Sets Uniquely Determined by Projections on Axes I. Continuous Case
Author(s): P. C. Fishburn, J. C. Lagarias, J. A. Reeds, L. A. Shepp
Source: SIAM Journal on Applied Mathematics, Vol. 50, No. 1 (Feb., 1990), pp. 288-306
Published by: Society for Industrial and Applied Mathematics
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?
SIAM J. APPL. MATH.
Vol. 50, No. 1, pp. 288-306, February1990
1990 SocietyforIndustrialand Applied Mathematics
017
SETS UNIQUELY DETERMINED BY PROJECTIONS ON
AXES I. CONTINUOUS CASE*
P. C. FISHBURNt, J. C. LAGARIASt, J. A. REEDSt,
AND
L. A. SHEPPt
fromtheirhyperplane
Abstract.This paper studies sets S in 1R"which are uniquelyreconstructible
integralprojectionsPi(xi; S) = f * fxs(xl, * ,,Xi, *, X,,)dxl ... dxi-l dxi+l ... dx,,onto the n coordinate axes of 11".It is shownthatany additiveset S = {x = (xl, -* , x,1): j7_I
f(x,) ? O}, whereeach fi(xi)
In particular,
balls areuniquelyreconstructible.
is uniquelyreconstructible.
is a boundedmeasurablefunction,
sets are additive.For n _ 3, Kempermanhas shownthat
It is shownthatin R2 all uniquelyreconstructible
sets in 11" of bounded measurethatare not additive.It is also noted for
thereare uniquelyreconstructible
n ? 3 thatneitherofthepropertiesof beingadditiveand beinga set of uniquenessis closed undermonotone
pointwiselimits.
A bad
A necessaryconditionfor S to be a set of uniquenessis thatS containno bad configuration.
is two finitesets of points T, in Int(S) and T2 in Int(Sc), whereS' =1R" - S, such that T,
configuration
and T2 have the same numberof pointsin any hyperplanexi = c for1 _ i '-n, and all c E 1R.We show that
foruniquenessforopen sets S in R2.
thisnecessaryconditionis sufficient
The resultsshow that priorinformationabout a densityf in R12to be reconstructedin tomography
f
(namelyiff is knownto have only values 0 and 1) can sometimesreduce the problemof reconstructing
can in principlebe of enormous
to knowingonlytwo projectionsof f Thus even meagerpriorinformation
value in tomography.
Key words.projections,uniqueness,additivesets,probability
AMS(MOS) subjectclassifications.05B, 53A, 60B, 62H
1. Introduction.Horn [4] asked in-connectionwith a problem of robot vision
whetherthe unit disk D in R2 iS uniquely determined(up to null sets) by its line
integralprojectionsonto the x and y axes. At firstit may seem unlikelythatonlytwo
projectionsof the density
fD (X
Y)
=XD(X,
)
1
if (x, y)
ED
answer
could determine
f(x, y) uniquely,butin factHorn's questionhas an affirmative
despitesuch tempting"counterexamples"as Fig. 1.1. Here is a simpleproof:Anytwo
sets havingthe same projectionson the x and y axes have the same area and moment
of inertia.The disk D has the minimummomentof inertiaamong all sets S having
the same area as D, so thatthe area and momentof inertiaof a disk D characterize
it uniquely(up to a null set).
problemin which
This problemis a special kind of tomographicreconstruction
a densityf(x, y), giventwoprojectionsand thepriorinformawe attemptto reconstruct
tionthatthedensity
densityf(x, y)
f(x, y) onlytakesthevalues 0 or 1. For an arbitrary
we generallyneed to knowinfinitely
f(x, y) uniquely.
manyprojectionsto reconstruct
thatthe densityf(x, y)
However,the disk example shows thatthe priorinformation
is an indicatorfunctioncan greatlyinfluencethe numberof projectionsneeded for
tomographicreconstruction.
a set S in Rn the
More generally,we consideras given data forreconstructing
hyperplaneintegralprojections
Pv(t; S) =
f
H, (v)
Xs(X) dHt,
-xc < t < oo
* Receivedby theeditorsAugust10, 1987; accepted forpublication(in revisedform)January20, 1989.
t AT&T Bell Laboratories,MurrayHill, New Jersey07974.
288
SETS
UNIQUELY
DETERMINED
289
BY PROJECTIONS
1
.0
FIG. 1.1. The soliddiskis extendedoutin quadrantsII and IV and in in quadrantsI and III. Thedashed
as theoriginal.
warpeddiskcannothave thesame projections
whereXs(x) is thecharacteristic
functionof S, Ht(v) denotesa hyperplanewithnormal
unitvectorv givenby
Ht(v) = {x: (v,x) = t}
and dHtis Lebesgue(n 1)-dimensionalmeasureon Ht.We call theset{P,(t; S): -oo
t <oo} the projectiononto the line L4= {tv: -oo < t<oo} and associate P,(t; S) to the
point tvon thisline.
This paper examinesthe problemof characterizing
sets S of finitevolumein RDn
thatare reconstructible
(up to null sets) fromtheirprojectionsonto the n coordinate
axes of RDn,i.e., from
(1.1)
Pi(xi; S) =
Xs (XI
...
.
X
xn) dxl.
dxil dxi+l
dXn
for-oo < xi < oo and 1 i _ n. We call theproblemof characterizing
setsof uniqueness
the continuousunique reconstruction
problem.We show that such sets S exist. The
numbern is the minimalnumberof directionsof projectionon lines forwhichthere
is a set S (not a null set) such thatunique reconstruction
of S is possible. To see this,
note thatif projectionsin n - 1 or fewerdirectionsare given,thenthereis a direction
orthogonalto all thesedirections,and we recoverno information
aboutthedistribution
of themass of S in thisdirection,so thatunique reconstruction
is impossible.Furthermore,this argumentshows that for unique reconstruction
fromprojectionsonto n
lines to be possible these lines mustlie in n linearlyindependentdirections.In this
case we may,withoutloss of generality,
reducethegeneralproblemto thespecial case
of
(1.1) wherethe lines are the coordinateaxes by an invertiblelineartransformation
coordinates.
There is an analogous discreteversionof the problemwhichwe call the discrete
unique reconstruction
problem.The discreteproblem is to characterizewhich finite
subsetsS of the integerlatticepointsZn givenby theirindicatorfunction
fS
if
(x1, ** , Xn)E S,
~~~~~~
f (Xi
*..
*Xn)
=
SXj{
tgC
290
FISHBURN,
LAGARIAS,
REEDS,
AND
SHEPP
are uniquelyreconstructible
fromthe projections{Pi: 1 ' i
Pi(xi)=
E
(y -'Yl)E
fs(Yl9
n}, where
Yn)-
y,= xi
The discreteproblemis studiedin a companionpaper [1].
The problem of characterizingsets of uniqueness was studied in the twodimensionalcase by Lorentz [9] in 1948 in an elegantbut apparentlylittleknown
paper. Lorentzgave the followingnecessaryand sufficient
conditionthata set S be
Let the projectionsP1(x), P2(y) of S onto the x-axis and
uniquely reconstructible.
y-axis in R2 be
Xs(x,y) dy;
P1(x) =
P2(y) =
Xs(x,y) dx
and definethe nonincreasing
rearrangements
pi of Pi on 0? u <oo by
pi(u) =
x(x: Pi(x) - u) dx
=Leb{x: -oo<x<oo
and Pi(x)?u}
so thatpi(u) decreases on 0 u < oo,whereLeb { -} denotes Lebesgue measure.
2 offinitemeasureis a set of uniquenessif and
THEOREM
(Lorentz). A set S in fR
onlyifPi and P2 are inversesof each otheron u _ 0, i.e.,
U,
PI(P2(U))
u _ O.
Lorentz'sresultcan be used to show thatthe unitdisk D is a set of uniqueness.
Indeed, it is verifiedeasily thatthe projectionsof the unitdisk centeredat (0, 0) are
PI (x) = P2(x) = 2( 1x2)
and thatthe nonincreasingrearrangements
u _ 0,
PI (u) = P2(u) =V/(4 u2)+,
are inversesof each otherforu ? 0.
In this paper we give a sufficient
conditionthat S be a set of uniquenessin R',
namelythatS be additive(as definedbelow). We also give a necessaryconditionthat
S be a set of uniquenessin R', whichis thatS have no bad configuration
(as defined
below).
A set S is said to be additiveif thereare bounded measurablefunctionsf for
n, where f(x) =f(xi) depends only on the ith coordinate xi of x=
(x1,
,
xJ), such that
x CZS<
(1.2)
n
i=l
f(xi) -i-
For example,the unitball is additiveas may be seen by takingf(x) = (1/n) _X2 for
1i
n.
measureis a set of
1.1. Any additiveset S in RWn
THEOREM
offiniten-dimensional
uniqueness.
Proof.Suppose therewere anotherset T in RWn
havingthe same projections
Pi(xi)=
{{.{
Xs(x) dxl
dxi-l dxi+I
* dXn
as S. Let gi(x) = gi(xi) be any integrable(over SU T) functionon RWn
whose values
SETS
UNIQUELY
291
BY PROJECTIONS
DETERMINED
depend only on the ith coordinateof x. Then
...J
gi(x)dxl...dxn=
(1.3)
gi(xi)Pi(xi) dxi
=J.j
gi(x)dx.
dxn,
over the n -1 coordinatesotherthan xi and usingthe factthatS and
by integrating
T have the same projections.Clearly,S and T have thesame n-dimensionalmeasure.
T in (1.3) we get
the integralover
Settinggi =fi in (1.3) and subtracting
(1.4)
{
sn
s finT{(xi)
dxl .
dXn=J
T...JJf(Xi)
dx,
dxn
and
...sT dxl..dXn
(1.5)
f
... n
dXl ..dXn
But summingover i in (1.4) we obtain
(1.6)
s
(X dx...
If f d(x)
dXn =j
fE
(xJ) dxI.l dxn
The integrandon the leftin (1.6) is nonnegativeby (1.2), and the integrandon the
T) =0 and
negative,also by (1.2). It followsthatLeb (T-sn
rightin (1.6) is strictly
O
then from(1.5) that Leb (S- S nT)= O, so S and T differby null sets.
of
our
version
notion
Kemperman[7] has subsequentlyprovedthata generalized
of additivityalso guaranteesuniqueness.
by showing
We can show that various simple sets are uniquely reconstructible
to tell
that theyare additive,as forthe unit n-sphere.However,it may be difficult
whethera generalset S is additive.
Thereis a simple necessaryconditionfora set S to be a set of uniqueness,which
we state next. A set of fourpoints z1= (xl, Yl), Z2= (x2, Y2), wI = (xI, Y2), and w2=
2 are a 2-bad configuration
(or bad
(x2, YI) formingthe cornersof a rectanglein fR
in
Int
(SC),
w2
are
in
S
and
and
Int
(S)
of
S
if
z1
interior
the
and z2 are
rectangle)for
w,
=
k-bad
(or
of
S.
a
configuration
More generally,
where sc
R2- S is the complement
,
where
*
forS in R8nconsistsof 2k vectorszl, ***, Zk and wl,
k-configuration)
Wk,
all thezj are distinctpointsin Int (S) and all thewj are distinctpointsin Int (SC), and
everycoordinateplane xi = c containsthe same numberof wj's as zj's. We definea
forS in RWnthe same way, exceptthatzl, ***, Zk need not
weaklyk-badconfiguration
be distinctpoints,and wl, ... , Wk need not be distinctpoints.We say thatS has no
forany k-' 2.
if S has no k-bad configuration
bad configuration
in
has
no bad configuration.
A
set
1.2.
S
RWn
THEOREM
of uniqueness
, Zk and
of pointsz1,
consisting
Proof.Suppose S has a k-bad configuration
the
set
8,
Cl consisting
W1, ... , Wk as describedabove. Then fora smallenoughpositive
oftheunionof k balls of radius 8 aroundthezi lies in Int (S), and theset C2 consisting
of theunion of k balls of radius 8 aroundthewi lies in Int (SC). Then T = S - Cl + C2
E
has the same projectionsas S so S cannotbe a set of uniqueness.
In the restof thispaper wvestudythe relationsbetweenthe conditionsthata set
We
S is additive,thatS is a set of uniqueness,and thatS has no bad configuration.
now summarizethe resultsobtained.
and "having
In ? 2 we firstshow thatthe concepts"havinga bad configuration"
for
in
all
coincide
dimensions.
sets
a weaklybad congfiuration"
open
292
FISHBURN,
LAGARIAS,
REEDS,
AND
SHEPP
THEOREM
2.1. Let S be an open set offinitemeasurein R'. The followingare
equivalent.
for any k ' 2.
(1) S has no k-badconfiguration
for any k ' 2.
(2) S has no weaklyk-badconfiguration
betweenthe continuouscase and
This theoremhighlightsan essentialdifference
the discretecase, forin [1] we show thatin the discretecase the concepts"having a
bad configuration"and "having a weakly bad configuration"are distinctin all
dimensionsgreaterthan or equal to 3.
In ? 2 we studythe two-dimensionalcase in detail. In the two-dimensionalcase,
additivityand uniquenesscoincide.
THEOREM
2.2. Let S be a set in R2 offinitemeasure.Thefollowingconditionsare
equivalent.
(1) S is a set of uniqueness.
(2) S is additive.
conditionsforbeinga set
Kuba and Volcic [8] give othernecessaryand sufficient
of uniquenessin the two-dimensionalcase.
thenthe"set ofuniqueness"conceptessentiallycoincides
If S is suitablyrestricted
withthe otherconceptsin the two-dimensionalcase.
THEOREM
2.3. Let S be an openset in R2 offinite
measure.Thefollowingconditions
are equivalent.
(1) Thereis no open set S unequal to S havingthesame projectionson thex-axis
and y-axis.
for all k ' 2.
(2) S has no k-badconfiguration
for all k ' 2.
(3) S has no weaklyk-badconfiguration
If inadditiontheboundary&S has measurezero,theseare also equivalentto thefollowing
condition.
(4) S has no 2-bad configuration.
It seems likelythat the hypothesisthat&S has zero measure is unnecessaryin this
theorem.
An interesting
questionconcernswhetheror not the conceptsof uniquenessand
additivitycoincide forthreeor more dimensions.In an earlierversionof this paper
we conjecturedthattheycoincide,but Kemperman[6] settledthatconjecturewitha
counterexampleof a set of uniquenessthatis not additive.Kellererfound a similar
it to us. Kemperman[7] then proposed a
counterexampleand kindlytransmitted
forcertainsets to be
generalizednotionof additivitythatis necessaryand sufficient
sets of uniqueness.
Section 3 studiesgeneralpropertiesof additivesets. We firstgive a criterionfor
a set not to be additive.
THEOREM
3.1. If S is a boundedmeasurablesubsetof Rn and ifthereare measures
p and P on RD withthesame projections(one-dimensional
marginals)on all n axes, and
on S and v is concentrated
on SC, thenS is notadditive.
,u is concentrated
Section 3 thenstudiesthe followingsubsetof the cube in lR3:
S0={(XI,X2,X3):0
X.?1
i=1,2,3,x3'
max{xI,x2}}.
Note that So is the limitas n -> oo of the additivesets
Sn ={(X,
X2,X3):
0- xi-' 1 fori = 1,2,3,xn >?x1 +
Xn}
withSI c S2 C S3 c *. Our studyof thisset was motivatedby our searchfora set of
factsabout So.
uniquenessthatis not additive.We provetwo interesting
is
not
additive.
THEOREM
3.2. SO
SETS
UNIQUELY
DETERMINED
BY PROJECTIONS
293
3.3. S0 is nota set of uniqueness.
These theoremsimplythatboththe additivitypropertyand the set of uniqueness
propertyare not closed under the operationof takingmonotonepointwiselimitsin
dimensionsgreaterthan or equal to 3.
Theorem 3.2 follows fromthe fact that thereare a pair of measures A and v
concentratedon S0 and S', respectively,having the same projections onto the
coordinateaxes. An alternativedirectproofof Theorem3.2 has been shownto us by
Kemperman.
We prove Theorem 3.3 by showing there exists a pair of measures A and v
concentratedon S0 and S', respectively,
havingdensitiestakingonlythevalues 0 and
1, whichhave the same projectionson the coordinateaxes. We actuallyconstructa
pair /i, v havingbounded densitiesand thenuse a generalresult([2]) whichasserts
thattheexistenceof a pair (pu,v) withbounded densitiesimpliestheexistenceof such
a pair havingdensitiestakingonlythe values 0 and 1.
The relatedproblemof decidingwhetherthereexistsa set S havinggivenprojections is not consideredin the paper. The existenceproblemis studied in the twodimensionalcontinuouscase in [5], [9], wherenecessaryand sufficient
conditionsfor
existenceare given.
THEOREM
2. Continuouscase: generalresults.Our firstresultin the continuouscase shows
thattheconceptsof"no bad configuration"
and "no weaklybad configuration"
coincide
foropensets S, in all dimensionsn. This differs
fromthe discretecase forn ?-3.
THEOREM
2.1. Let S be an openset in lR'. Thefollowingconditionsare equivalent.
(1) S has no weaklyk-badconfiguration
for all k- 2.
(2) S has no k-badconfiguration
for all k ' 2.
to show (1)X'(2). To do thiswe
Proof.It is clear that(2)X==(1),so thatit suffices
show thatif S has a weaklyk-bad configuration
Let
thenit has a k-bad configuration.
...
=
of
k
w'i)
,
in
theweaklyk-bad configuration
the
Int
consist
points
(w),
w(0)
(S),
and the k pointsz'i) = (z', ... , z(')) in Int (SC), for1 ' i ' k. Since the sets {w5)} and
{z ()} have equal projections(countingmultiplicities)foreach j from1 to n,thereare
permutations irj of {1, 2,
, k} such that z5' = wij')'. Clearly, there are k points w'()
in
close to the w(i) Int (S) whose coordinatesconsist of kn distinctreal
arbitrarily
numbers. For any such w'i), define
(zl)
by
4) = Wi) i'))
are all distinctpoints and the sets {Wi'} and
for all i and j. Then the z(i) and
have the same projections.
As w approaches wi, z(i) approaches z , so by choosing the w sufficientlyclose
i) Int (S) and zi) E Int (SC); hence we concludethat
to thewi we mayensureboth EV
thereis a k-bad configuration. D
The remainderof this sectionestablishesresultsforthe two-dimensionalcase.
THEOREM
2.2. Let S be a set offinitemeasurein R2. Thefollowingare equivalent.
is
a
set
(1) S
of uniqueness.
(2) S is additive.
Proof.The result(2)=X(1) is just Theorem 1.1. To prove (1)X'(2) we use ideas
of Lorentz[9]. We are givena set of uniqueness S. Suppose firstthatS is contained
in R = {(X1, X2): X1 0, X2- O} and thatthe projection
w
Pl(XI) = J
XS(Xi,
{z'i)}
X2) dX2
is nonincreasingforx -0, and thatthe projection
P)0
P2(X2) =
X0
S(XI,
X2) dXl
294
FISHBURN, LAGARIAS, REEDS, AND SHEPP
is nonincreasingforx2?. Note thatP1(xl) is finiteforeach x1> 0 since S has finite
measure.Lorentz'sconditionthatS be a set of uniquenessis
P2(P1(xl))
for x1 _.
(2.1)
x1 a.e. [Leb]
Using this it is easy to verifythat
S ={(x,x2):
x1
i O and0 x2
P1(xI)}
is identicalto S up to a null set, since the rightside of (2.1) produces the correct
projections.Using the factthatP1(xI) is nonincreasing,it is easy to verifythat
(2.2)
S'
= {(XI,
X2): fl(XI)
fA(x{)
P1(xI)
+f2(X2)
0}
0
where
(2.3)
00
if xI '-0
ifXI<0
and
(2.4)
f2(X2) ={_
ifx2 <O
This would show thatS is additiveiff] and f2 did not take the value -oo. We avoid
in the generalcase by a change of variable,as describedbelow.
thisdifficulty
Consider the generalcase and recall thatLorentzprovesthatassociated to each
projectionPi(xi; S) thereare measurablefunctionso iO -*>R+ = {x: x> 0} which are
one-one (almost everywhere)and onto (almost everywhere)such thatthe functions
pi(t) definedby
pi(0oi(xi))= Pi(xi)
i= 1,2
are nonincreasingon 0< t<oo. These are monotonerearrangements
of the Pi(xi) on
the xi-coordinates.Now defineS+ c R+ x R+ by
S+ = {(1(XI),
0-2(x2)):
(X1, X2) E S}.
Then S+ has projectionsp1(xl) and p2(x2),and it is easy to check thatS+ is a set of
S is a setofuniqueness,
uniquenessifand onlyifS is a setofuniqueness.Byhypothesis,
hence so is S+, and by the precedingargumentthereare functionsf1 and f2 for S+
satisfying
(2.2)-(2.4). Then (2.2) impliesthatS is additiveforft(x1) and f (x2), where
f*(x1) =f1(o1(x1)) and f2*(x2) =f2(0u2(x2)),
afternoting that f*(x1) and f*(x2) are finiteeverywherebecause o-1(xl)>0 and
o2(x2) > 0 forall x1, x2E R. This completesthe proof.
D
We now observethatforopen sets S of finitemeasureour conceptsessentially
coincide in the two-dimensionalcase.
THEOREM 2.3. Let S be an openset in R2 offinite
measure.Thefollowingconditions
are equivalent.
(1) Thereis no open set S unequal to S havingthesame projectionson thex-axis
and y-axis.
(2) S has no k-badconfiguration
for all k _ 2.
(3) S has no weaklyk-badconfiguration
forall k ' 2.
If in additionthe boundaryaS of S has measurezero, thentheseconditionsare also
equivalentto thefollowingone.
(4) S has no 2-bad configuration.
Proof.Condition (1)=>(2) by Theorem 1.2 and (2)X(3) by Theorem 2.1. We
prove (2)=> (1) in the contrapositiveformby the followinglemma.
SETS
UNIQUELY
DETERMINED
BY PROJECTIONS
295
LEMMA 2.4. SupposethatSI and S2 are disjointopensetsin R2 offinite
area having
thesame projectionson thex-axisand y-axis.Thentherearefinitesets T1 in SI and T2
inS2 havingthesame numberofpointson eachlinex = c and on eachliney= cforall c E DR.
Proof.We prove the strongerresultthatthe set T1 can include any givenpoint
vO= (xo, yo) in SI. To do thistake a small open square UOof side 8 centeredat vothat
lies entirelyin the open set SI. Since the projectionsof SI and S2 agree,we can find
pointsYi = (xo, Yi) and Y2= (x1, YO)withYi and Y2in S2. Since S2 is open,by decreasing
8 if necessary,we can findopen squares UO, U*, U1 of side 8 centeredat vo,Yi, and
Y2 withthe squares U1, U* being in S2. (See Fig. 2.1.) We say thatthe pointsin U*
cover those in UO in the "down" direction,by which we mean the projections
P1(x; U*) = P1(x; UO)forall x E R, and those in U1 coverthosein UOin the "across"
direction,by whichwe mean P2(y; U1) P2(y; UO). Note that UO, U1, and U* each
has area 62.
\~~~V
Y1
S2
(i)
~ U1ii ~
1 U
uji
FIG. 2.1. Initialopensquares in S1 and S2*
Now let E be a positiveconstantdependingon 8 thatwill be picked once and
forall. We claim thereexistsan open set U2 such that:
(i) U2 is in SI.
(ii) For each point w= (w1, w2) in U2 thereis some point (w1,y) in U1 having
the same firstcoordinate.
(iii) P1(x; U2)_ P1(x; U1) forall x.
(iv) J ocIP1(x; U2)- P1(x; U1)Idx < E, so thatArea (U)>82-e.
(v) U2 is a disjointunion of a finiteset of open rectangles.
To see this, we use the fact that since U1 is in S2, and S, and S2 have the same
projections,and since UOhas no x-coordinatein commonwith U1, then
P1(x; S1 - Co) ' P1(x; U1) forall x E ,
wherethebar denotesclosure.Now coverthe open set S, -C UOwitha meshof squares
of side '1 and let '1 -0O. Taking '1 sufficientlysmall we can approximate the Riemann
296
FISHBURN,
LAGARIAS,
REEDS,
AND
SHEPP
integralof SI - UOarbitrarily
well frombelow,and choosingan appropriatesubcollection of these squares yieldsa set U2 such that(i)-(v) hold.
We continueto constructa seriesof sets Ui such that:
(i) If i is odd, then Ui is in S2, and is in SI otherwise.
(ii) Each point w = (wI, w2) in Ui can be reached froma point (wI, y) in Uil
if i is even and froma point (x, w2)in Ui-l if i is odd.
(iii) PI(x; Ui) ' PI(x; Ui-1) forall x if i is even. P2(y; Ui) ' P2(y; Ui-1) forall y
if i is odd.
(iv) JfO
IPj(z; Ui) - Pj(z; Ui-1)Idz ' iE where]j 1 foreven i and j 2 forodd i,
so Area (Ui)_82-i%E
(v) Ui is a disjointunion of a finiteset of open rectangles.
(vi) UfnlUj=0 forOj <i.
This constructionis carriedout usingthe factthat(for i = 2j)
PI(X; SI-CO -C2
U2j-2)_ PI(x; U2j-1)
and (for i =2j+1)
P2(Y; S2 - CJI- 3-
* *- J2
21)
P2 (Y; U2j)
provided U2j+i n u*=0. Since SI has finitearea A, if we take E small enough,say
2(A/8),
=1/100A-284, then the sets Ui are all of area greaterthan 682for 1'i'
hence by pairwise disjointnessthis constructionmust halt for some i<2(A/8). It
followsthatthereis some i = 2j forwhich U2vhas a pointV2j = (X2j, Y2j) suchthatthere
is a point v = (x*, Y2j) in U*. By condition(ii) we can extendthis to a sequence of
pointsVk = (Xk, Yk) in Uk such thatXk= Xk+I if k is odd and Yk = Yk+I if k is even.Now
since U* and U1 are rectangles,v1= (xi, yo)E U1 and v2j+l = (xo, Y2j)E U*. Then the
sets of points T = {vO,V2, V4,
,V2j} in S, and T2= {vl, V3, V5,*
V2j+1} in S2 have
the requiredproperty. O
We now completethe proofof Theorem2.3. The implication(2)=X(4) is always
true.We prove (4)X=(2) underthe assumptionthatthe boundarydS = S - S of S has
Lebesgue measurezero. (Since open sets are Lebesgue measurable,it followsthatdS
is always measurable,but it may have positivemeasure;cf. [12, p. 59].) We show the
fork ' 3 thenS has
whichassertsthatif S has a k-bad configuration
contrapositive,
a 2-bad configuration.
has the form
We may suppose thatthe k-bad configuration
x(l) = (x|'M,y(I)) E Int(S),
x(2) = (xI),
x(3)= (x( 2, y( ))
X(4) -(X(2)
x(2k -)
(xk,
E
Int(),
)
E
It (),
Since these sets are open, forall y withIlyll<
(|2))E Int (Sc),
y(3)) E
Int (SC),
x(2k)= (xlk) y(l))
8
E
Int (SC).
and 8 small enough,the set
S(y) ={x(') + y: 1-' i ' 2k}
is a k-bad configuration.
Since dS has measure0, the set of y with Ilyll<8 forwhich
some pointx(')+y is in aS has measure0, hence we can choose y so thatall k2points
(XI(IyAi)) for1-i, j-' k are in Int (S) U Int (SC), i.e., none of themare on dS (=-dC).
Now the argumentof Lemma 1 in [1] in the discretecase can be applied to conclude
that there is a subset of four of these k2 points with {(xI, yI), (x2, Y2)} C S and
{(xI, Y2),(X2,Yi)}C Sc. Since these points are actuallyin Int(S) or Int (SC), this is a
2-bad configuration. O
SETS
DETERMINED
UNIQUELY
BY PROJECTIONS
297
FIG. 2.2. Staircase example.
Remarks.(1) Lemma 2.4 is false in two-dimensionsif SI and S2 are allowed to
finite.The infinitecheckerboard
area, even if projectionsare everywhere
have infinite
whereshaded open squares are in SI
staircasein Fig. 2.2 providesa counterexample,
and unshaded open squares are in S2.
analogue of Lemma 2.4 is false foropen sets SI and
(2) The three-dimensional
S2 of finitevolume.
LEMMA 2.5. Thereexisttwodisjointboundedopensets SI and S2 in 3 such that:
(1) S1 and S2 have identicalprojectionson all threeaxes.
for SI or S2 containedin SI U S2.
(2) Thereexistsnofinitebad configuration
This resultis proved in the Appendix.
3. Propertiesof additivesets. Recall that S is additiveif and only if thereare
f (xi) for1 i _ n such that,up to setsof measurezero,
bounded measurablefunctions
(3.1)
S =
~~~~~~n
0
x = (xI,
XX):
E f;(Xi)-0
i=1
to decide when a set is additive,thereis a simple necessary
Althoughit is difficult
conditionforadditivity.
and thereare
THEOREM
3.1. Suppose S is a boundedmeasurablesubset of RW
on S and
measuresg and v withthesameprojections
forwhichg and v are concentrated
and
SC, i.e., g(S)>O
g(SC)=
(3.2)
v(S)=0.
ThenS is notadditive,i.e., (3.1) holdsfor nof, i= 1,
Proof.If f existsatisfying(3.1) thenforeach i
..
(3.3)
fi (xi) du-(x) =
...* *fi(x)
,
n.
dv(x),
since thef are bounded measurableand depend onlyon one coordinatewhileg and
v have the same one-dimensionalmargins.Adding over i in (3.3) gives
I
s
{Ei1f;(xi) d(x)
{s
f
dv(x).
J(xi)
negative.
By (3.1), theleftintegrandis nonnegativewhiletherightintegrandis strictly
298
FISHBURN,
LAGARIAS,
REEDS,
AND
SHEPP
It followsthat v has measurezero and so g has measurezero since g and v have the
O
same marginals.But a (S)> 0, so f cannotexist.
in provingthata set is additiveor is one of
To illustratesome of the difficulties
uniqueness,we studythe followingparticularsubset SO of the cube in R3:
SO= {(x,
x2,x3): x3 'max {x1, x2}, 0?x
_?1fori = 1,2, 3}.
Theorem
We willshowthatSOis notadditivebyexhibitinga pair ofmeasuressatisfying
3.1 forSo.
(3.2) withS = SO.
3.2. SO is notadditive,i.e.,thereare g and v satisfying
THEOREM
Proof.Since we would naturallybelieve that max {xl, x2} cannot be writtenas
f3-(f(xI) +f2(x2)), the resultthatSO is not additiveis perhapsnot surprising.
The proofgives a g and v not absolutelycontinuouswithrespectto Lebesgue
measureA. The idea behindthe proofis thatif we can finda singleprobabilityspace
on whichrandomvariables XI, X2, Xl, X2 are defined,and if Xi and X' have the
same distributionfor i = 1, 2, and yet max {Xl, X2} is everywherestrictlyless than
max {Xl, X2}, i.e.,
XI - Xl, X2- X2, max {XI, X2} < max {Xl, X2},
thenwe may assume by change of scale that0 ? Xi c 1 fori = 1, 2 and set
(3.4)
=
X=
(max {Xl, X2} + max {X,
X}).
Then the measure g induced by (X1, X2, X3) is concentratedon SO since X3 ?
max{X1,X2} while the measure v induced by (Xl, X2, X3) is concentratedon SO
since X3 < max {Xl, X2}. The measuresg and v have the same projectionsonto the
firsttwo coordinateaxes since Xi - X' fori = 1, 2; theyalso have the same projections
onto
X3
since
X3 = X3.
We constructrandomvariablessatisfying
(3.4). Let X1 = X2 =,3 where
P(/3E du) =
,
0o cu.<
1,
and let (Xl, X2) be a uniformpoint on the union of the line segmentsfrom(0, 1) to
(,3,13)and from(3,1,3)to (1, 0). Clearlymax {X1, X2} =,3 < max {Xl, X2}. It remains
of 13.We have
onlyto show thatXl and X2 have the distribution
{
1
P(XI
E
du) =
[u
P(,l3 dv)P(X'
dv
--TNV P-)
JO
E
du 1,3
E dv)
du
[1
2(1 -vP)
dv
u P(- )
du
du
2v 7N/u( -0),
X2 Xl . Of coursesince g assignsmeasure1 to X1 =X2,
and therefore
Hence Xl 1,3
O
, is not absolutelycontinuous.
We now show thatSOis not a set of uniqueness,hence thatneitheradditivitynor
beinga set of uniquenessis closed undermonotonepointwiselimitsforn greaterthan
to show thatthereare measuresg and v satisfying
or equal to 3. To do thisit suffices
the conditionsof Theorem3.1 forSOwhichhave Radon-Nikodymderivative0 and 1
at all points. Indeed, if To is anotherset with the same projectionsas SO and if
A = Lebesgue measure on Rn', x(A) is the indicator function of A, and go and voare
the measureswithRadon-Nikodymderivatives
dp-L0
dA -X(so-son
To) and
then -o and vohave the same projections.
dv0
dA =X(To-son To),
SETS
DETERMINED
UNIQUELY
299
BY PROJECTIONS
THEOREM
3.3. S0 is not a set of uniqueness,i.e., thereare I.Land v withdensities
takingonlyvalues 0 and 1 satisfying
I.L(S') = P(So) = 0.
Proof.The idea of the proof is to constructg and v with bounded densities
satisfying(3.2), and then to use a generalresultthatthe existenceof Iu and v with
bounded densitiesimpliesthe existenceof such g and v havingdensitiestakingonly
values 0 and 1.
We now exhibitg and v withboundeddensitiesthatsatisfy(3.2). We willconstruct
a probabilityspace and randomvariablesas in (3.4). To thisend suppose q(y) > 0 on
I = (0, 1) and definethe measure -,Ton 12 by
(3.5)
dxdy, wherexvy=max{x,y}.
7r(dx,dy)=q(xvy)
We willdefinea mapping(x, y) -* (x', y') withx v y < x' v y' and thensetX1 = x, X2 = y,
X1 =x', X2=y' so that(3.4) holds. To do this,we choose +(y):I- I and q(y):I->I
and set
(3.6a)
y'=4+(y),
x
x' =-q(y)
forx<y
(3.6b)
x'=+(x),
y' =-q(x)
x
fory<x
y
where k and qfare to satisfy
(3.7)
y<4(y),
4(0)=0,
4(1)=1,
< a)
For (3.4) to hold we need -,T(x
v-(x'< a) so thatwe want
{ dx I dyq(x) + { dx{ dyq(y) { dxI dyq(x)X(d>(x)
ra
o38
4increases, 0<q+(y)<1.
x
ra
o
Ilr
x
o
rx
=
o
o
+{
(3.8)
dx
<
f
dyq(y)X(x
a)
(y)< a)
We will requireqfand q to satisfy
(3.9)
whereO< a < 1 is tobe chosen
qf(y)-a
and
ra
(3.10)
(3.11)
aq(a) > {q(y)
0
a
0
q(y)dy=j
Then (3.8) withb = b(a)=
(3.12) 2
xq(x)
o
dy,
yq(y)dy,
O< a < a
O<a<1.
q(a)>O,
471(a) maybe writtenas
dx+a{
q(y) dy=j
xq(x) dx+min{ a,
1}
yq(y) dy.
on a we obtain
Clearly b(0) = 0, b(1) = 1, and differentiating
(3.13)
(3.14)
aq(a)+J
q(y)
aq (a) +
dy--|
'1
yq(y) dy=bq(b)b'(a),
q (y) dy= bq(b) b'(a),
a > a.
a< a
300
FISHBURN,
LAGARIAS,
REEDS,
AND
SHEPP
Using (3.10) and (3.11) we see thatthe leftside of (3.13) is aq(a)-Jg q(y) dy> 0 so
thatb = b(a), definedby (3.12), has b'(a)> 0 (if q(b)> 0) fora < a. By (3.14), b'(a)> 0
fora > a. Thus b(a) = 4 -'(a) increasesfrom0 to 1 so that+(a) does also. Furthermore,
we musthave b = b(a)< a for0 < a < 1 because if b _ a in (3.12) thenthe rightside
is greaterthan the leftsince under (3.11) it is easy to see that
O
(3.15) 2 {xq(x)
dx+a
{a
q(y) dy<
{o
ato
xq(x) dx+min{-,
1} Jyq(y)
dy.
Thus under (3.9)-(3.11), (3.7) and (3.8) hold. Leaving aside the questionof findingq
to satisfy(3.10) and (3.11) forsome 0 < a < 1, suppose we thenset X3 = X3= z where
z=(1
- 0)(x
v y) + O(x'
v y')
and 0 is independentof (x, y) and uniformon I = (0, 1). Thus the measurespace on
which Xi and X' are definedis P= {(x, y,0)}. Let g be the measure induced by
(XI, X2, X3) = (x, y, z) and v be the measure induced by (Xl, X2, X3) = (x', y', z).
Clearly , and v satisfy(3.2) since x v y <z <x'v y' and g and v have the same
between(x, x') and (y,y').
projectionsby (3.8) and the symmetry
We now verifythatg and v have bounded densities.The densityof g forx <y
is from(3.5) and (3.6a) and the factthatP(0(y'-y) E dz -y) = dz/(y'-y),
p.
(dx, dy,dz)
P((1 - 0)y + O(x'v y') E dz)
q(y)
= q(y)
dz
(x'vy')-y
dxdydz
(3.16)
q(y)
4'(Y) -Y-
The densityof v forx'<y' is as follows.If (x', y') comes from(xo, yo) withxo<yo,
wherexo< yo denotesan eventAo withindicatorXo,thenfrom(3.5) and (3.6a),
(3.17)
ir(Aofn(dx', dy',dz'))
dx' dy' dz'
=
= xoq(yo)
dxodyoP((1 - 0)yO+Oy'Edz')
dz'
dy'
1
1
yo
k(yo) k'(yo) Y' - Yo
However,(x', y') mayalso come from(xl, Yl) withxl > Yi, an eventAl withindicator
X1-But
(3.18)
U A1)
v(dx', dy',dz')/dx' dy' dz'= i7r((AO
n(dx' dy' dz'))/dx' dy' dz'
since (x', y') comes fromeithera unique (xo, yo) withxo< yoor froma unique (xl, Yi)
withxl > Yi, or both.That is to say,if (x', y') also comes fromsay (x2, Y2), theneither
(X2, Y2) = (xO, YO) or (x2, Y2) = (xi, Yi) because 4 is strictlymonotonic.By (3.5) and
(3.6b)
-(Al
(3.19)
n(dx', dy',dz'))
dx, dy, P((1 - 0)xI + Oy'E dz')
dz'
dx' dy'
dx'dy'dz'
=XXq(xl)
1
4X>'X1
X1
-,(j
1
lX
By (3.18) and (3.17) and then (3.19) we have
2 v(dx', dy',dz')
d
-<q
(3.20) dx' dy'dz' q=y)
1
Yo
(yo)
k'
q(y0) (yo)
1
X
1
1
+ q(x,)
qf(XI) 4'(x1) y'l-XI'
-yo
kp(yoY
SETS
UNIQUELY
DETERMINED
301
BY PROJECTIONS
But
(3.21)
yl-x1= Yiq(x )-xl-'>O(xl)-xl
X1
since yf= Y' = (yl/xl)qf(xl)> xl = xI
(xi) and we are in case (3.6b). Thus v will
have a bounded densityif and only if thereis an M forwhich
(3.22)
y
q(y)
1
1
<
And,from(3.16), A willhave a bounded densityifand onlyifthereis an M forwhich
(3.23)
q(y)l(
(y) -y) c M
since both termson the rightin (3.20) have the same formafter(3.21) is used in the
denominatorof the last termin (3.20).
It remainsto show thatthereis 0< a<1 and q(y), O<y<l, forwhich (3.10),
(3.11), (3.22), and (3.23) hold forthe 4) determinedby (3.8), as in (3.12)-(3.15). We
choose in particular
a = 2/5,
(3.24)
q(y) = y(l
_y)2
and directlyverifythat (3.10) and (3.11) hold. It is clear from(3.13) and (3.14) that
the onlychance for4 (y) -y or 4>'(y) to be small is neary = 0 or y = 1. But near a = 0,
from(3.12), recallingb = b(a) =4-'(a), (3.11), (3.13), and (3.24),
{
b
a
2(1_x)2dx=2{2
a
xq(x)dx-a{
q(y)dy=
a3
6
a~~~~~4
a5b
6 203
3
b4 b5
Let - standfor"bounded above and below by a constantmultipleof." It followsthat
near a=0, 4)'(a)l1, I )(a)-a
q(a) and hence (3.22) and (3.23) hold near a-0.
Near a =1, from(3.12), subtractingfrom(3.12) witha = b = 1, and setting1 -a =
and 1-b=6=
X2(1-X)2
b
dx=2
a
xq(x) dx-a
x2(1-X)2
=2
{
a
q(y) dy
dx-(1-
a)
o
x2(1 _X)dx
o
a3 a4 3'ab5 3 b
3
1
12
20
3
2
b5
5
It followsthatnear a = 1,+'(a) = 1,+(a) - a = (1 - a)2 q(a) and so (3.22) and (3.23)
hold near a = 1. This completesthe proofthatA and v have bounded densities.
The factthatthereexistA and v withdensitiestakingvalues only0 and 1 follows
easily fromthe followingsimple,powerful,and apparentlynew result,provedin [2].
This resultsays thatiff is any densityin R' with0 ?f? 1, thenthereis a densityg
withvalues in {0, 1} thathas thesame projectionsas f This is provedin thecase n - 2
conditionfortheexistenceof a bounded
by observingthatthenecessaryand sufficient
densitywithgiven marginalsdue to Strassen[11] is the same as the necessaryand
sufficient
conditionforthe existenceof a set withgivenmarginalsdue to Lorentz[9]
for n = 2. The proofforgeneraln followseasily fromthe resultforn = 2.
0
302
FISHBURN,
LAGARIAS,
REEDS,
AND
SHEPP
Final remarks.The firstpart of the proof of Theorem3.2 shows thatthereis a
fuzzysetwiththesameprojectionsas SO,wherea fuzzysetis a functionfo
=fo(x) E [0, 1].
Indeed, if Juand v are as in the proofwithdensitiesbounded by 1 (multiplyby a
smallconstant),we can setfo= X(So) - dt/ dA+ dv/dA.Notethatthisis a muchdifferent
statementthanthe trivialone thatthereis a bounded densitygo(x) supportedon the
cube withthe same projectionsas y(SO), i.e., as S; merelytake
g0(x) = 9P1(xJ)P2(x2)P3(x3),
wherePi(xi) is the projectionof SO onto xi,
_x2
fori = 1,2;
2
Pi(xi) =
2
The difference
is that go is not a fuzzyset because go(x) > 1 at some pointssuch as
to notethatthereare analogous reconstruction
go(O,0, 1) = 9/4. It is interesting
problems in whicha measureexistssatisfying
a givenset of "moment"conditions,but no
set exists. For example, it is impossibleto finda set Ac [0, 1] with Leb (An I) =
2 Leb (I) foreveryintervalI, but therecertainlyis a measure(or fuzzyset), -A,with
thisproperty.Related resultsare foundin [10] (see also [3]), [5], and [11].
Appendix.Proofof Lemma2.5. The bounded open sets SI and S2 are constructed
recursively:each Si is an infiniteunion of open unit cubes of size 2-k for various
values of k.
The basis of the construction
is a configuration
whichfitsinside a 3 x 2 x 2 block
made up of sets T1 and T2 each of whichconsistsof threeunit cubes. Let the block
be B = {(x, y,z): 0 c x c 3, o c y _ 2, o c z c 2} and label each cube by itscornerhaving
the smallestcoordinatevalues. The configuration
is:
T, = {(O, O,), (1, 0, 0), (2, 1, 1)}
T2= {(0, 1,O), (2, 0, O), (1, 0, 1)}.
It is picturedin Fig. A.1, wherethe cubes in T1are labeled witha plus sign,and those
in T2 by a minussign.
has threekeyproperties:
The basic configuration
(1) The sets T1 and T2 have the same projectionson all threeaxes.
(2) T1 containsa 2 x 1 x 1 block in the lower leftcornerof the lower layerof B
and a 1 x 1 x 1 block in the upper rightcornerof the upper layerof B.
for T1 (or T2) in T1U T2mustcontaina pointin every
(3) Anybad configuration
cube of T1U T2.
=
y
1
o
x=0
1
1
=
2
z0=
1+1
1
x=O
z1
FIG. A.1. Basic configuration.
2
SETS
UNIQUELY
DETERMINED
BY PROJECTIONS
303
The firsttwo propertiesare clear. To prove the thirdproperty,we forman
undirectedgraphwhose verticesare thecubes in T1U T2and wherean edge represents
includesa pointin one vertex,thenit necessarily
a condition:"If a bad configuration
includes a point in the othervertex."Such conditionsare forcedby the requirement
thata bad configuration
has equal projectionsforthesubsetsof T1and T2. For example,
if a bad configuration
includesa pointin the cube (2, 1, 1) in T1, thenit mustinclude
a point in the cube (1, 0, 1) in T2, and vice versa, which is seen by examiningits
projectionin the z-direction.For a pointin (2, 1, 1) in T1 can onlybe matchedin the
z-directionby a pointin T2having1 < z < 2, and all such pointslie in thecube (1, 0, 1).
Proceedingin this way, we obtain the graph picturedin Fig. A.2, where edges are
labeled by the directionof projectionthatverifiestheiroccurrence.Since thisgraph
includeseveryvertex.
is connected,it followsthata bad configuration
(
(1,0,0) )
(
(10,19)
x
FIG. A.2. Graphassociatedto basic configuration.
Constructa shrunkenand rotatedcopy
We now beginthe recursiveconstruction.
first
the
firstaxis by 2 and then permutingthe
of the basic configuration
scaling
by
in
a
2 X x 2 block consistingof 1 x x 1
to
a
axes cyclically obtain configuration
x
x
2
block
that
its lowerleftcornerin its lowest
so
parallelepipeds.We place this2 X
Ix
1
lx
1,
T1. The lx I x 1 cube (2, 1, 1)
in
cube (2, 1)
level exactly overlaps the
correspondsto the union of two cubes labeled (0, 0, 0) and (1, 0, 0) in the shrunken
Let (T', TV)denotetheanalogues of T1and T2in theshrunkenconfiguconfiguration.
ration.Our second set is:
U' = T1U T'-{cube (2, 1, 1)}
Uf= T2U Tf -{cube (2, 1, 1)}.
Now we repeat this constructionone more time, shrinkingand rotatingthe basic
as a set of
to obtain a 2 x 1 x 2 block containingthe basic configuration
configuration
22
1X X
and addingitin'toobtain:
parallelepipeds,
U' = T1U TfU T'1'- {two overlappingcornercubes},
Uf'= T2U TfU T' - {two overlappingcornercubes}.
the compositeconfiguration.
We call this configuration
304
FISHBURN,
LAGARIAS,
REEDS,
AND
SHEPP
Now we describethecompositeconfiguration
explicitly.For conveniencewe scale
up the sets U', U2' so thatall cubes involvedhave side at least one, by expandingthe
y and z axes by a factorof 2. When we do so, the resultingconfigurationUl, U2 is
embeddedin a 5 x 6 x 7 block B'=-{(x, y,z): oc x c 5 O_ y c 6, 0o z-c7}. U1 consists
of a 2 x 2 x 2 cube A, threeI x I x 2 parallelepipedsB, C, D, and a I x 1 x I cube E:
A. (0, 0, 0), (0, 0, 1), (0, 1, 0), (0,
(1,0,0),
1, 1)
(1,0, 1), (1, 1,0), (1, 1, 1)
B. (2,4,2),(2,4,3)
C. (2,3,4),(2,3,5)
D. (3,2,2),(3,2,3)
E. (4,5,6).
The set U2 consistsof three1 x 2 x 2 parallelepipedsF, G, H and three1 x 1 x 1 cubes
I, J,K, whichare:
F. (0, 2, 0), (0, 3, 0), (0, 2, 1), (0, 3, 1)
G. (2, 0, 0), (2, 0, 1), (2 1,0), (2, 1, 1)
H. (1, 0, 2), (1, 0, 3), (1, 1,2), (1, 1, 3)
I. (3, 5, 4)
J. (4,4,5)
K. (3,4,6).
The compositeconfiguration
( U, U2) has threekeyproperties:
(1) The sets U1 and U2 have the same projectionson all threeaxes.
(2) U1 containsa 2 x 2 x 2 block in the lower leftcornerof its lowesttwo layers
and a 1 x 1 x 1 block in the upper rightcornerof its highestlayer.
(3) Any bad configuration
for U1 (or U2) in U1U U2 mustcontaina pointin its
highestcube (4, 5, 6).
Here (1) and (2) are easilychecked.Property(3) is proveddirectlyusingan argument
similarto thatgivenforthebasic configuration.
We use directedgraphswhosevertices
consist of several cubes, and whose directededges indicate: "A bad configuration
containinga pointin some cube in the enteringvertexmustcontaina pointin one of
thecubes in thevertexpointedto." The edgesin thegraphare constructed
byarguments
similarto thoseused forthebasic configuration.
FigureA.3 givessuch graphsshowing
that any bad configurationmust include a point in the cube E, thus proving
property(3).
To describethe recursiveconstructionof SI and S2, we forman infiniteset of
shrunkby factorsof 2 at each step,so thatthe
copies of the compositeconfiguration,
kth composite configurationlies inside a 5/(2k-1) x 6/(2k-1) x 7/(2k-I ) block Bk. These
blocks are stackedsuccessivelyso thatthe cornerE of the block Bk exactlyoverlaps
the cornerA of block Bk+l, as indicatedin Fig. A.4.
in block Bk. The sets SI, S2
Let U(k),U9k) denote the compositeconfiguration
are givenby:
SI
S
=
U( U(2k-))
00
k=I
(U
k=l
(00
k=I
U(2k-1)) U
If we let U1 denote U1 - {A}
Sl-{A}
00
U
-
U(2k))
00
U U (2k-1)) U U u(2k)
k=I
- {all overlappedcorners}.
then
{E}
(
k=I
U2k)) -{all overlapped
corners}
k=I
)
)
00
U U(2k-1))
U
k=I
00
k
U U(2k))
k=I
SETS
DETERMINED
UNIQUELY
3
= 2 0rDEIC
x
z2or3
G
H
305
BY PROJECTIONS
x3
Y=~~~5
I
c
z-4r
EIJ
rN4
FIG.
A.3. Directedgraphsassociatedto compositeconfiguration.
FIG.
of blocksBk.
A.4. Recursiveconstruction
It is easy to see thatthe sets S1 and S2 are bounded. Moreover,we show that:
(1) SI and S2 have the same projectionson all threeaxes.
forSI containedin SI U S2.
(2) There existsno finitebad configuration
To show that(1) holds, we observethat
A
(A
Uk-1))
u
(A
u2k))
u2=(
u(2k-1)) u
(A
u2k))
306
FISHBURN,
LAGARIAS,
REEDS,
AND
SHEPP
have the same projectionson all threeaxes. Indeed, thisholds by property(1) forthe
compositeconfiguration
and thefactthatall setsin theuniondefiningS1 (respectively,
S2) are disjoint.The sets SI and S2 are not disjoint,since theirintersectionV is:
V=SFnS2=
00
U
k=I
E'k),
where E(k) denotes the cornercube E in U(k), which is also the cornercube A in
U(k+l). Since S = S1- V and S2= S2- V,property(1) follows.
To prove property(2), let J(k) U(k) U U(k), and call U(k) the levelk pointsin
SI U S2. We firstprove the analogue of property(3) of the compositeconfiguration,
whichis:
(3') Any bad configuration
for SI in SI U S2 that containsa point in Uk) must
containa point in U(k+1)
argumentas in Fig. A.3, except
This is provedby exactlythesame graph-theoretic
to be "all pointsin SI U S2 in all levels
thatthe vertexE in thatgraphis interpreted
'k+ 1." The conclusionof property(3) thattheremustbe a point in E is replaced
at a higherlevel.
by the conclusionthattheremustbe a pointin thebad configuration
This pointmustbe on level k + 1 because pointson level k and pointson levelsgreater
thanor equal to k +2 do nothave anyprojectionsthatagree.So (3') is proved.Finally,
property(3') implies property(2), because repeated applications of property(3')
at all levels ' k. Hence the bad
produce at least one point in the bad configuration
containsinfinitely
O
configuration
manypoints.
-
We are gratefulto S. H. Kahan for introducingus to the
Acknowledgments.
problem,to D. D. Sleator,H. S. Witsenhausen,and P. G. Doyle fortechnicalobservations,and to J. H. B. Kemperman,H. Kellerer,and A. Kuba fortellingus about their
own results.
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Sets uniquely determined by
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