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Minimal Ratio Spanning Trees R. Chandresekaran University of Texas at Dallas Richardson, Texas
ABSTRACT
Given an undirected graph G: (N;E) with a node se t N and an edge s e t E and nwnbers Ce and De, e EE, we provide a poZy- nominaZ2y bounded algorithm t o solve the problem: ning tree T such that the ra t io
Find a span-
i s minimized. minimize such ra t io functions is {mediate. an algorithm that is "greedy," i n the sense of Ehonds [21, w i 2 2 not work for t h i s problem.
An extension t o finding bases i n matroids that I t i s sham that
INTRODUCTION
The minimal spanning tree problem is well-known and e f f i - c i e n t algorithms f o r solving it e x i s t ( [ l l ,[31,[81, [lo] 1. One of these algorithms has been ca l l ed a "greedy" algorithm by Edmonds [21 , who descr ibes t h e f u l l p o t e n t i a l of such an algo- rithm. In t h i s paper, w e consider t h e following spanning tree problem f o r a given undirected, connected graph G: (N;E) with node set N and edge set E.
Problem MR:
t r e e T such that r a t i o
Given numbers Ce and De, e e E , f i n d a spanning
is minimized.
Networks, 7: 335-342 @ 1977 by John Wiley & Sons, Inc. 335
We will assume that the denominator of (1) does not change sign for any tree T, and without loss of generality this sign will be assumed to be positive. In algorithms proposed for solving (11, we will require solution of the minimal spanning tree problem:
ProbZem MS f k ) : meter k, find a tree t such that
Given numbers Ce and De, e E and a given para-
He(k) (1' 1 e&T
is minimized, where
He(k) = Ce - k De. The paper proceeds as follows: We first show that a gen-
eralization of the "greedy" algorithm does not solve Problem MR. Then we provide a characterization of an optimal tree. This is an extension of a condition in [31 for the minimal spanning tree problem. This characterization not only provides an improvement routine, but also an algorithm that has some desirable features. This is followed by an algorithm whose growth is polynomially bounded. All the work in this paper generalizes readily to ma- troids in place of trees. When the assumption regarding sign of the denominator in (1) is relaxed, the problem becomes quite difficult, and this is discussed finally.
MODIFIED "GREEDY" ALGORITHM AND AN EXAMPLE
We will assume that the reader is familiar with the "greedy" algorithm. The modified algorithm is as follows: "At each step, pick an edge which (i) does not form a loop with the edges al- ready chosen and (ii) among such edges increases the objective function by the smallest quantity." The following example shows that such an algorithm does not work for this problem
MINIMAL RATIO SPANNING TRJ3ES 337
where the pa i r on edge e is (Ce,De) and n is taken t o be "suf-
f i c i en t ly large." The above algorithm picks a l l the edges ex- cept the one with weights (n,2n) and the value of the objective
10 function f o r t h i s solution is -. The optimal solution is one 18 i n which a l l edges except one of the edges with weights ( 1 , 4 )
a re included and the value of the objective is
value can be made as close t o 1/2 as we please and hence t h i s is the optimal solution fo r large values of n.
' W e do not know any modification t o the "greedy" algorithm f o r the minimal spanning t r ee problem which solves Problem MR. It is c lear t h a t the optimal ordering of a subset of edges is influenced by weights on edges outside the se t .
n + 9 This 2 n + 1 4
SOLUTION ALGORITHMS
The following resu l t s w i l l be used i n establishing algo- rithms for (1).
Lema I: (See [5], [61, [71). Let k* be the m i n i m value of (1) and suppose T * ( k ) solves MS(k) . Then
Proof:
<+ 3T* + ( 1 C e / 1 De) < k eeT* eeT*
<+ 3T* + 1 He(k) < o
<@ 1 H e ( k ) < 0 .
eeT*
eeT* (k)
c ) Equality par t of ( 3 ) follows from (a) and (b) .
338 CHANDRASEKARAN
Corollary I : timal tree of MR is optima2 for M S W ) and conversely.
Let k* be the m i n i m value of (I). Then any op-
The main thrust of the remaining par t of t h i s paper is t o show tha t one need evaluate only polynomially many values of k i n Lemma 1 t o find a t ree tha t minimizes (1).
r i ca l ly and i n the process prove tha t only a polynomial number of problems MS(k) need be considered. The following arguments establish th i s resul t .
The idea is t o solve the problem of solving MS(k) paramet-
For any pa i r e , f E E, the two l ines (C - kDe) and (Cf - kDe) , e
- co < k < m e i ther intersect not a t a l l , in tersect exactly a t one p o i n t or they are identical . L e t P be the s e t of unordered pairs ( e , f ) where e and f E E such tha t the two l ines given above intersect precisely once and l e t
2 There are a t most O ( l E l tha t there i s no simple way t o delete multiple edges i f they are present unlike i n t h e minimal spanning t ree problem. L e t the elements of K = (k (e, f 1 I (e, f 1 E PI be ordered and numbered as
such values. We would l i k e t o s t r e s s
- < k < kl < ... 0 < k < a (5) r
Lemma 2: interval I = ( k i , k i + l ) or ( - m , k o ) or ( k r , a ) . MS(k) for a l l k E closure of I .
Suppose T * ( b solves MS& for some 32 i n the_open Then T*(k) solves
Proof: A t ree T solves MS(k) i f f it s a t i s f i e s
f I the loop of { f ) UT includes e
L(e,T) = (7
(See [31 . I ? = [a, B 1 such tha t T* ($1 solves MS (k) for a l l k E I. This is seen using (6). tion t o the assumption tha t the above k
i n K. f 2 closure of I.
Since T*&) solves M S ( G 1 , there is a ciosed interval
If ? 3 I, then a or B E K which i s a contradic-
The essence of t h i s argument implies tha t 1 3 I and hence
are the d i s t inc t numbers i A
M I N I M A L RATIO SPANNING TREES 339
Thus a l l we need t o do is t o solve MS(k) for one va lue of k i n each of the above ( r + l ) open i n t e r v a l s . Note t h a t r is p l y n o m i a l l y bounded as a func t ion of IEI, t h e number of edges i n t h e graph. Then w e have an algori thm t h a t is polynomially bounded.
L e t Io= (-OD,k0), I . = [kj-l, k.1 j = 1 , ..., r and I = [ k , , ~ ) , 3 r where k . are def ined by (5). L e t IKI = r.
A lgorithm S t e p 0:
S t e p I:
7
L e t a. = 0, B = r and t = 0.
L e t S = {Ij lat zj I B t 1. i n t e r i o r of I
G o t o Step 1. 0 t t Pick a va lue of k i n t h e
t = [ k j ,kj+ll and l e t T* (k ) solve r at+Btl
2
eeT* (kt)
t MS(k 1 . I f A = 1 H ( k . ) > 0 and
B = 1 He (kj+l) > 0 then go t o Step 2. If both
these q u a n t i t i e s are < 0 then go t o Step 3 . Otherwise go t o S tep 4.
e l
eET*(k )
S t e p 2: Define a
S t e p 3: Define a = c1 and Bt+l = j and go t o Step 1.
= j + l and Bt+l = B, and go t o Step 1. t + l
t + l t S t e p 4 : 3k* E [kj ,kj+ll + 1 He(k) = 0. This i s given by
ksT* (k )
k* = k . + (kj+l-kj) , I 1AI I , 3 A + B
where A and B are as def ined i n S tep 1.
rithm; T*(k 1 i s the opt imal tree f o r MR and t h e opt imal va lue of t h e o b j e c t i v e func t ion is k* s a t i s f y i n g (8).
Stop t h e algo- t
THE ORDER OF COMPLEXIm
Since t h e c a r d i n a l i t y of So i s bounded by ! E l 2 , t h e e f f o r t
The num- 2
involved i n a r ranging t h e va lues k i s O ( I E I l o g l E l ) .
ber of spanning tree problems MS(k) solved are bounded by 2 log1 E l , and each of t h e s e problems is 0 ( 1 E l log log (V( ) [ l l and [ l O l ) . Thus, t h e o rde r of t h e above algori thm i s
0 ( IE1 log 1E 1 ) . b u t t h e a i m has been t o e x h i b i t a polynomial algorithm.
j
(see
2 There has been no attempt to minimize t h e o rde r
340 CHANDRASEKARAN
GENERALIZATIONS
A genera l iza t ion t o determine a basis B of a matroid such t h a t
(91
is minimized is immediate. W e a l s o would l i k e t o po in t o u t t h a t Lemma 1 is e a s i l y extended t o t h e following general proposit ion which is proved i n a manner similar t o t h a t of Lemma 1. This i s used elsewhere 191 t o produce a " f i n i t e " algorithm i f one s u i t - ably defines the set of operations.
Theorem 1: L e t S be any arbitrary bounded cZosed subset of Rn.
Let f a d g be continuous functions from S to R1 such that
Let g(x) 0 Y X E S
and
Then
Remark: It should be c l e a r t h a t i f w e have a f i n i t e (polynomial) algorithm f o r determining Z* (k) i n equation ( 1 2 1 , t h i s algorithm together with t h e theorem can be used t o provide an algorithm which is almost f i n i t e (polynomial) i n t he following sense. The algorithm determines an i n t e r v a l containing k* whose length can
be made less than E i n log '"1 i t e r a t i o n s of so lv ing (121, where
k is the length i n i t i a l i n t e r v a l . This i n i t i a l i n t e r v a l I. = [ao,Bol where (ao,B,l a r e given by:
0
0 2 1
1 g(x 1 = min g ( x )
g ( x 1 = max g ( x )
XE s 2
X E S
(14)
(15)
MINIMAL RATIO SPANNING TNES 341
Such algorithms will be called E-finite (€-polynomial). In the case when S is the set of spanning trees we were able to show the existence of an even more efficient algorithm, in the sense that it is polynomially bounded.
So far we have treated the case when the denominator does not change signs over the set of feasible solutions. When the
set S is convex, if there exists x , x2 E S such that g(x ) > 0,
g (x ) < 0 then there also exists xo such that g (x ) = 0. This, of course, causes the usual problem of division by zero. But this may not be the case when the set S is discrete, even if g(x)
1 changes sign. We may be able to partition S into two sets S and
S such that
1 1
2 0
2
1 2 and S U S = S
In what follows, we will show that the problem becomes quite difficult (NP-complete to be precise) when we allow the above kind of a generalization, at least for the case of matroids. In order to show this consider the following NP-complete problem:
find a subset Given 2n nonnegative real numbers a 1, a2, ..., a of n numbers that add up as nearly as possible to half the sum of all the 2n numbers. By defining another set af of numbers by the relations:
2n
then the above problem also detects the existence of a subset of n numbers that add to zero.
One feature of any algorithm must be the detection of the existence of x .) g(x) = 0. Thus, if a polynomial algorithm exists for solving the generalized problem, then this would contradict the NP- completeness of the problem of finding a subset of n number a' i that add up to zero. Lastly, it is easy to verify that the sets of cardinality n numbers do indeed define a matroid. Hence this generalization will, in general, be difficult.
Now consider the generalization proposed above.
342 CHANDRASEKARAN
SUMMARY
W e have considered a ratio-minimal spanning t r e e problem and have provided both an s-polynomial and a polynomial algo- rithm when the denominator does n o t change sign. W e have indi- cated the order of d i f f i c u l t y when t h e denominator does change sign. There has been no attempt t o reduce the order of com- p l ex i ty of any of t he algorithms.
ACKNOWLEDGMENT
I a m thankful t o D r . K. Truemper f o r he lp fu l discussions and t o a r e fe ree f o r pointing out references [ l l , [51 , [61, [71 and [ l o ] .
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Paper received May 20, 1976.