An Efficient Algori thm for Edge-Ranking Trees

Xiao Zhou and Takao Nishizeki *

Department of System Information Sciences Graduate School of Information Sciences Tohoku University, Sendai 980-77, Japan

Abs t rac t . An edge-ranking of an undirected graph G is a labeling of the edges of G with integers such that all paths between two edges with the same label i contain an edge with label j > i. The problem of finding an edge-ranking of G using a minimum number of ranks has applica- tions in scheduling the manufacture of complex multi-part products; it is equivalent to finding the minimum height edge separator tree. Deogun and Peng and independently de la Torre et al. have given polynomial- time algorithms which find an edge-ranking of trees T using a minimum number of ranks in time O(n 3) and O(n 3 logn) respectively, where n is the number of nodes in T. This paper presents a more efficient and simple algorithm, which finds an edge-ranking of trees using a minimum number of ranks in O(n 2) time.

1 I n t r o d u c t i o n

An edge-ranking of a graph G is a labeling of edges of G with positive integers such that all paths between two edges with the same label i contain an edge with label j > i [DP90, IRV91]. The integer label of an edge is called the rank of the edge. The minimum number of ranks needed for an edge-ranking of G is called the edge-ranking number and denoted by X~r(G). An edge-ranking of G using X~r(G) ranks is called an optimal edge-ranking of G. The edge-ranking problem is to find an optimal edge-ranking of a given graph G. The problem seems to be NP-complete in general [dlTGS93]. The problem has applications in scheduling the manufacture of complex multi-paxt products; it is equivalent to finding the minimum height edge separator trees of G. The edge-ranking of trees, in particular, can be used to model the parallel assembly for a product from its components in a quite natural manner [BDJ+94, dlTGS93, IRV88a, IRV88b, IRV91]. Fig. 1 depicts an optimal edge-ranking of a tree using four ranks.

An edge-coloring of a graph G is to color all the edges of G so that no two ad- jacent edges axe colored with the same color. By the definition of edge-rankings, two edges which are incident to the same vertex cannot have the same rank, and hence any edge-ranking of a graph is also an edge-coloring. This implies that the maximum vertex degree of a graph G is a lower bound on the edge-ranking number x~(G) of G [DP90, IRV91]. Furthermore it is known that log 2 n _< X~r(G) for any connected graph G of n vertices [IRV91].

* E_maih (zhoulnishi)@ecei.tohoku.a~.j p

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Fig. 1. An optimal edge-ranking of a tree T.

The node-ranking problem of a graph G is defined similarly [IRV88a, KMS88]. A node-ranking of G is a labeling of vertices of G with integers such that all paths between two vertices with the same label i contain a vertex with label j > i. The node-ranking problem is to find a node-ranking of a given graph G using the minimum number of ranks. The node-ranking problem is NP-complete in general [BDJ+94, Pot88]. On the other hand Schaffer has presented a linear algorithm to solve the node-ranking problem for trees [Sch89]. Very recently Bodlaender et al. have given a polynomial time algorithm to solve the node-ranking problem for graphs with treewidth at most k for fixed k [BDJ+94].

Deogun and Peng [DP90] have given the first polynomial-time algorithm to solve the edge-ranking problem for trees, which takes O(n 3) time. Their algo- rithm consists of the following two steps: (1) transform the edge-ranking prob- lem for trees into the equivalent node-ranking problem for line-graphs of trees which they call "clique-trees'; and (2) find a node-ranking of clique-trees with a minimum number of ranks. Recently de la Torre et al. have rediscovered a polynomial-time algorithm which solves the edge-ranking problem for trees by means of a complicated greedy procedure, but the time complexity is O(n 3 log n) [dlWGS931.

In this paper we give a more efficient and simple algorithm which solves the edge-ranking problem for trees in O(n 2) time. Our algorithm improves the known algorithms [DP90, dlTGS93] in time-complexity and simplicity. We di- rectly solve the edge-ranking problem for trees without transforming it to the node-ranking problem for cliques trees or using a complicated greedy procedure. More precisely, we find an optimal edge-ranking of a given tree T by means of "bottom-up tree computation" on T: we find an optimal edge-ranking of a subtree rooted at an internal node of T from those of the subtrees rooted at its children.

2 P r e l i m i n a r i e s

In this section we define some terms and present easy observations. Let T = (V, E) denote a tree with node set V and edge set E. We often denote by V(T)

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and E(T) the node set and the edge set of T, respectively. We denote by n the number of nodes in T. T is a "free tree," but we regard T as a "rooted tree" for convenience sake: an arbitrary node of tree T is designated as the root of T. We will use notions as: root, node, internal node, child and leaf in their usual meaning. An edge joining nodes u and v is denoted by (u, v). The degree of node v is denoted by d(v). A sum T + T t of two trees T and T ~ is a graph (V(T) U V(T') , E(T) U E(T')) .

The definition of an edge-ranking immediately implies the following lemma [DES0, IRV91].

L e m m a 1. Any edge-ranking o/ a connected graph labels exactly one edge with the largest rank.

Let ~ be an edge-ranking of a tree T. The number of ranks used by ~ is denoted by # ~ . One may assume w.l.o.g, that ~ uses the ranks in [1, #~]. An edge e of T and rank ~(e) are visible (from the root with respect to ~p) if all the edges (except e) in the path from the root to e have ranks smaller than ~(e). Define L(~) as

i ( ~ ) = {~(e) [ e e E is visible},

and call L(~) the list of an edge-ranking ~ of the rooted tree T. Obviously #V E L(~) and ~(e) E L(~) for every edge e incident to the root. The ranks in the list L(~) are sorted in decreasing order. Thus the edge-ranking ~ in Fig. 1 has the list L(~) = {4, 3, 1}.

We define the lexicographical order -~ on the set of decreasing sequences (lists) of positive integers as follows: let A = {a l , . . . , ap} and B -- { b l , . " , bq} be two sets (lists) of positive integers such that al > ..- > ap and bl > . . . > bq, then A -~ B if and only if there exists an integer i such that

(a) aj = b i for all 1 < j < i, and (b) either ai < bi or p < i < q.

We write A _ B if A = B or A -~ B. For integers c~ and/? , a < /? , we denote by [a, fl] the set of integers between c~ and/~, that is, [a, fl] = {c~, a + 1 , - . . , fl}. If c~ > /~, then [~,/?] = r Obviously the following (1)-(3) hold for any integer a > l :

i f A _ B, then A - [1,~] ~ B - [ 1 , c~]; (1)

if A _ B and A - [1, c~] = B - [1, ~], then A n [1, a] _ B f'l [1, cr]; and (2)

if A -~ B and A - [1, c~] = B - [1, c~], then A t3 [1, a] -~ B n [1, cr]. (3)

An edge-ranking ~ of T is critical if L(~) _ L(~/) for any edge-ranking of T. The optimal edge-ranking depicted in Fig. 1 is indeed critical. The list of a critical edge-ranking of T is called the critical list o] tree T and denoted by L* (T). Thus L* (T) corresponds to an equivalent class of optimal edge-rankings of T. Since any critical edge-ranking is optimal, it suffices to find a critical edge- ranking of T.

The subtree of T rooted at node v is denoted by T(v). Let e = (u, v) be an edge in T such that v is a child of u. Then the tree obtained from T(v) by adding

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e is called the subtree of T rooted at edge e and is denoted by T(e). We denote by T - T(e) the tree obtained from T by deleting all edges in T(e) and all nodes in T(e) except u.

For an edge-ranking ~ of tree T and a subtree T ~ of T, we denote by ~[T ' a restriction of ~ to E(T ' ) : let ~ ' = ~]T' , then ta'(e) = !a(e) for e e E(T' ) . By Lemma 1 T has exactly one edge e with the largest rank # ~ . Denote by T ( # ~ ) the subtree T(e) rooted at the edge e. Let T ' = T - T ( # ~ ) , then

L(~) = { # ~ } U L(~]T ' ) . (4)

If ~ is a critical edge-ranking of T, then by (2) and (4) ~ IT ' is a critical edge- ranking of T ~. Let T ~ be the tree obtained from T(e) by deleting e. I f ~ is an optimal edge-ranking, then

X'~(T'), X'~(T") < # ~ = X',(T). (5)

3 O p t i m a l E d g e - R a n k i n g

The main result of this paper is the following theorem.

T h e o r e m 2. An optimal edge-ranking of a tree T can be found in O(n 2) time where n is the number of nodes in T,

In the remaining of this section we give an O(n 2) algorithm for finding a critical edge-ranking of a tree T. Our algorithm uses the technique of "bo t tom-up tree computat ion." For each node u of a tree T, we construct a critical edge-ranking of T(u) from those of u's children. This can be done in t ime O(nd(u)), and hence the total running time is O(n2), as shown later.

We first have the following lemma.

L e m m a 3. If a tree T has two edge-disjoint subtrees Ta and Tb with X~r(Ta) = X~r(Tb) = c, then c + 1 < x~(T).

Proof. Clearly c < X'~(T). Suppose that X~(T) = c. Then by Lemma 1 T has exactly one edge e with rank c in an optimal edge-ranking of T. Remove the edge e from T, and let T ' and T H be the resulting two trees. Then by (5) X~r(T'), X~r(T '') < c. Since Ta and Tb are edge-disjoint, either Ta or Tb, say Ta, does not contain e. Then Ta is a subgraph of either T ~ or T ' , say T ~, and hence

! X~r(Ta) < X~(T ') < c, contrary to the assumption xr(Ta) = c. Q.C.~.

Let Vl,. " " , "Up be children of an internal node u of T, let el = (u, vl), 1 E i < p, and let T p = T(el) + . . . +T(ep). Our idea is that a critical edge-ranking of T(el) can be constructed from a critical edge-ranking of T(vl) and that a critical edge- ranking of T p can be constructed from those of T ( e l ) , . . . ,T(ep). ~Te have the following two lemmas.

L e m m a 4. If ~ is an edge-ranking of T p, then

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(a) L(~) = Ul<i<,L(~lT(e i ) ) , and (b) sets L(~IT~e~) , 1 <_ i < p, are pairwise disjoint.

Conversely, g tai, 1 <_ i < p, are edge-rankings of T(ei) and sets L(~i) are pairwise disjoint, then the labeling ~ of E ( T p) extended from ~i, 1 < i < p, is an edge-ranking of T p and L(~) = Ul<i<_p L(Vi)"

Proof. immediate from the definition of an edge-ranking and its list. Q.C.7:).

L e m m a 5. Let m = max{x'~(TP-1), x ' (T (ep) ) } , then x ' ( T p) = m or m + 1.

Proof. Clearly m < x" (TP). Therefore it suffices to prove that X'~(T p) < m + 1. Let ~ and r be critical edge-raakings of T p-1 and T(ep), respectively. Extend

and r to an edge-labeling ~ of T p as follows:

, ( e ) = { ~(e) if e E E(Tp-1) , r if e 6 E(T(ep)) - {ep}, and m + l i f e = e p .

Clearly q is an edge-ranking of T p and # y < m + 1. Therefore x ' ( T p) < m + 1. Q.E.~).

The following Lemma 6 gives a necessary and sufficient condition for x ' ( T v) = m .

L e m m a 6 . Let m -: m a x { x ' ( T P - 1 ) , x ' ( T ( e p ) ) } . Then x ' ( T p) = m if and only if the following (a) or (b) holds:

(a) L*(T , - 1 ) n L*(T(ep)) = r or (b) there exists fl 6 [~ + 1, m] such that ~ ~ L*(T p - l ) U i* (T(ep) ) where c~

is the maximum integer in L*(Tp-1) ~ L*(T(ep)).

V 1

6

1 2

V 7 v8 V 9 VlO

Fig. 2. Critical edge-raakings of T(el) and T(e2).

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Before presenting a proof, we illustrate the conditions (a) and (b) for the tree T in Fig. 1. Fig. 2 depicts critical edge-rankings of T(el) and T(e2). Let p = 2, then T p-1 = T(el), Xtr(T p-l) = 3, x~(T(ep)) = 3, m = 3, L*(T p-l) = {3,2}, L*(T(ep)) = {3,1}, L*(T p-l) Cl L*(T(ep)) = {3}, c~ = 3, and [a + 1, m] = r Thus neither (a) nor (b) holds.

Proof of Lemma 6. *== Let ~ and r be critical edge-rankings of T p-1 and T(ep), respectively. Then L(~) = L*(T p-l) and L(r = L*(T(ep)). From ~ and r we construct an edge-ranking ~/of T p with # ~ / = m.

Suppose first that (a) holds, that is, L(~) f? L(r = r Then by Lemma 4

,f ~(e) if e 6 E(TP-1), and 7/(e) = [ r if e 6 E(T(ep))

is an edge-ranking of T p and #~} = m. Suppose next that (b) holds. Let r be an edge-ranking of T(ep) such that

{ ~ ( if e = ep; r = e) otherwise.

Then fl is visible, but none of the ranks in L(r Cl [1, fl] is visible with respect to r in T(ep). Hence L(r = (L(r - [1 , fl]) U{fl}. Since L(~) n L ( r C_ [1, fl - 1],

L(~) N L(r = L(~) Cl L(r - [1, fl] = r

Hence by Lemma 4

{~(e) i fe 6 E(T p-l), and ~/(e) = r if e 6 E(T(ep))

is an edge-ranking of T p with #q < re,. Suppose that x~(T p) = m but neither (a) nor (b) holds:

L*(T p-l) N L*(T(ep)) 7 s r and

[c~ + 1,m I C L*(T p-l) U L*(T(ep))

(6)

(7)

where c~ is the maximum integer in L*(T p-l) fl L*(T(ep)). Note that m is the largest integer in L*(T p-l) U L*(T(ep)). Let q be a critical edge-ranking of T p, and let qa = qlT p-1 and ~/b = qlT(ep) �9 By Lemma 4

L(7}) = L(t}a) U L(yb), and

L(r/a) Cl L(r}b ) = r

(8) (9)

Clearly

LO?a) h L*(TP-1), and

L(qb) ~ L*(T(ep)).

(10)

(11)

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By (1) L(y~) - [1 , al _ L*(T p-i) - [ i , al and L(~lb) --[1, al _ L*(T(ep)) - [1 , al. We then claim

L(y~) - [1, a] = L*(T p-i) - [1, a], and (12)

L(yb) - [1, a] = L*(T(ep)) - [1, a I. (13)

If (12) or (13) does not hold, that is, L(y~) - [1, a] ~- L*(T p- i ) - [ 1 , a] or L(yb) --[1, c~] ~- L*(T(ep)) - [1 , a], then

L(~) - [1, a] = (L(ya) - [1, a]) U (L(~b) - [1, a]) by (8)) ~- (L*(T p-i) - [1 , a]) U (i*(T(ep)) - [1, a]) by (9)) = L*(T p-i) U L*(T(ep)) - [1, c~] = [a + 1, inf. by (7))

Thus L(~/) _ {m + 1} and hence # y _> m + 1, contrary to the supposition X'~(T p) = m. Thus we have verified (12) and (13).

By (2) and (10) - (13) we have

L(rla ) f'l [1, a] ___ L*(T p- i ) N [1, a] _ {a}, and (14)

L(yb) N [1, a] _ L*(T(ev) ) n [1, a] _ {a} (15)

since c~ e L*(T p-i) N n*(T(ev) ). However, by (9) we have either a tg L(ya) or a r L(yb) and hence either L(y~) n [1, a] -( {a} or L(yb) N [1, a] -~ {a}, contrary to (14) or (15). Q.E.T).

In order to find a critical edge-ranking of T p from those of T p-1 and T(ev) , we need the following four lemmas.

L e m m a T . Let vi, v2 and v3 be children of an internal node u, and let ei = (u, vi), i = 1, 2,3. I f L*(T(ei)) ~_ t*(T(e2)), then L*(T(ei)+T(e3)) ~_ i*(T(e2)+ T(e3)).

L e m m a 8. Let v be a child of an internal node u, and let ei = (u, v). Let ~ be a critical edge-ranking of T(v), and let 7 = min{l > 1 I I r L(~)}. Then

ire E E(T(v)), and = i f =

is a critical edge-ranking o fT(e l ) . Given ~, one can find such ~ in O(n) time.

Proof. One can easily observe that ~ is an edge-ranking of T(el) and

L(~) = (L(~) - [1, 7 - 11) U {7}-

Let ~' be any critical edge-ranking of T(ei) . Then

L(rl') = (L(~'lT(v)) -[1,~/ '(ei) - 1])U {~'(el)}.

(16)

(17)

Furthermore ~'(ei) ~ L(~'IT(v)); (18)

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otherwise 7}' would not be an edge-ranking of T(el ) . By (17) and (18) we have

L*(T(e l ) ) = L(q') >~ L(~I'[T(v)) ~ L(~).

On the other hand, since [ 1 , ~ - 1] C_ L(~) and 3~ r L(~) , by (16) L(q) is the lexicographically smallest one among all the lists ~- L(~) . Thus we have L*(T(e l ) ) ~_ n ( , ) . Q.E.Z).

L e m m a g . Let Vl and v2 be children of an internal node u, and let el = (u, vl) and e2 = (u, v2). Let ~ and r be critical edge-rankings of Ta = T(e l ) and Tb = T(e2), respectively. Let T~ ! = Ta(#~) and T~ = T~-T~' . Let m = m a x { # ~ , # r I f X~r(T 2) = m, m = # ~ , and 7}' is a critical edge-ranking of T~a + Tb, then

( ~(e) if e �9 E(T~a'), and ~}(e) = ;f(e) if e �9 E(T~ -{- Tb)

is a critical edge-ranking o f T 2. (Remember T 2 = T(e~) + T(e2) = T~ + Tb.)

Proof. We first claim tha t # ~ ' < m 1, tha t is, ! ! _ - xr(T~ + Tb) < m -- 1. Since ! T ~ x ~ ( T : ) = X ~ ( a ) m and Ta and Tb are edge-disjoint subtrees of T 2 by

L e m m a 3 we have

x~(Tb) _< m -- 1. (19)

By (5) we have ! !

x~(T~) < m = X'~(Ta). (20)

Suppose for a contradiction that X~(T~ + Tb) ---- m. Then by (19), (20) and Lemma 6 L*(T'~) n L*(Tb) ~ r and [~ + 1, m - 1] C_ L*(T') U L*(Tb) where is the max imum integer in L*(T~) n L*(Tb). Since L*(T~) = {m} U L*(T~) by (4), we have a �9 L*(Ta) N L*(Tb) and [~ + 1, m] C_ L*(T~) U L*(Tb). Hence by L e m m a 6 X!~(T 2) = m -{- 1, a contradict ion. Thus we have verified #7]' < m - 1.

Since #~/~ < m - 1, one can easily observe tha t q cons t ructed from ~1' and is an edge-ranking of T 2 and L(~/) = {m} kJ L*(T' a + Tb).

We finally prove tha t ~ is a critical edge-ranking of T 2, tha t is, L(~) ___ L(~) for any critical edge-ranking ~ of T 2. Since X!,.(T 2) = m, m = # ~ �9 L(~). Since L(~) is a critical list ofTa , L((]T~) ~_ n(~) ~ {m} and hence m �9 L(~]T~). Since T~ = T~ - Ta', by (4)

n (~ ) = {m} U L(~]T'~) ~ {m} U L*(T~).

Let ~' = ~[T~ and T' a" = T~ - Ta(#~') = Ta - Ta(#~]Ta). Let p be a crit ical edge-ranking of Ta'!~ and let

#(e) = (~(~) if e �9 E(To(#~')), p(e) if e �9 E(T~").

Then # is an edge-ranking of T~, and L(#) = {m} kJ n(p). Since ~ is a crit ical edge-ranking of Ta, L(~) -~ L(#). ~Ve therefore have

{m} U L*(T:) ~ L(~) ~_ n(#) = {m} LJ L(p) = {m} U L*(T~"),

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and consequently L*(Tta) ~_ L*(T~a '') by (2). Therefore by Lemma 7

L*(T a + Tb) ~ L*(Ta" + Tb). (21)

Clearly

Ta t' + Tb = T 2 - Ta(#~'). (22)

By (21) and (22) we have

L*(T~ + Tb) "g L*(T 2 - Ta(#~')) .

Thus we have

L(y) = {m} U L*(T a + Tb) _ {m} U L*(T 2 - T~(#~')) = L(().

Q.c.7).

Thus, since ~[T~ is a critical edge-ranking of T~, the problem of constructing a critical edge-ranking y of T 2 = Ta + Tb from critical edge-rankings ~ and r can be reduced to the problem of constructing a critical edge-ranking y' of a smaller tree Tta + Tb = T 2 - Ta" from critical edge-rankings ~[Ta ~ and r when X~ (T2) = m.

L e m m a 10. Let vi and v2 be children of an internal node u, and let et = (u, Vl) and e2 = (u, v2). Let ~ and r be critical edge-rankings of T(el) and T(e2), respectively. Let m = max{#~ , # r I f x~(T 2) = m + 1 and L(~) ~ L(r then

{ ~(e) ire 6 E(T(e l ) ) , ~(e) = r l i e 6 E(T(e2)) - {e2}, and

m + l f i e = e 2

is a critical edge-ranking of T 2.

We have the following two lemmas which imply Lemmas 9 and 10. We omit the proofs since they can be proved similarly as Lemmas 9 and 10.

L e m m a 11. Let Vl," " , vp be children of an internal node u, and let ei = (u, vi) for each i. Let ~ and r be critical edge-rankings of Ta = T p-1 and Tb = T(ep), respectively. Let m = max{#~ , # r and suppose that X~r(T p) = m.

(a) / f n ( ~ ) _ L(r then

f ~(e) if e e E(T~'), and , (e) = [ , ' (e) if e e E(T~ + Tb)

is a critical edge-ranking o f t p, where T~a ' = Ta(#~) , T~a = Ta - T~' and ~' is a critical edge-ranking o] Tla + Tb.

(b) q L ( r ~- L(~), then

)" r if e e E(T~'), and ~?(e) = [ y'(e) if e e E(Ta + T~)

is a critical edge-ranking of T p, where T~' = Tb(#1b), T~ = Tb -- T~' and ~' is a critical edge-ranking of Ta + T~.

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L e m m a 12. Let v l , . . . , vp be children of an internal node u, and let ei = (u, vi) for each i. Let ~ and r be critical edge-rankings of Ta = T p-1 and Tb = T(ep), respectively. Let m = m a x { # ~ , # r and suppose that x ' ( T p) = m + 1. Let L(~IT(ej) ) be the lexicographically largest list among L(~JT(ei)), 1 < i < p - 1.

(a) If L(~JT(ej)) -< L(r then

(p(e) if e 6 E(Ta), y(e) = r if e e E(Tb) - {ep}, and

m + l if e = ep

is a critical edge-ranking of T p. (b) If L(~IT(ej) ) ~_ L(r and rl' is a critical edge-ranking of T i = TP -

T(ej) , then

{~( e) ire �9 E(T(e j ) ) - {ej}, y(e) = y'(e) if e �9 Z(T~a + Tb), and

m + l i f e = e j

is a critical edge-ranking of T p.

By Lemmas 8, 11 and 12 above we have the following recursive algorithm to find a critical edge-ranking of a subtree T(u) rooted at a node u.

P r o c e d u r e Edge-aanking(T(u)); b e g i n

1 if u is a leaf t h e n r e t u r n a trivial edge-ranking: r ~ r 2 else 3 b e g i n 4 let V l , . . . , v p be the children of u, and let ei = (u, vi) for all i, 1 < i < p ; 5 for i := i to p do Edge-aanking(T(vi)); 6 for i := 1 t o p do

find a critical edge-ranking of T(ei) from a critical edge-ranking of T(vi) by Lemma 8;

7 for i := 2 to p do 8 find a critical edge-ranking of T i

from those of T I-1 and T(ei) by Lemmas 6, 11 and 12; 9 r e t u r n a critical edge-ranking of T(u) {T(u) = T p}

10 end end.

Clearly one can find a critical edge-ranking of a tree T by calling P r o c e d u r e Edge-Ranking(T(r)) for the root r of T. Therefore it suffices to verify the time- complexity of the algorithm. As we show the detail later, one execution of line 8 takes in O(n) time. Furthermore line 8 is executed at most d(u) - 1 times in the for-loop at line 7. Therefore the for-loop at line 7 can be done in O(d(u)n) time. Since P r o c e d u r e Edge-Ranking is recursively called for all nodes, the total running time is

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This completes the proof of Theorem 3.1. We finally give an implementation of line 8 of Edge-Ranking, which finds a

critical edge-ranking y of T i from a critical edge-ranking T of T i-1 and a critical edge-ranking r of T(ei) .

P r o c e d u r e Line-8(9% r ~/); beg in

1 L : = r { L = L ( , ) } 2 let Ta := T i-1 and Tb := T(ei); 3 let La := L(T) and Lb := L(r 4 whi le La N Lb # r do beg in 5 a := max{ /6 La VI Lb}; {or does not increase } 6 fl := min{l >_ a + 1 ] I r La U Lb}; { fl decreases } 7 if La U Lb - [1, fl] # r t h e n

{ x ' ( T a + Tb) = m where m = max{x'~(T~), x ' (Tb ) } since (b) in Lemma 6 holds }

S for each label l 6 L~ U Lb -- [1, fl] in decreasing order do 9 i f / 6 L~ t h e n beg in { cf. Lemma l l (a ) }

10 T" := Ta(#(gvlTa)); { T" is the subtree (of Ta) rooted at

11 ~/ITa" := ~]Ta"; 12 Ta := Ta - Ttat; 13 La := La - {l}; 14 L := L U {/} 15 end 16 else beg in { l 6 Lb, cf. Lemma 11(b) } 17 T~' := Tb(#(r 18 , ITS' := r 19 Tb := Tb -- T~'; 20 Lb := Lb -- {1}; 21 L := L U {l} 22 end;

23

24 25 26 27 28 29 30 31 32 33 34

the edge e with the largest label l = 9~(e) } { oIT~" has been decided }

{ prune Ta t } { La = L(~]Ta) }

{ 7/]Tg' has been decided } { prune ~-~' }

{ ib = i(r }

{ f l = m + l a n d x ' r ( T a + T b ) = m + l } let ej be the edge with the lexicographically largest list L(9~[Ta(ej))

among the edges ej, 1 _< j < i - 1, remaining in Ta; i f L(9~lTa(ej)) ~" Lb t h e n beg in { cf. Lemma 12 }

, [ (Ta(e j ) - ej) := ~l(Ta(ej) - ej); ,(e~) : = ~; La := La - L(~ITa(ej) ) ; To : = G - G ( e j ) ; L := L U {fl};

e n d e l s e beg in

, l (Tb - ei) :-- r - el); , (e i) : = ~; Lb := r

{ rl[Ta(ed) has been decided }

{ prune Ta(ed) }

{ i(~iTo(e~)) "< Lb }

( T1]T b has been decided }

129

35 36 37 38 39 40 41 42

let Tb be empty tree; L := L O {fl}

e n d end;

~IT~ := ~IT~; ~lTb := r L:= L U L a U L b

end;

{ L~ N Lb = r eL Lemma l l } { ~]JTa has been decided } { ~]JTb has been decided }

{ ,] has been decided }

Whenever lines 5 and 6 are executed, a does not increase and/3 always decreases. Furthermore fl > a. Thus lines 5-37 in the whi le- loop are executed at most n times. Noting this fact, one can easily observe that the book-keeping operations takes O(n) t ime in total. Thus P r o c e d u r e Line-8 takes O(n) time.

R e m a r k : A more sophisticated implementat ion leads to an O(nlogn) algo- rithm. We omit the detail for the sake of space.

References

[BDJ+94]

[dlTGS93]

[DP90]

[IRV88a]

[IRV88b]

[IRV911

[KMSS8]

[Pot88]

[Sch89]

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