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Hipparchus Schroder and Hough

Richard P Stanley

and Plutarch Plutarch was a Greek biographer and

philosopher from Chaeronea who was b orn b efore ad and died after ad

He is b est known for his Parallel Lives which inspired such Renaissance

writers as Montaigne Shakespeare Dryden and Rousseau His many other

works have b een gathered together under the name Moralia a collection of

comparatively short treatises and dialogues which cover an immense range

of sub jects literary ethical p olitical and scientic p Part of the

Moralia consists of the TableTalk a collection of dialogues purp orting to

repro duce the afterdinner conversation of Plutarch and his friends and rel

atives on various o ccasions p In the TableTalk VI I I

app ears the following statement

Chrysippus says that the of comp ound prop ositions that

can b e made from only ten simple prop ositions exceeds a million

Hipparchus to b e sure refuted this by showing that on the ar

mative side there are comp ound statements and on the

negative side

Chrysippus c bc came to around and b ecame a

leading Stoic philosopher Hipparchus was a Greek c after

bc from in now Iznik Turkey who sp ent much of his

life at Rho des He was p erhaps the greatest astronomer of antiquity He is

most famous for his discovery of the of the based on his

own observations and those of Timo charis years earlier For further infor

mation on the work of Hipparchus see Bo ok I E Hipparchus was an

excellent though for a contrary view see p he was

the rst p erson to make systematic use of and he was probably

the inventor of stereographic pro jection However for many centuries no one

was able to make sense of the statement of Plutarch For instance T L

Heath vol p a standard older authority on Greek

says of Plutarchs statement that it seems imp ossible to make anything of

these gures while the more recent authority O Neugebauer p

states that Plutarchs statement has however so far eluded a satisfactory

explanation Similarly W and M Kneale p authorities on the

history of logic remark that It is dicult to make any satisfactory sense of

the passage N L Biggs p notes the paucity of combinatorial com

putations by the ancient and referring to Plutarchs passage says that

the most interesting of them is also the most mysterious A number of em

inent and historians of mathematics such as M Cantor J

Tropfke S G unther and E Artin have attempted to understand Plutarchs

statement without success An attempt to reconstruct Hipparchus pro ce

dure app ears in though it will b e apparent from our discussion that this

attempt is incorrect Another incorrect sp eculation app ears in p

Schroder Friedrich Wilhelm Karl Ernst Schroder was a German

logician who was b orn in Mannheim on November and died in

Karlsruhe on June He passed the do ctoral exam at the University

of Heidelb erg in and had p ositions in Zurich at the Eidgenossische

Polytechnikum Karlsruhe Pforzheim and BadenBaden b efore accepting

a p ost as full professor at Karlsruhe in Schroder worked mainly on

the foundations of mathematics notably with the theory of

functions of a real variable and mathematical logic He was one of the rst

p ersons to accept Cantors ideas in set theory and was one of the developers

of mathematical logic in the second half of the nineteenth century Schroder

is b est known to combinatorialists for his pap er in which he discusses

four bracketing problems The rst two problems concern the bracketing or

parenthesization of a string of letters that we may assume to b e all identical

say the letter x The second two problems are analogues of the rst two

where the string of letters is replaced by a set of elements We will discuss

only the rst two problems here

The formal denition of a bracketing is the following First x itself

is considered to b e a bracketing Recursively dene a bracketing to b e a

B B B where k and each B is a bracketing We

k i

represent the bracketing B as a parenthesized string of xs Thus think of B

as a k ary pro duct B B B If some B is the single letter x then

k i

we remove the parentheses surrounding B for clarity of notation Thus for

i

example the bracketing

xxxxxxxxxxxxxx

t

t t t

LL LL LL

L L L

L L L

L L L

t t t t t t t t L L L

LL DD LL DD

L D L D

L D L D

L D L D

L D t t t t t t t t t L D

Figure A tree

represents a way of multiplying xs whose last op eration was a ternary

op eration B B B where B xx B xxxxxxxx and B

xxxx and similarly for B B and B There are exactly eleven bracket

ings of four letters namely

xxxx xxxx xxxx xxxx xxxx xxxx

xxxx xxxx xxxx xxxx xxxx

Note that the last ve of these are built up entirely from binary op erations

and are therefore called binary bracketings

There are three fundamental equivalent ways to represent a bracketing in

addition to a parenthesized string discussed ab ove as plane trees polygon

dissections and Lukasiewicz words We now briey describ e these alternative

representations If B is a bracketing then we rst dene the plane tree B

corresp onding to B If B consists of a single letter then B is a single

ro ot vertex If B B B then B consists of a ro ot vertex drawn

k

at the top with subtrees B B drawn in that order from left to

k

right Thus the key prop erty dening a plane tree is that the subtrees of

every vertex are linearly ordered For instance the plane tree corresp onding

to the bracketing of is shown in Figure Note that a binary

bracketing corresp onds to a binary plane tree ie a plane tree for which

every nonendp oint vertex has exactly two successors

Next we consider p olygon dissections Let P b e a convex p olygon A

dissection of P is obtained by drawing some diagonals that dont intersect

in their interiors Thus P is divided up into regions that are themselves

convex p olygons In particular if P has m sides and we draw m such

diagonals the maximum number p ossible then we obtain a dissection for

which every region is a such dissections are called We

now explain how to asso ciate a plane tree D with a p olygon dissection D

We asso ciate with the degenerate p olygon with just two vertices a single

ro ot vertex Now x once and for all an edge e of the p olygon P called the

root edge In a given dissecton D the edge e is contained in a unique p olygon

Q which is a region of D Let k b e the number of edges of Q If we remove

the edge e and the interior of Q from D then we are left with dissections

D D D of k p olygons some p ossibly with just two vertices reading

k

counterclockwise from e along the b oundary of Q such that D and D

i i

intersect at a single vertex for i k Dene recursively D to b e

the plane tree whose subtrees of the ro ot are D D in that order

k

Note that if P has n vertices then D has n endp oints Figure shows

the p olygon dissection corresp onding to the tree of Figure

Finally we consider Lukasiewicz words The letters of such words come

from the alphab et A fx x x g The weight x of a letter x is

i i

dened by x i A word y y y made of letters from A is said

i m

to b e a Lukasiewicz word if y y for j m and

j

y y Thus y x The set of all Lukasiewicz words

m m

is called the Lukasiewicz language Ch To obtain a Lukasiewicz

word from a plane tree do a depthrst preorder search through the

tree By denition this is a linear ordering v v v of the vertex

p

set of dened recursively by v where v is the ro ot

k

of and are the subtrees of v in that order Dene

k

x x x

deg v deg v deg v

1 2

k

where deg v denotes the number of successors or children of vertex

i

v For instance the Lukasiewicz word corresp onding to the plane tree of

i

Figure is

x x x x x x x x x x x x

Note that since our bracketings B do not allow unary op erations the plane

tree B has no vertices of degree one and the corresp onding Lukasiewicz

word do es not involve the letter x

e

Figure A p olygon dissection

The corresp ondences established ab ove are easily seen to yield the follow

ing result

Prop osition a Let sn denote the total number of bracketings of a

string of n letters Then sn is also equal to i the number of plane trees with

no vertex of degree one and with n endpoints ii the number of dissections

of a convex n gon and iii the number of Lukasiewicz words with no

x s and with n x s

b Let bn denote the number of binary bracketings of a string of n

letters Then bn is also equal to i the number of binary plane trees with n

endpoints and hence with n vertices ii the number of triangulations

of a convex n gon and iii the number of Lukasiewicz words with n

x s and n x s and with no other letters such words usual ly with the

last x deleted are sometimes called Dyck words

We are now ready to explain the contribution of Schroder to these brack

eting problems Schroders rst problem asks for the number bn of binary

bracketings of a string of n letters Using a argument

Schroder derives the formula stated slightly dierently

n

bn

n n

Thus bn is just the Catalan number C for which an enormous literature

n

exists For some further information and references see A list of

ab out fty combinatorial interpretations of Catalan will app ear in

Exercise and is available on the World Wide Web at httpwww

mathmitedu~rstanecechtml

Schroders second problem asks for the total number sn of bracketings

of a string of n letters Schroders main result on his second problem is the

generating function

X

p

n

snx x x x

n

He also gives the values with the typographical error for s

s s

Perhaps the quickest way to obtain equation is the following Let y denote

the lefthand side The recursive denition of bracketing is equivalent to the

formula

y

y x y y y x

y

Multiplying by y yields the

y xy x

One of the solutions is spurious and the other one is just the righthand side

of

The numbers sn are now called Schroder numbers Schroder do es not

mention any other combinatorial interpretations of Schroder numbers nor

do es he give a single outside reference Let us p oint out some additional ref

erences The problem of counting the triangulations of a convex p olygon was

raised by Segner and solved anonymously by Euler The connection

b etween bracketings and plane trees was known to Cayley The bijection

b etween plane trees and p olygon dissections app ears in Etherington with

a sequel by Erdelyi and Etherington in The bijection b etween brack

etings and Lukasiewicz works is essentially the reverse Polish notation or

parenthesisfree notation developed by the Polish logician Jan Lukasiewicz

He came up on the idea of this notation in and rst pub

lished it in as explained in p fo otnote The connection

b etween reverse Polish notation and enumerative combinatorics app ears in a

pioneering pap er of George Raney

There is now a considerable literature on Schroder numbers and related

numbers To get into this literature see p Let us also

mention that it is easy to obtain a simple p for

the Schroder numbers which allows them to b e computed rapidly Namely

0

dierentiate with resp ect to x and solve for y to obtain

x y x y

0

y

y x x x

the latter equality a consequence of the quadratic equation Hence

0

x x y x y x

Expanding the lefthand side in a p ower in x and setting the co ecient

n

of x equal to yields

n sn n sn n sn n

No direct combinatorial pro of of this formula was known until D Foata and

D Zeilb erger after reading an earlier version of this pap er found such a

pro of

Hough The stage is now set for the denouement The astute reader

may have already anticipated it by comparing Plutarchs cryptic statement

with the values of the Schroder numbers In January David Hough

a graduate student at George Washington University who decided

only in that he would pursue a career in mathematics noticed that

the mysterious number of Plutarch ie the number of comp ound

prop ositions that can b e formed from ten simple prop ositions is just the

tenth Schroder number Hough learned ab out Plutarchs statement from

Exercise Houghs discovery strongly suggests that Hipparchus was

carrying out a calculation equivalent to the mo dern calculation of the number

of bracketings of a string of ten letters However it remains to determine

exactly what Hipparchus and Plutarch meant by a comp ound prop osition

In Stoic logic comp ound prop ositions are built up from simple ones using

such connectives as and or and if then Ch I I I This do es

not seem like enough information to pinp oint precisely what Hipparchus had

in mind

We can also ask how Hipparchus computed the number As

noted in p this number is much to o large to have b een computed

by a direct of all the cases Moreover it is highly unlikely that

Hipparchus was aware of the sophisticated recurrence More probable is

that Hipparchus used the obvious recurrence equivalent to equation

X

si si n sn

k

i i n

1

k

where the sum ranges over all ways to write n as an ordered sum of k

p ositive The sum on the righthand side of equation in the

case n has terms There are only essentially dierent terms

corresp onding to the partitions of into a least two parts ie the

ways to write as an unordered sum of at least two p ositive integers If the

terms of the sum are group ed according to the partition of to which they

corresp ond it is still necessary to count the number of ways of ordering each

partition For instance the partition has orderings of

its terms thus contributing the amount ss s to the sum We

cannot but admire Hipparchus ability to compute the Schroder number s

at a distant when not even a remotely similar accurate computation is

known For further information ab out combinatorics in ancient see

The number in Plutarchs statement ie the number of com

p ound prop ositions that can b e formed from ten simple prop ositions on the

negative side remains an enigma Many p ossible variants of plane trees

have b een lo oked at without success Moreover Neil Sloane has veried that

the numbers and do not app ear any

where in the valuable tables Thus the mystery of Plutarchs statement

remains at most half solved

Acknowledgment The research was partially supp orted by NSF grant

DMS I am grateful to Judith Grabiner and four

anonymous referees for providing invaluable suggestions and references

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Math

Department of Mathematics

Massachusetts Institute of Technology

Cambridge MA

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