Robert Baier, NiraDyn and ElzaFarkhi
Abstrat. Westudypropertiesofabinaryoperationbetweentwoom-
pat sets depending on a weight in [0;1℄, termed metri average. The
metriaverage isused inspline subdivision shemes for ompat setsin
IR n
,instead oftheMinkowskionvexombination ofsets,toretain non-
onvexity[3℄. SomepropertiesofthemetriaverageofsetsinIR,likethe
anellationproperty,andthelinearbehavioroftheLebesguemeasureof
the metri average with respet to the weight, are proven. We present
analgorithmforomputingthemetriaverageoftwoompatsetsinIR,
whihareniteunionsofintervals,aswellasanalgorithmforreonstrut-
ing one of themetriaverage's operands, given the seondoperand,the
metriaverageandtheweight.
x1. Introdution
Inthispaperwestudypropertiesofabinaryoperation,termedmetriaverage,
between twoompatsetsA;BIR.
For two ompat sets A;B in IR n
and a weight t 2 [0;1℄, the metri
averageof AandB withweight tis givenby
A
t B=
tfag+(1 t)
B
(a) : a2A [
t
A
(b)+(1 t)fbg : b2B ;
where
A
(b)isthesetofalllosestpointstobfromthesetA,andtheaddition
aboveisthe Minkowskiadditionofsets.
Themetriaverageisintroduedin[1℄forpieewiselinearapproximation
of set-valuedfuntions. It is usedin spline subdivisionshemes for ompat
sets, toreplae the average between numbers [3℄. With this binary average,
the limitset-valued funtionof aspline subdivisionshemeoperatingon ini-
tialdataonsistingofsamplesofaunivariateLipshitzontinuousset-valued
funtion, approximatesthesampledfuntionwitherrorofthe orderof O(h),
for samples h distane apart [3℄. Thus the limitset-valued funtions of the
spline subdivision shemes retain the non-onvexity nature of the approxi-
matedset-valuedfuntions,whileif we usethe Minkowskiaverageinstead of
ApproximationTheoryX 1
CharlesK.Chui,LarryL.Shumaker,andJoahimStoekler(eds.),pp.1{13.
Copyrighto
2001byVanderbiltUniversityPress,Nashville,TN.
ISBN0-8265-xxxx-x.
Allrightsofreprodutioninanyformreserved.
the metri average, any limit is onvex, and the spline subdivision shemes
failtoapproximateset-valuedfuntionswithnon-onvex images[4℄.
The metriaveragehas manyimportantproperties [1℄,[3℄. Itis asubset
of theMinkowskiaveragetA+(1 t)B,generallynon-onvex, reovering the
set Afor t=1, andB fort=0. Here weonsider themetri averageas an
operationbetween ompat setsinIR. Inthis settingthe metriaverage has
several more important properties, suh as the anellation property whih
guarantees that for a given weight t, if the metri average and one of its
operandsare known, thenthe seond operand is determined uniquely. Suh
a property is valid for Minkowskisums, only for onvex sets, and not valid
for non-onvex ones. Whileredundantonvexifyingparts mayappearin the
Minkowskiaverageofnon-onvexsets,therearenoredundaniesinthemetri
average of ompats in IR. In this sense the metri average of sets in IR is
optimal.
Wealsoshowthattheomputationofthemetriaverageisnotostlyand
doesnotrequiretheomputationofdistanes. Bypresentinganalgorithmfor
thealulationofthemetriaverage,weprovethatthenumberofoperations
required is linear in the sum of the numbers of losed intervals in the two
operands,independentlyofthe weightparameter.
ThemetriaverageforsetsinIRisimportantinthereonstrutionof2D
setsfromtheirross-setions,andmoregenerally,inapproximatingset-valued
funtions withimagesinIR.
Here is an outline of the paper: Denitions and notation arepresented
in Setion 2. Properties of the metri average for 1D sets are presented in
Setion3,without proofs. Analgorithmfor alulatingthemetriaverage is
giveninSetion4. Theanellationpropertyisderivedfromanalgorithmfor
the reonstrutionof the set Afrom the sets B;C and the weight t2 (0;1)
whenC =A
t
B. Thisis doneinSetion5,whereaentraltheoremfor the
validityoftheanellationalgorithmisstated. Themainproofsarepostponed
tothe lastsetion.
x2. Denitions and Notation
DenotebyK (IR n
)the setofallompat,nonempty subsetsofIR n
, byC(IR n
)
thesetofallompat,onvex,nonemptysubsetsofIR n
andbyK
F
(IR)theset
of all ompat, nonempty subsets of IR whih arenite unionsof nonempty
intervals.
TheLebesguemeasureofthesetAisdenotedby(A). TheHausdordistane
between the sets A;B2 K (IR n
) ishaus (A;B). The Eulideandistane from
a point a to a set B 2 K (IR n
) is dist(a;B) = inf
b2B
ka bk
2
. The set of all
projetionsof a2IR n
onthe setB2K (IR n
)is denotedas
B (a):=
b2B : ka bk
2
=dist(a;B) :
The setdierene ofA;B2K (IR n
)isAnB=fa : a2A; a62Bg.
AlinearMinkowskiombinationoftwosetsAandBis
for A;B2K (IR )and;2IR.
TheMinkowskisumA+BorrespondstoalinearMinkowskiombinationwith
==1. ThelinearMinkowskiombinationwith;2[0;1℄,+=1,is
termedMinkowskiaverageorMinkowskionvexombination.
Asegmentis denotedby[;d℄=f+(1 )d : 01g,for;d2IR n
.
Denition 1. Let A;B 2 K (IR n
) and 0 t 1. The t-weighted metri
average ofAandBis
A
t B=
tfag+(1 t)
B
(a) : a2A [
t
A
(b)+(1 t)fbg : b2B (1)
where thelinearombinationsin (1)areintheMinkowskisense.
The setsA;B2K
F
(IR)aregivenas
A= M
[
i=1 [a
l
i
;a r
i
℄ B=
N
[
j=1 [b
l
j
;b r
j
℄: (2)
Eahintervalisproper,i.e.theleftendpointis notbiggerthanthe rightone,
equalityispossiblewhihstandsforaso-alledpoint(ordegenerate)interval.
The intervalsareorderedinan inreasingorder, i.e. a r
i
<a l
i+1 andb
r
j
<b l
j+1
for allrelevanti;j.
Weextendthe representation(2)toaommonlosedintervalontaining
theonsideredsetsbyaddingpointintervalstotheleftandtotherightofthe
sets:
A= M+1
[
i=0 [a
l
i
;a r
i
℄; B= N+1
[
j=0 [b
l
j
;b r
j
℄; (3)
where
x
min
=a l
0
=a r
0
=b l
0
=b r
0
<minfa l
1
;b l
1 g ja
l
1 b
l
1
j; (4)
x
max
=a l
M+1
=a r
M+1
=b l
N+1
=b r
N+1
>maxfa r
M
;b r
N g+ja
r
M b
r
N j: (5)
ThishoieguaranteesthatC=A
t
Bishangedonlybytheadditionofthe
pointintervalsfx
min
gandfx
max
gforanyt2[0;1℄. DenoteX=[x
min
;x
max
℄.
The \holes"of eahset, namelythemaximalopenintervalsinX,whih
donotintersettheset,playanimportantrole,aswellastheirenters. Denote
the holesby
H A
i :=(a
r
i
;a l
i+1
); i=0;:::;M; H B
j :=(b
r
j
;b l
j+1
); j =0;:::;N; (6)
andtheirentersby
a
i :=
a r
i +a
l
i+1
i=0;:::;M; b
j :=
b r
j +b
l
j+1
; j=0;:::;N: (7)
The dualrepresentationof AandB is:
A=Xn
A; where
A= M
[
i=0 H
A
i
; (8)
B=Xn
B; where
B= N
[
j=0 H
B
j
: (9)
The t-weighted metriaverage A
t
Bis denotedbyC,
C =A
t B=
L+1
[
k=0 [
l
k
; r
k
℄; (10)
where l
0
= r
0
=x
min ,
l
L+1
= r
L+1
=x
max
, andthe enter of the k-th hole
H C
k
=( r
k
; l
k+1
)is denotedby
k .
x 3. Properties of the MetriAverage
The followingpropertiesof themetriaverageareknown[3℄:
Let A;B;C2K (IR n
)and0t1,0s1. Then
1. A
0
B=B; A
1
B=A; A
t
B=B
1 t A:
2. A
t
A=A.
3. A\BA
t
BtA+(1 t)Bo(A[B).
4. haus (A
t B;A
s
B)=jt sjhaus(A;B):
The followingpropertiesarevalidforsetsinIR.
Proposition 2. Let A;B2K (IR),C;D2C(IR),t2[0;1℄. Then
(a) C
t
D=tC+(1 t)D,
(b) (A
t
B)=t(A)+(1 t)(B).
() (A
t
B)=t(
A)+(1 t)(
B).
The proofof the rstassertionfollows triviallyfrom the denition. The
third assertion is proven inthe last setion. The seond one follows diretly
from thethird.
In the following we present two properties of the metri average whih
arevalidfor setsinK
F (IR).
Proposition3. LetA;B2K
F
(IR),andletH(A)denotethenumberofholes
of A. Then foreveryt2(0;1)
(a) H(A
t
B)H(A)+H(B)
(b) The number of operations neessary for the alulation of A
t B is
O(H(A)+H(B)).
The rstassertion is proven in the lastsetion. The seond one follows
from thealgorithmpresented inthesequel.
Proposition 4. Let A 0
;A 00
;B2K
F
(IR). Thenfor anyt2(0;1)
A 0
t B=A
00
t
B=)A 0
=A 00
: (11)
The proof of this laim follows from the onsiderations in setion 5. It
an be extended to sets in K (IR ), sine all relevant statements arevalid for
setsonsistingof aninnitenumberof ompatsegments.
Tounderstandthenatureofthe metriaverageoftwosetsAandB,we
distinguishfourtypesofholes inAwithrespettoB, andvieversa:
Denition5. LetH A
i
beaholeofA. Aordingtoitspositionwithrespet
toB,H A
i
isalled:
1. pairedwithaholeofB,ifthereisaholeH B
j
ofB,suhthata
i 2H
B
j and
b
j 2H
A
i .
2. pairedwith apointin B,ifthe enter a
i 2B.
3. leftshadow ofaholeof B,ifthereisaholeH B
j
of B, suh thata
i 2H
B
j ,
andb
j a
l
i+1 .
4. rightshadowofaholeofB,ifthereisaholeH B
j
ofB,suh thata
i 2H
B
j ,
andb
j a
r
i .
Clearly, eah holeof Abelongstoexatly oneof theaboveategories of
holes withrespettoB,andvieversa.
NotethatH A
0
ispairedwithH B
0
bythehoieofx
min
,andsimilarly,H A
M
is pairedwithH B
N
by thehoieofx
max .
WhenAisaveragedwithBandt2(0;1℄,eahholeofAreatesa\hild"
holeof C,whihinherits thetypeofitsparentwithrespettoB,asisstated
below.
Proposition6. LetH A
i
andH B
j
beholesofAandB,respetively,t2[0;1℄
andC =a
t B.
1. IfH A
i
andH B
j
arepaired,thenthe interval
H C
=tH A
i
+(1 t)H B
j
isahole ofC, pairedwithbothH B
j
andH A
i .
2. IfH A
i
ispairedwithapointofB,thenfor t>0theinterval
H C
=tH A
i
+(1 t)fa
i g
isahole ofC, pairedwiththepointa
i 2B.
3. IfH A
i
isaleftshadowof H B
j
,then fort>0theinterval
H C
=tH A
i
+(1 t)fb r
j g
isahole ofC, andaleftshadow ofH B
j .
4. IfH A
i
isarightshadowof H B
j
, thenfort>0the interval
H C
=tH A
+(1 t)fb l
g
isahole ofC, andaright shadowofH B
j .
The proof ofthispropositionis postponed tothelastsetion.
InterhangingtherolesofAandBandreplaingtwith1 tinProposition
6, weget that for t2(0;1)some holes of C are generatedby holes of A,or
respetively,by holes of B, by thefour ways presented above. The following
proposition, proved in the last setion, states that every hole of C has this
property.
Proposition 7. Let H C
=( 0
; 00
)beaholeofC =A
t
B, t2[0;1℄. Then
H C
isobtained eitherfrom aholeof A,byone ofthe fourwayspresented in
Proposition6,orfromahole ofB, inasymmetriway.
Inthe next examplewehaveplottedthe one-dimensionalsetsA,B and
thesetC
t
=A
t
Binonepiture,givingBatthey-oordinate0,Aaty=1,
andC
t
aty= tfor t= 1
4
; 1
2
; 3
4
(seeFigure 1).
The linesonneting the boundary points of A topoints of B and vie
versa,show whih holes of Aareonneted withwhih holes orpointsof B,
aording to their type with respetto B, and similarly for the holes of B.
These linesgivetheholesof C
t
whenrossedwiththeliney=t.
C = B 0 C 1/4 C 1/2 C 3/4 C = A 1
1 3 4 5 6.5 14 16
0 1 5 6 7.5 8 9 10 11.5 14
Fig. 1. ThesetsA,BandC
t
ofExample8.
Example 8. Considerthe twosets
A=[0;1℄[[5;6℄[[7:5;8℄[[9;10℄[[11:5;14℄;
B=[1;4℄[[5;6:5℄[[14;16℄:
Forthese twosets apossibleX is [ 2;20℄. The metri averageC
t
=A
t B
is C
t
=Xn
C
t , where
C = t( 2;0)+(1 t)( 2;1)
[ t(1;5)+(1 t)f3g
[ tf5g+(1 t)(4;5)
[ t(6;7:5)+(1 t)f6:5g [ t(8;9)+(1 t)f6:5g
[ t(10;11:5)+(1 t)(6:5;14)
[ t(14;20)+(1 t)(16;20)
:
The end points of X and of A;B;C
t , x
min
= 2; x
max
= 20, are not
present inthepiture.
The holesof thesetAarerelated tothesetBasfollows:
Thehole(1;5)ispairedwithapointinB,eahoftheholes(6;7:5),(8;9)
is aleftshadow ofaholeof B,thehole (10;11:5)is pairedwithaholeof B.
The holesof thesetB arerelated tothe setAasfollows:
The hole (4;5)is aright shadow of ahole of A andthe hole (6:5;14) is
pairedwithahole ofA.
SeeFigure 1forthe waytheseholes indueholes inC
t
=A
t B.
x4. Algorithmfor Computingthe Metri Average
In this setion we propose an algorithm for alulating the metri average
C =A
t
B, oftwogivensetsA;B2K
F
(IR),andt2(0;1).
Relying on Propositions 6 and 7, we onstrut the holes of the set C
onsidering the generating holes of A andB, in an order from left to right,
determining thetypeofeah hole.
Algorithm for alulating C=A
t B
Given aret2(0;1); A;B2K
F
(IR)of theform (3).
1. H C
0 :=tH
A
0
+(1 t)H B
0
; i:=1; j :=1; k:=1:
2. WhileiM andj N,
(a) IfH A
i
is arightshadowof H B
j 1 ,then
H C
k :=tH
A
i
+(1 t)fb l
j
g; k:=k+1; i:=i+1.
(b) Else,ifH B
j
isarightshadowof H A
i 1 , then
H C
k
:=tfa l
i
g+(1 t)H B
j
; k:=k+1; j:=j+1.
() Else,ifH A
i
is aleftshadowof H B
j ,then
H C
k :=tH
A
i
+(1 t)fb r
j
g; k:=k+1; i:=i+1.
(d) Else,ifH B
j
isaleftshadowof H A
i ,then
H C
k
:=tfa r
i
g+(1 t)H B
j
; k:=k+1; j :=j+1.
(e) Else,ifa
i
<b r
j ,then
H C
k :=tH
A
i
+(1 t)fa
i
g; k:=k+1; i:=i+1.
(f) Else,ifb
j
<a r
i ,then
H C
k :=tfb
j
g+(1 t)H B
j
; k:=k+1; j :=j+1.
(g) Else(H A
i
andH B
j
arepaired)
H C
k :=tH
A
i
+(1 t)H B
j
; k:=k+1; i:=i+1; j:=j+1.
End ofthe loop.
3. L:=k 1; C=Xn
L
[
k=0 H
C
k
:
Eah hole of A,B belongs exatlytoone of the ases desribed inStep
theotherset,isonnetedtoasinglepointoftheothersettogenerateahole
of C (ases (a)-(f)of Step2). Notethatthe ondition(e)(resp. (f))heked
aftertheondition(a)(resp.(b))yieldsthattheholeH A
i
(resp. H B
j
)ispaired
withapointof B(A).
Notealsothattheorderoftheholesfromthelefttotherightyieldsthat
alltherightshadowholesofagivenholeareonsideredafterit. Thatiswhy,
in the ases (a),(b) of Step 2 we hek for right shadows of the previously
onsidered holesH A
i 1 ,H
B
j 1 .
x5. CanellationProperty
Toprovethe anellationproperty(11),wepresentan algorithmwhihom-
putesthe setA,ift2(0;1); B andC(=A
t
B)aregiven.
The followingpropositionis thebasisforour anellationalgorithm.
Proposition9. Givenaret2(0;1),C=A
t
BwithholesH C
k
(0kL),
andB withholesH B
j
(0j N).
1. Let H C
k
and H B
j
bepaired anddene a 0
=a 0
k
= 1
t
r
k +(1
1
t )b
r
j
; a 00
=
a 00
k
= 1
t
l
k+1 +(1
1
t )b
l
j+1 . Ifa
0
<a 00
, then(a 0
;a 00
)XnA.
2. LetH C
k
bepairedwithapointofB,anddenea 0
=a 0
k
= 1
t
r
k +(1
1
t )
k ,
a 00
=a 00
k
= 1
t
l
k+1 +(1
1
t )
k
. Then(a 0
;a 00
)XnA.
3. LetH C
k
bealeftshadowoftheholeH B
j
,anddenea 0
=a 0
k
= 1
t
r
k +(1
1
t )b
r
j , a
00
=a 00
k
= 1
t
l
k+1 +(1
1
t )b
r
j
. Then(a 0
;a 00
)XnA.
4. Let H C
k
be arightshadow hole ofH B
j
, anddene a 0
=a 0
k
= 1
t
r
k +(1
1
t )b
l
j+1 , a
00
=a 00
k
= 1
t
l
k+1 +(1
1
t )b
l
j+1
. Then(a 0
;a 00
)XnA.
Denition 10. A hypotheti hole (a 0
;a 00
) of A is any properopen interval
(a 0
;a 00
)onstrutedinoneofthefourwaysdesribedintheaboveproposition.
Let C =A
t
B; t2 (0;1). ByProposition9, every hypotheti hole of
Ais asubsetofsome (real)holeof A. Thusthesetof allhypothetiholes is
ontainedinthesetof holesof A.
Onthe otherhand, byProposition6,everyholeof Ageneratesa\hild"
hole of C of the same type with respet to B. The proedure desribed in
Proposition9guaranteesthateveryholeof Awillbereoveredbyits \hild"
holeofC. ThusthesetofallholesofAisontainedinthesetofallhypotheti
holes of Aonstrutedfrom the holes of C. Therefore the setof all holes of
Aisequal tothesetof allhypothetiholes.
Theorem 11. Let J =fk : 0kL; a 0
k
<a 00
k
g,wherea 0
k
;a 00
k
aredened
inProposition9. ThenA=Xn
[
k2J (a
0
k
;a 00
k )
.
NotethatPropositions6,9andTheorem11 remaintrue whenB andC
are inniteunionsof ompat segments, sine their proofs donot use essen-
tially the nite number of segments. Thus the anellation property is true
Given twosets B;C 2 K
F
(IR), and a weight t 2 (0;1), we propose the
followingalgorithmfor reonstrutingA2K
F
(IR),ifC=A
t B.
Canellation Algorithm
1. J :=;; k:=0.
2. WhilekL,
(a) Computea 0
k
;a 00
k
aordingtoProposition9.
(b) Ifa 0
k a
00
k
, thenJ :=J [fkg.
3. A=[b l
0
;b r
N+1
℄n
L
[
k =0
k=2J (a
0
k
;a 00
k )
:
x6. Proofs
First weprovepropositions6,7, whih arethen usedinthe proof of Proposi-
tions2()and3(a).
Proof of Proposition 6:
1. Let H A
i
be paired with H B
j
. Denote 0
= ta r
i
+(1 t)b r
j ,
00
= ta l
i+1 +
(1 t)b l
j+1
. Toprove that H C
= ( 0
; 00
) = tH A
i
+(1 t)H B
j
is a hole
of C = A
t
B, we rstprovethat H C
\C =;. Supposethat there is
2H C
\C. Then=ta 0
+(1 t)b 0
,whereeithera 0
2A;b 0
2
B (a
0
),or
a 0
2
A (b
0
);b 0
2B. Supposethatb 0
2
B (a
0
),wherea 0
2A,anda 0
a r
i .
Then sine a 0
< b
j
, it follows that b 0
b r
j
, and = ta 0
+(1 t)b 0
ta r
i
+(1 t)b r
j
= 0
, i.e. 2= H C
, a ontradition. Similarly one gets
ontraditionsif a 0
a l
i+1
;b 0
2
B (a
0
), orif a 0
2
A (b
0
), where b 0
2 B
satises bb r
j
orbb l
j+1 .
Thuswehaveproven that H C
XnC. To verify that H C
isa hole of
C, wehavetoprovethat its endpoints areelementsof C. This follows
trivially from the denition of H C
and the fat that, for the left end
points, either b r
j 2
B (a
r
i ), or a
r
i 2
A (b
r
j
), and similarly, for the right
endpoints, eitherb l
j+1 2
B (a
l
i+1 ), ora
l
i+1 2
A (b
l
j+1 ).
TheproofthatH C
ispairedwithAandBistrivialandfollowsfromthe
relation
0
+ 00
2
=ta
i
+(1 t)b
j 2H
A
i
\H B
j .
2. LetH A
i
bepairedwithapointofB,i.e. a
i
2B. Thena r
i
;a l
i+1 2
A (a
i ),
hene 0
=ta r
i
+(1 t)a
i 2C,
00
=ta l
i+1
+(1 t)a
i
2C. Toprovethat
H C
=( 0
; 00
)is ahole ofC, wehavetoprovethatH C
\C =;. If there
is 2H C
\C,then=ta 0
+(1 t)b 0
,where eithera 0
2A;b 0
2
B (a
0
),
ora 0
2
A (b
0
);b 0
2B. Supposerstthata 0
2A;b 0
2
B (a
0
)anda 0
a r
i .
Then sine a
i
2 B, it follows that b 0
a
i
. Thus = ta 0
+(1 t)b 0
ta r
i
+(1 t)a
i
= 0
, i.e. 2= H C
, aontradition. Similarlyoneproves
the otherthreeases.
Thus( 0
; 00
)isaholeofCwith
0
+ 00
2
=a
i
2B,heneH C
ispairedwith
apointof B.
3. Let H A
i
be a left shadow of H B
j
. Denote 0
= ta r
i
+(1 t)b r
j
; 00
=
ta l
+(1 t)b r
, andH C
=( 0
; 00
). Clearly, b r
2
B (a
l
), andeither
b r
j 2
B (a
r
i ),ora
r
i 2
A (b
r
j
). Thus 0
; 00
2C. IfH C
\C =;,itistrivial
toshowthatH C
isaleftshadowof H B
j .
It remains to show that H C
\C = ;. Suppose that 2 H C
\C, i.e.
=ta 0
+(1 t)b 0
,whereeithera 0
2A;b 0
2
B (a
0
),ora 0
2
A (b
0
);b 0
2B.
Suppose rstthat a 0
2 A;b 0
2
B (a
0
)anda 0
a r
i
. Then sine a r
i
<b
j ,
it follows that b 0
b r
j
. Hene = ta 0
+(1 t)b 0
ta r
i
+(1 t)b r
j
= 0
,
i.e. 2=H C
, aontradition. The otherthreeases areprovensimilarly.
4. The asethat H A
i
is arightshadowof H B
j
issymmetritothe previous
ase andweomittheproof.
Proof of Proposition 7:
Let 0
= ta 0
+(1 t)b 0
, where either b 0
2
B (a
0
) for some a 0
2 A, or
a 0
2
A (b
0
) for some b 0
2 B, and let 00
= ta 00
+(1 t)b 00
, where either
b 00
2
B (a
00
)forsome a 00
2A,ora 00
2
A (b
00
)for someb 00
2B.
First weprovethat both inequalitiesa 0
a 00
; b 0
b 00
holdand atleast
one of them is strit. Clearly, ifa 0
a 00
and b 0
b 00
, then 0
00
, whih is
impossible. Next we show that a 0
> a 00
; b 0
< b 00
is impossible. We use the
inequality
maxfjb 0
a 0
j;jb 00
a 00
jg>maxfjb 0
a 00
j;jb 00
a 0
jg; (12)
whihis proven attheendofthe present proof.
Suppose, e.g. that jb 0
a 0
j = maxfjb 0
a 0
j;jb 00
a 00
jg. It follows from
(12) that a 0
= 2
A (b
0
) and b 0
= 2
B (a
0
), a ontradition. Similarly we get a
ontraditionifjb 0
a 0
jjb 00
a 00
j.
Thusa 0
a 00
; b 0
b 00
andatleastone of theseinequalities isstrit. To
provethat(a 0
;a 00
)XnA,supposethatthere existsa2(a 0
;a 00
)\A. Then
a belongstoone ofthe followingranges(some ofthem mightbeempty):
1. If b 0
ab 00
, there isb2
B (a)\[b
0
;b 00
℄,hene ta+(1 t)b2( 0
; 00
),
aontradition.
2. If a < b 0
, then there exists a(b 0
) 2
A (b
0
)\(a 0
;a 00
)\(a 0
;b 0
), suh that
ta(b 0
)+(1 t)b 0
2( 0
; 00
), aontradition.
3. The aseb 00
<ais symmetritothe previousone.
Thus(a 0
;a 00
)XnA. Similarlyone provesthat(b 0
;b 00
)XnB.
Next,weprovethattheintervals(a 0
;a 00
),(b 0
;b 00
)satisfytheonditionsof
one ofthe fourasesof Proposition6.
Assumethat(a 0
;a 00
),(b 0
;b 00
)arenon-degenerate,i.e. a 0
<a 00
andb 0
<b 00
.
We will prove that they are paired. If a
= a
0
+a 00
2
= 2 (b
0
;b 00
), for instane
a
b 0
, then either a
b 0
a 00
, implying that a 00
2
A (b
0
) and 0
<
ta 00
+(1 t)b 0
<
00
, a ontradition, or a 00
<b 0
, implying that a 0
= 2
A (b
0
),
and therefore b 0
2
B (a
0
), from whih it is onluded that in the interval
(a 0
(b 0
a 0
);a 0
+(b 0
a 0
))thereareno pointsofB. Thusb 0
2
B (a
00
)and
0
< ta 00
+(1 t)b 0
<
00
, a ontradition. The ase a
b 00
is symmetri.
Similarlyoneprovesthat b
= b
0
+b 00
2 2(a
0
;a 00
). Therefore (a 0
;a 00
),(b 0
;b 00
)are
paired.
0 00 0 00 0 00
Ifb 00
<a
,thena 00
= 2
A (b
00
)andb 00
2
B (a
00
). Hene intheinterval
(a 00
jb 00
a 00
j;a 00
+jb 00
a 00
j) there areno points of B. Let b 000
=minfb 2
B; ba 00
g,then(a 0
;a 00
)is aleftshadow of(b 00
;b 000
).
Similarly,ifb 00
>a
,wegetthat(a 0
;a 00
)isarightshadowofaholeofB.
Ifb 00
=a
,thenobviously(a 0
;a 00
)is pairedwitha
=b 00
2B.
In a similarway, if a 0
=a 00
, then(b 0
;b 00
) is ashadow of ahole inA, or
pairedwithb
=a 00
2A.
Proof of (12):
Theinequality(12)followseasilyfromthefatthatinthetrapezoidwith
verties (a 00
;0), (a 0
;0), (b 00
;1), (b 0
;1), the large diagonal is longer than the
sides, and the Pythagorean theorem. To prove the abovegeometri fat, it
is suÆienttoprovethat ifoneofthe sidesBC;AD ofthe trapezoid ABCD
(ABkCD),isnotlessthanoneofthediagonalsofABCD,thenitislessthan
the other diagonal. Suppose, for instane, that BC BD. We prove that
BC<AC. Inthe triangleBCD, theinequality ofthe sidesyieldsinequality
of the angles,
6
BCD
6
BDC. Continuing to ompare the angles, sine
ABkCD, it follows that
6
BAC =
6
ACD <
6
BCD. On the other hand,
6
BDC =
6
ABD <
6
ABC. Thus we get
6
BAC <
6
ABC, hene, in the
triangle ABC,BC<AC,whih ompletesthegeometri proof of(12).
Proof of Proposition 2()and Proposition 3(a):
As was proven in Proposition 7, every hole in C is generated either by
a holeof A onneted toasingle pointof B (if the holeof A is ashadow of
some hole of B oris paired with apoint of B), or, symmetrially,by a hole
of Bonnetedtoasinglepointof A,orbytwopairedholesofAandB. By
Proposition6,dierentholesof A(orofB)produedierentholesinC,and
the onlyase whentwo holes,oneof Aandoneof B,produeonehole of C
is thease ofpairedholes. Thisyieldsthelaim ofProposition3(a).
Denote by I
A
(respetively I
B
) the set of indies of holes in A (resp.
B) whih are onneted to a single point in B (resp. A). Sine for every
i2=I
A
thereexistsauniquej(i)2=I
B
suhthatH A
i
ispairedwithH B
j(i) ,then
Proposition6implies
(
C)= X
i2I
A t(H
A
i )+
X
j2I
B
(1 t)(H B
j )+
X
i2IA=
t(H A
i
)+(1 t)(H B
j(i) )
=t(
A)+(1 t)(
B):
Proof of Proposition 9:
1. LetH C
k
bepairedwithH B
j
. SineC=A
t
B,byPropositions6,7,the
onlypossibility for H C
k
is that H C
k
=tH A
i
+(1 t)H B
j
, where H A
i is a
hole ofA,pairedwithH B
j
. ThenlearlyH A
i
=(a 0
;a 00
).
2. Let H C
k
be pairedwithapoint of B, i.e.
k
2B. Then a 0
= 1
t
r
k +(1
1
t )
k
; a 00
= 1
t
l
k+1 +(1
1
t )
k
. Supposethata2A\(a 0
;a 00
). Then,sine
ais loserto
thana 0
anda 00
,there isapointa
0 2(a
0
;a 00
)\
A (
).
Hene ta
0
+(1 t)
k
2C\H C
k
,aontradition.
3. Let H C
k
be aleft shadow of H B
j
; a 0
= 1
t
r
k +(1
1
t )b
r
j
; a 00
= 1
t
l
k+1 +
(1 1
t )b
r
j
. Dene also the point a 0
s b
r
j
suh that ja 0
s b
r
j j = ja
0
b r
j j
(possiblya 0
s
=a 0
).
Weperform theproofinseveralsteps.
Step1 Firstweprovethat (a 0
;a 0
s
)\A=;.
Assume (a 0
;a 0
s
) 6= ;, i.e. a 0
< b r
j
< a 0
s
, and suppose that there is
a2A\(a 0
;a 0
s
). Thenfor b r
j
there isa(b r
j )2
A (b
r
j )\(a
0
;a 0
s
). Thus
ta(b r
j
)+(1 t)b r
j 2(
r
k
; l
k+1
)\C,whih is aontradition.
Step2 Weprovenow that(a 0
;a 00
)\[a 0
s
;b
j
℄\A=;.
Supposethatthereisa2A\(a 0
;a 00
)\[a 0
s
;b
j
℄. Thensineab
j ,it
followsthatb r
j 2
B
(a).Heneta+(1 t)b r
j
<ta 00
+(1 t)b r
j
= l
k+1 .
On the other hand, ta+(1 t)b r
j
> ta 0
+(1 t)b r
j
= r
k
. Thus
ta+(1 t)b r
j 2(
r
k
; l
k+1
)\C, whih isaontradition.
Clearly,ifa 00
b
j
, theproof isompleted. In allnext steps wesupposethat
b
j
<a 00
.
Notethat thepoint l
k+1
2C isobtained eitherby
l
k+1
=ta(b)+(1 t)b; where b2B; a(b)2
A
(b); (13)
orby
l
k+1
=ta+(1 t)b(a); where a2A; b(a)2
B (a
00
): (14)
In Steps3and4wesupposethat(13) holdsandprovethat[b
j
;a 00
)\A=;,
whihimplies(a 0
;a 00
)\A=;. InStep5weshowthat(14)isimpossiblewhen
b
j
<a 00
.
Step3 Weprovethat[b
j
;a 00
)\A=;,inasebb r
j
in(13).
Indeed, sine 1
t
> 1, then a(b)= 1
t
l
k+1 +(1
1
t )b
1
t
l
k+1 +(1
1
t )b
r
j
=a 00
. ThisyieldsthattherearenoelementsofAintheinterval
[b;a 00
)[b
j
;a 00
).
Step4 Weprovethat[b
j
;a 00
)\A=;,inasebb l
j+1
in(13).
Assumebb l
j+1
. Deneb s
j
>
l
k+1
suh that b s
j
l
k+1
= l
k+1 b
r
j .
SuhapointexistssineH C
k
isaleftshadowofH B
j ,i.e. b
r
j
<
l
k+1
<
b s
j
<b l
j+1 .
Denotea 00
s
= 1
t
l
k+1 +(1
1
t )b
s
j
. Thenbythedenition ofa 00
weget
a 00
l
k+1
= l
k+1 a
00
s
=( 1
t 1)(
l
k+1 b
r
j
). Sinebb l
j+1
>b s
j and
a(b)= 1
t
l
k+1 +(1
1
t
)b, it follows that a(b) <a 00
s
. Sine a(b) is a
projetionofb,itfollowsthattherearenopointsofAintheinterval
I=[b (b a 00
s
);b+(b a 00
s )℄[b
s
j (b
s
j a
00
s );b
s
j +(b
s
j a
00
s )℄=I
0
.
Sinea 00
s
<
l
k+1
<b
j andb
s
j +(b
s
j a
00
s )=2
l
k+1 b
r
j +
1
t (
l
k+1 b
r
j )=
a 00
+(b s
j b
r
j ) >a
00
, it is easy to see that [b
j
;a 00
) I 0
I. Thus
therearenopointsof Ain[b
j
;a 00
).
Step5 Let l
k+1
=ta+(1 t)b(a),wherea2A,b(a)2
B
(a). Sinea2A,
itfollowsbySteps1,2thata62(a 0
;b
j
℄\(a 0
;a 00
),heneeitheraa 00
,
ora>b
,oraa 0
.
If a> a 00
, thensine a 00
>
l
k+1
>b r
j
, it follows that b(a)b r
j and
ta+(1 t)b(a)>ta 00
+(1 t)b r
j
= l
k+1
, aontradition.
Ifa>b
j
,thenb(a)b l
j+1
andta+(1 t)b(a)>tb
j
+(1 t)b l
j+1
>b
j ,
aontradition.
If a minfa 0
;b
j
g, then sine a b
j
, b(a) b r
j
. Hene l
k+1
ta 0
+(1 t)b r
j
= r
k
<
l
k+1
,aontradition.
Thustheonlypossibility fora2Aisa=a 00
, hene b(a)=b r
j . But,
sineb
j
<a 00
, weobtainthatb r
j 62
B (a
00
),whihis aontradition.
Thisompletesthe proof of3.
4. The asethat H C
k
is arightshadowof H B
j
issymmetritothe previous
ase andis proven similarly.
Aknowledgments. Theseondandthirdauthorswerepartiallysupported
by theIsrael Siene Foundation {Center ofExelleneProgramandby the
HermannMinkowskiCenterforGeometry atTel AvivUniversity.
Referenes
1. Artstein,Z.,Pieewiselinearapproximationsofset-valuedmaps,Journal
of Approx. Theory56(1989),41{47.
2. Dyn,N., Subdivision shemes in Computer-Aided Geometri Design, in
AdvanesinNumerialAnalysis,W.Light(ed.),Vol.II,Wavelets,Subdi-
visionAlgorithmsandRadialBasisFuntions,ClarendonPress,Oxford,
1992,36{104.
3. Dyn,N.andE.Farkhi,Splinesubdivisionshemesforompatsetswith
metriaverages,inTrendsinApproximationTheory,K.Kopotun,T.Ly-
he andM.Neamtu(eds.),VanderbiltUniv.Press,Nashville,TN,USA,
2001,95{104.
4. Dyn,N.andE.Farkhi,ConvexiationrateinMinkowskiaveragingpro-
esses,preprint.
RobertBaier
ChairofApplied Mathematis
UniversityofBayreuth
D-95440Bayreuth,Germany
Robert.Baieruni-bayr eut h.de
NiraDyn and Elza Farkhi
Shool ofMathematialSienes
SaklerFaultyofExat Sienes
TelAvivUniversity
TelAviv69978,Israel
niradynpost.tau.a.i l, elzapost.tau.a.il
