algorithm proposed

in [LIM92]. To preserve optimality, it is necessary to merge the vertices that represent

terminals that are the endpoints of nets that are to be routed using no tracks as in Figure 7.

The resultant constraint graph is also acyclic.

It is easy to see that the number of vertices and edges in the constraint graph is O(n)

where n is the total number of terminals. Furthermore, the graph can be constructed in

O(n) time given the number of routing layers and the number of tracks in each channel. The

constraint graph described by us is identical to that of [LIM92] except in the way channels

with zero tracks are handled and in that our graph is defined for 1 > 1 routing layers while








procedure Heuristicl ;
begin
for i := 1 to c 1 do
begin
determine the minimum area join of each pair of adjacent cells ;
select the pair that has minimum area and replace it with its
minimum area join ;
end;
end;

Figure 8: Heuristic 1

that of [LIM92] is only for 1 = 1.
The length of the longest path from the source vertex of the constraint graph to each of the
remaining vertices can be computed in O(n) time by doing this in topological order [HOR094,
Section 6.5]. It is easy to see that if each terminal is placed at a vertical position given by
the longest path length from the source, then all nets can be routed in the given number of
tracks (as the conditions of Theorem 2 are satisfied in each routing channel). Furthermore,
Lim [LIM92] has shown that such a positioning of terminals results in a stretched layout of
minimum height for the given channel widths. As a result, when channel widths are known,
cells can be stretched to minimize area in O(n) time. The channel widths that result in
minimum area can be determined in O(n(n/c)c-1) time where c is the number of cells by
trying out all possible channel widths [LIM92]. Since this is feasible only for small c, we
propose several heuristics in the next section.



4 Heuristics to Minimize Area

We formulate three greedy heuristics to obtain the minimum area join of a row of c compacted
cells that have a total of n terminals.


4.1 Heuristic 1

The heuristic is described in Figure 8.
In each iteration of the for loop we examine every pair of adjacent cells. For each pair,
the minimum area join is found using the algorithm of [LIM93] extended to the multilayer
case as discussed in Section 2. The pair which has the minimum area join is replaced by a









single cell that represents this join. So following each iteration of the for loop the number
of cells decreases by one. When the for loop terminates, we are left with a single cell that
represents the join of all c cells. The time needed to determine the minimum area join
of a pair of cells with n; nets between them is O(n2). The time to do this for all pairs
of adjacent cells is O(E n-;) = O(n2). So, the for loop iteration with i = 1 takes O(n2)
time. On subsequent iterations, only the two pairs that include the cell introduced in the
previous iteration need to have their minimum area join computed. Since each cell pair
being considered includes at least one composite cell, the minimum area join is computed
by considering the portion of the constraint graph that represents all the basic cells in
the cell pair. Channel widths for channels within a composite cell are not changed while
obtaining the minimum area join of the cell pair. However as different channel widths for
the channel between the two (composite) cells being joined are tried, the constraint graph
is used to determine the minimum height of the combined cell. So, the time to combine two
(composite) cells with n; terminals in the channel between them is O(nnr). Hence the time
for the remaining c 2 iterations is O(n ZC n;) = O(n2). The overall complexity of Heuristic
1 is therefore O(n2). In case the terminals are uniformly distributed over the cells, n; =
O(n/c) for all i. The time for the first iteration of the for loop is now O(n2/c) and that for
each of the remaining iterations is O(n2/c). The overall time is O(n2).

4.2 Heuristic 2

In this heuristic, we begin by assigning each channel the number of tracks needed to route the
channel with no cell stretching. This number can be determined in O(n;) time for a channel
with n; nets as described in Section 2. The time taken to do this for all c- 1 channels is O(n).
The configuration obtained in this way is the maximum width layout. Starting from this
configuration, we reduce the total number of tracks available across all c 1 channels by one
on each iteration. For this, the effect of a one track reduction is computed for each channel.
The minimum layout height is determined by computing the length of the longest path in
the constraint graph of Section 3. The track reduction is done in the channel that results in
the smallest layout height (hence the minimum area for the given number of tracks). The
algorithm is stated more formally in Figure 9.
When the algorithm terminates, A is the area of the minimum area join found by the









procedure Heuristic2 ;
begin
for each channel determine the number of tracks, ti,
needed to route with no stretching, 1 < i < c ;
c y-1 /
t := Flti;
set up the constraint graph using ti tracks in channel i, 1 < i < c;
compute layout area, A ;
for tracks := t downto 1 do {reduce by 1}
begin
for i := 1 to c 1 do
begin
reduce the number of tracks in channel i by 1 ;
modify the constraint graph to reflect this ;
determine the length of the longest path in the graph
and from this the layout area, a ;
end ;
select j such that aj = min{ ai } ;
reduce the number of tracks in channel j by 1 ;
A = min{ A, a } ;
end;
end;

Figure 9: Heuristic 2


heuristic. To reconstruct the layout, it is necessary to store the tracks per channel each time
A is updated in the statement A = min{ A, aj }.
For the time complexity, we see that the steps that precede the outer for loop take O(n)
time. Each iteration of the outer loop takes O(nc) time. Hence this loop contributes a total
of O(nct) to the time. Since t = O(n), the overall time complexity of Heuristic 2 is O(n2c).


4.3 Heuristic 3

Unlike Heuristic 2 which attempts to minimize the layout height for each value of t, the
total number of tracks, Heuristic 3 attempts to minimize the width (i.e., total number of
tracks) for each choice of layout height. The heuristic begins with a layout height, ht, equal
to the height of the tallest compacted cell. At each iteration, the next layout height to use
is computed as described later. During each iteration, cells are combined in groups of at
most k (k > 1 is a parameter to the heuristic). Each group of combined cells is replaced by
its minimum area join subject to the constraint that the height of the join does not exceed








procedure Heuristic3;
begin
ht := height of the tallest cell;
repeat { minimize width subject to height < ht }
repeat { do this by combining k cells at a time }
select k adjacent cells such that the minimum height cell is selected
and the height of the tallest selected cell is minimum
(if there are fewer than k cells, then select all of them) ;
obtain the minimum area layout for the selected cells under
the constraint that the layout height does not exceed ht ;
during the preceding step record the next value of ht
that is possible for a layout ;
until one cell remains ;
compute the area of the remaining cell and record it
if it is less than the minimum area found so far ;
if there is no next height then terminate;
ht := next height ;
until false;
end ;

Figure 10: Heuristic 3

ht. This joining of < k cells at a time continues until only one cell remains. Its area is
computed and recorded. The minimum area obtained over all heights tried is then reported
as the best. Heuristic 3 is given in Figure 10.
In our implementation of heuristic 3, the minimum area join of k cells is found by consid-
ering the portion of the constraint graph for all the basic cells included in these k cells. So,
for this purpose composite cells are not handled as single cells. Rather, as in Heuristic 1, the
basic cells they are composed of are considered and channel widths previously assigned to the
associated channels are not changed. Track assignment is done only for the k 1 channels
between the k composite cells. We found this to give better results than when composite
cells were regarded as atomic. For the case k = 2, the minimum area is determined by a
binary search over the number of tracks in the single channel. This takes O(n log n;) time
where n; is the number of nets in channel i. Thus the time needed for the inner repeat loop
when k = 2 is O(cn log n) (for uniform terminal distribution it is O(cn log(n/c)). During
the binary search, the heights corresponding to channel widths that require height > ht are
recorded. The minimum of these heights yields the next value of ht.
When k > 2, all track combinations for the k 1 channels are tried as in Section 3.









Again, each composite cell is broken up into its basic cells. As different track combinations
are tried, we record the minimum height > ht that results from any track combination. This
gives the next value of ht. The time for the inner repeat loop is O((c/(k 1))n(n/k)k-)
(or O((c/(k 1))n(n/c)k-1) when terminals are uniformly distributed).
In all our experiments, the outer repeat loop was iterated fewer than (k 1)n times. To
ensure that the number of iterations is O(kn), one may adopt the following scheme. When
the number of iterations first reaches (k 1)n, compute a set of at most n new heights by
beginning with the current constraint graph. This uses the current assignment of number of
tracks in each channel. Heuristic 2 is next used to reduce the total number of available tracks
by one and determine the height needed to complete the routing with the reduced number of
tracks. This process gives us at most n new heights hi < h2 < ... < hp. Heuristic 3 is now
resumed with hi as the next height. Only two iterations are performed. Then Heuristic 3
is resumed with max{ h2, ht } as the next height. Again two iterations of the outer repeat
loop are done. Next the heuristic is resumed with max{ h3, ht } as the next height. This
continues until we have gone through p resumptions of the heuristic. With this scheme to
limit the number of iterations, the complexity of Heuristic 3 becomes O(cn2 log n) when
k = 2 and O((c/(k 1))kn2(n/k)k-1) = O(cn2(n/k)k-1) when k > 2. For the case when the
n terminals are uniformly distributed over the c cells, the complexity is O(cn2 log(n/c)) when
k = 2 and O((c/(k 1))kn(n/c)k-1) = O(cn(n/c)k-1) when k > 2. One may verify that
since Heuristic 3 tries the maximum useful height (i.e., the height needed when no routing
tracks are available), it generates optimal solutions when k = c.



5 Experimental Results

We programmed our three heuristics as well as the heuristic Fang [CHEN89] in C and ran
tests on a single KSR processor. Optimal solutions for instances with upto nine cells were
obtained using the corrected version of the exhaustive search algorithm of [LIM92]. Our
test set consisted of instances that had a number of cells, c, equal to one of the numbers
in the set {3, ..., 9, 10, 20, 50, 100}. For each value of c, there were twenty instances and
the results were averaged over these instances. An instance with c cells had c 1 routing
channels. The number, t, of terminals on either side of each routing channel was equal to c









for 3 < c < 9 and was 10 for the other values of c. In addition, when c = 100, we also had
instances with 20 terminals on either side. In our experiments, we considered only single
layer and two layer routing.
Table 1 gives the average percentage by which the area of the single layer solutions
generated by each of the heuristics exceeded the area of the single layer optimal solution.
As is evident, each of the heuristics proposed in this paper gave noticeably better solutions
than did Fang. This table is only for the cases 3 < c < 9 as for c > 9 the optimal algorithm
of [LIM92] required too much time to complete.


t = number of terminals on each side of each routing channel


Table 1: Error rate ( ) over optimal, 1 = 1

In table 2, we have used the single layer solution produced by Fang as the benchmark
against which the solutions obtained by our three heuristics are compared. This table gives
the average percentage by which the area of the solutions produced by our heuristics is less
than that of the solutions produced by Fang. Our solutions have area 9 to 1i' less.
Table 3 compares the computing time requirements of the various algorithms for the case
of one layer. The optimal algorithm is useful only for small values of c (say upto 7). While
Fang is significantly faster than the heuristics proposed here, the quality of the solutions
generated by our heuristics is superior.
Table 4 is the analog of table 1 for the case of two layers. Again, our heuristics performed
considerably better than did Fang. Table 5 gives the improvement in area due to increasing
the number of routing layers from one to two. This is influenced somewhat by the width of
cells which in our case ranged from 5 to 30 times the track separation. With narrower cells,


cells t Fang Heuristic Heuristic Heuristic3
1 2 k=2 k=3 k=4
3 3 5.9 0.5 0.2 0
4 4 10.0 0.9 0.1 0 0
5 5 11.0 3.7 0.3 0.3 0.1 0.1
6 6 12.7 2.4 0.3 0.2 0.2 0.0
7 7 16.1 3.5 0.3 0.1 0.1 0.1
8 8 17.9 2.7 0.4 0.3 0.3 0.1
9 9 18.8 3.5 0.3 0.4 0.3 0.1




















Table 2: Improvement ( .) over Fang, 1 = 1


the impact of the second layer would have been greater and with wider cells, it would have
been less. Also, the impact of the second layer is more when more routing tracks are needed.
For the smaller instances of table 4, for example, the optimal solutions with 1 = 2 required,
on average, only 1.t' less area than when 1 = 1.
Table 6 is the analog of table 2 for the case of two layers. The results are similar to those
in table 2. Table 7 gives the average computing times for the two layer instances. These are
less than for the one layer case as the constraint graph has fewer edges.
For large c, we recommend the use of heuristic 2 or 3 (with k = 2) and for small c we
recommend using heuristic 3 (with k = 3 or 4).



6 Conclusion

We have developed necessary and sufficient conditions for multilayer river-routing. In addi-
tion, two simple algorithms to do multilayer river routing using the fewest number of tracks
were developed. Next, we considered the problem of joining a row of compacted cells and
developed heuristics to stretch cells and river-route the nets so that the layout area is min-
imized. Our proposed heuristic was compared, experimentally, with Fang [CHEN89] and
found to produce layouts with less area. However Fang is faster. We recommend the use of
our Heuristic 3 with k = 3 or 4 in practice.


cells t Heuristic Heuristic Heuristic3
1 2 k = 2 k =3 =4
10 10 9.4 14.0 14.0 14.2 14.3
20 10 10.5 15.4 15.4 15.6 15.6
50 10 10.6 16.1 16.1 16.2
100 10 9.1 16.5 16.5 16.6
100 20 9.3 18.3 18.0










cells t Fang Heuristic Heuristic Heuristic3 Optimal
1 2 k =2 k=3 '=4
3 3 0.0 0.01 0.02 0.02 0.01
4 4 0.0 0.02 0.02 0.04 0.09 0.05
5 5 0.0 0.03 0.04 0.09 0.36 1.2 0.77
6 6 0.0 0.03 0.08 0.21 0.66 3.7 13
7 7 0.0 0.04 0.14 0.50 3.4 27 278
8 8 0.0 0.07 0.24 0.84 4.4 34 1.8t
9 9 0.0 0.09 0.42 1.8 16 72.3 47t
10 10 0.0 0.14 0.78 3.1 19 334
20 10 0.01 0.32 5.8 16 117 1196
50 10 0.01 1.4 93 127 896
100 10 0.03 4.4 782 590 4873
100 20 0.06 11 3022 4087

Times are in seconds.
S: Times are in hours.


Table 3: Time taken, 1 = 1

References

[BOYE88] Boyer, D. G., Symbolic Layout Compaction Review, 25th Design Automation
Conference, June 1P'"' pp. 383-:;'1.

[CHEN89] Cheng, G and A. Despain, Fang: A Joiner for Compacted Cells, in VLSI 89,
1989, pp. 455-463.

[HORO94] Horowitz E. and S. Sahni, Fundatamentals of Data Structures in Pascal, 4th
Edition, W. H. Freeman and Company, 1994.

[LEIS83] Leiserson, C. E. and R. Y. Pinter, Optimal Placement for River Routing, SIAM
Journal on Computing, 1'-" ; pp. 447-462.

[LIM92] Lim, A., Efficient Algorithms for CAD in VLSI, PhD Dissertation, Univ. of Min-
nesota, 1992.

[LIM93] Lim, A., S. Cheng, and S. Sahni, Optimal Joining of Compacted Cells, IEEE Trans-
actions on Computer, Vol 42, No 5, May 1993, pp. 597-607.










cells t Fang Heuristic Heuristic Heuristic3
1 2 k = 2 k=3 3 =4
3 3 6.3 0.5 0 0
4 4 9.8 1.1 0.1 0 0
5 5 11.0 2.7 0.1 0.2 0.1 0.0
6 6 13.0 1.4 0.1 0.0 0.0 0.1
7 7 15.9 2.3 0.1 0 0 0
8 8 16.8 1.7 0.2 0 0.1 0.0
9 9 18.4 2.2 0.1 0.1 0.1 0.0


Table 4: Error rate ( .) over optimal, 1 = 2


cells t Fang Heuristic Heuristic Heuristic3
1 = k =2 k= 3 /k=4
10 10 4.1 4.7 3.3 3.3 3.0 3.0
20 10 4.1 5.8 3.4 3.4 3.3 3.2
50 10 4.6 5.2 3.4 3.2 3.2
100 10 4.7 6.3 3.5 3.4 3.4
100 20 7.1 10.4 4.8 5.2


Table 5: Improvement ( ) over 1 = 1 cases


[MCCRB,] McCreight, E. M., Priority Search Trees, SIAM J. Comput., Vol 14, No 2, May
1'-.K pp. 257-276.

[MIRZ87] Mirzaian, A., River Routing in VLSI, Journal of Computer and System Sciences,
Vol 34, 1987, pp. 43-54.

[PINT82] Pinter, R. Y., On routing two-point nets across a channel, 19th Design Automation
Conference, June 1982, pp. 894-902.

[PINT83] Pinter, R. Y., River Routing: Methodology and Analysis, Third Caltech Confer-
ence on VLSI, March 1'i" ;

[WEST81] Weste, N., Virtual Grid Symbolic Layout, 18th Design Automation Conference,
June 1981, pp. 225-233.
























Table 6: Improvement ( ) over Fang, 1


Times are in seconds.
t : Times are in hours.


Table 7: Time taken, 1 = 2


cells t Heuristic Heuristic Heuristic3
1 2 k = 2 _k =3 =4
10 10 10.0 13.3 13.2 13.2 13.3
20 10 12.2 14.8 14.8 14.9 14.8
50 10 11.2 15.1 14.9 15.0
100 10 10.7 15.4 15.4 15.4
100 20 12.5 16.3 16.3


cells t Fang Heuristic Heuristic Heuristic3 Optimal
1 2 k =2 k=3 '=4
3 3 0.0 0.01 0.02 0.02 0.0
4 4 0.0 0.02 0.02 0.03 0.06 0.04
5 5 0.0 0.02 0.02 0.07 0.25 0.75 0.60
6 6 0.0 0.02 0.04 0.13 0.38 2.0 11
7 7 0.0 0.04 0.06 0.29 1.9 14 223
8 8 0.0 0.06 0.12 0.53 2.5 19 1.5t
9 9 0.0 0.06 0.20 0.96 7.5 34 39t
10 10 0.01 0.09 0.32 1.7 9.9 153
20 10 0.01 0.24 2.7 9.9 65 651
50 10 0.01 1.1 42 Si 586
100 10 0.02 3.7 357 472 3069
100 20 0.05 7.9 1377 3166




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