EXAMENSARBETEN I MATEMATIK

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1 EXAENSARETEN I ATEATIK ATEATISKA INSTITUTIONEN, STOCKHOLS UNIVERSITET Parh atrces and Permutaton Statstcs av Chrstopher Yamba No 8 ATEATISKA INSTITUTIONEN, STOCKHOLS UNIVERSITET, 069 STOCKHOL

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3 Parh atrces and Permutaton Statstcs Chrstopher Yamba Examensarbete matemat 0 poäng, fördupnngsurs Handledare: ats Oldn och Yshao Zhou 006

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5 Abstract In ths paper I ntroduce a permutaton statstc, the Parh matrx statstc, whch counts the number of permutatons wth a gven Parh matrx and prove ths s equdstrbuted wth the descent statstc, when we consder permutatons n S n Frst I wll ntroduce a generalzaton of the classcal Parh vector and show a matrx completeness for Parh matrces of words n S n When we consder S n, ths Parh matrx completeness mples that the Parh matrx statstc s equvalent to what can be defned as a Parh vector statstc of order In order to prove the equdstrbuton between the Parh matrx statstc and the descent set statstc I mae use of results from RP Stanley on descent set statstcs I defne analogous nclusve and exclusve Parh vector statstcs The desred result follows from the Prncple of Incluson- Excluson

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7 Contents Notatons 4 Introducton 5 Parh atrces, asc Defntons and Theorems 0 3 The Generalzed Parh appng 5 4 Descent Statstc 5 Comparng the Parh Vector appng to the Descent Vector appng 6 6 Yamba s Theorem 9 7 ( v) ρ for bnary words 35 8 Summary, Conclusons and Suggeston of Further Research 37 3

8 Notatons P( S ) the power set of S P ( ) S the set of all multsets wth the elements from S [n] the set {,, n } n the number of -combnatons of [n] wth repetons S ( ) n the set of all permutatons of length n of elements from a multset D( π ) the descent set of π α ( S) the nclusve descent set statstc β ( S) the exclusve descent set statstc * w u an alphabet the set of all words wth letters from the alphabet the number of occurrences of the word u as a scattered subword n w a, the word a, aa + a = of consecutve letters from an ordered alphabet = { a < a < < a n } + the set of upper trangular ( + ) ( + ) -matrces ( m, ), + : the Parh matrx mappng gven an alphabet of letters * ψ + ψ : N * + the Parh vector mappng of order ϑ ( v) the nclusve Parh vector statstc of order ρ ( v) the exclusve Parh vector statstc of order ϑ (( m ) + ) the nclusve Parh matrx statstc over an alphabet of letters,, ρ (( m ) + ) the exclusve Parh matrx statstc over an alphabet of letters,, D φ ( π ) the descent vector of π β ( S) the descent vector statstc φ ( π ) the number of nversons n a permutaton π 4

9 Introducton The Parh vector mappng s an mportant tool n the theory of formal languages, ntroduced 966 by RJ Parh n [6] Gven an ordered alphabet wth letters = { a < < a} and a word * w, where * denotes the set of all words wth letters from the alphabet, the * Parh vector mappng s a morphsm ψ : N, defned by ψ ( w) = ( w,, w ), where a a w denotes the number of occurrences of the letter a a n the word w and N denotes the set of nonnegatve ntegers One mportant result concernng the Parh vector mappng, found n [6], s that the mage by the Parh vector mappng of a context-free language s always a semlnear set The Parh vector contans numercal propertes of a word expressed as a vector For example, consder the ordered 3-letter alphabet 3 = { a < b < c} and the word w = abccbac Then we get the Parh vector ψ ( w) ( abccbac) ( w, w, w ) ( abccbac, abccbac, abccbac ) (,,3) 3 = ψ = = = N a b c a b c The Parh vector mappng s not nectve snce two words can have the same Parh vector Thus, much nformaton of the words s lost n ths mappng and therefore the natural contnuaton of ths topc was the development of an extenson of the Parh vector mappng Ths new tool, the Parh matrx mappng, contans more nformaton about a word than the Parh vector mappng The Parh matrx mappng was ntroduced several decades later n [], qute recently 00 It contans nformaton about the number of occurrences of certan subwords Here subwords means scattered subwords defned n the followng way: Defnton A word u s a subword of a word w f there exst words x,, x n and y,, 0 y n, some possbly empty, such that u = x xn and w = y0x y xn yn Let w denote the number of occurrences of u as a scattered subword of w For example, u tae the word w abccbac * = and consder the subword u abc = Then w means the u number of occurrences of u = abc as a scattered subword of w = abccbac We have 5

10 abccbac abccbac followng 4 occurrences 3 abccbac 4 abccbac Thus w = 4 When worng wth Parh matrces we u wll repeatedly encounter subwords denoted a,, where,,, denotes the ndces of the letters poston n the gven alphabet So n other words we are dealng wth subwords consstng of consecutve letters from an alphabet For example we mght be n the ordered alphabet = { a < < a5} The notaton a,4 means that we are consderng the word a,4 = aa3a4 If we let w = aaaa 3a4 then we have that w denotes the number of occurrences of a,4 = aa3a4 as a scattered subword n a,4 w = aaa a3 a4 w = aaaa3a 4 We have the followng occurrences w = a a a a a 3 4 Thus w = a,4 The Parh atrx mappng s a morphsm * ψ : + where + s a collecton of ( + ) - dmensonal upper trangular matrces wth nonnegatve ntegral entres and unt dagonal The classcal Parh vector wll appear as the second dagonal (the dagonal above the man dagonal) n the Parh matrx The other entres above the second dagonal contan nformaton about the order of the letters (n the word n examned) and are of the form w, a, < The man dagonal entres are s and 0 s below t Gven the alphabet = { a < < a }, we have ψ * ψ : + w w w a, a, a, 0 w w a, a, ( w) = 0 0 w w a, a, w a, 0 0 0, defned by (I should pont out that ths defnton s equvalent to Defnton 8 on page ) In ths thess I wll examne words n S, the set of all permutatons of the elements of the set [ ] = {,, } The Parh matrces we deal wth n ths paper wll have entres of the form 6

11 w, Consder the classcal Parh vector mappng of ψ ( w) = ( w, w,, w ), for w S Snce the letters of words belongng to S occur exactly once, ( gven the alphabet = { < < } ) we have that all the words of S have the same Parh vector, namely ψ ( w) = (,,) So, gven 5 = { < < 3 < 4 < 5} what s the Parh matrx of w = 345 S? 5 Frst of all note that snce we are worng wth S 5, the value of w s ether 0 or In general the Parh matrces of words from S wll have entres from Z In our case we have w = 345 S5 and ψ 5 ( w) ψ (345) w w w w w w w w w w w w = = = w w w The Parh matrx mappng s not nectve Ths can easly be seen If we consder the case studed n ths thess, that s, ψ : S +, we have a surectve mappng from a set wth! elements to a set wth less than! elements, namely elements Ths mght not be obvous to the reader at ths moment, but t wll be seen later, when we show that the Parh matrx of a word n S s completely determned by ts thrd dagonal The thrd dagonal can be seen as a bnary strng of length and we wll show that t s the same as the Parh vector of second order It can easly be realzed f one consders for example the subword 34 can only exst f the word 3 and 34 exsts Generalze ths dea and the result s clear Ths s proved n the Parh atrx Completeness Theorem So as a result we wll focus on ths thrd dagonal It s very specal n ths thess and wll be referred to as the Parh vector of second order Let s examne the entres of the above Parh matrx a bt closer The man (frst) dagonal s, by the Parh matrx Theorem (Theorem 0), m, =, + So we may thn of t 7

12 as the unt vector (Corollary ) (,,) The second dagonal ( w,, w ) s ust the classcal Parh vector + Let w = 345 S5 Then n ths case as we establshed earler we have ψ ( w) = ( w,, w ) = (,,) 5 Now t seems natural to contnue ths procedure, nvestgatng 5 the entres by gong through the dagonals n ncreasng order Let us contnue The thrd dagonal of the Parh matrx n ths case s ( w, w, w, w ) = (,0,,) The general case would be the + dagonal corresponds to ( w, w,, w ) 3 ( ) ( ) + + Now, why not call ths a Parh vector of :th order, f we let the classcal Parh vector be the Parh vector of frst order In the same way we have Corollary, we can establsh a Corollary such that the thrd dagonal corresponds to the Parh vector of second order Ths follows drectly from the Parh matrx theorem (Theorem 0) n the same way we get Corollary We wll get a generalzed Parh vector of order,where and our attenton wll be drected toward =, the Parh vector of second order Two words wth the same (classcal) Parh vector wll n many cases have Parh matrces that are dfferent I should also menton that Parh mappng s a morphsm from λ (where denotes * (,, ) concatenaton) to ( N, +,(0,,0)) and the Parh matrx mappng s a morphsm from λ to consstng of the set of ( + ) -dmensonal upper trangular matrces as * (,, ) defned above together wth the operaton multplcaton of matrces and wth as ts unt the unt matrx of dmenson +, (,,() ) I have now provded some bacground of Parh matrces Now I wll provde some bacground on permutaton statstcs What s permutaton statstcs? Permutaton statstcs s a branch of enumeratve combnatorcs dealng wth enumeraton of permutatons wth respect to propertes such as the descent set of a permutaton 8

13 Defnton The descent set of a permutaton π = π π S s denoted D( π ) and s defned by D( π ) = { : π > π + } [ ] Sometmes t s useful to thn of t as a mappng D : S P([ ]), where P([ ]) denotes the powerset of [ ], that s the set of all subsets of [ ] = {,, } The permutaton statstc that counts the number of permutatons wth a gven descent set s called the descent (set) statstc and has been extensvely studed The descent set statstc s denoted by β ( S) = #{ π S : D( π ) = S}, S [ ] In ths thess we wll consder descent set statstc of words that belong to S Consder the powerset of [ ], P([ ]) It wll be very useful to represent each subset S [ ] as a bnary strng correspondng to S For example f we consder S = {,3, 4} [4] = {,,3,4}, may represent ths by lettng poston, 3, and 4 n the bnary strng of length 4 correspond to a and the rest correspond to 0 Formally ths s a mappng φ, where stands for bnary, defned as follows: Defnton 3 Let P([ ]) denote the powerset of [ ] Defne φ : P([ ]) Z, by φ ( S ) = v, such that v = v v and v v = 0 f S = v = f S Ths s easly seen to be a becton Havng ths way of representng sets we also have a way of representng descent sets as bnary strngs I sometmes denote descent sets of permutatons as ( π ), where the set s mapped to ts bnary strng (or alternatvely φ ( D ( π )) ) For example we have D φ (345) = 0 (Note that ψ (345) = 0 ) D φ y ths pont you mght have begun to suspect what I m amng for I want to nvestgate possble connectons between Parh vectors of second order of words n S (they are bnary strngs of length ( ) ) and descent sets of words n S I wll do ths n ths by ntroducng a new permutaton statstc by thnng of the Parh matrx mappng as the property n queston To be more specfc, I am nterested n countng the number of permutatons mappng to a gven Parh matrx, and as a consequence of the Parh atrx Completeness 9

14 (Theorem 35) Ths turns out to be the same as determnng the number of permutatons mappng to a certan Parh vector of second order, snce every Parh matrx of a word n S s completely determned by ts thrd dagonal, whch s the dagonal correspondng to the Parh vector of second order The number of permutatons mappng to a gven Parh vector v of second order s denoted by ρ ( v) The number of permutatons mappng to a gven descent vector v s denoted by β φ (see the secton of Notatons below) Later I wll prove that ρ ( ) ( ) v = β φ v (see Theorem 6) In other words that the Parh statstc of order s equdstrbuted wth the descent statstc Ths may be consdered the man result of ths thess and possbly a new contrbuton connectng the theory of formal languages wth the permutaton statstcs Ths opens up a vast number of nterestng enumeratve problems to explore further Parh atrces asc Defntons and Theorems We have been nformally ntroduced to what Parh matrces represents, namely t contans nformaton of the number of certan subwords (of consecutve letters) of a word, gven an ordered alphabet In ths secton I wll more formally go through the theory as found n [], before gong further and ntroducng my own results concernng Parh statstcs and descent statstcs Frst of all we need some defntons Defnton [] An ordered alphabet s a totally ordered fnte set denoted by a a = { < < } In ths thess, worng n S, we wll mostly consder the ordered alphabet where the letters are a fnte set of ntegers = { < < } Defnton A fnte multset s a fnte set where repettons of elements are taen nto consderaton: Let N = {,,, } be a set A fnte multset wth respect to N of cardnalty n 0

15 b b s denoted = {,, }, element # = b = n, where b denotes the number of repettons of = We wll consder words w of length n that are permutatons of elements of multsets Ths s denoted by w S ( ) n Example 3 Let N = {,,5} and consder the multset = 3 0 {,,3, 4,5 } words w S ( ), we are consderng words that are permutatons of length n If we consder # = = 7 The number of such words s equal to the multnomal coeffcent 7 7! = = 80 One such word s w = 3533 S5( ),,3, 0,!!3!0!! Defnton 4 [] Let = { a < < a } be an ordered alphabet The Parh vector mappng s a morhpsm ψ N defned by ψ ( ) (,, ) w = w w, where : * a a w a denotes the number of occurences of the letter a n w Gven ψ ( w * w ) = ψ ( w ) + ψ ( w ), where * denotes concatenaton and * w, w, we have * denotes the set of all words wth letters from Example 5 Consder the ordered alphabet 7 = { < < < 7} and the words w = 34 * Σ and * w = Then we have w w = and ψ ( w w ) = ( w w, w w, w w, w w, w w, w w, w w ) = (,,,,,,) ψ ( w ) = ( w w w w w w w ) = (,,,,0,0,0) ψ ( w ) = ( w w w w w w w ) = (0,0,,0,,,) So we have ψ ( w w ) = (,,,,,,) = (,,,,0, 0, 0) + (0, 0,, 0,,,) = ψ ( w ) + ψ ( w )

16 In ths thess, worng n S, we have that the Parh vector mappng ψ ( ) (,, ) w = w w a a s ψ ( w) = ( w,, w ) = (,,), snce each letter of the -letter alphabet n a permutaton n S of length occurs exactly once The followng defnton concerns how the subwords are notated Defnton 6 [] Consder the alphabet = { a < < a } Then the a, denotes the word a, = aa + a of consecutve letters from the alphabet In partcular a, = a As we wll wor wth words belongng to S, the above subwords wll have the form ( + ), where and are letters from the ordered alphabet = { < < } Defnton 7 [] The number of occurrences of a, as a scattered subword of w s denoted by w N a, Agan, n our case S, we wll have w Z, and snce each subword n a permutaton can occur at most once, We have nformally defned the Parh matrx mappng but we wll do t more formally here Defnton 8 [] Let = { a < < a } be the an ordered alphabet The Parh matrx mappng s a morphsm * ψ : +, where + denotes the set of upper trangular ( + ) ( + ) matrces ( m, ), +, such that m, N ψ s defned on by ψ ( al ) = ( m ),, ( +, where we set m ), =, m l, l + = for + and for all other entres n ψ ( a ) we set m, = 0 l + has a unt whch s the dentty matrx I For * w wth w = a a, s defne ψ ( w) = ψ ( a ) ψ ( a ) where multplcaton s matrx multplcaton s a,

17 We have ψ ( a a ) = ψ ( a ) ψ ( a ), a, a r t r t r t Example 9 Let 3 be the ordered alphabet { < < 3} and assume that w = 3 Then ψ ( w) s a 4 4 trangular matrx and can be computed as follows: 3 ψ (3) = ψ () ψ () ψ () ψ () ψ (3) = = However as we mentoned earler although two words have the same Parh vector they can have dfferent Parh matrces For nstance, tae u = 3 gven the same alphabet as above u = 3 has the same Parh vector as 3 but t can easly be verfed that ψ (3) = () ψ () ψ () ψ () ψ ψ (3) = = ψ (3) Theorem 0 (Parh matrx mappng) [] Let = { a < < a } be the an ordered alphabet wth and propertes: ) m = 0, < +, ) m, =, + * w, then the matrx ψ ( w) ( m, ), + = has the followng 3) m + = w, +, a, 3

18 Proof [] Clearly property ) and ) follows drectly from the defnton Let s prove property 3) by nducton Assume that w = n If n the asserton s clearly true Assume that 3) holds for all words of length n and let w be of length n + u w = ua, where = n and a wth It follows that ψ ( w) = ψ ( ua ) = ψ ( u) ψ ( a ) Assume that ψ m, m, + 0 m, + ( u) = y the nducton hypothess ψ ( u ) has m, property 3) From Defnton 8 we have that ψ ( a ) = 0 All the elements n the matrx ψ ( a ) are 0, except the man dagonal, that s all :s and entry m, + = So we have ψ ( w) = ψ ( ua ) = ψ ( u) ψ ( a ) = m, m, + 0 m, + m, = N 0 The resultng matrx N has the followng property: n, + = m, + m, + for all and for all ndces, np, q = mp, q Gven the above theorem we can easly get the followng corollary, snce ths follows drectly from property 3) m + = w, +, n the above theorem:, a, Corollary [] Let = { a,, a } < be an ordered alphabet The Parh matrx ( w) = ( m), has as ts second dagonal, the Parh vector of w That s ψ + ( m,, m,3,, m, + ) = ( w,, w ) = ψ ( w) a a 4

19 Example Gven the alphabet = { < < 3} and w = 3 we have ψ (3) = the second dagonal of the matrx We can see that the Parh vector ψ (3) = (,,) equals Now I wll proceed and present some of the new results on Parh matrces n connecton wth permutatons statstcs that I have developed 3 The Generalzed Parh appng Ths secton s the part where I start develop the Parh matrx subect further to establsh new connectons wth exstng permutaton statstcs Some of the deas have been mentoned nformally n the ntroducton Here I wll attempt to gve a more formal foundaton for these deas I am gong to show a matrx completeness (Theorem 35) for Parh matrces of words n S n We shall see that the Parh matrx of words n S n s completely determned by ts thrd dagonal Therefore t s natural to generalze Corollary so that t connects each +:th dagonal wth a certan Parh vector of order n the followng way: Defnton 3 (Generalzaton of the Parh Vector appng) Let = { a,, a } < be an ordered alphabet Defne the Parh vector mappng of order as the mappng ψ : * + Z, defned by ψ ( w) = ( w, w,, w ) a a a,, + +, In partcular = gves the classcal Parh vector mappng If we set a, = a, then ψ ( w) = ( w, w,, w ) = ( w, w,, w ), correspondng to the second dagonal a, a, a, a a a 5

20 Analogcally, whch s easy to verfy by usng Theorem 0, we get a generalzaton of Corollary as follows: Corollary 3 Let = { a,, a } < be an ordered alphabet The Parh matrx ( w) = ( m), has as ts :th dagonal, the Parh vector of order of the word w That s ψ ( m +, m +,, m + ) = ψ ( w) = ( w, w,, w ),,, a a a,, + +, Proof Agan ths follows drectly from property 3) n Theorem 0, same as for Corollary and by the defnton of the generalzed Parh vector mappng Example 33 (Generalzed Parh vector and the Parh matrx dagonals) Let 5 = { < < 3 < 4 < 5}, = {,,3,4,5 } and consder the word w = S7( ) w w w w w w w w w w w w Then we have ψ ( w) = = w w w y Corollary 3 we have the followng: The (trval) Parh vector of order = 0 corresponds to the man dagonal That s ψ ( w) (,,,,,) 0 = corresponds to the underlned man (frst) dagonal n The (classcal) Parh vector of order = corresponds to the second dagonal That s 6

21 ψ ( w) = ( w, w, w, w, w ) = (,,,, ) corresponds to the underlned second dagonal n The Parh vector of second order ( = ) corresponds to the thrd dagonal That s ψ ( w) = ( w, w, w, w ) = (,,, ) corresponds to the underlned thrd dagonal n etc The followng result wll be useful when provng the atrx Completeness Theorem (Theorem 35) for words n S The atrx Completeness Theorem smplfes the tas of fndng the connecton between descent statstc and Parh statstcs because we can consder bnary strngs n both cases nstead of bnary strngs and bnary matrces Proposton 34 Consder the Parh matrx mappng on the set of permutatons of the elements of [] Defne ψ : S +, by ψ π π π π π 0 π π π 0 0 π ( π ) = π π π ( ) ( )( ) ( )( ) ( ) = m,, + ( ) Then for <, we have 7

22 ) If π = ( ) ) = ) If π = 0 ( ) ) = 0 3) If π = 0 ( ) ) = 0 4) If π = 0 0 ( ) ) = 0 Proof Let π Then the entres of ψ ( π ) are exclusvely 0 s and s, snce for each word S S π every letter of = { < < } occur exactly once and by consequence the subwords of π S can occur at most once, thus π s ether 0 or We wll use ths property of S n ths proof ) For <, assume that π = and π ( + ) ( ) ( + ) ( = Ths means that the )( ) words ( + ) ( ) and ( + ) ( )( ) both exst as subwords of π and therefore, snce we are worng n S, ( + ) ( ) must also exst as a subword of π Thus π = ) For <, assume that π = 0 and π ( + ) ( ) ( + ) ( = Ths means that the word ) ( + ) ( ) does not exst as a subword of π Then ( + ) ( ) cannot exst snce ( + ) ( ) s a subword of ( + ) ( ) Thus π = 0 3) For <, assume that π = and π = 0 Then by a smlar reasonng as ( ) ( + ) n ) we get that π = 0 4) For <, assume that both π = 0 and π = 0 Then by the same ( ) ( + ) reasonng as n ) we get that π = 0 In concluson the only way to get π = s f both when π = and π = ( ) ( + ) otherwse π = 0 We can thn of ths as a rule when computng the Parh matrx ψ ( w ) of a word π S All entres of the man dagonal n a Parh matrx are s, by the Parh matrx Theorem (Theorem 0) As we mentoned earler about wordsπ S, all the entres of the second dagonal of the Parh matrx (the classcal Parh vector, Corollary ) also conssts exclusvely of s y Corollary 3, the thrd dagonal s the Parh vector of second order 8

23 ψ ( π ) = ( π, π,, π ) The entres π 3 ( ) ( + of ths vector depends on whether or not ) ( + ) exst as a subword of π If ( + ) exst as a subword of π then π ( + = If ( + ) does not exst as a subword of ) π then π 0 ( + = ) y the above proposton we can automatcally compute the entres n the fourth dagonal from the thrd dagonal and the ffth dagonal from the fourth dagonal etc, untl we have completed the computaton of the matrx Now havng establshed ths result we have the tools for provng the atrx completeness Theorem 35 (Parh atrx Completeness Theorem) Consder the Parh vector of second order ψ : Z, ψ ( π ) = ( π,, π ), for ( π S Then the Parh matrx ) S mappng ψ : S +, ψ ( π ) = ( m), s completely determned by ψ ( π ) Proof y the Parh matrx Theorem (Theorem 0) we have ψ π π π π π 0 π π π 0 0 π ( π ) = π π π ( ) ( )( ) ( )( ) ( ) All the entres below the man dagonal are always 0 (by Theorem 0) All the entres n the man dagonal are always (by Theorem 0) Snce we are worng n S, all the entres of the second dagonal are always (by Corollary (classc vector)) y Corollary 3 we have that the thrd dagonal corresponds to the Parh vector of second order so the Parh vector of second order determnes the thrd dagonal n the Parh matrx y Proposton 34 we have, for <, that the values of π and π determnes the value of ( ) ( + ) π Therefore the Parh vector of second order π π π π 3 ( )( ) ( ) (,,,, ) 9

24 determnes π π π π 3 34 ( 3)( )( ) ( )( ) (,,,, ) whch s the Parh vector of thrd order y Corollary 3 ths corresponds to the fourth dagonal of the Parh matrx In the same way we get by Proposton 34, that the ffth dagonal follows from the thrd dagonal n the same way We get all the dagonals by the same reasonng In concluson the Parh vector of second order s enough nformaton to compute all the entres of the Parh matrx, when we consder words that belong to S The Parh matrx completeness for permutatons n S sgnfcantly speeds up the computatons of Parh matrces To demonstrate ths, here s a smple example Example 36 Let 5 = { < < 3 < 4 < 5} and consder the words n S 5 Consder the followng 3 permutatons π = 534, π = 534, π 3 = 534, π 4 = 345, π 5 = 543 Then we have: ψ ( π ) = ( π, π, π, π ) = (,0,,0) ) ψ ( π ) = ( π, π, π, π ) = (,,,0) ) ψ ( π ) = ( π, π, π, π ) = (0,,,0) 3) ψ ( π ) = ( π, π, π, π ) = (,,,) 4) ψ ( π ) = ( π, π, π, π ) = (0,0,0,0) 5) Let us ust compute the dagonals resultng from these (thrd dagonals) Parh vectors of second order as descrbed earler To mae t easer to vsually follow the computatons I have removed the all entres below the thrd dagonal snce they are the same for all Parh matrces over words n S Then we get: 0

25 ) ) 3) 4) 5) ψ ψ ψ ψ ψ ( π ) = ( π ) = ( π ) = ( π ) = ( π ) = It s also vsually more apparent f vewed as arthmetcal trangles where and bottom row corresponds to the thrd dagonal and the next row to the fourth dagonal etc Entres follow the rules from Proposton 34 Then we get

26 0 0 0 ) ) 0 0 3) ) ) These arthmetcal trangles have the form ψ 5( π ) ψ 4( π ), by Corollary 3 ψ 3( π ) ψ ( π ) A consequence of ths result s that f we are able to show somethng for Parh vectors ψ ( π ), π S, t can also be used for Parh matrces ψ ( π ), π S For example f we are countng the number of permutatons mappng to a gven Parh vector we are countng what s also equvalent to the number of permutatons mappng to a gven Parh matrx Smlar to the descent set statstc, I wll defne an analogous permutaton statstc for the Parh mappngs Consder the Parh vector of order I wll defne a permutaton statstc by countng the number of permutatons that maps onto a specfc vector of order y Theorem 35, the Parh vector of second order of a word w Sn determnes the Parh matrx of ths word As a consequence, countng the number of permutatons mappng to a specfc Parh vector of order s equvalent to countng the number of permutatons mappng to a gven Parh matrx Now let s revst the descent sets 4 Descent Statstc Permutaton statstcs s an ongong branch of enumeratve combnatorcs and s currently under a lot of development Computng the descent set statstc has been done by several The methods I have chosen to present n ths paper can be found n Stanley s enumeratve combnatorcs and s based on the Prncple of Incluson-Excluson Frst let s go through the permutaton statstcs of nterest n ths thess, the descent set statstc We have already been

27 ntroduced to t nformally n the ntroducton Now I wll go through materal n a more formal manner Defnton 4 Let π = ππ π = S Defne the descent set mappng by D : S P([ ]), where D( π ) = { : π > π + } = S [ ] The followng functon wll be useful because t maps each descent set of a permutaton to a correspondng bnary strng and thus we can compare t to the bnary Parh vector of a permutaton Defnton 4 Let φ : P([ ]) Z be a functon defned by φ ( S) = v = ( v, v,, v ) Z, such that v 0 f S = f S, where S [ ] It s easy to see that φ n the above defnton s a becton D φ We can consder the composton φ D : S P[ ] Z, where π v We may call the vector v mapped by ths composton functon the descent vector v of π and denote ths by Dφ ( π ) = v Example 43 Consder the permutaton π = ππ π 3π 4π 5π 6 = 4563 S6, then D( π ) = {4,5}, snce we have the followng descents π 4 = 6 > 3 = π 5 and π 5 = 3 > = π 6 and all other are ascents ( π < π + ) The correspondng descent vector s D φ ( π ) = φ({4,5}) = 00 So, how many permutatons n S6 has got the descent set D( π ) = {4,5}? How s ths computed? In fndng a formula for the descent set statstc we shall fnd the Prncple of Incluson- Excluson very useful I want to use bnary strngs as representatves for descent sets, snce I wll show the connecton to Parh vectors of second order later n the thess Therefore I wll use a functon φ between sets and bnary strngs 3

28 Frst let s defne the nclusve descent statstc Ths s an mportant constructon n ths thess ecause we can use the Prncple of Incluson-Excluson on t to establsh the exclusve descent statstc formula Later we wll also defne an analogue for Parh vectors of second order and n the proof of the an Theorem 6 we wll eventually arrve at ths formula Defnton 45 [3] Gven S [ n ] let α( S) denote the number of permutatons π Sn whose descent set s contaned n S That s α( S) = #{ π S : D( π ) S} n n Proposton 46 [3] Let S = { s,, s} < [ n ] Then α( S) = s, s s, s3 s, n s Proof ([3] see page, proposton 3) Consder the permutaton π = aa an wth D( π ) S, we may frst choose a < a < < as n n s ways Proceed by choosng as < as < < as n + + n s ways etc From ths we get s s α( S) n n s n s n s n = = s s s s3 s n s s, s s,, n s as desred Now let s defne the exclusve descent statstc Defnton 47 [3] Gven S [ n ], the exclusve descent statstc s defned as β ( S) = #{ π S : D( π ) = S} n It s clear that α( S) = β ( T ) Ths holds the ey for applcaton of the Prncple of T S Incluson-Excluson to get a formula for β ( S) I wll not go through ths here but from Stanleys enumeratve combnatorcs [3] (by applcaton of the Prncple of Incluson-Excluson) we have the formula: 4

29 S T n (48-9) β ( S) = ( ) α( T ) = ( ) s, s s,, n s T S < < <, where n α( S) = s, s s,, n s S = { s < s < < s } [ n ] and S T = #{ S T }, Example 40 Gven n = 7 and S = {,5} We get S = { s =, s = 5} {,,,7} β ( S) = ( ) α( T ) = T S S T = = 0,8 5,8 5,5,8 5 0 ( ) ( ) ( ) 7 In Stanleys enumeratve combnatorcs [3], page 69, the descent statstc s also wrtten n an alternatve form as a determnant It can be shown that n βn ( S) = ( ) = n!det[/( s s )] s, s s,, n s < < < + We have ( s s )! ( s s )! ( s s )! β n ( S ) = ( s s )! ( s s)! ( s+ s )! n! ( s s )! + ( s s )! ( s s )! ( s s )! + So we can express the computaton of the sum n Example 40 as! 5! 8! β ( ) 8! 7 4! 7! n 0 3! 5

30 5 Comparng the Parh Vector appng to the Descent Vector appng Ths secton s smply dedcated to provdng some computatons showng of the dstrbuton of the Parh Statstc and the Descent statstc The computatons ndcate that they seem to be equdstrbuted We do ths by computng the Parh vector mappng of second order and then for the same value of n = 5 we compute the Descent vector mappng (by usng the functon φ mappng descent sets to ther correspondng bnary strngs) Computatons 5 Consder the Parh vector mappng of second order on S 5 We have ψ ( π ) = ( π, π, π, π ) = v 4 ψ : S5 Z, where (For convenence I wll notate v as a word For example 0 nstead of (,,,0) ) 4 Z S ,435,435, ,534,534, ,543,543, ,345,345, ,543,543,543,543, ,435,345,345,345, ,453,435,453,453,435,453,453, ,354,534,354,354,534,354,354, ,345,345,345,345,345,345,345, ,543,543,543,543,543,543,543, ,435,453,435,435,435,453,453,435,435, ,534,354,534,534,534,354,354,534,534, ,354,354,354,354,354,354,354,354,534,534, 534,534,534,534, ,453,453,453,453,453,453,453,453,435, ,435,435,435,435 6

31 4 Computaton 5 Consder ρ ( v) = #{ π S : ψ ( π ) = v Z } 5 ρ () = ρ (0000) = ρ (0) = 4 ρ (000) = 4 ρ (0) = 4 ρ (000) = 4 ρ (00) = 6 ρ (00) = 6 ρ (000) = 9 ρ (0) = 9 ρ (000) = 9 ρ (0) = 9 ρ (00) = ρ (00) = ρ (00) = 6 ρ (00) = 6 7

32 D φ Let s chec the descent set dstrbuton Gven φ D : S P([ ]) Z, where π S v For = 5, we have Computaton 53 (Descent vector mappng, n = 5) 4 Z S ,453,345, ,543,534, ,345,435, ,354,543, ,453,354,453,543, ,534,435,435,543, ,453,435,435,543,534,534,543, ,453,354,345,453,354,345,534, ,354,453,543,354,453,543,435, ,354,534,534,345,435,435,345, ,354,354,453,453,435,534,543,534,543, ,453,453,354,354,534,435,345,435,345, ,435,453,435,453,435,534,543,534,543,435, 534,345,354,345, ,534,543,435,453,354,534,543,435,453,453, 453,354,354,345,345 4 Computaton 54 Consder β ( v) = #{ π S : D ( π ) = v Z } φ 5 φ β φ () = β φ (0000) = β φ (0) = 4 β φ (000) = 4 8

33 β φ (0) = 4 β φ (000) = 4 β φ (00) = 6 β φ (00) = 6 β φ (000) = 9 β φ (0) = 9 β φ (000) = 9 β φ (0) = 9 β φ (00) = β φ (00) = β φ (00) = 6 β φ (00) = 6 We see that for S 5 and v Z, ρ ( ) ( ) v = β φ v In the man theorem I wll prove that ths s 5 true for all v Z when we are consderng ψ and D φ over S 6 Yamba s Theorem We have seen that for v Z, ρ ( ) ( ) v = β φ v I conecture that ths s true for all 5 v Z and for all when we are consderng ψ and consdered as the man result of ths paper D φ over S The theorem can be The proof of the exclusve descent set statstc formula s based on the Prncple of Incluson- Excluson There are several methods of computng ths statstc but I have chosen the methods found n [3] (See Proposton 46 and Theorem 4) It s acheved by frst defnng the descent set nclusvely n terms of subsets of a descent set and nvertng the formula by dentfyng that t holds the settng of the Prncple of Incluson- Excluson To get the analogue thnng for the Parh vectors of second order I wll mae use 9

34 of subvectors (defned n Defnton 6) nstead of subsets The becton φ, between subsets and bnary vectors motvates the use of ths Ths s useful later when we defne the nclusve descent set statstc Out of smplcty we wll denote the bnary vectors as bnary words Defnton 6 Consder the set of vectors of Z Let u = u u Z and v = v v Z Defne a relaton on the set Z by u v f u v,, u v If u v we may call u a sub vector of v and v a super vector of u Example 6 Let = { < < 3 < 4 < 5} and consder w = 345 S5 and z = 354 S5 Then we have ψ ( w) = ( w, w, w, w ) = u = ( u , u, u3, u4) = (,0,,) and ψ ( z) = ( z, z, z, z ) = v = ( v, v, v, v ) = (,0,,0) We have v = = u v v v = 0 0 = u = = u 3 3 = 0 = u 4 4 Thus v u, so we have that v s a sub vector of u and u s a super vector of v Descent statstc s orgnally formulated n terms of sets but one should eep n mnd the becton φ to be aware of the connecton between the descent statstc and the Parh statstc Defnton 68 Let = { a < < a } and v Z The nclusve Parh vector statstc of second order s defned byϑ ( v) = #{ w S : ψ ( w) v} Defnton 69 Let = { < < }, and v Z The exclusve Parh vector statstc of second order s defned by ρ ( v) = #{ w S : ψ ( w) = v Z } 30

35 We wll now defne the Parh matrx statstc ρ (( m, ) + ), countng the number of permutatons wth a gven Parh matrx ( m, ) + y the Parh atrx Completeness Theorem (Theorem 35), we get that ρ (( m, ) + ) = ρ( v), where v s a Parh vector of second order correspondng to the thrd dagonal n ( m, ) + (Corollary 3) Defnton 6 Let = { < < }, consder ψ ( S ) and let ( m), + ψ ( S ) The Parh matrx statstc s defned by ρ (( m ) ) = #{ w S : ψ ( w) = ( m) ψ ( S )}, +, + Now I wll show that the defned Parh vector statstc s n fact equdstrbuted wth the descent set statstc, when we consder words n S The proof gven for ths s nspred by Stanley s proof of the nclusve descent statstc It can be smplfed by observng the smlarty of reasonng Choosng a permutaton wth a gven nclusve descent corresponds to choosng ts elements of [n] satsfyng a gven ascent pattern In the Parh case we have a gven subword pattern and choose among [n] ncdes n the same way as n the descent case To compare the smlarty n reasonng see [3], proposton 3 an Theorem 6 Let φ be the functon from Defnton 43 For S [ ] and v = φ ( S) Z consder ρ ( v) = #{ w S : ψ ( w) = v Z } and ( v) = #{ w S : D ( w) = v Z } β φ φ Then ρ ( ) ( ) v = β φ v (The Parh vector statstc of second order s equdstrbuted to the descent statstc when we consder words n S ) The proof follows the same deas as n [3], page Proof Consder φ : P([ ]) N, φ ( S ) = v, S [ ], Let ϑ ( v) be the nclusve Parh statstc ϑ ( v) = #{ π S : ψ ( π ) v Z } and ρ ( v) be the exclusve Parh statstc ρ ( v) = #{ γ S : ψ ( γ ) = v Z } 3

36 Clearly ϑ ( v) ρ( u) = (n the same way α( S) β ( T ) u v formula for ϑ ( v), we can nvert t, n analogue wth = =, [3], page ), so f we can fnd a T S S T α( S) β ( T ) β ( S) = ( ) α( T ) to get the desred result T S T S Let v = v v Z, such that φ ( v) = S = { s,, s } [ ] To create a π S that has the Parh vector of second order ψ ( ) v π, frst let s tae a closer loo at v We have, for, Ths means that v = v v v v where v = 0 and v l = for all other s s s s ψ ( π ) =, π,, π,, π, (,0,,,, 0,,,, 0,,) = v s ( s + ) s ( s + ) s ( s + ) s s s Now to create a π S that has the Parh vector of second order ψ ( π ) v s l Frst place the subword ( s ) s among avalable ndces of π Ths can be done n s ways Then place ( s + ) ( s ) s among the s postons Ths can be done n s s s ways If we contnue ths process we see that s s s ϑ ( v) = = s s s s s3 s s, s s, s3 s,, s Now recall that we have ϑ ( v) = ρ( u) Hence by the Prncple of Incluson-Excluson (the u v formula) and Theorem 4 we have t ρ( v) = ( ) = β ( S) = β ( v),,, t s s s s φ < < < t Let S = { s,, s } [ ] Then to choose a permutaton π = a a S, where D( π ) S We can do ths by frst choosng the elements a from [ ] a < < a Ths can be done n s ways and ths corresponds to placng the subword ( s ) s among avalable s postons 3

37 I clam that Dφ π = = ψ π ut frst let s defne what the nverse permutaton ( ) v ( ) π s Defnton 63 Gven a permutaton π = aa a = S a a a a a a The nverse permutaton s defned as π = S There s a one-to-one correspondence between a permutaton and ts nverse permutaton To get a permutaton of the same form as π = aa a the ndces a n π are placed n ncreasng order we get π = bb b S Example 64 Consder π = 345 = S = = = 534 S π 5 Then Theorem 65 Letπ, let = { < < } be an ordered alphabet and let D φ ( π ) = v Z Then for S exst we have that Dφ ( π ) = v = ψ ( π ) Proof Consder π = a a S such that D( π ) = S = { s, s,, s } [ ] Ths means a < a < < a > a < < a > a < < a > a < < a Let s use an alternatve s s s s s s s notaton for π : s s + s s + s s + a < a < < a > a < < a > a < < a > a < < a π = s s + s s + s s + s + Wth standard notaton we have s s + s s + s s + π = a a as a s + as a s + as a s a + s + 33

38 The frst row s ndces of π, the second row s the letters a of π = a a S wth nequaltes The thrd row s the descent vector Dφ ( π ) = φ ( S) = v If we let ndces change places then we get (wth an alternatve notaton) a < a < < a > a < < a > a < < a > a < < a π = s s + s s + s s + Wth standard notaton we have π s s + s s + s s + s + a a a a a a a a a = s s + s s + s s + s s s + s s + s s + s + Now s must exst as a subword of π snce ndces a < a < < a s s ( s + ) does not exst as a subword of π snce ndces > means the letter s + s a a s + s to the left of s n,,,,,,,0, = π Therefore we have ( π π π 3 s ( s + ) ) s If we contnue the same reasonng for s, we wll attan all the entres of what s the Parh vector of second order We have ( ) ψ ( π ) = π, π,, π, = = ( π ) 3 s ( s + ) s s s I wll now use ths result for an alternatve proof of Theorem 64 D φ Proof of Theorem 6 Snce there a one-to-one correspondence between the set of permutatons and ther nverse permutatons and by Theorem 65, ψ ( π ) = D φ ( π ), we have for v Z that ρ( v) = #{ γ S : ψ ( γ ) = v} = #{ γ S : D( γ ) = v} = β φ ( v) For each permutaton γ S, where ψ ( γ ) = v, there s exactly one permutaton γ, where S D( γ ) = v So βφ ( v) = ρ( v) Corollary 66 Let = { < < < }, let v ( m ) Z and let, + be a Parh matrx of a word belongng to S, where v corresponds to the thrd dagonal Then ρ (( m, ) + ) = β φ ( v) 34

39 Proof As mentoned earler by the Parh atrx Completeness Theorem (Theorem 35), we get that ρ (( m, ) + ) = ρ( v), where v s a Parh vector of second order correspondng to the thrd dagonal n ( m, ) (Corollary 3) y the an Theorem + ρ ( ) ( ) v = β φ v so t follows that ρ (( m, ) + ) = ρ( v) = β φ ( v) 7 ρ ( v) for bnary words In ths secton I wll present some results n computng the number of bnary word mappng to a gven parh vector of second order Defnton 7 Let ( b, b ) denote the nverson of the letters b and b n the word w = bb bn Sn( ), (where s a fnte multset), f < and b > b Let ( w ) denote the number of nversons of w = bb b S ( ) Defnton 7 The number of words wth nversons s denoted nv( w) = #{ w S ( ) : ( w) = }, where 0 n n n n A generatng functon for ths would have the form ( w) q w S n ( ) Let s say we want to count #{ w Sn( ) : ψ ( w) = ( w ) = ( )}, where 0 n and { a b =, }, # = a + b = n can be attaned by observng that countng the number of subwords n w That s, w corresponds to countng the number of nversons n a permutaton and the fact that we are dealng wth bnary words Also, snce we are worng wth bnary words, symmetry gves us thatψ ( w) ( w) =, where w s the (bnary) complement word (for example w = w = ) And snce we have, by 35

40 a a a Proposton 37 ([3] page 6), that for = {,, m m }, wth # = n ( w) q = w S ( ),,, n a a a m q n We get n partcular a a {, } =, wth # = n ( w) q = q = w S ( ) ( ), n w S n a a q, where ψ ( w) n = Smlar observatons have also been made by D Foata and G-N Han n [4], pages 4-3 Defnton 73 (nary words) [4] Let W ( N, n ) denote the set of all words of length ( N + n) havng exactly N letters equal to and n letters equal to 0 If x = xx x N + n s such a word, the nverson number nv( x ), s defned as the number of scattered subwords, 0 of the word x Example 74 We can also wrte the number of s that appear to the left of each letter equal to 0 For the word x = we have ( x ) = = 0 In [4] t s shown that x W ( N, n) q ( x) N + n = n q I wll not go through ths here n more detal (t can also be found n [3], proposton 37, page 6), but snce we have shown that ( x) = ψ ( x) we can mae use of ths result to get a generatng functon for Parh vector statstc of second order on bnary words Proposton 7 [4] N + n = n q x W ( N, n) q ( x), where N + n n q s the q bnomal coeffcent) As a consequence we get N + n ( x) q q n = =, where = ψ ( x) q x W ( N, n) x W ( N, n) 36

41 8 Summary, Conclusons and Suggeston of Further Research Theorem 6 and Theorem 65 are the man result of ths thess wth consequences for Formal language theory (and future Parh matrx theory) and permutaton statstcs The result connects formal languages wth permutaton statstcs n such a way that the results establshed for descent set statstcs (and nversons statstcs) s transferred to Parh matrx theory and thus motvates future research concernng, for example, lattce theory of Parh matrces and the permutaton statstcs of varous Parh vector mappngs (generalzed Parh vectors) For nstance a generalzed formula for the Parh vector statstc ρ n n = = N, where = {,, }, = {,, }, such that ( v) #{ w Sn( ) : ψ ( w) v } # = n = n would be desrable It would also be nterestng to study algorthms for lstng = the words wth a gven Parh vector/matrx The next step would be to nvestgate the Parh vector statstc ρ for arbtrary words w Sn ( ) The general case, ρ ( v) #{ w Sn ( ) : ψ ( w) v } < = = N, gets more complcated 37

42 References [] Alexandru ateescu, Arto Salomaa, Ka Salomaa, Sheng Yu, On an Extenson of the Parh appng, TUCS Techncal Report No 364, 003 [] Alexandru ateescu, Arto Salomaa, Sheng Yu, Subword Hstores and Parh atrces, Turu Centre of Computer Scence, TUCS Techncal Report No 44, 004 [3] Rchard P Stanley, Enumeratve Combnatorcs Volume, Cambrdge Unversty Press,997 [4] Lothare, Combnatorcs on words, Cambrdge Unversty Press, 00, [5] D Foata, Guo-Nu Han, The q-seres n Combnatorcs; Permutaton Statstcs (prelmnary verson), Aug 7, 004 [6] Parh RJ, On Context-Free Languages, JAssoc Comp ach, 3 (966)