obl i gati ons rul e of mi ni mum c ost spanni ng
ext ensi on probl ems
by
Vin cent Feltkamp 1
S tef Tij s
CentER and Ec on ometric sDe partment, Ti l burgUni versity,
P.O.Box90153,5000L ETil burg, Th eNetherl ands. Emai l : V.Fel tkamp@ku b.nl.
Shigeo Muto 2
Facul tyof Economic s,Tohoku Uni versi ty,K awau chi,
Aoba-ku ,S endai 980,Japan.
December 1, 1994
Abst ra ct
Min imum c ost spann ing e xte nsion prob lems are ge neral i zati ons of mi nimum
cost spanni ng tree probl ems i n wh ich an existi ng ne twork has to b e exten ded to
conne ct users to a source. This p aper generali ze s the de n iti on of i rreduci bl e
core to min imum c ost sp an nin g extensi on probl ems and intro duce s an al gorithm
generatingal lele ments ofthei rred uci bl ecore. Moreove r,thee qualremai ni ngobli
-gati on s rul e, a one-point re nement of th ei rreduci bl e core i s p re sente d. Final ly,
thepaperch aracteriz esth ese solu ti ons axi omati call y. Thecl assical Bird tree al
lo-cationofmi ni mu mcostspanni ngtre eprobl emsi s obtai ned asa parti cu larcasei n
ouralgori thm forth ei rred uci bl ecore.
1 Introduct io n
Co nsid er a g roup o f villag es , ea ch of which needs te b e co nnected directly o r via other
villa ges to a so urce. Such a conn ectio n need s costly links. Each villag e co uld co nnect
itself directly to the source, but by co o p erating co sts mig ht be reduced. This cost
minimization pro blem is a n old problem in Op erations Research , and Bor
u vka (192 6)
1
This authoris spons oredby t heFoundation forthePromot ionofRe se ar ch inEconomic Scienc es ,
whichispar toftheDutchO rgan iz ationfor Sc ie nti cRe search(NWO )
2
The thir d author wis hes to acknowledge the Canon Foundation in Europ e Visiting Rese arch Fe
came up with a lg orithms to cons truct a tree co nnecting everybo dy to the so urce w ith
minima ltota lco st. L ater,Krus kal(19 56 ),P rim(1 957 )a ndDijkstra(19 59 )fou ndsimilar
algo rithms. A histo ric overview o f this min imum cost spann ing tree (hencefo rth mcst)
prob lemca nbefo und inG ra hama nd Hell(198 5).
In thispaper,wea nalizea mo reg eneral problem, inw hicha partial networkof links
existsa lreadya ndhasto b eextendedtoanetworkconnectingeveryplayertothesource.
However, ndingaminima lcostspanningextensio nisonlyparto fth epro blem: ifthe
costofthisextens io nhastobeb o rnebythevillag es ,thenacostalloca tio npro blemhasto
b ea ddresseda swell. Clausand Kleitman(197 3)intro d ucedthisco sta llo cationproblem
forthe originalminimumco stsp anning trees etting,whereupon Bird(1 976 )treatedthis
prob lemw ith g ame- theo retic methods a nd propos ed a co st a llo cation closelyrelated to
the Prim-Dijkstra algo rithm. Grano t and Hub erma n (198 1)proved that this allocation
is a n extremal point of the co re of the ass o ciated min imu m co st s panning tree g ame.
This ga meisde neda sfo llows : th eplayersa rethe villag esa nd theworthof aco alition
isthe minima lcos tofconnecting this co alitionto the source vialinksb etweenmemb ers
of this coa litio n. Aarts (19 92 ) fo und other extreme points of the co re in case the mcst
prob lem h as a n mcst tha t is a chain, i.e. a tree w ith o nly two leaves. Kuipers (199 3)
inves tiga tedth ecoreofinfo rma tio ngames. T hes ea rega mesa rising fro mmcs tp roblems
in which th eco sts o f a con nection a re either one o rzero .
Th e outlineof this paper is as fo llows.
Sectio n 2 presents a formal model of the minimum co st spa nning extension (mcse)
prob lem in w hich an exis ting network ha s to b e extended to a span ning netwo rk, i.e. a
netwo rk conn ectin g every village to th e so urce. An a lg orithm to nd a minimum cost
spa nnin g extension and a ssociated set of co st allocatio ns is p resented . In co ntra st to
earlierwork(s eeFeltka mp,Tijs andMuto (19 94 a)),this extens io nandsetofallocatio ns
are not genera ted by a n a lg orithm a la P rim-Dijks tra, but by an algo rithm s imilar to
Kruska l's (1 95 6) alg orithm. It is proved thatthe seto f allo cations g eneratedisa subset
of the co reof the a ssocia tedmcseg ameand that isis independent o fthe extension that
is constructed.
Sectio n3g eneralizesthede nition oftheirredu cibleco re,propos edinBird(19 76)for
minimum cost sp anning tree problems, to minimum co st spa nning extension p roblems
and proves that the setof a llo cations genera ted by the algorithmin s ectio n 2coincides
withthe irreduciblecore. Aco rolla ryistha t Bird 'streeallocations(seeBird(19 76)) for
minimumcost spann ing tree p ro blemsa re also g enera ted by o uralgo rithm.
Sectio n 4intro du ces the equalremain ing o bliga tion sva lu e,a one-p o int re n ement of
the irreduciblecore. It is obtained as a s p ecialca seo f the algo rithm for the irredu cible
core presented in sectio n 2 . Like the irreducible core, the E RO value is indep endent of
the extension cons tructed. This in co ntrast with Bird's tree a lloca tio n rule, which is
dep endent on th etree constructed.
Sectio n 5 axio matically ch aracterizes the irredu cible core a nd the equal remaining
obligations va lue. Amo ng others,axio ms we use a reeciency, consistencyan d converse
consistency.
Finally, section 6 concludes with so me remarks a nd sug gestions fo r further research.
Prelimina ries a nd nota tions
We recall so me standa rd d e nitio ns from g raph theory w hich can be found in a ny
elementary textbo o k o n g raph theory to show the nota tio nal conventio ns. A g raph
G =< V;E > co nsists of a set V o f vertices and a set E o f edges. An edg e e incident
with two vertices ia nd j is identi ed with fi;jg 3
. Fo r a gra ph G=< V;E > a nd a set
W V,
E(W):=fe2E jeWg
is the seto f ed ges linking twovertices in W and fora subset E 0
o f E,
V(E 0
):=fv2V jthereexis ts a nedge e2E 0
w ith v2eg
is the seto f vertices incident with E 0
.
Th e co mpleteg raph on avertex setV isth e g raph K
V =< V;E V >, w here E V :=ffv ;w gjv ;w2V a nd v6=w g:
A path fro m a vertex i to a vertex j in a g raph < V;E > is a sequence (i =
i 0 ;i 1 ;:::;i k
= j) of vertices s uch tha t fo r all l k, the edg e fi
l01 ;i
l
g lies in E. A
cycl e is a path of w hich the beg in and end p o ints co incide. Two vertices i;j 2 V are
conn ectedin agrap h<V;E >if thereis apathfro mi to j in<V;E >. A su bsetW of
V isconnec tedin<V;E > if every two verticesi;j 2W a reconn ected inthe subg raph
< W;E(W)>. A connected s et W is a connec ted compone nt of the gra ph <V;E > if
nos up erset ofW isco nnected. Wewillusua llys ayco mponentw henwemea nco nnected
comp o nent. If W V, th e set of con nected comp o nents of the g raph <W;E(W) > is
denoted W=E. Aconn ected graphisag raph<V;E >withV conn ectedin<V;E >. A
treeisaconnectedg raphtha tcontainsno cycles. Aconnectedcomponento fag raphwill
b e denoted by the letter C, a nd for a vertex v of the gra ph, the co nnected comp o nent
containing v is denoted C
v .
Th e eco nomic situations in the sequ el invo lve a set N of users of a source 3. For a
coalition S N, wed eno te S[f3g by S 3 . For two vectors x2IR S a nd y2IR T , w here
S a nd T are two disjointco alitions , wedenote (x;y ) the vecto rwith comp o nents
(x;y) k = ( x k if k 2S y k if k 2T:
Furthermo re,fortwocoa litio nsS T an d avectorx2IR T ,wed eno tex S the restriction of x to S. Th esymb o l 1 S
is used to deno te the vecto rin IR N with coordinates 1 S ;k = ( 1 if k 2S 0 if k 2N nS:
For a ny co alition S,the simp lex 1 S is de nedby 1 S :=fx2IR S + j X i2S x i =1g: 3
Withma nyecono micsituationsinw hichcos tshaveto b edividedoneca na ssocia tea
cost game (N;c) cons isting ofa nite setN =f1;:::;ng ofplayers,a nd acharac teristic
func tionc:2 N
!IR,withc(;)=0. Herec(S)representsthemaximalcos tforcoa litio nS
if it s ecedes , i.e. if p eop le o f S co o p erate and do no t count up o n help from peo ple
outs ideS.
Th e core Co re(c) of a co stg ame (N;c), is de ned by
Core(c)=fx2IR N j X i2N x i =c(N) and X i2S x i c(S)fo r all S Ng:
Th e ca rdina lity o f a setS will b e d eno ted by jSj.
2 Mcse pro blems : a solution
In this s ection we fo rma lly present minimum cos t spa nning extens io n pro blems, mcse
ga mesanda na lg orithmwhichforanymcsepro blemcomputesaminimumcos tsp anning
extensionand ana ssocia tedsetofalloca tio ns,w hichappearsto beco nta inedinthecore
of the mcsega me.
A minimum cost spannin g ext ension probl emconsistso f asetN o f users whohaveto
extendanexis tingnetworkinordertob econnectedtoas ource,deno ted3. T helinksare
costly, and the users have to pay for the extens io n. Such a pro blemis repres ented by a
completegra ph<N 3 ;E N 3 > onth esetN 3
containinga llusersand thesource,tog ether
with a set E E
N 3
o f a lready cons tru cted links an d a weightfunctio n w:E
N 3
! IR
+ .
The co sto fconstructing an edg eeis g ivenbythe p o sitiveweightw(e)>0of this ed ge.
Becaus e the g raph o f pos sible ed ges is always the complete g ra ph, we d eno te a mcse
prob lem with set of u sers N, so urce 3, weight function w and existin g edge set E by
<N;3;w ;E >. If th e set of existing edges E is empty, the mcse pro blem b ecomes the
cla ssica lminimumco stspan ning treeproblem, andin stea dof writing <N;3;w ;;>,we
will write <N;3;w>.
Mcseproblemsca nb esp litupintotwosub problems,anOpera tio nsResearchproblem
of connecting allus ers to the s ource ina extended gra ph< N 3
;E[E 0
> suchtha t the
cost o f the extensio n E 0
is minimal an d acost a lloca tio n problemo f allocatin gthis cost
to the users ina rea sona ble way.
In the specialcase of mcst problems, an mcst isco nstructedby Kruska l's algorithm,
whichis de n ed as fo llows :
Algorithm 2.1 (Kruskal 1956)
input : a n mcs t co nstructio npro blem <N;3;w>
output : a n edge setE 0
o f a minimu mco st spa nnin g tree
1. sta rt with the empty setE 0
=;.
2. Find an edg e e o f minimum co st such that the graph < N 3
;E 0
[feg > does not
contain a cycle.
3. Join this edg eto the setE 0 (E 0 :=E 0 [feg). 3 0
5. E 0
is the edge setwe weres eeking .
For mcs e pro blems, a generalization of Krus kal's a lg orithm is demo nstrated in
exam-ple2.2.
Exa mple 2.2 L et N = f1 ;2 ;3;4;5g, a nd let the weights o f the edges a nd the g raph
whichisa lready co nstructed b eas in g ure1. T hecostsof ed ges th ata re no tindicated
are$2 00 . Firs tco nstructthe edgef1;2 g,itisthechea p esto newh ichdo esno tintroduce
u 3 $ 100 u 2 A A A A A A $ 10 $2 0 u 1 u 3 1 1 1 1 1 1 $ 40 u 4 H H H H H H $7 0 u 5 Th eweights o f links. u 3 u 2 u 1 u 3 u 4 H H H H H H u 5 f4;5 g isalrea dyconstructed.
Figure 1 : A simple mcsepro blem
acycle. Th esamereas oning picks f2;3 g as secondedge, as third edge f3;4g and nally
as la st edge, f2;3g is cons tructed.
The a lg orith md emons trated is the follow ing :
Algorithm 2.3 (Kruskal generalized to mcse problems)
input : a n mcs e co nstruction problem<N;3;w;E >
output : a n mcs e
1. Given M<N;3;w;E >, de ne
t = 0 the initia ls tage,
= jN
3
=Ej01 the numb er of stag es ,
E 0
= E the initia ledg e set.
2. t:=t+1 .
3. Ats tage t,givenE t0 1
,chooseacheapes tedgee t
suchthat theg raph<N 3 ;E t01 [ fe t
g> doesno t contain more cycles than the graph <N 3 ;E t0 1 >. 4. De neE t :=E t01 [fe t g.
6. E
is th e extensionwe wereseeking. Deno tethe sequenceof edg es cons tructed by
E =(e 1
;:::;e
).
In lemma 7.1 it is proved tha t the exten sio n cons tructed is ind eeda mcse. Notethis
algo rithm cons tructs the edg es of a minimum cos t spa nning extension in the order of
non -decreas ing costs. It is eas yto seetha t any sequ enceE =(e 1
;:::;e
) wh ich co nsists
of the edgesof a minimumco sts panning extensionorderedby non-d ecreasing costscan
b e con structed with algorithm2.3 by a suitable choice of the edges chosen ins tep 3.
Bird(1 97 6) a ssociatesa treea llo ca tion withevery minimumcost s panning treeinan
mcs tpro blem. InFeltka mp ,Tijsa ndMuto(19 94 a),weprovethatthistreeallo cationcan
b ea ssociated withthe sequenceof edgeswhichP rimandD ijkstra'salgo rithmgenerates
when genera ting this mcst. T his sug gests looking for a lloca tio ns associa ted with the
sequen ces g en era ted by Krus kal's alg orithm.
For a nmcse pro blemM<N;3;w;E >, the minimal cos ts panning extensionsE 0
have on e edge less tha n the number o f co mponents jN 3
= Ej (if mo re edg es were bu ilt,
a cycle wo uld be intro duced, which ca nnot be minimal in cost, as the weights o f the
edges a re p o sitive). Hence, de ning := jN 3
=Ej01 , we a ssocia te an allocatio n w ith
everysequenceE =(e 1
;:::;e
)o fedgeswhichdoesno tintro ducen ewcyclesinth emcse
prob lem M. Note th at a ny such s equence connects a ll p layers to the s ource. The idea
b ehind the allocation is tha t at each succes sive stag e t, the co st of the edge e t
which
is cons tructed at s tage t is sha red among the p layers in N acco rdin g to a share vect or
f t
21 N
. Three ru les have to be observedw hen allocatingthe cos tof e t
:
Atstag et,theedgee t
con nectstwocomp on entsofthegraph<N 3 ;fe 1 ;:::;e t0 1 g>
crea ting a comp o nent C t o f the g raph < N 3 ;fe 1 ;:::;e t g >. Only players in C t
contribute to the cost o f e t
.
Theco mponentC t01
3
of thesourceinthe g raph<N 3 ;fe 1 ;:::;e t01 g>cons tructed
b efores ta get doesno t co ntribute to the co st o f e t
.
Furthermo re, summing over all edges in the sequen ce, every comp onent of the
originalgra ph<N 3
;E >whichdoesno tco ntainthe sourcepaysfractionsofedges
to ato tal of o ne.
Hence, de ne the set V E
(M) o f sequ ences o f share vectors val id for the s equenceE in
Mby V E (M):= 8 > > > > > > > > > > < > > > > > > > > > > : (f 1 ;:::;f )2[0;1] jNj j X k 2C t f t k = 1 for a llt; and X k 2C X s= 1 f s k = 1 8C 2N 3 =E with 362C ; and f t k = 0 if k 2C t0 1 3 : 9 > > > > > > > > > > = > > > > > > > > > > ;
For a sequen ce F (f 1
;:::;f
)valid forE in M,de nethe allocatio n
x E;F (M) := X t=1 f t w (e t )2IR N (2:1)
and de ne the set D E (M) by D E (M) :=fx E;F jF 2V E g: (2:2)
If no co nfusio n can o ccur, we dro p the a rgumentM.
Lemma 2.4 Fo r a ll mcse problems M, for all sequences E = (e 1 ;:::;e ) su ch that E 0 := fe 1 ;:::;e
g isa n mcse of Ma nd allF validfor E,the allocatio n x:=x E;F (M) is ecient: P i2N x i =c M (N).
Proof : Validity of F implies every edg e e t
2E
nE is paid fo r by the co mponent C t
it co nstructs. Moreover,E 0
is a minimalcos tspan ning extension of <N 3 ;E >,h ence X i2N x i = X e2E 0 nE w (e)=c M (N): 2
Note that b eca use the set of valid sequ ences of share vecto rs V E is convex a nd the map x E :F 7! x E;F
is linear, the setD E
isa lso convex, fo r any sequence E.
Insteadof rstco nstructingtheedgesa ndlaterallocatingtheirco st,onecoulda llo cate
the cos to f the edge e t
immediately, b eca use the validity of a sequence of sha re vectors
can be checked stag e by s tage : as equencef 1 ;:::;f is va lidfor e 1 ;:::;e in Mif and
only if at everysta ge t it satis es
the comp on ent C t
co nstructed at stage t pays the co st o f the ed gee t
,
for every comp o nent inthe original gra ph<N 3
;E >,the to tal ofthe sh ares paid
up to sta ge t does not exceed 1 ,
the p eop leou ts ide C t
do not pay a nything,
the p eop lein the comp on ent C t0 1
3
of the s ource do n ot pay a nything.
In formula ,this g ives
8 > > > > > > > > < > > > > > > > > : X k2C t f t k = 1; X k2C t X s= 1 f s k 1 fo r allC 2N 3 = E; f t k = 0 if k 62C t ; f t k = 0 if k 2C t0 1 3 (2:3)
Exa mple 2.5 For a nmcstpro blemT <N;3;w>,P rimand Dijkstra'salgorithm(cf
Prim(19 57)orDijkstra (19 59 ))constructsasequenceE =(e 1 ;:::;e jNj )ofedgesleading to a nmcst<N 3
;T >as fo llows : ateverysta ge t,e t
isan edg ewh ichcon nectsap layer
with the co mponent of the so urce in the gra ph < N 3 ;fe 1 ;:::;e t01 g > a nd which h as
minima l co st a mo ng all s uch edg es . Without los s o f generality, we numb er the players
in N such that for every t, the edge e t
con nects player t with the comp o nent of the
so urce, f3;1 ;:::;t01g. Hence, the edge e 1
co nnects player 1 to th e so urce and using
the s ys tem 2.3 , we see player 1 h as to pay the co st of e 1
and the o ther players do not
contribute. In the seco nd sta ge, player 2 is connected to the comp o nent o f the source,
whichnowequa lsf3;1g. The rstequationinsystem2.3impliesplayers1an d 2paythe
cost ofedgee 2
. The fou rthimpliestha t p layer 1,whoisinthe componento fthe source,
does not contribute, hen ce, player 2 is as sig ned the cos to f edg e e 2
. The third equation
implies the other players do no t co ntribute. Th e inequality is sa tis ed, b ecause up to
now, every comp o nent in the o rig ina l g raph (i.e. every player) paid eith er o ne ed ge or
no ed ges. By ind uction, we see that a t every stag e, the comp o nent C t
cons ists of the
comp o nentC t01
3
a ndth enewly co nnectedplayert. Beca use theco mponentoftheso urce
doesno tcontribu tetothe co sto fe t
,the uniquevalidallocationofthe cos tofthisedgeis
to allocate it co mpletely to player t. Hence, D E
(T) con sis ts o f o neallo cation, inwhich
eachplayeri isallocatedthe costofthe edgeincidentto i onth eu niquepa thinthe tree
from i to the so urce. This allo cation is precisely Bird 's tree a lloca tio n as sociated w ith
the mcs t<N 3
;T >,deno ted T
(cf Bird (19 76)).
Exa mple 2.6 Co mputing the extreme points of the s et o f va lid sha re vectors fo r the
sequen ceE co nstructedinexa mple2 .2showsth at inthis caseD E
(M)istheco nvexhull
of the vecto rs
(1 0;20 ;4 0;1 00;0 ); (10;2 0;40 ;0 ;100 ); (10 ;2 0;10 0;40 ;0); (10 ;20;1 00;0 ;40);
(1 0;40 ;2 0;1 00;0 ); (10;4 0;20 ;0 ;100 ); (10 ;1 00 ;2 0;40 ;0); (10 ;100 ;20;0 ;40);
(2 0;10 ;4 0;1 00;0 ); (20;1 0;40 ;0 ;100 ); (20 ;1 0;10 0;40 ;0); (20 ;10;1 00;0 ;40);
(4 0;10 ;2 0;1 00;0 ); (40;1 0;20 ;0 ;100 ); (10 0;10 ;2 0;40 ;0); (10 0;10 ;20;0 ;40):
Inspired by Bird (1 976 ), we as sociate a minimum cos t spa nning extens io n ga me
(N;c M
) with an mcseproblem M <N;3;w ;E > a s follows . E ach coa litio n S N,
if it canno t count o n the p layers in its complement, ha s to so lve a problem s imilar to
the problem of the gra nd co alition, na mely, extending the existin g gra ph to a g raph
connecting all users in S to the sou rce. The cost of this extensio n is the worth c M
(S)
of co alition S inthe mcse ga me.
When co mp uting the co sto f a coa litio nS, severa l questions a rise. Ca n the co alition
use a ll o r so me o f the edges which a re alrea dy present? Is it a llowed to use vertices
outs ide S? We o pt fo r the fo llowing an swers : a coa litio n S is a llowed to use all edges
which are initially present, but ca n on ly us e those vertices which lie in aco mponent of
<N 3
;E> wh ich conta ins memb ers o f S or the so urce. L et us cons ider an example to
u 3 u 1 u 2 H H H H H Hu 3 $4 J J J J J J J J J J J J $ 3 $1 $3 8 8 8 8 8 8 $ 1
Figu re 2 : f1;2 g is allowedto use the edge f2 ;3 g, b utf1gis n ot.
Exa mple 2.7 In the pro blem depicted in gure 2, edge f2;3 g is alrea dy co nstructed .
Co alition f1 ;2 g is allowed to use the edg e f2;3ga nd can con nect itself by building the
edges f1;2g an d f3 ;3g, so c(f1;2 g)=1+1=2. Co alition f1gis no tallowedtouse the
edge f2 ;3g beca use the co mponent f2 ;3 g does not have a ny vertices in commo n w ith
f1go rthe source,h encec(f1g)=3 . Theother worthsare c(f2g)=c(f3 g)=1 (co nnect
player2 via player 3),and nally c(f1;3 g)=c(N)=2.
In general,th e fo rmula beco mes
c M (S):=min 8 < : X e2E 0 w (e)j S C E 0 3 a nd E 0
containso nly edg es b etween
comp o nents co ntain ing membersof S 3 9 = ; fora llS N,whereC E 0 3
isth ecomp o nent ofthe so urce3 inthe gra ph <N 3
;E[E 0
>.
Th enext theo rem sta tes a nallocationas sociated with asequence of edges g enerated
by algorithm2.3 isa core element o f the mcse g ame.
Theorem 2.8 Fo ranymcs eproblemM,fo ranysequenceo f cho icesE =(e 1
;:::;e
)in
thealgo rithm2.3appliedtoMandanysequenceoffra ctionsF validfo rE theallocation
x E;F
,asde nedinequation2 .1 ,isaco re- alloca tio nofthemcsegame(N;c M
)a ssociated
with M.
Th eproofofthistheo remislengthya ndtechnica l,andcanbefou ndinthe a pp endix.
An immed iate con sequenceis
Corollary 2.9 For any sequen ce E lead ing to a minimum co st s panning extension for
an mcsepro blem Mwith as sociated mcse ga me(N;c M ), D E (M)C ore(N;c M ): Proof : Fo r any x 2 D E
(M), there is a sequence F which is valid for E with x =
x E;F
(M)2Core(N;c). 2
A questio n wh ich arises is, how does the set D E
Proposition 2.10 For a nymcseproblemM,fo r anyE a nd e
E cons tructed by the a lgo
-rith m 2 .3 ap plied to M, D E (M)=D e E (M):
A proof can b e found in the append ix.
Beca use D E
is independent o f the s equence E of edges, as lon g as this s equence is
constructed by the algo rithm2.3, wede ne for a nmcs eproblemM :
D G K
(M):=D E
(M)
for any sequ ence E o btained by the a lgo rithm 2.3 ap plied to M. (T he su p erscript GK
sta nds fo r generalizedKruskal).
3 The irreducible core o f mcse pro blems
In th issectio nwegeneralizeth econceptofirreduciblecore, known frommcstproblems ,
tomcseproblemsandproveth atou rsetofallocationsD G K
coincidesw iththisirredu cible
core.
De nition 3.1 Given an mcse problem <N;3;w;E >, we de ne the a ssociated mcst
prob lem<N E ;3 E ;w E >as fo llows : N E
consistso fth ecomp o nentso f<N 3
;E >which
do n ot contain the s ource 3, th e new so urce 3
E
is th e comp o nent o f <N 3
;E > which
contains the o rig ina ls ource 3 and w
E
is de ned by
w
E
(C;D):=minfw(i;j)ji2C ;j 2Dg
forallcomp on entsC andD ofthegrap h<N 3
;E >. Theintuitiveidea isto s hrinkea ch
comp o nent not containing the so urce into a s ing le player, and to sh rink the comp o nent
of the source into an ewsource.
Furthermo re, fo r an edg ee=fi;jg2E
N 3,de n e e E :=fC i ;C j g (3:1)
and for a seto f edg es F E
N 3, de ne F E :=ffC i ;C j gjfi;jg2Fg; (3:2) where C i and C j
a re the comp o nents of <N 3
;E > co nta ining players i and j, resp
ec-tively.
Itiseas ytoseethatifF isa nmcs eofth emcsepro blem<N;3;w ;E >,th enthe tree
<N 3
E ;F
E
> is a n mcsto f the ass o cia ted mcstpro blem <N
E ;3 E ;w E >. Conversely, if <N 3 E
;T >isa n mcsto f the as sociatedmcstp ro blem,then thereexists an mcseF w ith
F
E
=T. Th is co rres p on dence,thoug hposs iblynot one to one, tran sfers thewell-known
structureo fthe co llectionofmcstreeso fa nmcstproblemonto theseto fmcsextensio ns
Zumsteg (1 992 )de ned twoplayersi;j in agame (N;c) to bema rio nettes if
c(S[fig)=c(S [fjg)=c(S[fi;jg)
for all S N. Cons iderin g players to b e marionettes of themselves turns b eing
mar-ionettes into a n equiva lence relation, we d eno te it by . Fo r any player i, the s et of
marionettes of iis deno ted by S
i .
De nition 3.2 Fo ra ga me(N;c), the marionette-red ucedga me (N 0 ;c 0 ) isthe gamein which N 0 =fS i
j i 2Ng, and wh ich sa tis es c 0 (C) =c( S S2C S) for all C 2 N 0 . Hence
a player inthe ma rio nette-reduced ga me con sis tsof all marionettes o f one p layer inthe
originalg ame.
Equivalently,o neco uldobtainthe marionette- reducedga mea sas ubga meoftheo rig ina l
ga me : fo r each player U 2 N 0
, ta ke o ne representa tive player j
U 2 U an d d e ne T =fj U jU 2N 0
g. D e ne the s ubga me(T;c T
) by
c T
(U)=c(U)
for all U T. Fo r every player i in T, there is a unique player S
i in N 0 sa tisfying j S i
=i and fo r every player U in N 0
, there is a un ique player j
U
in T satisfying S
j
U =
U. Furthermore this b ijection between the players of T a nd N 0
turns out to b e an
isomorphism betweenthe g ames (N 0 ;c 0 ) and (T;c T ) : c 0 (C) = c( [ S2C S) = c(fj S jS 2Cg) = c T (fj S jS 2Cg) c T (U) = c(U) = c( [ i2U S i ) = c 0 (fS i ji 2Ug)
for all coa litio nsC N 0
a nd U T.
Lemma 3.3 If (N;c)is agame, and(N 0
;c 0
)isitsmarionette- redu cedga me,their cores
are rela ted as follows :
1. if x2Co re(N;c), theny 2Core(N 0 ;c 0 ), where y2IR N 0 is de nedby y S = P i2S x i for a ll S 2N 0 . 2. if y 2 C ore(N 0 ;c 0 )\IR N 0 +
, then x 2 Core(N;c), for all x 2 IR N + s atisfying y S = P i2S x i for all S 2N 0
. Moreover,such an x exists.
Proof : The proof o f part 1 is trivia l. To prove pa rt 2,take y2 Core(N 0
;c 0
). We rst
provea nxwhichsatis estherequirementsexists . Fo rallS 2N 0
,cho o searepresenta tive
playeri S 2S an d as sig n x i := ( y S if i=i S for an S 2N 0 ; 0 otherwise:
Becaus e y is non- neg ative, so is x. Now x(N) = x(fi S j S 2 N 0 g) = y (N 0 ) = c 0 (N 0 ) =
c(N) a nd for any co alition T N, there exis ts a subset T 0 o f T, such that S i2T S i = S i2T 0S i
, where the rig ht-hand sideis adisjointunion. Hence,
c(T) = c( S i2T S i ) = c( S i2T 0S i ) = c 0 (fS i ji2T 0 g) P i2T 0y S i = P i2T 0x(S i ) = x( S i2T 0S i ) x(T);
whichimplies x sa tis es the requirements .
Nowta keany x2IR N + sa tisfyingy S = P i2S x i for all S2N 0 . T hen x(N)= X S2N 0 x(S)= X S2N 0 y S =y(N 0 )=c 0 (N 0 )=c(N)
andfo r a ny co alition T,take ag ainthe subsetT 0 su chthat S i2T S i = S i2T 0 S i ,wherethe
right-hand sideisadisjointunion. Thenbecaus exisno n-nega tiveandy 2Co re(N 0 ;c 0 ), x(T) X i2T 0 X j2Si x j = X i2T 0 x(S i ) = X i2T 0 y Si c 0 (fS i ji2T 0 g) = c(T 0 ) = c(T) Hence,x2Co re(N;c): 2
Lemma 3.4 Fo r a g iven mcse problem <N;3;w ;E > with mcse g ame (N;c M
) and
as sociated minimum cost sp anning treep roblem <N
E ;3
E ;w
E
>, the mcs t game asso
-cia ted with <N
E ;3
E ;w
E
>co incides with the ma rio nette-reduced ga me o f (N;c M
).
Proof : It isclear tha t two playerswhich are inthe s amecomp o nent of <N 3
;E >are
marionettes in the mcseg ame : if either one is con nected to the s ource, so is the o ther,
so the cost of conn ectin g o neis the cost of connecting both. Hence, the p layers in N
E ,
b eing components of <N 3
;E >,are coalition sof ma rio nettes.
On the o ther ha nd, if two players are marionettes, it means tha t conn ecting one h as
the sa me cost a s connecting the other or co nnecting both. Beca use the co stof a ll edges
is p o sitive,it follows that b oth playersmus t liein the same comp o nent. 2
We now de nethe irreducible core o f a n mcse problem. It is a stra ig htfo rward
De nition 3.5 Given an mcse pro blem M = <N;3;w ;E > and an mcse E 0
, d e ne
the irreducible core IC (M;E 0
) of M with respect to E 0
as follow s : cons ider the set
Va r(M;E 0
)o f all mcse problems o btainedfromM by va rying theweightw (e) of edges
e62 E 0
, that still have E 0
as mcse. Now IC(M;E 0
) is the inters ectio n of the cores of all
mcs e g ames as sociated w itha n mcs e problem inVar(M;E 0 ), i.e. IC (M;E 0 ):= \ Core(N;c M 0 )jM 0 2Var(M;E 0 ) :
If the set E of initially present edges is empty, the present de nition co incides with the
de nition o f irreducible co re of an mcs t pro blem in Bird (1 97 6). Fo r mcstp ro blems, it
is already known that the irred ucible core is indep endent ofthe mcst used tode ne it.
Equivalently,one co uldde neth eirreducib lecore asfollow s : givenan mcseproblem
M=<N;3;w ;E > and an mcse E 0
, fo r any two players i;j 2N, let P
ij b e a path in the gra ph < N 3 ;E [E 0
> fro m i to j. It is pos sible that this path is not un ique, but
the part of the pa th in E 0
is . De nea new weightfunction
w (i;j):=ma xfw(e)je2P
ij \E
0
g: (3:3)
Then the fo llowing ho lds:
Lemma 3.6 The irreducib le core IC(M;E 0
) of a mcse pro blem M = <N;3;w ;E >
coincideswith the core o f the ga me (N;c) =(N;c
< N;3;w ;E >
).
Proof : It is easy to s ee that E 0
is a n mcse o f the pro blem w ith red uced weights
< N;3;w;E >. Hence, the irreducible co re of M is included in the core of the ga me
(N;c). Conversely, if E 0 is an mcse o f a problem M 0 2 Va r(M;E 0
), then the weight
of a ny edge e in the pro blem M 0
has to b e larg er tha n the weig ht w(e). Hence, the
prob lem < N;3;w ;E > is the problem with the s ma lles t weig hts o f a ll pro blems in
Va r(M;E 0
). This implies that C ore(N;c) is included in the Core(N;c M 0 ), for a ny M 0 2Var(M;E 0
)and hencetha t Core(N;c)IC (M;E 0
). 2
Th e next propositio n states the rela tio n b etween the irreducible core o f a n mcse
prob lemand th e associa tedmcstproblem.
Proposition 3.7 For a n mcs e problem M= <N;3;w;E > and a n mcse E 0
, the
irre-ducib lecore IC(M;E 0 ) s atis es IC(M;E 0 )=fx2IR N + jy2IC(<N E ;3 E ;w E >) where y S := X i2S x i 8S 2N E g
Proof : T hisfollows ea silyfro mlemma ta 3.3and3.4and thefa ctthatanmcs eistra
ns-formed into an mcst by the tra nsitio n from mcse pro blem to a ssocia ted mcst pro blem.
2
Corollary 3.8 Theirredu cibleco reofa nmcsep roblemisindep endentofthemcseused
Accordingly, we willdeno te the irreducible co re of a nmcs e p roblemM by IC(M).
Having g ivenso me propertieso f the irreducib leco re,wenowproceedto prove th atit
coincideswith the s et o f allocationsgenerated by algorithm2.3.
Lemma 3.9 Let M<N;3;w ;E > be an mcseproblem. T hen D GK
(M)IC(M).
Proof : In section 2 we sta ted that for a n mcs e problem M<N;3;w ;E >, the set
D GK
(M)o fallocatio nsg eneratedbythe algorithm2.3app liedtoMarecoreallocatio ns
oftheass o ciatedga me(N;c M
). Sincetoproveth isinsection7 ,weo nlyus etheweights
of edges in the mcse E 0
, which is an mcse in all mcse problems M 0 2 Var(M;E 0 ), it holds tha t D G K
(M) is a su bset o f th e core of any mcseg ame a ssociated with an mcse
prob lemin the set Va r(M;E 0
), de nedabove. Hence,
D G K
(M)IC (M): (3:4)
2
In o rder to prove the revers e inclus io n,we need the following lemma.
Lemma 3.10 Let T =<N;3;w> b e a nmcstpro blem a nd let <N 3
;T >bean mcst
for T. Then Bird's treeallocatio n T
lies inth es et D G K
(T).
Proof : The number of ed ges in T equals n := jNj. Co nsid er any sequen ce E =
(e 1
;:::;e n
)o f edges o btained by o rdering the edg es o f T by non- decreasingcost. D e ne
F =(f 1 ;:::;f n ) by f t i := ( 1 if e t
is the rst edg e on the pa th in <N 3
;T >from i to the source,
0 otherwis e. It follows that F 2V E (T) a nd tha t T =x E;F 2D E =D G K (T): 2
Theorem 3.11 Let T b e a nmcs t problem. Then D GK
(T)=IC (T).
Proof : It follows from lemma 3.9 tha t we only have to prove IC(T) D G K
(T).
Bird(19 76 )p rovedIC(T)is theco nvexhull ofthe setofallBirda lloca tion sof the mcst
prob lem e
T, with reduced weight fu nction de ned by equa tio n3 .3 . By pro p o sition 3.10 ,
these Bird a lloca tio ns lie in D G K ( e T). Now D GK ( e T) equals D G K
(T), b ecause the set
D GK
is o btained by o nly co nsid ering the weights of ed ges in an mcst, and these edges
have the sa me weight in e
T as in T(cf. Aa rts and Driess en (19 93)). Mo reover D GK
(T)
is convex, h ence
IC(T)=conv hullf T jT isan mcst of e TgD G K ( e T)=D GK (T): 2
Corollary 3.12 Let Mbe an mcseprob lem. ThenD GK
(M)=IC(M).
Proof : L et x 2 IC(M), and let T =< N
E ;3
E ;w
E
> b e the mcst problem a ssociated
with M. We knowby pro p os ition 3.7 th atth evecto ry2IR NE ,de nedby y S = P i2S x i forallS 2N E ,liesinIC(T)(=D G K
(T)). Hence,thereexistsasequenceE =(e 1
;:::;e
)
of edges lea ding to a n mcst of T an d a sequence F = (f 1
;:::;f
) of fractio n vectors
valid for E, such that y = x E;F
. Now fo r each edge e t = fC i ;C j g, there exists an edge e~ t
with same weight in the weighted gra ph < N 3
;E
N 3;
w >, which con nects the
comp o nents C i and C j . Hence, e E =( ~e 1 ;:::;e~
) isa sequencelea din g to an mcseof M.
De ne e F =( ~ f 1 ;:::; ~ f ) by ~ f t i = ( f t Ci x i y C i if y C i >0 0 if y C i =0:
for all t and all i 2N, where C
i
is the comp onent containing i. Then e F is valid for e E. Moreover, fo r a ny player i, x e E; e F i = 0 if y C i = 0 , but then a ls o x i = 0, a nd if y C i 6= 0 , then x e E; e F i = X t= 1 f t C i x i y C i = x i y C i X t= 1 f t C i =x i y C i y C i =x i : Hence,x=x e E; e F 2D GK
(M), which co mpletes the pro o f. 2
4 The equa l rema ining obligations value
In most ca ses, the irreducible co re of a n mcs e problem M co nta ins a co ntinuum of
allocations . If the objective is to choo se a division o f the cos t, one mig ht be b etter o
with a one-point solutio n.
Th eequal re main ing obl igation svalue (henceforth EROvalue), sugg estedbyJos
Pot-ters, isaon e- p ointre nementof theirredu cible corea nd iscons tru cted by thefo llowing
extension of a lg orith m2 .3 .
Algorithm 4.1 (Equa l remainin g obligations solution)
input : a n mcs e p roblem<N;3;w ;E >
output : a n mcs e a nd the EROva lue
1. Given M<N;3;w;E >, de ne
t = 0 the initia ls tage,
= jN
3
=Ej01 the numb er of stag es ,
E 0
= E the initia ledg e set.
2. t:=t+1 .
3. Ats tage t,givenE t0 1
,chooseacheapes tedgee t
suchthat theg raph<N 3 ;E t01 [ fe t
g> doesno t contain more cycles than the graph <N 3
;E t0 1
4. If C t = C t0 1 i [C t0 1 j
is the co nnected comp o nent ju st formed by co nnecting the
comp on ents C t01 i an d C t0 1 j o f the gra ph < N 3 ;E t01 > w ith th eedg e e t = fi;jg,
de ne the vectorf t =(f t k ) k2N of fra ctions by f t k = 8 > > > > > > > > > > > > > < > > > > > > > > > > > > > : 1 jC t01 i j 0 1 jC t j if k2C t0 1 i and362C t ; 1 jC t01 j j 0 1 jC t j if k2C t0 1 j and362C t ; 1 jC t01 i j if k2C t0 1 i and32C t01 j ; 1 jC t01 j j if k2C t0 1 j and32C t01 i ; 0 o therw is e: 5. De neE t :=E t01 [fe t g. 6. If t< , g oto step 2 . 7. E
isthemcs ewesoug ht. Asbefo re,d eno te E =(e 1
;:::;e
)thesequenceo f edges
constructed. 8. De neERO E (M):= X t= 1 f t w (e t ):
Applied to the mcse problem o f example 2.2 , this algo rithm g enerates s uccessively
edge f1 ;2g,of which players 1 and 2 each pay 10 1
2 =
1
2
, edge f2 ;3 g,of which player 3
pays 10 1 3 = 2 3
and players1and 2ea ch pay 1 2 0 1 3 = 1 6
,edgef3 ;4g,o fwhichplayers1,2
and 3 each pay 1 3 0 1 5 = 2 15
w hile p layers 4 and5 pay 1 2 0 1 5 = 3 10 and n ally ed ge f2;3g,
to which eachplayercontributes 1
5
. T his yields the allocation
E RO(M)= 1
3
(10 1;10 1;11 6;96 ;9 6):
G enerica lly,thecho iceofedgeinstep3isu niquebutevenintheca sethatthesequence
E isn otuniquelyde ned,this algorithmyieldso nly onea llo cation,ind epend ently ofthe
cho iceofed gesma de. T hisincontrastwithBird'streeallocationrule,thatmaya ssociate
a di erent allo cation withea ch mcst o f an mcstpro blem.
Proposition 4.2 For any two sequences of edges E and e
E chos en by th e algo rithm4.1
app lied to a nmcs eproblem M,
ERO E (M)=ERO e E (M):
Thepro o fissimila rtotheproofofpropositio n2 .1 0,sowedonotgiveit. T hispro p os ition
allows us to d e ne
ERO(M):=ERO E
(M)
for a ny sequence E con structed by algorithm4 .1 . Clea rly, the fra ctions con structed are
To see tha t the ERO value deserves its n ame, d e ne fo r an mcse p roblem M the initial ob ligation o i of aplayer iby o i := ( 1 jC i j if 362C i 0 if 32C i (4:1) where C i
is the comp o nent o f <N 3
;E >co nta in ing player i.
For asequenceF =(f 1
;:::;f
)of fra ction vecto rs,a fterasta ge t ap layeri2N
hasp aid P s t f s i
,whiletheinitialoblig atio nwa so
i
. Hencep layeri'sre main ingoblig ation
o t i satis es o t i =o i 0 X s t f s i =o t0 1 i 0f t i : (4:2)
Theorem 4.3 The algo rithm4 .1 hasthe pro p erty th ata fter ea chs tage t,ineach
com-p o nent C of th e g raph <N 3
;E t
>, every player k in the co mponent C h as the sa me
rema ining oblig atio n
o t k = ( 1 jCj if 362C ; 0 if 32C : (4:3)
Proof : The proof goes by indu ction o n the sta ge t.
1. After s ta ge zero, fork ina comp o nent C of <N 3 ;E >, o t k =o k 00= ( 1 jCj if 362C ; 0 if 32C :
2. Supposeequa tion4 .3ho ldsa fterstag et01. L etC t =C t01 i [C t01 j b etheco nnected
comp on ent fo rmed a tsta ge t by co nnecting the co mp onentsC t01 i a nd C t0 1 j of the gra ph<N 3 ;E t0 1
>withtheedg ee t =fi;jg. L etC t k a ndC t01 k
bethecomp on ents
of p layer k in <N 3 ;E t > a nd <N 3 ;E t0 1 >. Then o t k = o t01 k 0f t k = ( 1 jC t01 k j 0f t k if k 62C t01 3 000 if k 2C t01 3 = 8 > > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > > : 1 jC t01 i j 0( 1 jC t01 i j 0 1 jC t j ) if k 2C t01 i a nd 362C t 1 jC t01 j j 0( 1 jC t01 j j 0 1 jC t j ) if k 2C t01 j a nd 362C t 1 jC t01 i j 0 1 jC t01 i j if k 2C t01 i a nd 32C t01 j 1 jC t01 j j 0 1 jC t01 j j if k 2C t01 j a nd 32C t01 i 1 jC t01 k j if k 62C t [C t0 1 3 0 if k 2C t01 3
= 8 > > > > > > > > > > > < > > > > > > > > > > > : 1 jC t j if k 2C t 633 0 if k 2C t01 i a nd 32C t0 1 j 0 if k 2C t01 j a nd 32C t0 1 i 1 jC t01 k j if k 62C t [C t0 1 3 0 if 32C t01 k = 8 < : 1 jC t k j if 362C t k 0 if 32C t k :
Henceequa tio n 4 .3holds a fters tage t a swell. This completes the pro o f. 2
5 Axiomatic characterizatio ns
Ins ections2a nd4weintroducedtheirreducibleco rea ndtheequalremainingoblig atio ns
value for mcseprob lems . We axiomatica lly chara cterize these rules inthis sectio n.
In contrast with the chara cteriza tion s in Feltka mp, T ijs a nd Mu to (19 94a ) and
Feltkamp, Tijs and Muto (19 94b), here, a s olutio n con sis ts only o f a s et of allo cations .
This is p o ssible b eca use both the irreducib le core a nd the ERO ruleare indep endent of
the set o f edg es constructed.
An allocat ion of a n mcse problem<N;3;w ;E > is a vector x 2 IR N + w hich satis es P i2N x i c M
(N). In e ect,a n a lloca tio nis a vector thata lloca tesat leas tthe cost of
a minimumco sts panning extensio nto the players.
Pro p erties that a n allo cationx of an mcsepro blem <N;3;w ;E >can s atisfy are
De nition 5.1 E x is ec ient if X i2N x i =c M (N):
MC x has the min imal contribut ion prope rt yif everycomponent th atdoes not contain
the s ourcecontributes a tleast the co stof aminimum co st ed geth at co nnectstwo
comp on ents. In formula : fo r ea ch comp o nent C 2 N 3
= E that does not contain
the so urce,
X
i2C x
i
minfw (e)j e connects two comp o nents of <N 3
;E >g:
FSC x ha s the free for sou rce compon ent property if x
i
= 0for alli in the comp o nent
of th es ource inthe grap h<N 3
;E >.
lea stthecosto fa nedg e,a ndthecomp o nentofthesources houldnotco ntribute,b ecause
it is alrea dy connected.
A solution o f mcse problems is a map a ssigning to every mcse problem a set of
allocations .
De nition 5.2
NE A solution is sa id to benon- empty if
(M)6=; fo r allM:
We will s ay a solutio n is ecient, satis es th e minima l co ntributio n pro p erty or the
freeforso urceco mp onentpropertyiffo rallMa llelementsoftheso lution (M)sa tisfy
the corresp o nding pro p erty.
De nition 5.3 Givena nmcseproblemM<N;3;w ;E >andanedgee=fi;jg62E
that co nnectstwo comp on entso f <N 3
;E >,de ne the edge -reduced mc se prob le m
M e
=<N;3;w ;E[feg >:
Notethat theed ge-reducedpro blemisindeed asimp lerproblem tha nthe o rig ina lpro
b-lem: les sedg es haveto b e cons tructed. However,it h asthe sa me numb er o f playersas
the original pro blem.
Th e next three properties us e edg e- reduced mcse problems, w ith as extra edg e an
edge w hich co nstructs a n ew comp o nent if it is adjoined to the gra ph <N 3
;E >, and
whichhas minimumco sta mo ng a ll edges w iththis property.
De nition 5.4
Lo c is l ocal if for all M, for all x 2 (M), for all minimum cos t edge e that when
add ed to <N 3
;E > cons tructs a new comp o nent C, there existsa n x~2 IR C s uch that (~x;x NnC )2 (M e ):
In e ect, this axio m requires tha t when a minimum cost edg e is a dded , this h as no
in uence on the allocatio n to players that are not in the comp o nent constructed by
add ing this edge.
ECons is minimum cost edg e consiste nt if fo r all M <N;3;w;E >, for all x 2
(M), fo reachminimumcost edg ee tha t w hen add edto <N 3
;E >constructs a
new comp o nent C,for each 21 C satisfying w(e)x C , it h old s tha t x0x e; 2 (M e ); where x e; :=( w (e);0 NnC ).
This axio m mea ns tha t w hen a minimum cos t edge is added, the savings obta ined by
not having to constru ct this edg e can be allo cated arbitrarily over the players that are
CEC ons is conv erse min imu m cost edg e con sisten t if fo r a ll M <N;3;w ;E >,
for every min imu m cost edge e tha t w hen a dded to <N 3
;E > co nstructs a new
comp on ent C,fo r everyx2IR N th atsa tis es a MC, b FSC , c w (e)x C imp lies x0x e ; 2 (M e ) fora ll 21 C , it holdstha t x2 (M):
Thisa xiomrequiresthatifaddinga nallocationto thesolutio ndoesnotdestroytheMC ,
FSC a nd ECo ns pro p erties, then its hould b e pa rt of the solutio n. In e ect, it requires
the so lution to bethe larg est so lution th atsa tis es the other pro p erties.
Lemma 5.5 Theirreducibleco resa tis es thepro p ertiesNE, MC ,E , FSC,E Cons and
CEC ons.
Proof : Beca use o f the coincidence o f the irreducible co re with the set D GK
, the
irre-ducib le core is n on-empty fo r a ny mcse pro blem : there are always va lid s equences of
fractio n vectors for a ny s equenceof edges con structed by the a lg orith m2 .3 . It satis es
the minimumco ntributionpro p erty becaus eeveryco mp onentthat doesno tco nta inthe
so urce has to pay for fractions of edges that tota l on e, so it co ntributes at leas t the
minimumcost ofa ned gethatconnectstwoco mponents. That itisecient isprovedin
lemma 2 .4. By cons truction, itis clear th atD G K
sa tis es FS C.
To prove edge cons istency, ta ke an mcse prob lem M a nd sup p os e x 2 IC (M) =
D GK
(M). Fo r ea ch minimum cost edge e th at when added to <N 3
;E > constructs
a new comp o nent C, there exists a sequen ce E = (e = e 1
;e 2
;:::;e
) s tartin g with the
edgee,tha tisconstructedbythealgo rithm2.3. BecausethesetD G K
(M)isindependent
ofthesequenceo fedg esco nstructed,th ereexistsasequenceF =(f 1 ;:::;f )2V E s uch that x=x E;F . For a ny 21 C satisfying w (e)x C ,de ne 1 21 C by 1 i :=f 1 i for alli2C. Then 8 > > > > > < > > > > > : X i2C i 0 1 i = 0 ; ( 0 1 )w (e 1 ) X t=2 f t w (e t ); f t w (e t ) 0 fort 2 :
Hencethere exist vectors 2 ;:::; 2IR C satisfying 8 > > > > > < > > > > > : t i f t i fo rt 2and fora ll i2C ; X i2C t i = 0 for t 2; ( 0 1 )w (e 1 ) = X t w (e t ): (5:1)
E.g. take t
theprojectiono f f t
on the hyperplane withcoo rdina tes zero,along theline
throu gh the points P s=2 f s w (e s ) and( 0 1 )w (e 1 ), i.e. t =f t 0 P i2C f t i P i2C P s= 2 f s i w (e s ) ( X s= 2 f s w (e s )0( 0 1 )w(e 1 ))
for all t 2 . Rew riting th elas t equa tio nof system(5 .1 ),we o btain
w (e)= X t=1 t w (e t ): Itfollow stha tx 0x e ; = P t= 2 (f t 0 t )w (e t
). The rsttwoequa tio nso fsystem(5.1)imply
that th es equence(f 2 0 2 ;:::;f 0 )is validfo r (e 2 ;:::;e ). Hence, x0x e; 2M e .
Beca use ed geco nsistency imp lieslocality,theirreducibleco resatis eslo ca litya swell.
To prove that the irreduciblecore satis es CEC ons, ta ke a n mcse problem M, take
a minimumco stedg ee tha t when a dded to <N 3
;E >cons tructs anewco mp onent C,
take an x2IR N tha t sa tis es a MC , b FSC , c w (e)x C implies x0x e ; 2IC(M e ) fo r all 21 C . Wehave to proveth at x 2IC (M): D eno te by C 1 an d C 2
the two components th at are joined by e. T he a llo ca tion x
sa tis es FSC , hence if o ne of these co mponents (say C
1
) contains the so urce, x
i = 0 for i 2 C 1 a nd a n 2 1 C with x C sa tis es i = 0 fo r i 2 C 1 . For s uch an
(which exists), there exis ts a s equence (e 2
;:::;e
) cons tructed by the a lg orith m 2.3
app liedto the pro blemM e a nd as equence(f 2 ;:::;f )2V (e 2 ;:::;e ) suchthat x0x e ; = x (e 2 ;:::;e );(f 2 ;:::;f ) : So with f de nedby f k := ( k if k 2C 0 oth erwise
for all k 2N,it ho lds that (f;f 2
;:::;f
)is va lidfor the sequence (e;e 2 ;:::;e )a nd x=x (e;e 2 ;:::;e );(f;f 2 ;:::;f ) 2 (M):
Ifneitherofthetwocomp o nentscontainthesource,thenbytheminimalco ntribution
property, b oth comp o nents co ntribute a t lea st w(e) in the allocation x. On the other
han d, both together contribu te x(C),s o there exists an a 1 2[0;1 ], suchtha t ( x(C 1 ) = a 1 w (e)+(10a 1 )(x(C)0w(e)) x(C 2 ) = (10a 1 )w (e)+a 1 (x(C)0w(e)) De ne 21 C ,by i = ( a 1 x i = x(C 1 ) if i2C 1 ; (10a 1 )x=x(C ) if i2C :
Then, w (e)x C
. Hence,thereexistsamcsefe 2 ;:::;e ga ndasequence(f 2 ;:::;f )2 V (e 2 ;:::;e ) (M)s uchtha tx 0x e; = P t=2 f t w (e t ). De nea 2 1 ;:::;a 1 bya t 1 =(10a 1 ) P i2C f t i and a 2 2 ;:::;a 2 by a t 2 =a 1 P i2C f t i . Then 8 > > > > > > > > < > > > > > > > > : a t 1 +a t 2 = X i2C f t i fo r allt 2 a 1 + X t=2 a t 1 = a 1 +(10a 1 ) X t=2 X i2C f t i = 1 10a 1 + X t=2 a t 2 = 10a 1 +a 1 X t=2 X i2C f t i = 1 De ning g 1 =( ;0 NnC )a nd g t i = ( a t 1 x i =x(C 1 ) if i 2C 1 a t 2 x i =x(C 2 ) if i 2C 2 we seethat (g 1 ;:::;g )2V (e;e 2 ;:::;e ) (M). Furthermore, x(C 1 )0a 1 w(e) = (10a 1 )(x(C)0w (e)) = (10a 1 ) X t=2 X i2C f t i w (e t ) = X t=2 a t 1 w (e t ):
This implies that for i2C
1 , x i = x i x(C 1 ) x(C 1 ) = x i x(C 1 ) (a 1 w (e)+ X t=2 a t 1 w (e t )) = g 1 i w (e)+ X t= 2 g t i w (e t ):
Simila rly for i2C
2 . Hen ce, x=x (e;e 2 ;:::;e );(g 1 ;:::;g )
is inthe irredu cible co re of M. This
implies that the irred ucible core s atis esconverseedge cons istency. 2
Lemma 5.6 If a so lution of mcse problems sa tis es MC, FSC an d E Cons , a nd a
so lution of mcse prob lems sa tis es NE , FSC an d CEC ons, then (M) (M) for
each mcsepro blem M.
Proof : We prove the lemma for any mcse pro blem M by induction o n the numb er
of comp o nents o f the gra ph <N 3
;E >. First, cons ider rst an mcs e pro blem M
<N;3;w ;E >, where the gra ph <N 3
;E >is connected. Then by FSC, x =0 for a ny
Now suppose the lemma holds for every mcse problem Mwith p01 co mponentsin
theg raph<N 3
;E >. Ta kea nmcseproblemMs uchthat<N 3
;E >ha spcomp on ents
and ta kex 2 (M). By E Cons o f , fo r a ny minimum co stedg ethat constru cts a new
comp o nent C when a dded to <N 3
;E >, for ea ch 2 1 C
s atis fying w(e) x C , it holds that x0x e; 2 (M e ) (M e );
where the inclusion holds by th e indu ction hypothesis . Beca use x sa tis es MC,it holds
that x2 (M) by CECo nso f . Hence (M) (M): 2
Theorem 5.7 Theuniquesolutionofmcsepro blemstha tsa tis esNE,MC,FSC,ECo ns
and CECo ns is th e irreducible co re.
Proof : By p rop o sition 5.5, the irreducible core ha sthese p ro p erties. By lemma 5 .6 , if
two so lutions have thesepro p erties, theycontaineach othera nd hencetheycoincide. 2
To cha racterisethe EROva lue,weneedsome oth erprop ertiesthat as olu tion can have.
De nition 5.8
ET aso lution satis esequalt reatmentifforeverymcsep roblemM,forallx2 (M),
foreachco mponentCoftheoriginalgraph<N 3
;E >,an d fo rallpa irso fplayersi
and j 2C, x i =x j :
IPC ons Aso lution isin versel yproportionalcon sist entiffo reverymcsepro blemM
<N;3;w ;E >, for every minimum cos t ed ge e that wh en added to <N 3
;E >
connects two co mponents C
1
and C
2
, neither of wh ich co nta ins the s ource, fo r all
x2 (M), there existsan x~2 (M e ) su ch tha t jC 1 j X i2C 1 (x i 0x~ i )=jC 2 j X i2C 2 (x i 0x~ i ):
Proposition 5.9 The uniqu e s olutio n of mcse prob lems that satis es NE, FSC, Loc,
E ,ETa ndIPCon sistheE ROvalue. Here,theEROva lueisidenti edwiththesolution
that ass ign sthe singleton fERO(M)gto every M.
Proof : First we prove tha t the ERO value sa tis es the required properties. T hat the
ERO va lu e sa tis es the properties NE, MC, FSC, L o c and E is a co nsequence o f its
b eing are nement o f th eirreducible co re. Tha t is sa tis es equaltreatment is also easy
to see. To prove it satis es IP Cons , take a n mcse problem M, a nd a minimum cost
edge connecting the comp o nentsC
1
a nd C
2
,neithero f whichconta ins the source,into a
comp o nent C. Then there exists a sequence o f edg es E =(e =e 1 ;:::;e ) a sequence of fractio nsF =(f 1 ;:::;f
)constructedbythea lg orithm4.1suchthatE RO(M)=x E;F
Moreover, by de nition of the algo rithm, x (e 2 ;:::;e );(f 2 ;:::;f ) = ERO(M e ). Beca use e
connectstwocomp o nents that d o no tcontain the so urce,
x E;F k 0x (e 2 ;:::;e );(f 2 ;:::;f ) k = 8 > > > > < > > > > : w(e)( 1 jC1j 0 1 jCj ) if k 2C 1 ; w(e)( 1 jC2j 0 1 jCj ) if k 2C 2 ; 0 if k 62C : Then X k 2C1 (ERO(M)0E RO(M e ))=10 jC 1 j jCj = jC 2 j jCj
and simila rly,
X k 2C 2 (ERO(M)0ERO(M e ))= jC 1 j jCj : Hence jC 1 j X k2C 1 (ERO k (M)0ERO k (M e ))= jC 1 jjC 2 j jCj =jC 2 j X k 2C 2 (ERO k (M)0ERO k (M e )):
To prove un iquenes s, suppose a solutio n satis es these s ix properties. We prove
(M)=fERO(M)gbyinductiono nthenumb erofcomp on entso fth eg raph<N 3
;E >.
L et <N 3
;E> have one co mponent. By FSC,x
i
(M)=0=ERO
i
(M) for alli 2N
and all x2 (M).
Sup p os e (M) = fERO(M)g fo r a ll mcs e problems M such that <N 3
;E > h as
less than p co mponents. C onsider an mcse pro blem M such tha t <N 3
;E > has p
comp o nents. Ta ke a minimum cos t spa nning edge e that connects two comp o nents C
1
and C
2
into a new comp o nent C in<N 3 ;E >. By ET of ap plied to M and M e , for allx2 (M), fo r all x~2 (M e ), for a ll i;j 2C 1 we have x j 0x~ j =x i 0x~ i =: 1 (x;x)~ and for a lli;j 2C 2 we have x j 0x~ j =x i 0x~ i =: 2 (x;x):~ Moreover,by FSC of , if C 1
co nta ins the sou rce,then
1
(x;x)~ =0 and by loca lity and
eciency,thereexis tsa nx~2 (M e )s uchthat 2 (x;x)~ = w (e) jC 2 j . IfC 2
co nta insthesource,
then s imilarly o neproves that
1
(x;x)~ a nd
2
(x;x)~ a reuniquely determined. If neither
of C
1
a nd C
2
contain the so urce, then by IP Cons ,th ereexists an x~2 (M e ) suchtha t jC 1 j X k2C1 1 (x;x)~ =jC 2 j X k 2C2 2 (x;x);~
and by locality andeciency,
X i2C 1 (x;x)~ + X i2C 2 (x;x)~ = X i2C (x i 0x~ i )=w (e):
Hence, 1 (x;x)~ = w(e) jC 1 j 2 and 2 (x;x)~ = w (e) jC 2 j 2 : So whether C 1 o r C 2
contain the so urce o r not, the numb ers
1
(x;x)~ and
2
(x;x)~ are
uniqu ely determined a nd indep endent o f x and x.~ Now th e ERO va lue als o satis es
the six properties and so ha s these sa me numbers
1
(x;x);~
2
(x;x)~ cha racterizing the
di eren cebetweenE RO(M)a nd ERO(M e
). The inductio nhyp othesis th en implies
x0 1 (x;x)1~ C1 0 2 (x;x)1~ C2 =x~=ERO(M e )=E RO(M)0 1 (x;x)1~ C1 0 2 (x;x)1~ C2 and so x=ERO(M): 2
6 Co ncluding remarks and sugg estions for f urt her
research
New in th is pap er is the integ rated appro ach : we solve the problem of constructing
an o ptimal n etwork a nd at the sa me time a llo ca te the cos ts. Beca use o f this integrated
viewo fthe problem,thedi erentalgo rithmsto co nstructaminimumco stspa nning tree
sug gested s evera lclos ely related a lg orith msfor so lutions to the cos ta llo ca tion pro blem.
For in stance,Prima nd D ijkstra'salgo rithmappea redto beclosely linked to Bird's tree
allocations . This s uggested to to lo o k fo r allocation rules that a rerelated to Kruska l's
(19 56 )algo rithmforco nstructinga nmcst. Inthe presentpaperwerela tetheirredu cible
core to Kru skal's algo rithm, an d moreover provide a on e-point re nement, the equa l
rema iningoblig ationssolution. Athirdalgo rithmfo rconstru ctinganmcstisknown,and
inFeltkamp, Tijsan d Muto(1 994 b) weass o ciate an alloca tio n rulewiththis algorithm.
Second,ins tea do f lo o kingonlya tthe extremecase where noedg esa repres entatthe
b eg inning, and a spa nnin g n etwork h as to b e co nstructed, we consider pro blems w here
so me network ca n be present a lready, construct a minimum cost spa nning extension
and prove that the cos t o f this extension ca n b e a lloca ted in a a sta ble way. This h as
two adva nta ges. The math ematica l a dvantag e is that a half-so lved pro blemis ag ain in
the sa me cla ss o f problems, which allows for a recursive so lution, the advantage from
an a pplied p oint o f view is that no t o nly prob lems in which all edges have yet to b e
constructed are treated, but als o extens io ns of n etworks can be solved. If the o rig ina l
prob lem wa s su ggested by among others electri ca tion of Moravia at th e begin ning of
the century,by nowthe p roblemis mo rehowto extenda na lread ypresent network and
allocate the cos tof the extens io n.
Th ird, we provide axio ma tic characterizations of the irreducible core and the equa l
rema ining o bliga tio ns so lution. These axio ma tic cha racterizations enab le one to s elect
an allocation rule,ba sed o nthe properties the ruleshou ld h ave.
Anotherwayofa pproa chingmcseproblemsistode netheco stofacoalition Sasthe
minima lcos to f an extension con necting S to the source,without a ny restriction on the
more op p ortunities to save cos ts. Hence,in generalthe core o f this variantis co nta ined
in the core o f the mcs e game we de ned. However, su ch a mono tonic game a ssociated
with an mcse problem <N;3;w ;E > ca n be cons idered a s an mcs e ga me according to
our de nitionas sociated to th emcseproblem<N;3;w 0
;E >where theweig hts oflinks
havebeen reduced to satisfy the trianglein equality
w 0 (fi;jg)w 0 (fi;kg)+w 0 (fk;jg) fo r alli;j;k 2N 3 :
So this d o es not introduce new games. Moreover,core elementsof the mono tonicga me
can be co mputed by a pplying algorithm2.3 to the prob lem withred uced weig hts.
A second p o ssible va ria tio nis no t to allow a co alition to use any players in its
com-plement when connecting to the so urce. This would yield a g ame with the sa me cost
for th e grandco alition,but la rger costs forthe o ther co alitions. Th eco reo f this va riant
contains the co re o f our mcse g ame and so all a lgo rithms presented in this paper yield
core elements o f the variant.
Beca use the set D G K
a nd the ERO value a re not de ned in functio n of a ga me but
ratherfro mthemcsepro blemits elf,theyareindep endentofth eg amewhicho nechoos es .
Moreover,for both va ria nts ,the irreducible co reco incid es with the set D GK
.
All alloca tio ns introduced here lie in the irreducible core of the mcse problems. In
ordertogetthewholeco reo ftheco rres p on dingga mes,o neneedstouseweig htsofedges
thatare notus edina ny min imu mcos tspan ningextension. Ingenera l,itisstillanop en
prob lem to compute the w hole core o f a n mcse g ame directly from the weights of the
edges, even if the attention isres tricted to mcstg ames .
Finally, it would be interesting to nd non-coopera tive games of which equilibria
sus tain the co o p erative so lutions pres ented here.
A cknow ledgements : Insp iring discussions with Ha rry Aarts , Peter Borm, Ruud
Jeuris sen,Micha elMaschler,Gert-Ja nOtten,JosPottersa ndHansReynierseaboutthe
subject of the pa p er are g ratefully acknow ledged.
7 Appendix
In this a pp endix, we prove theorem 2.8 and pro p o sition 2 .1 0. First, we need a few
lemmas.
Lemma 7.1 Fo r all sequen ces E cho sen by algo rithm 2.3 the constructed extension
<N 3
;E
> isa n mcs e fo r the prob lemM<N;3;w ;E >.
P roof : There are +1 = jN 3
=Ej comp o nents in <N 3
;E >, a t each stage two
comp o nentsare co nnected,s oa fters tage ,the resultinggra phisconnectedandn onew
cycles have beenintroduced. Ass ume tha t the extens io n<N 3
;E
> con structed is not
minima lin cost, i.e. there exis ts a set ofedg es ~ E co nta ining E, su ch th at <N 3 ; ~ E > is
a co nnected g raph, and
X e2 ~ EnE w (e)< X e2E nE w (e): (7:1)
Letthes equence e E =(~e 1 ;:::;e~
)co nsistof theedgesin ~
EnE orderedby non-decrea sing
weight. Equa tion7.1impliesthereexistsasmallestt suchthate t 0 =e~ t 0 fo r1t 0 <t and e t 6= ~e t . Beca use e t
is a minimum weight edg e that does n ot introduce a cycle
in < N 3 ;E t01 >=< N 3 ;E [ f~e 1 ;:::;e~ t01 g > a nd e~ t
does no t introduce a cycle in
<N 3 ;E t0 1 >,itfollows thatw (e t )w (~e t
). Co nsider theend pointsia ndj ofe t . T hey haveto b eco nnected in<N 3 ; ~
E >,h encethere exis tsa pa th fro mito j in <N 3
; ~
E >.
But no t all edges in this pa th can b e present in the graph < N 3
;E t0 1
>, o therwis e e t
would introduce acycle. Hence thereis an edge e2 ~
EnE inthis paththat comes la ter
in e E tha ne~ t ,so ecos ts a tlea stw( ~e t
),w hichisa tlea stw(e t ). NowE 0 :=( ~ Enfeg)[fe t g
is a s panning exten sio nof <N 3
;E > such that E 0
nE does no t co st more than ~
E nE,
and E 0
has o ne edge more in commo n w ith E
. Repea ting th is process eno ugh times
show sth atE
doesno tco stmo rethan ~
E. Thisisacontradiction, hencethe a ssumption
that the algorithm2.3 d o es n otlea dto an mcseis wro ng. 2
In ord er to prove tha t D E
is a subs et of the core of the as sociated mcs e game if E
is constructed by algo rithm 2.3, we n eed to compare the outco me of the algorithm2.3
app lied to relatedmcsep ro blems.
Sup p os e we have a n mcse pro blem M <N;3;w ;E > a nd an ed ge e = fi;jg
connectingthecomp o nentC
3
o ftheso urcewiththecomp o nentC
j
ofsomeplayerj inthe
gra ph<N 3
;E >. De ne ~
E :=E[feg. Cons iderthemcs eproblem f
M=<N;3;w ; ~
E >.
Distinguishthe graph s,comp o nents,edg es a nda llo cationsused inalgorithm2.3applied
to th e pro blems M an d f
M by g ivin g tho se in the latter problem a tilde. W ith this
setup, we prove twolemmata and acoro llary,whichweneed to prove theorem2 .8 .
Lemma 7.2 Fo r every s equence o f choices E = (e 1
;:::;e
) in the algo rithm 2.3 a
p-plied to M, o ne can n d a n s such that the s equence e E (~e 1 ;:::;e~ 0 1 ) := (e 1 ;:::;e s0 1 ;e s+ 1 ;:::;e
), obtained by deleting the edge e s
fro m E, is a sequence of
edges tha t can be obta ined by algo rithm 2 .3 applied to f Ma nd tha t satis es 1. (N 3 =E t )nfC t i ;C t j g=(N 3 = ~ E t )nf ~ C t i g and C t i [C t j = ~ C t i for a ll t<s,
that is, as lo ng as t < s, the gra phs < N 3 ;E t > and < N 3 ; ~ E t > have the sa me
comp on ents, except for the comp o nents o f i and j in < N 3
;E t
>, which are
connected to each other in<N 3 ; ~ E t >. 2. N 3 =E t =N 3 = ~ E t01 fora ll t2fs;:::; g,
thatis,a fterstag e s,the comp o nentsof<N 3
;E t
>co incidewithth ecomp on ents
of <N 3 ; ~ E t01
> at the previous stag e.
Proof : We prove the sta tements by inductio n on t. For t =0 : ~ E 0 = ~ E =E [feg = E 0 [feg, hence ~ C t i =C t i [C t j and (N 3 =E t )nfC t i ;C t j g=(N 3 = ~ E t )nf ~ C t i g.
If ca se 1 ho lds at sta ge t01 , look at the e ect of add ing e t
to ~
E t0 1
. Two ca ses can
occur.
a e t
6= e and a ddin g the edge e t
does not intro d uce a cycle in the g raph
<N 3 ; ~ E t0 1 >. Now e t
is a cheapest edg e w hich does not introduce a cycle in
3 t0 1 3
~ t01
does no tintroduce a cyclein <N 3 ;E t01 >. Hence e t
is a lso a ch ea p estedg e that
does not introduce a cycle in <N 3 ; ~ E t01 >. T his means e t
is a leg itimate choice
for ~e t
. Co nsequ ently, case 1 still holds at sta ge t.
b e t
=eoradding theed gee t
doesintroduceacycleinthegra ph<N 3 ; ~ E t01 >. This means e t
connects the comp o nents C t01 i a nd C t0 1 i of < N 3 ;E t0 1 >. Then C t i =C t01 i [C t0 1 j = ~ C t0 1 i
and the o ther comp onents are unchang ed,so (N 3 =E t )n fC t i g=(N 3 =E t01 )nfC t01 i ;C t01 j g=(N 3 = ~ E t01 )nf ~ C t01 i g. HenceN 3 = E t =N 3 = ~ E t0 1 ,
and ca se2 holds fo r stag e t.
Suppose cas e 2 ho ld s for stag e t01 . Then the edg e e t
= fk;l g is a legitimate choice
for ~e t0 1
(itd o es not introducea cycle,a nd has minima lcost a mo ng the edgessa tisfying
this). Hence,C t k =C t0 1 k [C t01 l = ~ C t02 k [ ~ C t0 2 l = ~ C t0 1 k ,w hichimpliesN 3 = E t =N 3 = ~ E t0 1 ,
and case 2 holds fo r stag e t as well.
Hence, if the rst stag ea t which case 2ho ldsis calleds, wes ee thatcas e1 holds for
t<s a nd cas e 2ho lds for ts. Note tha t N 3 =E =fN 3 g=N 3 = ~ E t01
, hence cas e2ho ldsat s tage , so s . 2
Lemma 7.3 Let M a nd f
M b e as ab ove, let E b e a sequence o f cho ices ma de by
al-go rithm 2 .3 applied to M, and let F b e va lid for E. L et the sequence e
E and the
sta ge s b e a s de ned in lemma 7.2. T hen there exis t ( ~
t
k )
k 2N
, s uch tha t the
se-quence e F =( ~ f 1 ;:::; ~ f 01 )de ned by ~ f t k := 8 > > > > > > > > < > > > > > > > > : f t k if t<minf ~ t k ;sg X t 0 = ~ t k f t 0 k if t= ~ t k <s f t+1 k if ~ t k ts 0 if t> ~ t k
fo r allt2f1;:::; 01 g and allk 2N
is valid for e E. Info rmula : e F 2V e E ( f M). Proof : For a ll k 2N,de ne ~ t k
:=minftjthere exists apath from k to 3 in <N 3 ; ~ E t >g; where ~ E t
is the edg e set resulting in sta ge t in th e algo rithm 2.3 ap plied to f M. Note that ~ t k =0for all k 2C j
, the comp onent of <N 3 ;E > connected to C 3 by the edg e e. Hence ~ f t k
= 0for a ll sta ges t if k 2 C
j
. This mea ns the players in C
j
do n otcontribute
to any ed ge. We nowp rove the lemmain three step s.
1. For t 01, let e~ t = fk;l g and let ~ C t = C t01 k [C t01 l
b e the comp o nent in the
gra ph <N 3 ; ~ E t
> fo rmed by the a ddition o f ~e t . Then P m2 ~ C t ~ f t m = 1. To prove
this, we d istinguish severa l ca ses :
Sup p os e ~e t is not incident to ~ C t01 3 , the co mp onent of3 in <N 3 ; ~ E t >. Then ~ t = ~ t >t for allm 2 ~ C t , hence
{ if t <s then ~ f t m =f t m fo rm 2 ~ C t . Moreover, ~ C t =C t
, the comp o nent in
th eg raph <N 3
;E t
> formed by the additio n o f ~e t =e t . Hence, X m2 ~ C t ~ f t m = X m2C t f t m =1 by the a ssumptio n o nF. { if t>s then ~ f t m =f t+1 k for m 2 ~ C t . Moreover, e~ t =e t+1 and ~ C t =C t+ 1 ,
th eco mponentintheg raph<N 3
;E t+ 1
>fo rmedbytheadditio no fe t+ 1 . Hence, X m 2 ~ C t ~ f t m = X m2C t+1 f t+1 m =1 by the a ssumptio n o nF. Sup p os e that ~e t is incident to ~ C t01 3
. Then o neo f k and l ,say k,lies in ~ C t0 1 3 . Th is means ~ t m = ~ t l =t fo r allm 2 ~ C t01 l and ~ t m <t for allm 2 ~ C t0 1 k . Hence, { if t<s,th en ~ f t m = P t 0 = t f t 0 m fo rm2 ~ C t0 1 l and ~ f t m =0for m2 ~ C t0 1 k . This implies X m 2 ~ C t ~ f t m = X m2 ~ C t01 l ~ f t m = X m2 ~ C t01 l X t 0 =t f t 0 m = X m2 ~ C t01 l ( X t 0 = 1 f t 0 m 0 t0 1 X t 0 =1 f t 0 m ): Now ~ C t01 l =C t0 1 l
isaunionofanumb er,sayp ,ofcomp o nentsC
1
;:::;C
p
o f the gra ph<N 3
;E >. Rememb ertha t for allq 2f1;:::;p g :
X m2C q X t 0 =1 f t 0 m =1 ; h ence X m2 ~ C t01 l X t 0 =1 f t 0 m = p X q = 1 X m2Cq X t 0 =1 f t 0 m =p (7:2) a nd a s C t01 l = S p q = 1 C q
contains exactly thos e players that contributed
to the p01 edgesin fe 1
;:::;e t0 1
g tha t co nnect the comp o nents (C
q ) p q =1 into C t01 l , X m2 ~ C t01 l t01 X t 0 = 1 f t 0 m =p01 : (7:3)
E qua tio ns7 .2and 7.3 imply
X m 2 ~ C t ~ f t m =p0(p01 )=1 :
{ if ts, then ~ f t m =f t+1 m , e~ t =e t+ 1 and ~ C t =C t+1
, the component inthe
g raph <N 3
;E t+1
> fo rmed by the addition o f e t+ 1 . Hence, X m 2 ~ C t f t m = X m2C t+1 f t+1 m =1 by the a ssumptio n o nF. 2. For ea chco mponentC 2(N 3 = ~
E) that do es not contain3 : C is alsoa comp o nent
of the g ra ph <N 3 ;E > (b eca use <N 3 ;E > a nd <N 3 ; ~ E > d i er only in the
comp on ent o f the sou rce). Moreover, ~ t k = ~ t l for allk;l 2C a nd if ~ t k <s for a ll k2C, then X k 2C 01 X t= 1 ~ f t k = X k2C ~ t k X t=1 ~ f t k = X k2C ( ~ t k 0 1 X t=1 ~ f t k + ~ f ~ t k k ) = X k2C ( ~ t k 0 1 X t=1 f t k + X t= ~ t k f t k ) = X k2C X t=1 f t k = 1 :
Th e rs tequalityfollow sb ecause ~ f t k =0fo rt> ~ t k
,the thirdequality b ecause
o f the de nition of ~ f ~ t k k
, a nd the fthby the as sumptio ns o nF.
if ~ t k sfo rk 2C,thenCisn otconnectedto3in<N 3 ; ~ E s01 >=<N 3 ;E s >.
Hen ce, a cco rding to F, n ob ody in C contributes to e s , which is an edge incidentto C s 3 . This impliesf s k
=0fo r allk 2C. But then
X k 2C X t=1 ~ f t k = X k 2C ~ t k X t= 1 ~ f t k = X k 2C ( s0 1 X t=1 f t k + ~ t k X t=s f t+1 k ) = X k 2C ~ t k + 1 X t=1 f t k = 1: 3. Furthermo re, ~ t k tift 01andk 2 ~ C t0 1 3 ,thecomponento f3in<N 3 ; ~ E t01 >. Hence, ~ f t k =0 by de nition. Steps 1 , 2a nd 3 imply e F 2V e E ( f M). 2
Corollary 7.4 Let M and f
M b e a s a b ove, let E b e a s equence of cho ices made by
algo rithm2 .3applied to M, and let F b e va lid fo r E. Then
x E;F k (M)x e E; e F k ( f M) for a ll k 2N; where e E is as de nedin lemma7.2 a nd e F in lemma7 .3 . Proof : x e E; e F k ( f M)= P ~ t k t= 1 ~ f t k w (~e t ). If ~ t k =0 then ~ t k X t=1 ~ f t k w( ~e t )=0x E;F k (M) If 0< ~ t k
<s (w ith s as de ned in lemma 7.2) then
~ t k X t= 1 ~ f t k w (~e t ) = ~ t k 0 1 X t=1 ~ f t k w (~e t )+ ~ f ~ t k k w (~e ~ t k ) = ~ t k 0 1 X t=1 f t k w (e t )+( X t= ~ t k f t k )w(e ~ t k ) ~ t k 0 1 X t=1 f t k w (e t )+ X t= ~ t k f t k w(e t ) = X t= 1 f t k w (e t ) = x E;F k (M): :
The second equation fo llows fro m the de nitio n of e
F an d the inequality holds
b ecause E iso rdered by no n-decrea sin gweig hts .
If ~
t
k
s thenk isno tco nta ined in ~ C s0 1 3 =C s 3
,so k is nota llowedto co ntributeto
e s ,i.e. f s k =0. Hence, ~ t k X t= 1 ~ f t k w (~e t ) = s01 X t= 1 f t k w (e t )+ ~ t k X t=s f t+1 k w (e t+ 1 ) = s01 X t= 1 f t k w (e t )+ X t= s+1 f t k w (e t ) = X t= 1 f t k w(e t ) = x E;F k (M): 2
Th is nowenab les usto p rove theorem2.8 .
Theorem 2.8For anymcsepro blemM,fora ny s equenceo f cho icesE =(e 1
;:::;e