The Number of Three-Dimensional Convex Polyhedra
Author(s): Edward A. Bender
Source: The American Mathematical Monthly, Vol. 94, No. 1 (Jan., 1987), pp. 7-21
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2323498 .
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THE NUMBER OF THREE-DIMENSIONAL CONVEX POLYHEDRA
1987]
The NumberofThree-Dimensional
ConvexPolyhedra
EDWARD A. BENDER*
Department
ofMathematics,
University
ofCaliforniaat San Diego,La Jolla CA 92093
Abstract
A convexpolyhedron,or polytope,is theboundedintersection
of closed half-spaces.
The problemsof determining
thenumberof threedimensionalconvexpolyhedraas
a functionof the numberof facesor edges or both have been aroundforover 150
years. Except for Steinitz'sconversionof polyhedrato "planar maps", littlewas
done on the problem until the work on "rooted" planar maps in the 1960's.
Recentlythe original(unrooted)questionshave been answeredasymptotically.
We
will retracethe stepsthatled to thisresult.
1. Introduction
Convex polyhedra(also calledpolytopes)are theanaloguesto convexpolygonsin
of
higherdimensions.We can definea convexpolyhedronas a boundedintersection
we could definea convex polyhedronto be the
closed half-spaces.Alternatively,
convex hull of a finiteset of points. We will be concerned exclusivelywith
three-dimensional
polyhedra:thosethatlie in 3-spacebut do not lie in a plane. In
the future,"polyhedra" will always mean "three-dimensional
convex polyhedra."
Cubes, tetrahedra,and prismsare all examplesof polyhedra.In an obvious way,
but we
polyhedrahave vertices,edges,and faces.(These can be describedformally,
need not do so here.)We will use the notationP0, P1, and P2 forthe numbersof
vertices,edges, and faces,respectively,
of a polyhedronP. Euler's famoustheorem
(1752) statesthat
P0 - P, + P2 = 2.
Suppose that v, e, and f are positiveintegerswith v - e + f = 2. Does there
exist a polyhedronP with P0 = v, P1 = e, and P2 = f? The answer is No in
general;however,it is easy to givenecessaryand sufficient
conditions:
THEOREM
1. Thereis a convexpolyhedron
P with
P0 = v,
ifand onlyif
v-e
+
P1 = e,
and P2=
f
f = 2, and 4 < v < 2e/3, and 4 < f < 2e/3.
EdwardA. Bender:Workingin thearea wherematrixtheoryand numbertheorymeet,I receivedmy
at Harvardand a memberof the
Ph.D. underOlga Tausskyat Caltechin 1966. I was a PeirceInstructor
researchstaffat the CommunicationsResearchDivision of the InstituteforDefense AnalysesbeforeI
joined the Universityof Californiain 1974. My major professionalinterestat presentis asymptotic
enumeration.I flirtwith other areas of "concrete" mathematics,population biology,and computer
science.
*Researchsponsoredby Departmentof ComputerScience,University
of Georgia at Athens.
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8
EDWARD A. BENDER
[January
Proof. The firstconditionis Euler'sTheorem.Since the"smallest"polyhedronis
thelowerbounds on v and f are necessary.An edge correspondsto
a tetrahedron,
an unorderedpair of vertices,called its ends. Everyvertexmustbe the end of at
least threeedges,and each edge has twoends. If t is thetotalnumberof ends,then
3v < t = 2e and so v < 2e/3. Similarly,since each face is borderedby at least
threeedges and each edge lies on two faces,f < 2e/3.
Here's a sketchof theconverse.If thetriple(v, e, f ) satisfiestheconditions,then
thereare nonnegativeintegersx and y such that
(v, e, f ) - x(l, 3,2) -y(2, 3,1)
equals one of (6, 10,6), (6,9, 5), (5, 9, 6), and (4,6,4). (Prove it by inductionon e.)
whichwe'll call
Each of the fourtriplesjust listedcan be realizedby a polyhedron,
irreducible.Adding (2,3, 1) can be realizedby slightlyadjustingthe edges meeting
at a vertexand replacingthe vertexby a triangularface. Adding (1, 3, 2) can be
facesin place of two adjacentedges
two triangular
realized similarlyby introducing
one of thefourirreduciblepolyhedracan
of some face. (See Figure1.) By iterating,
be builtup to realize(v, e, f).L
FIG. 1. Buildingup to a specified(v, e, f). Additionsare shownheavy.
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1987]
THE NUMBER OF THREE-DIMENSIONAL CONVEX POLYHEDRA
9
Since the existencequestionwas easily settled,we proceed to the next typeof
question:
Q. How many"distinct"convexpolyhedraP have P0 = v and P2 = f ?
onlyone of
There is no need to specifyP1 sinceit equals v + f - 2. By restricting
thePk's, we areled to askfork = 0,1 and2.
Qk.
How many"distinct"convexpolyhedraP have Pk = n?
To answerthequestions,we need to say whatwe mean by distinctpolyhedra.Given
two polyhedraP and Q, theremaybe a one-to-onemappingm of thefacesof P to
the faces of Q that preservesincidence;i.e., if F1 and F2 are faces of P that
intersectin exactlyone edge (resp., vertex),then m(Fl) and m(F2) intersectin
If no suchmappingexists,P and Q
exactlyone edge (resp.,vertex),and conversely.
This will be whatwe mean by "distinct."
distinct.
are called combinatorially
Steinerposed Q2 in 1832 and Kirkmanstatedin 1878 thathe saw no hope of
answeringQ with the presentpower of mathematics.Shepard (1968) asked for a
close approximationto the answerto Q0.
by placinga vertexin each face
The dual, P* , of a polyhedronP is constructed
of P * and joining two such verticesby an edge if and only if the corresponding
faces of P share an edge. The facesof P* correspondto theverticesof P. One can
show that P** = P. Thus dualityis a bijection,Q0 and Q2 have the same answer,
and the answerto Q remainsunchangedif thevalues of v and f are switched.
Q (resp.,Qk)
For specificvalues of v and f (resp.,k and n), thecorresponding
can be answeredin a finitelengthof timesincean algorithmexistsforconstructing
This methodis not an acceptableanswer.What
all polyhedrawith-givenparameters.
is an acceptable answer? One possible definition,suggestedby the theoryof
algorithms,is: a way of calculatingthe number,whichrequiresan amountof time
thatis a polynomialin v and f (or n). It is quite likelythatno answerin thissense
of an answer?One wayis to allow moretime
exists.How can we relaxthedefinition
forcalculatingthe formula.If thisis relaxedtoo much,one can use the algorithm
all polyhedra.
alluded to earlierforgenerating
Another way to adjust the notion of answer is by requiringonly a good
approximateformulathatcan be computedquickly.Whatis a goodapproximation?
go to zero as v and f
We will requirethatthepercentageerrorin theapproximation
about how numbersbehave
(or n) get large. Such answers,whichgiveinformation
as the parametersget large, are called asymptoticformulas.All the questions
Q, QO, Q1, Q2 above have now been answeredasymptotically.
In thispaper we willretracethepath to theanswers.The firststepwas takenby
Steinitz(1922), who convertedthe questionsto problemsabout countinggraphsin
the plane. Nothing furtherwas done until Tutte developed methods for planar
enumerationin the 1960s. As a result,Mullin and Schellenberg(1968) obtaineda
"generatingfunction"for"rooted" polyhedrawithgivennumbersof verticesand
faces. This led to an explicitbut messyformulaforrootedpolyhedra.Benderand
functionto obtainan asymptoticformulathat
Richmond(1984) used thegenerating
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10
EDWARD A. BENDER
[January
(1985)combined
this
is validforpartoftherangeof v and f. BenderandWormald
answerto Q or Qk forrooted
withvariousestimates
to showthatan asymptotic
The entirerangewas coveredby Bender
polyhedra
givesan answerforpolyhedra.
and Wormald(to appear)as a resultofworkon thepaperyouarenowreading.
If all proofshadbeenincluded,
thisarticlewouldhavebeena smallmonograph.
I havereplacedmostproofswithbroadsketches.
Ifyouareinterested
in
Therefore,
thedetails,consulttheoriginal
articles.
of
information.
For a variety
Federico(1975) was thesourceof thehistorical
questionsconcerning
see Shepard(1968).
polyhedra,
2. The graph-theory
problem
A graphconsistsofa setofvertices
withedgesjoiningsomepairsofthevertices.
to every
The vertices
and edgesarenotlabeled.If thereis a pathfromeveryvertex
A loopis an edge
othervertexalongtheedges,thenthegraphis calledconnected.
withbothendpoints
thesame.If theendsof e1 arethesameas theendsof e2, we
a multiple
edge.A (planar) mapis a connected
graph
saythatel and e2 constitute
drawnon theplaneso thatno edgescross.Themaximal
regions
containing
no edges
are called faces.The unboundedfaceis calledthe external
face. Two maps are
thesameifonecanbe converted
intotheotherbystretching,
considered
contracting
theplane.
and/orreflecting
A mapis calledk-connected
iftheredoesnotexista positive
integer
j < k and a
partition
oftheedgesintotwosetsE1 and E2 suchthateachsetcontainsat leastj
Herearesome
endpoints.
edgesand theedgesin E1 n E2 containat mostj distinct
simpleusefulobservations
on k-connectedness.
01. "Connected"is thesameas "1-connected."
ifand onlyifit containsno loops
02. A mapwithat least2 edgesis 2-connected
and no vertexis encountered
morethanonce as we walk arounda face
boundary.
if and onlyif it
03. A 2-connected
map withat least 4 edgesis 3-connected
in
containsno multiple
edgesand everypairof facesthathavetwovertices
common,sayv and w,also havetheedge(v, w) in common.
To prove02, notethatif v is encountered
Exercise 1. Provetheobservations.
thegraph.To prove03, notethatremovalof v
twicethenitsremovaldisconnects
bothv and w. If (v, w) is nota
ofeachfacecontaining
and w splitstheboundary
thegraph.You mayfindit easierto
commonedgeof thesefaces,thisdisconnects
ifyoudrawthemap so thatone of thosefacesis external.
see whatis happening
to a mapas follows.
Selecta faceF. Removeall
A polyhedron
maybe converted
Placea planeparallelto F on the
of thepolyhedron
excepttheedgesand vertices.
near
fromF. Placea lightoutsidethepolyhedron
oppositeside of thepolyhedron
noneof theshadowsthattheedges
thecenterof F. If thelightis placedcarefully,
of F will
cast on theplanewillcrosseachotherand theshadowof theboundary
bound the externalface of a map formedby the shadows.The facesof the
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1987]
THE NUMBER OF THREE-DIMENSIONAL CONVEX POLYHEDRA
11
polyhedroncorrespondto thefacesof themap. See Figure2. This pictureis called a
Schlegeldiagramforthepolyhedron.
FIG. 2. A polyhedronprojecteddown onto a plane by thelightL givesa Schlegeldiagram.
ifand
THEOREM 2. (Steinitz) A map is theSchlegeldiagramofa convexpolyhedron
onlyif it has at least 4 verticesand is 3-connected.
Exercise 2. Provethenecessityby using03.
For a proof,see Chapter13 of Griinbaum.The lack of a
The converseis difficult.
correspondingresultin higherdimensionsseriouslyhampers attemptsto count
convex polyhedrain thosedimensions.El
thereare generallymanySchlegeldiagramsforeach polyhedron.
Unfortunately,
This leads us to thenotionof rootedmaps and polyhedra.A polyhedronis rooted
by choosing an edge (called the rootedge), one vertexon the edge (called the root
vertex)and one face adjacentto therootedge (called the rootface). A 2-connected
map is rooted if an edge on the externalface (also called the root face) is
degreeof a map or polyhedronis thenumberof edges
distinguished.The root-face
on the root face.
COROLLARY2.1. There is a one-to-onecorrespondence
betweenrooted convex
mapswithat least4-vertices.
polyhedraand rooted3-connected
Proof. The directionof a rootedge willbe such thattherootvertexis thetail of
the rootedge. Arrangethepolyhedronso thatwhentherootfaceis viewedfromthe
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12
EDWARD
A. BENDER
[January
outside and traversedin the directionof the root edge, the traversalis clockwise.
thepolyhedron.)Now place thelightnear thecenterof
(This may requirereflecting
the root face. This givesa one-to-onecorrespondence.
O
functionfor3-connectedmaps.
In thenextsectionwe willdiscussthegenerating
In Sections 4 and 5 we will see how thatleads to asymptoticsforrootedSchlegel
diagramsand, hence,forrootedpolyhedra.If all theways of rootinga polyhedron
P weredistinct,it would have P1 *2 *2 = 4P1 rootedversions.In Section6 we will
see that formost polyhedraall rootingsare distinct.This providesus withasymptoticanswersto thequestionsQ and Qk.
ofrootedmaps
3. The exactnumber
Generating
functions.If an is a sequence definedfor n > 0, thenthe generating
functionforthe sequenceis
00
=
A(x
YEanxn
n=O
We will adopt the conventionof using a lower-caseletterfor a sequence element
and the corresponding
upper-caseletterforthe corresponding
generatingfunction.
These ideas extend to multiplyindexed sequences; for example, the generating
functionforthe triplyindexedsequence bi, k is
00
B(x, y, z)
=
00
E
i=O j=O
00
E bi , kX yjZk.
k=O
All theinfiniteseriesthatwe use convergewhenthevariablesare sufficiently
small.
Suppose thatwe are givena generating
functionforsome sequence,say, a, j. By
Taylor's Theoremforfunctionsof twovariables,thecoefficients
of thepowerseries
forA(x, y) are uniquelydeterminedand so mustbe the sequence a i1. Thus, if we
somehowexplicitlyexpand A(x, y) in a powerseries,we will obtain a formulafor
the sequence ai j
There are a varietyof rooted maps that one mightenumerate.Each of these
problemsis approachedby describinga methodforconstructing
maps out of other
maps. When this descriptionis translatedinto a statementinvolvinggenerating
functions,the resultis a functionalrelationshipamong the generatingfunctions.If
the constructioninvolvesonlythe typeof map we are counting,thenthefunctional
equation involvesonlythe generatingfunctionwe are interestedin. Thus it can be
solved, at least in principle,forthatgenerating
function.
2-connected
maps
Brownand Tutte(1964) enumerated2-connectedrootedmaps. Since theirresult
is centralin latercalculations,we'll look at theirmethod.
Suppose M is a rooted2-connectedmap thatis not a singleedge. The root edge
of M belongs to two faces, the exteriorface and some interiorface, which are
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1987]
THE NUMBER OF THREE-DIMENSIONAL CONVEX POLYHEDRA
~
MO~
13
M
FIG. 3. Decomposing a 2-connectedmap.
shown by heavylines in Figure3. By splittingthe map into pieces at all verticesv
such that v lies on bothfaces,we obtaina sequenceof maps MO, .., Mt,whereMo
is theroot edge. By 02, each Ml can be seen to be 2-connected.Each can be rooted
by rootingthe edge of Mi firstencounteredwhen followingthe externalface in a
clockwise directionstartingfromthe root edge. This decompositionis reversible
provided we specifywhich edge on the root face of each Mi is the last edge
we have a
encounteredon therootface of M in our clockwisetraversal.Therefore,
unique method for buildingup a 2-connectedmap from2-connectedmaps with
fewerfaces.
we need to keep trackof the numberof
In order to describethisnumerically,
vertices,the numberof faces,and the degreeof the root face. The last quantityis
needed because an M, withroot-facedegreek has k - 1 possible choices forthe
last vertexon the rootface of M.
Let i,j k be the numberof rooted2-connectedmaps withi + 1 vertices,j + 1
faces, and root-degreek, except that the map consistingof a single edge is not
counted. Since the root-facedegreedoes not interestus, we want F(x, y, 1), the
forfi j = Ekfi,j k. It can be shownthattheaboveconstruction
function
generating
is equivalentto the equation
F(x, y, z) =yz
f(
t=1
E
fij,XkXY (Z + Z2 +
i, j, k
?,. +Zk-l)
+
(3.1)
Aftera littlemanipulationwe get
00
F(x, y, z) = yz E ((zF(x, y, 1)
t=O
-
F(x, y, z))/(l
-
z) + xz)t - yz.
and writingF forF(x, y, z) and
Aftersummingthe geometricseries,rearranging,
F1 forF(x, y, 1), we obtain
F2 + ((1
-
z)(1
-
xz) + yz - zFi)F
-
yZ2(X
-
xz + F1) = 0.
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(3.2)
14
EDWARDA. BENDER
[January
Since (3.1) is a directtranslationof a construction
whichbuilds up maps out of
maps with fewerfaces,(3.2) and the initial conditionsfiO, k = 0 must determine
F(x, y, z) uniquely.Since both F and F1 appear in (3.2), it is not clear how to
extractF or F1 from(3.2) (settingz = 1 simplyleads to the equation 0 = 0). One
approach is to use educatedguessing,as done by Brownand Tutte.Thereis a more
systematicapproach (Brown, 1968), but it can lead to a morass of algebra:
Complete the square in (3.2) to obtainan equationof the form
(F + stuff)2= G (x, y, z, F1).
(3.3)
Let z = Z(x, y) stand forthe value of z forwhichthe leftside of (3.3) vanishes.
Since theleftside of (3.3) is a square,its derivativewithrespectto z also vanishesat
Z. Applyingthisto therightside of (3.3) we obtainthe two equations
G(x, y, Z, F1) = 0 and G,(x, y, Z, F1) = 0
in the two unknownsZ and F1. These are rationalequations,and theycan be
"solved" forF1.
THEOREM 3. (Brownand Tutte) The number
maps with
fi j of rooted2-connected
+
i 1 verticesandj + 1 faces is givenby
F(x, y)
=
uv(1
-
U)2,
-
u - v),
whereu and v are givenimplicitly
by
X = U(I
-
V)2,
y = v(1
and u(0,0) = v(0,0) = 0.
The explicitvaluesare
(2i +j - 2)!(2j + i - 2)!
i!j!(2i -1)!(2j - 1)!
Proof. Since F(x, y) = F(x, y,1), we could solve G = G, = 0 for F1. This is
x and y as statedin thetheorem.The explicitvalue forf
done by parameterizing
is obtained by a techniqueknownas Lagrangeinversion.
That techniqueexpresses
the coefficientsof any functionH(u(x, y), v(x, y)) in termsof derivativesof H
and of u(I - v)2 and v(I - u)2 with respectto u and v. See Section 5 for a
discussion.O
Quadrangulations
A quadrangulation
is a map such thateach internalface is a quadrilateral.
Thereis a connectionbetweenrooted2-connectedmaps and rootedquadrangulations with quadrilateralexternalfaces,discoveredby Brown (1965). Let M be a
rooted map withvertexset V and face set F. Place a vertexin the middleof each
face to forma new set of verticesV*. Definea set of edgesby connectingv and v*
to v*. This givesa map Q. Let r
if and onlyif v is a vertexof thefacecorresponding
be the root vertex of M, e = (r, v) the root edge of M and v* the vertex
correspondingto theexternalfaceof M. The rootvertexof Q is v and therootedge
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1987]
THE NUMBER OF THREE-DIMENSIONAL CONVEX POLYHEDRA
15
JIf~~~~~~t
.00.
~
~ ~
0
FIG. 4. Heavy dashed linesare theedgesof theoriginal2-connectedmap. Solid linesare theedgesof the
correspondingquadrangulation.
is (v, v*). The edgesjoining v* are drawnso thatno edges of Q lie outsideof the
regionbounded by (v, v*), (v*, r) and e. See Figure4.
is a bijectionbetweenrooted2-connected
THEOREM 4. The above correspondence
withquadrilateral
graphs with more than one edge and rootedquadrangulations
the originalgraph is 3-connected
if and onlyif each
externalfaces. Furthermore,
is a face.
4-cycleof edgesin thequadrangulation
Proof. The bijectionis due to Brown (1965, Sec. 7) and the last part of the
theoremis due to Mullinand Schellenberg(1968, Sec. 5).
Exercise 3. Constructproofsusing02 and 03.
Z
Call a quadrangulationthatcorrespondsto a 3-connectedmap simple.
Exercise 4. Show that the verticesof a quadrangulationcan be partitioned
uniquelyinto twosets,called red and green,suchthatedgesconnectonlyverticesof
differentcolors, the red verticescorrespondto the verticesof the corresponding
2-connectedmap and thegreenverticesto themap's faces.
withrootdegree4, i + 1 red vertices
Note thatthe numberof quadrangulations
the numberof 2-connectedrootedmaps with
and j + 1 greenverticesequals f,,J,
i + 1 verticesand j + 1 faces.Let p, j be thenumberof thosethatare simpleand
have at least 8 vertices.By Theorem4 and Corollary2.1, the functionP(x, y)
counts rooted polyhedra.(The condition on verticesin p,1 eliminatessmall
3-connectedmaps thatdo not correspondto polyhedra.)
A quadrangulationhas a diagonalif thereare externalverticesv and w and an
internalvertexx so that(v, x) and (w, x) are edges.Let n,, count thenumberof
quadrangulationscountedby f,,X that have no diagonals and let N(x, y) be the
correspondinggeneratingfunction.
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16
[January
EDWARD A. BENDER
Everyquadrangulationwithrootdegree4, morethan4 verticesand no diagonals
can be built froma simplequadrangulationhavingat least 6 vertices.This is done
by replacing the internalfaces of the simple quadrangulationwith arbitrary
(1968, Sec. 6) showthat
quadrangulationsof rootdegree4. Mullinand Schellenberg
function
and correspondsto thegenerating
thisconstructionis uniquelydetermined
equation
(3.4)
(xy/F)P(F/y, F/x) = N(x, y) - xy.
The quadrangulationscountedby fij can be brokeninto threedisjointclasses:
(i) no diagonals,countedby N(x, y);
(ii) diagonal at theroot,countedby, say, R(x, y);
(iii) diagonal not at theroot,countedby, say, D(x, y).
Thus,
F(x, y) = N(x, y) + R(x, y) + D(x, y).
shownin Figure5 it followsthat
By the typeof construction
R(x, y)
A
=
(N(x, y) + D(x, y))F(x, y)/x.
B
FIG. 5. Buildingup R (x, y). QuadrangulationA has no rootdiagonalbut B may have any number.
the rolesof red and greenvertices(see Exercise4) gives
Interchanging
D(x, y) = (N(x, y) + R(x, y))F(x, y)/y.
By performingsome algebraicmanipulationson the last threeequations,one can
show that
N = ((1 + Fly)
+ (1 + Fx)-
(3.5)
1)F.
By settingX = F/y and Y = F/x and combining(3.4) and (3.5):
P(X, Y) = ((1 +
X)?
-
+ (1 + Y)
_
1)XY-F(x,
-
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y).
1987]
THE NUMBER OF THREE-DIMENSIONAL CONVEX POLYHEDRA
17
Exercise 5. Using this with Theorem 3 and settingr = u/(1 - u - v) and
s = v/(1 - u - v), show
forthenumberof
function
THEOREM 5. (Mullin and Schellenberg)The generating
distinctrootedconvexpolyhedrawithi + 1 verticesandj + 1 faces is givenby
P(X, Y) = ((1 + X)
+ (1 + Y)-
1)XY
-
F,
where
F = rs/(r + s +
1)3
by
and r and s are givenimplicitly
r = X(s
+ 1)2,
s =
Y(r +
1)2
r(O,O) = s(O,O) = 0.
Mullin and Schellenbergapplied Lagrangeinversionto obtaina formulaforp, j.
it is a double summationwithalternatingsigns,so it seemshard to
Unfortunately,
see how p, , behaves except by computingspecificvalues. I'll say more about
Lagrangeinversionand an exact formulain Section5.
in X and Y. This
Note that the generatingfunctionforP(X, Y) is symmetric
result also follows immediatelyfromdualitywithoutever seeing the generating
functionsfolloweasilyfromP(X, Y). Here are
functions.Various othergenerating
of yk in P(1, Y) is the numberof rooted convex
two examples. The coefficient
of xiy' in P(xy, y) is
polyhedrawith k faces. By Euler's theorem,the coefficient
the numberof convexpolyhedrawithi + 1 verticesand n edges.
numberof rootedpolyhedra
4. The asymptotic
We need some notationforwritingasymptotics.
g(n) meansthatf(n)/g(n)
f(n)
f(n)
=
--
O(g(n)) meansthatf(n)/g(n)
1 as n -o;
->
0 as n -s 0o.
By convention,f(n)/g(n) = 1 when f(n) = g(n) = 0. For functionsof two (or
is moreinvolved.Let R be a regionin thexy-plane
more) variables,theterminology
large.We say that
containingintegerpoints(x, y) withmin(x,y) arbitrarily
in R
f(m, n) g(m, n) uniformly
if
supIf(m, n)/g(m, n) - 11 -O 0 as k -o,
wherethe supremumis takenover all (m, n) in R withmin(m,n) > k. We define
in R in a similarfashion.
f(m, n) = o(g(m, n)) uniformly
forrootedmaps.
There are severalpossibleapproachesto obtainingasymptotics
is to work with a simple formulalike that forf, in
The most straightforward
Theorem3 togetherwithStirling'sformula,
n!- (2 n)1/2 (nle
n
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(4.1)
18
EDWARDA. BENDER
[January
This idea can also be adapted to certaintypesof sums,but seldom to those with
alternatingsignsunless the initialtermsdominatethe sum. In cases like P(X, Y),
one tries to work directlywith the generatingfunction.For an introductionto
see Bender(1974).
asymptoticsin combinatorics,
We now turnto P(X, Y). Thinkof it as a functionof two complexvariablesX
and Y. Such functionshave places at whichtheymisbehave,called " singularities."
A
littlebit of knowledgeabout the natureof the singularities
closestto the originis
to provideinformation
of thepowerseriesfor
oftensufficient
about thecoefficients
the function.It would take too much space to definesingularities
and discuss the
connectionbetweentheirnatureand thecoefficients
of thepowerseries.In thisway
Bender and Richmond (1984) obtained messy asymptotic formulas from
P(X, Y), P(1, Y), P(xy, y), etc. The resultforpi j was valid for i -o co provided
1/2 + c < i/j < 2 - c. This resultleaves a gap at each end because i/j is constrainedto stayaway from1/2 and 2 whilev/fcould approacheither1/2 or 2 as f
gets large.The extremeends werefilledin by
THEOREM 6. (Tutte,1962) Let ti be thenumberof rootedconvexpolyhedrawith
i + 1 verticesand all faces triangular.Then
2(4i - 7)!
(3i -4)! (i -1)!
3(i
256
16(6ri)/
27
The same formulaholdsif ti countsrootedconvexpolyhedrawithi + 1 faces and all
verticesof degree3.
5. Reconsideration
While writingthis paper, I simplifiedthe messy asymptoticformulafor pi,
mentionedearlier.Using thisresultand Theorem6 as a guide,I conjectured
THEOREM
as min(i,j) s o
7. (Benderand Wormald,to appear) Uniformly
P i,
35ij (j
+ 3
j + 3)
Proof. No highpoweredtools are needed. By applyingLagrangeinversion(see
different
below) to Theorem5 in a slightly
way thanMullin and Schellenbergdid, a
singlyindexed summationis obtained.To make asymptoticcalculationseasier,the
sum is transformed
twiceby usingthePascal triangleidentity
(ck
k ?
(k)
(k-
1)
and rearranging
terms.
What is Lagrangeinversion?Suppose we want the coefficient
of x' in f(g(x))
wherethe power seriesforf(y) is knownbut thatforg(x) is not known.Instead,
we only know thepowerseriesfortheinversefunctiong-1(z). Lagrangeinversion
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1987]
19
THE NUMBER OF THREE-DIMENSIONAL CONVEX POLYHEDRA
tells us how to computethe answer.There are variousgeneralizationsto functions
of severalvariables.See (S. A. Joni,1977) fora discussionand also a proofof the
following.
THEOREM 8. (I. J.Good) Supposethatf(zl,..., Zk) and H(z1,..., Zk), 1 < i < k
are powerseriessuchthattheHi's have nonzeroconstantterms.Let h1= ziHi. Then
thereexist uniquepowerseries gj(Xl,.
satisfying
the set of equations
.., Xk)
...
the
x1l ...
Also,
coefficient
of
hi(gll
gk)= xi.
Z
zZk in
coefficient
ofz"
det(dhj/dzj)f/(Hjh+1
in f(gl
Xlk
...
equals the
... ., gk)
k+1
This can be applied to Theorem5 with
(Z1,
Z2)
=
(r,
s),
(X1, X2) =
f = -rs/(r + s + 1)3,
(n1, n2) = (i,
(X, Y),
h1 = r/(s +
j),
h2 = s/(r +
1)2,
1)2.
Exercise 6. Show thatfori, j > 1, pi,j is thecoefficient
of r'sJin
(r + s + IY3(S ?
+ 1)2J'3(3rs
1)2i-3(r
If we write
(r ? s +
=
)
t, u
-
r- s
-
1).
(
-k)(t)rus-u,
u
t
thenMullin and Schellenberg'sformulais obtained.If we write
(r + s+
=
(r +
r/(l
(+
(-k
ri)(1
+
s)
S))
-k-j
(5.1)
then Bender and Wormald'sformulais obtained.You may wish to carryout these
calculations.If so, write
(1 + r + s)3 (3rs - r - s
-
1)
=
3rs(r + s + i)
3-
(r + s +
1>)2
and use (5.1) withk = 2 and k = 3.
6. The asymptotic
numberof polyhedra
In thissectionwe willdiscusstheproofand applicationof thefollowingtheorem.
THEOREM 9. (Bender and Wormald,1985) Let ui j be the numberof unrooted
convexpolyhedrawithi + 1 verticesand j + 1 faces. There are constantsA and
0 < c < 1 such thatforall i andj,
1
4(i +)ui,
/pi,j< 1 + c',
where0/0 is interpreted
as 1.
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20
[January
EDWARD A. BENDER
The theoremsays that ui,j approaches pi 1/4(i + j) veryquickly.Thus Theorems7 and 9 answerquestionQ. AnsweringQk involvesestimating
1
E
4*35
i(i(+j)
/ ~2iir 2j j
()+ 3 i+3
where the sum rangesover appropriatevalues of i and j. This can be done by
standardmethodsas discussedin Bender(1974) or by usingresultsin Benderand
Richmond(1984). The answersare
I
Q
=A(i, j), say;
(21)(21)~~
A( j,sy
972iU(i+ j) (i + 3)(i + 3
Qo - ('nn(4 + Th)/4V7)"/2A(n
Q, -
-
1,(n
-
1)(3 +
?
)/4);
4 A(n/2, n/2);
4
Q2= Qo,
where fractionalfactorialsin binomialcoefficients
are approximatedby Stirling's
formula(4.1). Tutte(1963) conjecturedand Richmondand Wormald(1982) proved
the asymptoticformulaforQ1.
How is Theorem9 proved?As noted at the end of Section2, thereare 4(i + j)
distinctways to root a convexpolyhedronwith i + j edges and no symmetries.
If
thereare symmetries,
thenumberof rootingsis less. That accountsfortheleft-hand
inequalityin Theorem9.
The right-handinequalityis based on estimatesfor rooted polyhedra with
Thereare threedifferent
symmetries.
typesof symmetries
possiblefora polyhedron.
One typepreservesthe orientationof the polyhedronand is essentiallya rotation.
The othertwo typesreversetheorientationand are distinguished
by whetheror not
the symmetry
maps any vertexor edge intoitself.If it has such an invariant,it can
in a plane; otherwise,a reflection
in a point.
essentiallybe viewedas a reflection
For each typeof symmetry,
if you are given(i) a connectedpiece cut out of the
polyhedronwhose imagesunderthe symmetry
coverthe entirepolyhedronand (ii)
the nature of the symmetry,
thenyou can reconstructthe entirepolyhedron.The
piece cut out foryou mayinvolveedgesand facesthathave been cutin half.The cut
may also runthroughsomevertices.Connectall thoseverticesto a singlenewvertex
v and extend the cut edges to v. If the originalcut is chosen carefully,then the
resultingfigurewill be 3-connected.Since the numberof verticesand edges in the
new 3-connectedgraphis less than thatin the originalpolyhedron,Theorem7 can
usuallybe used to obtaina crudebut adequate upperbound.Variousadaptationsof
this idea are needed to handle all the cases thatarise. If you wish details,see the
originalpaper.
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1987]
COMBINATORIAL AND FUNCTIONAL IDENTITIES
21
Special cases of Theorem9 were proved by Tutte (1980) and Richmondand
Wormald(1982).
REFERENCES
E. A. Bender,AsymptoticMethodsin Enumeration,SIAM Rev., 16 (1974) 485-515 and (1976) 292.
E. A. Bender and L. B. Richmond.The AsymptoticEnumerationof Rooted Convex Polyhedra,J.
Combin. TheorySer. B, 36 (1984). 276-283.
E. A. Bender and N. C. Wormald,The Number of Rooted Convex Polyhedra,Canad. Math. Bull.,
submitted.
, AlmostAll Convex Polyhedraare Asymmetric,
Canad. J. Math.,37 (1985), 854-871.
W. G. Brown, Enumerationof QuadrangularDissections of the Disk, Canad. J. Math., 17 (1965),
302-317.
, An Algebraic Technique for SolvingCertain Problemsin the Theoryof Graphs, Theoryof
Graphs: Proc. Colloquium,Thiany,Hungary,1966 (P. Erdos and G. Katona, eds.), Academic
Press,New York (1968), 57-60.
W. G. Brown and W. T. Tutte,On the Enumerationof Rooted NonseparablePlanar Maps, Canad. J.
Math., 16 (1964), 572-577.
P. J. Federico,The Numberof Polyhedra,PhillipsRes. Repts.,30 (1975), 220-231*.
B. GrUnbaum,Convex Polytopes,Interscience,
New York, 1967.
S. A. Joni,Lagrange Inversionin HigherDimensionsand Umbral Operators,Linear and Multilinear
Alg., 6 (1978), 111-121.
R. C. Mullin and P. J. Schellenberg,The Enumerationof c-netsVia Quadrangulations,J. Combin.
Theory3 (1968), 259-276.
L. B. Richmond and N. C. Wormald,The AsymptoticNumber of Convex Polyhedra,Trans. Amer.
Math. Soc., 273 (1982), 721-735.
G. C. Shepard,TwentyProblemson Convex Polyhedra,Math. Gaz., 52 (1968), 136-147, 359-367.
W. T. Tutte,A Census of PlanarTriangulations.
Canad. J. Math.,14 (1962), 21-38.
, A Census of Planar Maps, Canad. J. Math.,15 (1963), 249-271.
, On the Enumerationof ConvexPolyhedra,J. Combin.TheorySer. B, 28 (1980), 105-126.
Combinatorial
and FunctionalIdentities
in One-ParameterMatrices
DAN KALMAN
26565 MazurDr., RanchoPalos Verdes,CA 90274
ABRAHAMUNGAR
Department
ofMathematics,
NorthDakota State University,
Fargo,ND 58105
1. Introduction.A one-parametermatrixis one whose entriesdepend on a
parameterin such a way thatmatrixmultiplication
correspondsto performing
an
operation,e.g., additionor multiplication,
on the parameter.More formally,if a
of
Dan Kalman: I studiedfirstat HarveyMudd College,thendid mygraduateworkat theUniversity
Wisconsin. After holding facultypositions at Lawrence University(Appleton, WI), Universityof
Wisconsin-GreenBay, and AugustanaCollege (Sioux Falls, SD), I acceptedmy presentpositionas an
applied mathematicianin theaerospaceindustryin Los Angeles.
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