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Title: Upgrading vertices in trees, series-parallel digraphs and general series-parallel digraphs to bound path length
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Title: Upgrading vertices in trees, series-parallel digraphs and general series-parallel digraphs to bound path length
Alternate Title: Department of Computer and Information Science and Engineering Technical Report
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Creator: Sahni, Sartaj
Paik, Doowon
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University of Minnesota
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Copyright Date: 1991
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Upgrading Vertices In Trees, Series-Parallel Digraphs And General

Series-Parallel Digraphs To Bound Path Length+



Doowon Paik Sartaj Sahni

University of Minnesota University of Florida

Abstract

We consider trees, series-parallel digraphs, and general series-parallel digraphs that have vertex

weights and delays. The length/delay of a path is the sum of the delays on the path. We show

that minimal weight vertex subsets X such that the length of the longest path is bounded by a

given value 6 when all vertices in X are upgraded to have delay 0 can be found in pseudo polyno-

mial time. In case all delays are unit or all weights are unit, our algorithms have a quadratic com-

plexity. For the case of trees with unit weights and unit delays, we develop a linear time algo-

rithm.






Keywords And Phrases

Vertex upgrading, bounding path length, trees, series-parallel digraphs, general series-parallel

digraphs


+ Research supported, in part, by the National Science Foundation under grant MIP 86-17374.









2



1. Introduction

Directed acyclic graphs (dags) can be used to model circuits. Each vertex of the dag has a delay

associated with. The delay of any signal path is the sum of the delays of the vertices on the path.

The delay of a circuit is the length of the longest path in the dag. In an effort to reduce circuit

delay, we may attempt to upgrade some of the dag vertices. Upgrading a vertex reduces its delay

but comes at some cost. In this paper we consider an extreme form of upgrade that results in the

vertex delay being negligible (i.e., zero) relative to the delay before upgrading. For each vertex v

of the dag G = (V, E) we use d(v) to denote its delay before upgrade and w(v) to denote the cost of

the upgrade.

Let d(G) be the length of the longest path in G. Let GX be the graph that results when the

vertices in X are upgraded. Let d(GX) be the length of the longest path in G[X. The dag vertex

upgrade problem (DVUP) is that of finding a minimal weight vertex set X such that d(G )) < ,

where 6 is an input parameter (Figure 1).

(5,3) (0,3)

(2,2)_ .Q (2,2)O
u1 05 1 05

(3,4) (3,4)
2 4 2 4
(1,3) (3,2) (1,3) (0,2)


(a) G with 6 = 6 (b)X = {3, 4} is the minimal solution
d(GC) = 6


Figure 1: DVUP example (the first(second) coordinate is the delay(weight) of the vertex).














Throughout this paper, we assume that the weights, delays, and 6 are nonnegative integers.

In [PAIK91], we showed that:

(1) DVUP is NP-hard for directed chains.

(2) DVUP can be solved in O(n3logn) time for unit weight unit delay dags (n is the number of

vertices in the dag).

(3) DVUP is NP-hard for dags with unit weight but non unit delays when 6 2 2 and is solvable

in O(n2) time when 6 = 1.

In addition, we proposed a backtracking algorithm and several heuristics for general dags.

The related problems of vertex splitting and vertex deletion to bound path lengths were studied

by Paik, Reddy, and Sahni in [PAIK90ab].

In this paper we consider the DVUP for trees, series-parallel dags (SPDAGs) and general

series-parallel dags (GSPDAGs). The results we obtain are:

(1) An O(n) time algorithm for unit weight unit delay trees (Section 2).

(2) Pseudo polynomial time algorithms for trees, SPDAGs and GSPDAGs with general weights

and delays (Sections 2, 3, 4). These algorithms have complexity O(n2) when either 6 = O(n),

all weights are unit, or all delays are unit (note that if all delays are unit, then d(G) < n and

so 8 < n for the problem to be nontrivial).














2. Trees



2.1. Trees With Unit Weight And Unit Delay

When the dag is a rooted tree T such that w(v) = d(v) = 1 for every vertex, the minimum weight

vertex subset X such that d(T[X) 6 can be found in O(n) time by computing the height, h, of

each vertex as defined by:

( 1 v is a leaf
h(v) = 1+ max {h(u) | u is a child of v}, otherwise


X is selected to be the set:


X= {v I h(v) > 6}


The vertex heights can be computed in O(n) time by a simple postorder traversal of the tree T

[HORO90]. The correctness of the procedure outlined above is established in Theorem 1. Note

that when all vertices have unit weight, the weight of a set, Y, of vertices is simply its cardinality

\Y\.


Theorem 1: For any tree T let h(v) be the height of vertex v. The set


X= {v I h(v) > 6}


is a minimum cardinality vertex set such that d(TX[) < 6.

Proof: The fact that d(T[X) 6 is easily seen. The minimality ofX is by induction on the

number, n, of vertices in T. For the induction base, we see that when n = 0, JX = 0 and the

theorem is true. Assume the theorem is true for all trees with < m vertices where m is an arbitrary

natural number. We shall see that the theorem is true when n =m + 1. Consider any tree with n =

m + 1 vertices. Let X be the set of all vertices with height > 8. If AIX = 0, then X is clearly a














minimal set with d(T[Y) < 6. Assume |AX > 0. In this case the root, r, of T has h(r) > 6 and so is

in X. First, we show that there is a minimal vertex set W that contains r and for which d(TIW) <

6. Let Z be a minimal vertex set for which d(TIZ) I 6. If r Z, then let r, u1, u2, ..., h(r)- be a

longest root to leaf path in T. Since h(r) > 6, at least one of the u,'s is in Z. Let u, be any one of

the u,'s in Z. Let W = Z + {r} {u,}. Clearly, | W = |Z|. Since all root to leaf paths that include

u, also include r, the length of these paths in TIW is the same as in TIZ. The length of the remain-

ing paths in TIW is no more than in TIZ. So, W is a minimal cardinality vertex set such that

d(TIW) < 6 and furthermore W contains the root r.

Let A(v), A e {X, W}, denote the subset of A that consists only of vertices in the subtree,

T(v), rooted at v. Since d(T(v)X[(v)) 6, d(T(v)IW(v)) 6, and I T(v)l m for each v that is a child

of r, it follows from the induction hypothesis that |X(v)| = I W(v)| for each v that is a child of r.

Hence, A = 1 + | IX(v)l= I W. D
v is a child of r



2.2. General Trees

Since a chain is a special case of a tree and since DVUP for chains with arbitrary weights and

delays is known to be NP-hard [PAIK91], we do not expect to find a polynomial time algorithm

for general trees. In this section we develop a pseudo polynomial time algorithm (i.e., one whose

complexity is polynomial in the number of vertices and the actual values of the vertex delays and

weights). We modify the definition of height used in the preceding section to account for the ver-

tex delays. We use H to denote this modified height.

( d(v) v is a leaf
(v) d(v) + max {H(u) | u is a child of v}, otherwise


For each vertex v, let L(v) be a set of pairs (1, c) such that there is a subset X c T(v) such

that d(T(v)[Y) = 1 and I w(u) = c. Let (l/, cl) and (12, c2) be two different pairs such that 1i <
u X















12 and cl, c2. In this case, pair (li, cl) dominates (12, c2). Let S(v) be the subset of L(v) that results

from the deletion of all dominated pairs. Let S(r) be this set of dominating pairs for the root r of

T. Let (1', c') e S(r) be the pair with least cost c'. It is easy to see that the least weight vertex set

W such that d(TIW) < 6 has weight c'. We shall describe how to compute S(r). Using the back-

trace strategy of [HOR078, cf chapter on dynamic programming] we can compute the W that

results in d(TIW) 6 and Y w(u) = c' in less time than needed to compute S(r) (however, S(r)
u W

and some of the other S's computed while computing S(r) are needed for this).

For a leaf vertex v, S(v) is given by:

{ (0, w(v)) } d(v)> 6
S(v) ( 0, w(v) ), (d(v), 0) d(v) <6


For a non leaf vertex v, S(v) may be computed from the S(u)'s of its children ul, ... uk

First, we compute U(v) as the set of nondominated pairs of the form (1, c) where 1 = max { /, 12,
k,
k } and c = C c, for some set of pairs (l,, c,) e S(u,), 1 < i < k_. Let V(v) and Y(v) be as
S=1

below:

(v)= { (1, c+w(v)) (1, c) e u(v)}

Y(v) = { (l+d(v), c) | l+d(v) < 6 and (1, c) e U(v) }


Now, S(v) is the set of nondominated pairs in V(v) u Y(v). Since S(v) contains only nondom-

inated pairs, all pairs in S(v) have different / and c values. So, IS(v)l min { 6, o } where ( =
n
I w(u). Using the technique of [HOR078], S(v) can be computed from the S(u)'s of its children
u=1

in time O(min{6,mc*k,). To compute S(r) we need to compute S(v) for all vertices v. The time

needed for this is O(min{6,m} *k,) = O(miin{6,(o}*n)














Note that for unit delay trees, 6 < n and for unit weight trees (o = n. So in both of these

cases the procedure described above has complexity O(n2).



3. Series-Parallel Dags

A series-parallel digraph, SPDAG, may be defined recursively as:

1. A directed chain is an SPDAG.

2. Let s, and t, respectively, be the source and sink vertices of one SPDAG G, and let s2 and t2

be these vertices for the SPDAG G2. The parallel combination of G, and G2, G1/G2, is

obtained by identifying vertex s, with s2 and vertex tj, with t2 (Figure 2(c)). G/1G2 is an

SPDAG. We restrict G, and G2 so that at most one of the edges < s,t>, < s,t> is present.

Otherwise, G//G2 contains two copies of the same edge.

3. The series combination of G, and G2, GIG2, is obtained by identifying vertex t1 with s2 (Fig-

ure 2(d)). GIG2 is an SPDAG.

The strategy we employ for SPDAGs is a generalization of that used in Section 2.2 for trees

with general delays and weights. Let s and t, respectively, be the source and sink vertices of the

SPDAG G. Let D(l,Y,G) be a minimum weight vertex set that contains the vertices in Y, Y c {s,

t}, and such that d(GID(1,Y,G)) 1 and let f(G) be as below:

f(G)= {(l,c,Y) | 0 l 6, c = w(u)}.
ueD(I,Y,G)

Let (1 ,cl ,Yi) and (12 ,c2 ,Y) be two different triples in f(G). (i ,cl ,Yi) dominates (2 ,c2 ,Y2)

iff /, < 12, c, c2 and Yi = Y2. Let F(G) be the set of triples obtained by deleting all dominated tri-

ples off(G). If (l',c',Y') is the least weight triple (i.e., the one with least c) in F(G) then the least

weight W such that d(Gii ) < 8 has weight c'. We shall show how to compute F(G) and hence

(l',c',Y'). The actual W may be obtained using a backtrace step as described in [HOR078].















S1 t 2 G, t

0< G, >o~ C< G o G




(a) G, (b) G2 (c) G1/G2


s' G2l t2






(d) GG2


Figure 2: Series-parallel digraph.


3.1. G Is A Chain

Consider the case when G has only two vertices s and t. F(G) is constructed using the code:

F(G) := { (0, w (s)+w(t), {s, t}) }

if d(s) < 6 then F(G) := F(G) u { (d(s), w(t), {t}) }
if d(t) < 6 then F(G) := F(G) u { (d(t), w(s), {s}) }

if d(s)+d(t) < 6 then F(G) := F(G) u { (d(s)+d(t), 0, 0) }

When G is a chain with more than two vertices, it may be regarded as the series composition of

two smaller chains G, and G2. In this case F(G) may be constructed from F(GI) and F(G2) using

the algorithm to construct F(G1G2) described in the next section.














3.2. G is of the form G1G2


The following lemma enables us to construct F(G G2) from F(GI) and F(G2).




Lemma 1: If(l, c, Y) e F(G1G2), then there is an (i ,cl ,Y1) e F(Gi) and an (2 ,c2 ,2) e F(G2) such

that

(a) D(llY ,1 ,) = D(I,Y,GIG2) rn V(G1)


(b) D(12,Y2 ,G) = D(l,Y,GIG2) n V(G2)


(c) D(I,Y,GG2) =D( ,Y1 ,G) u D(12 ,Y2 ,G2)


(d) c = I W(U), c1 = I W(U), C2 = C W(U)
u e D(1,Y,GIG2) u e D(1,YT,GI) u e D(1,YT,G2)

Proof: Let A = D(l,Y,GIG2) rn V(Gi) and B = D(l,Y,GIG2) n V(G2). Since V(GiG2) = V(Gi) u

V(G2), D(1,Y,G G2) = A uB. Let s, and t,, respectively, denote the source and sink vertices for G,, i

e { 1, 2}. s and t are the corresponding vertices for GiG2. We see that s = sl, t = t2, and t1 = s2.


Case 1 : [ t D(1,Y,GG2) ]

Let 11 = d(GiIA), Y1 =A r {sj}, and c, = w(u). We shall show that (li ,c ,Y1) e F(Gi). Sup-
u A

pose it is not. Then it must be dominated by a triple (l'j ,c' ,Y1) that is in F(GI). Let C =

D(l'I,Y1,G,). It follows that c', = w(u) and that (l',+d(G2IB)-d(s2), c'+ I w(u), Y) dominates
ueC ueB

(li+d(G2IB)-d(s2), ci+ I w(u), Y) (l, c, Y). So, (1, c, Y) V F(GIG2). This contradicts the
ueB

assumption on (1, c, Y). Hence, (1i ,c ,Yj) e F(G1). Similarly, (/2 ,c2 ,Y2) F(G2). (a) (d) are an

immediate consequence as D(l1 ,Y, ,Gl) = A and D( ,Y2 ,G2) = B.














Case 2 : [ tj E D(l,Y,G1G2)

This is similar to case 1 D

Lemma 1 suggests the following approach to obtain F(GIG2) from F(GI) and F(G2):

Stepi: Construct a set Z of triples such that F(GIG2) c Z. This is obtained by combining

together all pairs of triples (1i ,cl ,Y) e F(Gi) and (/ ,c2 ,Y) E F(G2).

Step2: Eliminate from Z all triples that are dominated by at least one other triple of Z.

The triples (/i ,cl ,Y1) and (12 ,c2 ,Y) are compatible iff( tj e Yi and s2 e Y2) or ( tj v Yi and

s2 Y2 ). Only compatible triples may be combined. Assume that we are dealing with two com-

patible triples. We first obtain the triple (1, c, Y) as below:

if t1 E Y1

then (1, c, Y) := (11+12, CI+c2-w(tl), Y1uY2-{t1})

else (1, c, Y) := (l,+12-d(tl), c1+c2, Y1uY2-{t })

Next, (1, c, Y) is added to Z provided I 6.


3.3. G = G1/G2

When G = G1/G2 we use Lemma 2 which is the analogue of Lemma 1.



Lemma 2: If (, Y) e F(G1/G2), then there is an (1i ,cj ,Y1) e F(Gi) and an (2 ,c2 ,2) e F(G2)

such that

(a) D( ,Y1 ,Gi) = D(I,Y,GiG2) n V(Gi)


(b) D(12,Y2,G2) = D(l,Y,GIG2) rn (G2)















(c) D(l,Y,GG2) = D(l ,Y, ,G1) U D(12 ,Y2 ,G2)


(d) c = I W(u), C1 = I w(u), c2 = w(u).
u e D(I,Y,GIG2) u e D(1,YI,GI) u e D(12,Y2,G2)

Proof: Similar to that of Lemma 1. D

To obtain F(G//G2) from F(G1) and F(G2) we use the two step approach used to compute

F(G1G2). For step 1, we compute the triple (1, c, Y) obtained by combining (1l ,c1 ,Yl) E F(G1) and

(12 ,c2 ,Y2) E F(G2). The triples are compatible iff Yi = Y2. Again, only compatible triples may be

combined. For compatible triples, (1, c, Y) is obtained as below:

S:= max {11, 12}

1 := C + C2 + W(c)
u E Yl

Y:= Y


Next, (1, c, Y) is added to Z.



3.4. Complexity

The series-parallel decomposition of an SPDAG can be determined in O(n) time [VALD79]. By

keeping each F(G,) as four separate lists of triples, one for each of the four possible values for the

third coordinate of the triples, F(GIG2) and F(G//G2) can be obtained in O( F(G1)| + F(G2)| ) time

from F(GI) and F(G2). Since F(GI) (F(G2)) contains only non dominated triples, it can contain at

most four triples for each distinct value of the first coordinate and at most four for each distinct

value of the second coordinate (these four must differ in their third coordinate). Hence, IF(GI)|

4*min { 6 + 1, 1 w(u) } and |F(G2)| < 4*min { 6 + 1, w(u) }. So, we can obtain F(G)
u e V(GI) u e V(G2)

for any SPDAG in time O(n*min {6, i w(u)} ). For SPDAGs with unit delay or unit weight,
u e V(G)


this is O(n2).














4. General Series Parallel Dags

General series parallel dags (GSPDAGs) were introduced in [LAWL78, MONM77, SIDN76]. A

linear time algorithm to determine whether or not a given dag is a GSPDAG was developed in

[VALD79]. This paper also contains a linear time algorithm to obtain a series-parallel decompo-

sition of a GSPDAG. The definitions and terminology used in this section are derived from

[VALD79].

A transitive dag is a dag G = (V, E) such that < ij> e E whenever there is a path from i to

j. The transitive closure, G+ = (V, E+) of the dag G = (V, E) is the transitive dag with E c E+ and

IEl is minimum. An edge < ij> of the dag G is redundant iff there is an i to j path in G that

does not include . A dag is minimal iff it contains no redundant edges. The transitive

reduction, G- = (V, E-) of the dag G = (V, E) is the minimal dag that has the same transitive clo-

sure as G and such that E- c E.

The class of minimal series parallel dags (MSPDAGs) is defined recursively as below:

1. If l0 = 1 and I|E = 0, then G is an MSPDAG

2. If G, = (V1, E1) and G2 = (V2, E2) are MSPDAGs, then their parallel composition Gi//G2 = (V

u V2, El u E2) (Figure 3 (a)) is an MSPDAG.

3. If G, and G2 are as above, then their series composition GIG2 = (V 2 u V2, E u E2 u E3)

where E3 = {< ij> I i V, dout(i) = 0, j e V2, d n(j) = 0 }, dout(i) and d'n(j) are, respec-

tively, the out and in degrees of vertex i (Figure 3(b)) is an MSPDAG.

Finally, a GSPDAG is a dag whose transitive reduction is an MSPDAG. The dag of Figure 4 is a

GSPDAG as its transitive reduction is the MSPDAG of Figure 3(b). It is interesting to note that

every dag which is a tree is an MSPDAG. However the only dag trees that are SPDAGs are

chains.






















G, G2


(a) G1//G2


Figure 3: Parallel and series combination of G1 and G2.


Figure 4: An example GSPDAG.

The following lemma shows that the DVUP solutions for a GSPDAG G and for its transi-

tive reduction MSPDAG G- are the same.



Lemma 3: Let G = (V, E) be a GSPDAG and let G- = (V, E-) be its transitive reduction (hence

G- is an MSPDAG). LetX be a subset of V. Then, d(G K) < 6 iffd(G-C ) < 6.

Proof: If d(G X) < 6 then d(G-CX) 6 as E- c E. If d(G[X) > then there is a path P = vv2 ... Vq

in G such that its delay in GKX is A = I d(v,) and A > 6. From the v, to vq path P of G we
1 < < n,v,EX

can construct a vi to v, path Q in G- by replacing each edge < v,,v+,,> V E- by a path from v, to

v,,, in G-. Since G- is the transitive reduction of G, such a path exists. When all edges < v,,v,+1>


(b) G1G2














SE- have been so replaced, we obtain the desired path Q. Clearly, the delay of Q is 2 A. Hence,

if d(G[Y) > 6 then d(G-Y) > 6. This completes the proof. D

As a consequence of Lemma 3 the DVUP for GSPDAGs can be solved using the steps:

Stepi: Compute G-, the transitive reduction of the GSPDAG G.

Step2: Let X be the minimum weight vertex subset such that d(G-~X) < 6.

Step3: OutputX.

Since the transitive reduction of a GSPDAG G can be obtained in time linear in the number

of vertices and edges in G [VALD79], we need be concerned only with step 2. Our strategy for

this is similar to that used in Section 3 for SPDAGs. However, since the source and sink vertices

of G, and G2 remain distinct following a series or parallel combination, we can deal with tuples

(1, c) rather than with triples (1, c, Y). Corresponding to D(1 ,Y ,G) of SPDAGs, we define D(1 ,G)

for an MSPDAG G to be a minimum weight vertex set for which d(GX[) 1. We use the series

parallel decomposition of G and begin with the non dominated tuple sets F(G) for each of the

vertices in G. Then using the series and parallel combinations that result in G we obtain F(G).


4.1. G Is A Single Vertex

If G is the single vertex v, then F(G) = { (0, w(v)) } when d(v) > 6 and { (0, w(v)), (d(v), 0) }

when d(v) < 6.



4.2. G Is Of The Form GiG2

The following lemma is the analogue of Lemma 1.


Lemma 4: If(l, c) e F(G1G2), then there is an (1i ,c) e F(Gi) and an (12 ,C2) e F(G2) such that














(a) D(l1 ,G1) =D(1,G1G2) r V(G,)

(b) D(12 ,G2) = D(1,G1G2) n V(G2)


(c) D(1,G1G2) =D(1 ,G) U D(12,G2)

(d) c = I W(u), c1 = I W(u), c2 = W(u).
u e D(I,Y,GIG2) u e D(I1,YI,GI) u e D(12,Y2,G2)

Proof: Similar to that of Lemma 1. D

For the case of MSPDAGs, all pairs (li ,cl) e F(GI) and (12 ,c2) E F(G2) are compatible pairs

for combination. The pair (1 ,c) that results from combining these two pairs is (1+12 ,c1+c2). This

is added to Z (see step 1 of the procedure for G1G2 in Section 3.2) provided 1+1/2 6.



4.3. G = G//G2

The analogue of Lemma 2 that applies to MSPDAGs is given below.



Lemma 5: If(l, c) e F(G1/G2), then there is an (1i ,c) e F(G1) and an (2 ,c2) e F(G2) such that

(a) D(1 ,G,) =D(l,G1G2) r V(G,)

(b) D(12 ,G2) = D(1,G1G2) n V(G2)

(c) D(1,G1G2) =D(1 ,G) u D(12 ,G2)

(d) c = W(U), C1 = I W(u), C2 = W(M).
u e D(1,Y,GIG2) u e D(1,Y,,GI) u e D(12,Y2,G2)

Proof: Similar to that of Lemma 1. D

Once again, all pairs (l1 ,cl) e F(G1) and (2 ,c2) E F(G2) are compatible pairs for combina-

tion. The result of combining these two pairs is the pair (1 ,c) = (max {l1,12, c1+c2 ).














4.4. Complexity

While the algorithm for MSPDAGs is simpler than that for SPDAGs, its asymptotic complexity

is the same. Note that for trees with unit weight and/or unit delay the complexity is O(n2) which

is asymptotically the same as the algorithm of Section 2.2 for unit weight or unit delay trees but

inferior to the algorithm of Section 2.1 for unit weight and unit delay trees.



5. Conclusions

We have obtained pseudo polynomial time algorithms for DVUP for trees, series-parallel dags,

and general series-parallel dags. These algorithms have quadratic complexity when all vertices

either have unit delay or unit weight. For 6 = O(n), the complexity is O(nS). For the case of trees

with unit weights and unit delays we have developed a linear time algorithm.



6. References

[HOR078] E. Horowitz, and S. Sahni, "Fundamentals of Computer Algorithms", Computer

Science Press, Maryland, 1978.

[LAWL78] E. L. Lawler, "Sequencing Jobs To Minimize Total Weighted Completion Time

subject to precedence constraints", Annals ofDiscrete Math. 2, 1978, 75-90.

[MONM77] C. L. Monma and J. B. Sidney, "A General Algorithm For Optimal Job Sequenc-

ing With Series-Parallel Constraints", Technical Report No. 347, School of Opera-

tions Research and Industrial Engineering, Cornell University, Ithaca, N.Y., July

1977.

[PAIK90] D. Paik, S. Reddy, and S. Sahni, "Vertex Splitting In Dags And Applications To

Partial Scan Designs And Lossy Circuits", University of Florida, Technical

Report, 90-34,1990.














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