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Hipparchus, Plutarch, Schr Oder and Hough

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Hipparchus, Plutarch, Schr Oder and Hough Hipparchus Plutarch Schroder and Hough Richard P Stanley Hipparchus 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 number 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 Athens around and b ecame a leading Stoic philosopher Hipparchus was a Greek astronomer c after bc from Nicaea in Bithynia 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 precession of the equinoxes 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 mathematician though for a contrary view see p he was the rst p erson to make systematic use of trigonometry 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 mathematics 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 Greeks and referring to Plutarchs passage says that the most interesting of them is also the most mysterious A number of em inent mathematicians 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 combinatorics 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 sequence 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 plane 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 equation 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 triangle such dissections are called triangulations 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 degree 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 generating function 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 numbers will app ear in Exercise and
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