Counting pairs of lattice paths by intersections

Transcription

1 Coutg pas of lattce paths by tesectos Ia Gessel 1, Bades Uvesty, Waltham, MA , USA Waye Goddad 2, Uvesty of Natal, Duba 4000, South Afca Walte Shu, New Yo Lfe Isuace Co., New Yo, NY 10010, USA Hebet S. Wlf 3, Uvesty of Pesylvaa, Phladelpha, PA , USA Lly Ye 4 Uvesty of Wateloo, Wateloo, Otao, Caada N2L 3G1 Abstact We cout the pas of wals betwee dagoally opposte coes of a gve lattce ectagle by the umbe of pots whch they tesect. We ote that the umbe of such pas wth oe tesecto s twce the umbe wth o tesecto, ad we gve a bjectve poof of that fact. Some pobablstc vaats of the poblem ae also vestgated. 1 Itoducto Cosde a ) plae lattce ectagle, ad wals that beg at the og south-west coe), poceed wth ut steps ethe of the dectos east o oth, ad temate at the oth-east coe of the ectagle. Fo each tege we as fo, the umbe of odeed pas of these wals that tesect exactly pots. The umbe of pots the tesecto of two such wals s defed as the cadalty of the tesecto of the two sets of vetces, excludg the tal ad temal vetces. The fgue below shows a pa of such wals whee = 9, = 17, ad = 5. N, FIGURE OMITTED It s well ow that N, 0 = ) ) 1. 1 Ideed, the umbes N, 0 /2 ae ow as the Naayaa umbes [5] ad cout the umbe of stacase polygos, whch ae well studed. Naayaa [6] showed that the value N, 0 /2 s also equal to the umbe of plae tees wth vetces ad leaves. A poof of ths usg lattce paths s gve [2]. At the othe exteme, f = 1 the the two wals cocde so that N, 1 = ). 1 Suppoted by the Natoal Scece Foudato 2 Suppoted by the Offce of Naval Reseach 3 Suppoted by the Offce of Naval Reseach 4 Suppoted by a Fellowshp of the NSERC, Caada 1

2 I sectos 2 ad 3 below, we establsh two explct fomulas fo the umbes N,. These fomulas ae N, = 2 + 1) 1 ) ) ) + 1 1, 0 2), 1a) ad N, 2 + 1) = 1) ) ) 2 ) 1 ) ), 0 2). 1b) I secto 2 we pove the valdty of fomula 1a), ad secto 3 we pove the equalty of fomulas 1a) ad 1b) the commo age of valdty. These fomulas eveal the teestg fact that N, 1 = 2N, 0,.e., that exactly twce as may pas of wals have a sgle tesecto as have o tesecto. Such a elatoshp clealy mets a bjectve poof, ad we supply oe secto 4 below. I sectos 5 ad 6 we dscuss a umbe of elated esults. I the fst of these sectos we cout the pas of otesectg wals statg at the og but edg at two dffeet specfed pots. I the secod we cout the pas of otesectg wals egadless of whee the two wales ed up. Related esults o so-called vcous wales have bee obtaed by Fshe [3] ad othes; patcula the umbes of pas of otesectg wals the d-dmesoal lattce have bee studed by Guttma ad Pellbeg [4]. Fally, secto 7 we dscuss a vaato whch we fd the pobablty that two depedet wales o a gve lattce ectagle do ot meet, ude a dffeet hypothess. I ths vaato we assume adom moto of the wales ad put a bae alog the les x = ad y = s. Hee the wales stat at the pots 0, 1) ad 1, 0), wal east o oth each step, gog east wth some pobablty p, j) depedet o the cuet pot, j), except that whe a wale eaches the bae at x = esp. the bae y = s) the all futue steps ae costaed to be oth esp. east) utl the pot, s) s eached. We fd see the case z = 1 of Theoem 5 below) that f the pobablty p, j) that a step fom, j) wll go oth depeds oly o + j, the: The pobablty that the two wales do ot meet utl they each the temus, s) s the same as the pobablty that a sgle wale who stats at the pot 0, 1) ad taes + s 2 steps wthout a bae, fshes at the pot 1, s). 2 Devato of the fomula 1a) Oe ca use a geeatg fucto to appoach the poblem. The fst step s to fd the equste geeatg fucto ad the secod step s to extact the coeffcets fom the geeatg fucto. 2

3 If we sot out the pas of wals that have tesectos accodg to the pot q, m q) of the last.e., most oth-eastely) tesecto, the we see the ecuece N, = N m,q 1 N m, q 0. 2) q,m Itoduce the geeatg fucto u x, y) =, N, x y. The equato 2) says smply that u = u 1 u 0, fo 1. Thus u x, y) = u 0 x, y) +1 fo = 0, 1, 2,... Now, u 0 s well-ow, havg bee calculated by Naayaa [5] ad othes. But oe ca also fd u 0 by obsevg that the coeffcet of x y the sum u x, y) s the total umbe of pas of wals, sce evey pa has some umbe of tesectos. The umbe of all pas of wals s 2 ), so we have 1 1 u 0 x, y) = u 0 x, y) 0 = 1 + u x, y) =, = 0 = ) 2 x y x y 1) P y + 1 y 1 ) 1 1 2xy + 1) + x 2 y 1) 2, whee the P s ae the Legede polyomals, ad the classcal geeatg fucto has bee used. Thus u 0 x, y) = 1 1 2xy + 1) + x 2 y 1) 2. 3) It follows that the umbe of pas of wals that have exactly tesectos s the coeffcet of x y u x, y) = ) xy + 1) + x 2 y 1) 2. 4) Now, we eed to extact the coeffcets of ths geeatg fucto. Fo ths we use the Lagage Iveso Fomula. Ths eques us to efomulate thgs slghtly. Note that N, s the coeffcet of y z u z, y/z). Fom equato 3) t follows that f we wte u 0 z, y/z) = y + z + 2f, the f satsfes the equato f = y +f)z +f), ad the umbe N, of y z y + z + 2f) +1. s the coeffcet 3

4 To extact the coeffcets of y+z+2f) +1 we use the Lagage Iveso Fomula the followg fom. See e.g., [7] eq ).) If F satsfes F = xgf), the fo ay fomal powe sees φ ad ay postve tege m, moeove, [x m ]φf) = 1 m [tm 1 ]φ t)gt) m ; 5) [x 0 ]φf) = φ0). Now we toduce a auxlay vaable x ad cosde the equato F = xy + F)z + F). We wll use equato 5) to expad y + z + 2F) +1. Settg x = 1 wll the gve us y + z + 2f) +1. Asde: the quadatc equato fo F has two algebac solutos, but oly the oe we wat has a powe sees expaso.) Let φt) = y + z + 2t) +1 ad gt) = y + t)z + t). The epeated use of the bomal theoem yelds φ t)gt) m = 2 + 1)y + z + 2t) [y + t)z + t)] m = 2 + 1)[y + t)z + t)] m ) y + t) z + t) = 2 + 1) ) y + t) m+ z + t) m+ ) [ ) ] m + m + )z s t m+ s = 2 + 1) ] [ y t m+ s s = 2 + 1) ) ) ) m + m + y z s t 2m+ s. s,,s The coeffcet of t m 1 ths sum comes fom the tems whch s = m Thus by equato 5), togethe wth φ0) = y + z) +1, we have y + z + 2F) +1 = y + z) +1 x m ) ) ) m + m ) y z m+ +1. m m m=1, By settg x = 1 ad mag the chage of dex of summato to m = 1, we obta y + z + 2f) +1 = ) + 1 y z +1 + ) ) ) y 2 + 1) z. 1 >+1 Thus the esult follows. 4

5 3 Equalty of 1a) ad 1b) Ths secto s devoted to the poof of the equalty of the two fomulas 1a), 1b), the age 0 2. Note that we adopt the usual coveto that ay tem wth the factoal of a egatve tege ts deomato s cosdeed to be zeo. Theoem 1 Fo 0 2 we have ) ) ) = + 1 1) j j j ) j j ) j 2 ) j 1 ) 1 j 1 j 2 j ). 6) Poof. We obta the two sdes of 6) by evaluatg two ways the double sum S = j,l 1) j + 1)! 2)! j 1)!! )! j! l! j l 1)! 2j l)! + j + l 1)!, whee the sum s ove all oegatve teges j ad l wth j 1. We shall eed two foms of Vademode s theoem: ) ) ) a b a + b = m m ad ) ) ) a c c a 1) =. 8) m m Fst we sum o l. We have ) ) 1 j + 1)! j 1)! 2j 2 S = 1)! )! j! 2j)! l j l 1 j=0 l = ) j + 1)! j 1)! 2j 2 1) by 7),! )! j! 2j)! j 1 j /2 ad ths s easly see to be equal to the ght sde of 6). Next we set l = j S ad sum o j. Ths gves S =,j = = 1) j + 1)! 2)! j 1)!! )! j! j)! 1)! j)! + 1)! 1 + 1)! 2)! 1) j j 1)!! )! 1)! + 1)! j! j)! j)! j=0 + 1)! 2)! 1)!! )! 1)! + 1)! )! 5 7) ) ) j 1 1) j j j j=0

6 = + 1)! 2)! 1)! ) + 1! )! 1)! + 1)! )! by 8), ad ths s easly see to be equal to the left sde of 6). It may be oted that the theoem ca also be obtaed fom fomula 1), p. 30 of [1] by tag the lmt as c the settg m = 1, a =, w = 1, ad b =. Fally, we ema that Zelbege s algothm see [8]) s capable of vefyg that the two sdes of 6) satsfy the same ecuece elato. I fact, though, the ecueces ad poof cetfcates that oe obtas ae vey upleasat ad cosdeably loge tha the huma poof we have gve above. 4 A bjecto The fomula 1b) shows that 2N, 0 = N, 1 = ) 1 1 Hee we gve a bjectve poof of the asseto that 2N, 0 = N, 1. Fo coveece, we defe s = so that we deal wth a s ectagle. The dstace at x betwee two lattce paths P 1 ad P 2 s d x P 1, P 2 ) = m{ y 2 y 1 : x, y 1 ) P 1, x, y 2 ) P 2 }. ). Defto of the map. We defe ou map φ fom pas of paths wth o tesecto to pas of pas of paths wth oe tesecto. Hece let P, Q) be a pa of paths that do ot tesect. We may assume that P s oth of Q. Thee ae two cases: 1. d x P, Q) 2 fo all 1 x 1. The φ maps P, Q) to P, Q) ad P, Q ) as follows. To obta P fom P, fst taslate P dow by 1 ut, the delete ts fst oth edge, ad the cocateate a oth edge to the last vetex of the ew path. The pa P, Q) tesects at, s 1) ad oly thee. To obta Q fom Q, fst taslate Q up by 1 ut, the delete ts last oth edge, the adjo a oth edge -boud to ts fst vetex. The pa P, Q ) tesects at 0, 1), ad oly thee. 2. d x P, Q) = 1 fo some x, 1 x 1. Let x 0 be the smallest such x, ad let y 0 = max y {x 0, y) Q}. Fst suppose x 0, y 0 ) 1, 0). The lowe by 1 ut the poto of P fom 0, 0) to x 0, y 0 +1), 6

7 ad move the fst oth edge of P to jo x 0, y 0 ) ad x 0, y 0 + 1), to obta the ew path P. The pa P, Q) tesects at x 0, y 0 ) ad oly thee. To get aothe pa of paths that tesect at x 0, y 0 ), techage the uppe path wth the lowe path betwee x 0, y 0 ) ad, s), P, Q). Fally, suppose x 0, y 0 ) = 1, 0). The we fst poduce P, Q) exactly as the pevous paagaph so that P, Q) tesects at 1, 0) ad oly thee. To poduce a secod pa that tesects at 1, s) ad oly thee, the double-east edge fom 0, 0) to 1, 0) P, Q) s, ths case, moved to the otheast coe as aothe double-east edge. The the esultg pa s taslated 1 ut westwad so the paths beg at 0, 0), ed at, s), ad tesect at 1, s) ad oly thee. Ivetblty of the map. We patto the collecto of all pas of paths that tesect exactly oce to goups of two as follows. ) If P, Q) tesects at 1, 0) the pa P, Q) wth P, Q ), whee P, Q ) tesects at 1, s) ad the emoval of the double-east edges fom both pas P, Q), P, Q ) esults the same pa of otesectg paths o a 1) s ectagle. ) If P, Q) tesects at 0, 1) the pa P, Q) wth P, Q ), whee P, Q ) tesects at, s 1) ad the emoval of the double-oth edges fom both pas P, Q), P, Q ) esults the same pa of otesectg paths o a s 1) ectagle. ) Suppose P, Q) tesects at a pot x, y) wth 0 < x < ad 0 < y < s. If P s oth of Q fom 0, 0) to the tesecto the pa P, Q) wth P, Q ) whee the two paths have bee techaged fom the tesecto pot to, s), so P s always oth of Q. Now we ca defe the vese mappg ψ = φ 1. Gve two pas of paths P, Q), P, Q ) fom goup ) above, ψ loos oly at P, Q). It emoves the tesecto by lftg oe of the double edges up by 1 ut P, Q), ad t moves the secod edge of P, whch s a east edge, to jo 0, 0) ad 0, 1). Fo P, Q), P, Q ) fom goup ) above, ψ loos at the pa P, Q ) whch tesects at, s 1). It lfts the uppe path, except fo ts fal oth edge, by 1 ut, the moves that oth edge to jo 0, 0) ad 0, 1). Fally, fo P, Q), P, Q ) fom goup ) above, ψ loos at P, Q) such that P s always oth of Q, ad suppose they meet at x 0, y 0 ). The t lfts the poto of P that pecedes the tesecto up by 1 ut, deletes the oth edge of P fom x 0, y 0 ) to x 0, y 0 + 1), ad adds a oth edge to jo 0, 0) ad 0, 1). 7

8 5 Temato at dffeet edpots I ths secto we exted the fomula fo N, by cosdeg pas of wals whee the two wales stat at the same pot but ed at dffeet pots. Say the wales both stat at 0, 0), ad the fst wale temates at, ) ad the secod at t, t). The, fo < t, let M,,t deote the umbe of uodeed) pas of these wals that tesect exactly pots, ot coutg the statg pot; fo = t, let M,, = N, 1. Theoem 2 Fo t, the umbes M,,t ae gve by 2 ) j t j u) 1 2u 1) u 1 j 2u 2u + 1 j j + t ) ) ) 1 j 2u 1 j 2u t j 1 2u ) ) ). 9) j j t j The secod expesso couts the umbe of pas that tesect each of the fst steps; the fst expesso couts the emag pas. Fo = 0 we ecove the famla: ) M,0,t = t ) t Poof outle. The poof cossts of fou steps. We wll expla the fou steps but most of the detals ae omtted. Let f,,, t) be the clamed fomula fo M,,t. 1. Show that the clamed fomula s coect fo = t. That s, show that the fomula fo f,,, ) smplfes to N, 1. Ths taes seveal les of mapulatos whch we omt, ad cludes a poof of the fact that 1 j=0 1) j 2 2) j 2 2 j 1 ) ) 2 1) j 2 1) ) 2 1) ) 2 j 1). j 1 ) = 1) ) ) ). 2. Notce a ecuece elato fo < t. By cosdeg the last steps of both wals, we obta: M,,t M,,+1 = M 1,,t + M 1, 1,t + M 1,,t 1 + M 1, 1,t 1, t > ) = M 1,,+1 + M 1, 1,+1 + N 1, 1 + M 1, 1,. 11) 3. Show that the clamed fomula satsfes the ecuece fo < t. Sce f,,, ) = N, 1, we eed oly show that f,,, t) satsfes the ecuece gve equato 10) fo t + 1 ode to pove that t satsfes both 10) ad 11). Note that s costat. We use the combatoal detty a b m ) m ) m a b = a b m 1 m 1 ) m a b ) + a 1 b m 1 m m 1 a 1) b 8 ) + a b+1 m 1 m 1 ) m ) a b 1 + a b m 1 m 1 ) m a 1 b 1).

9 The fst expesso of f,,, t) s a lea combato, wth coeffcets depedet of, t ad, of the left sde of the above detty wth a = t j, b = 1 2u ad m = 1 j 2u ad hece obeys the ecuece 10). The secod expesso s a smla lea combato wth a = t j, b = j ad m = ad hece also obeys that ecuece. 4. Show that the clamed fomula satsfes the bouday codtos. The values of M ae uquely detemed by the ecueces ad the bouday values fo < t. The bouday values ae M +1,,t = ) δ,t 1, 0 < t + 1. The secod expesso of f + 1,,, t) s equal to ) δ,t 1 sce the oly ozeo value occus whe j = ad t = + 1. It the taes seveal les of calculatos ad a vaat of Vademode s fomula to show that the fst summad of f + 1,,, t) s 0 whe < t. 6 Futhe emas I ths secto we dscuss the case whee the edpots of the wals ae ot pescbed. Theoem 3 Let f ) deote the umbe of odeed pas of wals that beg at the og, whch ed at the same pot, whch tae steps each of whch s oth o east, ad whch tesect each othe teally exactly pots. The f ) = )! + 1)! 1)!. Poof. Fom equato 4) above, we have that f )x = u x,1) = 1 1 4x) +1 = 2x) +1 m 0 The expaso s equato ) [7].) + 1)2m + )! m!m + + 1)! xm. Sce thee ae 2 ) pas of wals that stat at the og ad ed at the same pot, we ca tepet ths pobablstcally as: If two depedet wales stat at the og, ad each taes steps, each step beg oth o east wth equal pobablty, ad f they fsh at the same pot, the the pobablty that the teo of the paths tesect exactly pots s p, ) = )2 2)!! 1)!2)! 1; 0 1). It s teestg to ote that p, 1) = 2p, 0) f > 1. The followg theoem s pobably ow, though we fd o efeece. 9

10 Theoem 4 Let g ) deote the umbe of odeed pas of wals that beg at the og ad poceed wth oth o east steps, whch each have steps, ad whch tesect each othe exactly pots excludg the og. The g ) = 2 2 Poof. Let f ) be defed as Theoem 3, ad let F x) ad G x) be the geeatg fuctos fo f ) ad g ). The ). g ) = j f 1 j)g 0 j). 12) If we sum ove we fd that 4 = 2j ) j j g0 j). Ths sequece of equatos solves to g 0 ) = 2), ad so G0 x) = 1 4x) 1 2. Now fom equato 12), sce F x) = 1 1 4x) +1, we have G x) = 1 1 4x) 1 4x = 2 j ) 2j x j, j hece g ) = 2 2 ). The last expaso s equato ) [7].) Hece the pobablty that two such wals do ot tesect at all s 2) /4. Fo aothe poof of the theoem oe ca loo at the dffeece of the two wals. If x, y ) ad x, y ) ae the coodates of the wales o the two wals, put x, y) := x x, y y ). The x, y) wals to x, y) wth pobablty 1/2, to x+1, y 1) wth pobablty 1/4, ad to x 1, y + 1) wth pobablty 1/4. Thus the dffeece wal taes place etely o the le x + y = 0. The statstcs of tesectos of the ogal pa of wals ae detcal wth those of etus to 0 of a sgle oe-dmesoal wal of twce as may steps, ad ae gve ay boo o adom wals. Stlg s fomula ad some mapulato yelds: Coollay 1 [3] The aveage umbe of tmes that two depedet adom wals of steps, begg at the og, coss each othe s 2 + 1)! 4! 2 1 = 2 π 1 + o1). 7 Notesectg wals wth a bae I ths secto we cosde a vaato o the ogal poblem. Let, s z > 0. We cosde two wales, U ad L, that stat at the espectve lattce pots 0, z) ad z, 0) sde the ectagle wth the baes x = ad y = s. They depedetly move ethe oth o east utl they each ethe the bae x = o the bae y = s, 10

11 whee they ae costaed to move alog the bae to the temus, s). At each lattce pot of A = {, j) : <, j < s } the pobablty of movg east s p, j) ad the pobablty of movg oth s 1 p, j). We wll fd the pobablty B, s, z) that the fst tme the wales meet s at the temus. We say such a pa of wals s vald. Let us stop the wals afte + s z 1 steps. The a wale eds at ethe the pot 1, s) o the pot, s 1). The codto that a pa of wals s vald s thus equvalet to all thee of the followg holdg: 1) the wales eve meet A, 2) U eds at 1, s), ad 3) L eds at, s 1). Now, we clam that the pobablty that 2) ad 3) ae tue but 1) s false s equal to the pobablty that U eds at, s 1) ad L eds at 1, s). Fo, f we have a pa P of wals whch satsfy 2) ad 3) but tesect, we ca ceate a ew pa of wals P as commo wal poblems) by techagg the segmets fom the begg to the fst tesecto tal segmets ) of the two wals P. The esultat P s a pa of wals whch U eds at, s 1) ad L eds at 1, s), whee P has the same pobablty of occuece as does P. Smlaly, a pa P of wals whch U eds at, s 1) ad L eds at 1, s) must tesect somewhee; f we techage the tal segmets we obta a pa P of wals whch satsfy 2) ad 3) but ot 1). Hece f u deotes the pobablty that U eds at 1, s), ad l deotes the pobablty that L eds at, s 1), the the pobablty that a pa of wals s vald s B = ul 1 u)1 l) = u + l 1. Now, to smplfy the above expesso we poceed as follows. We emove the baes ad exted p, j) to the pots whee o j s abtaly. The the pobablty u s the pobablty that a sgle wale statg at 0, z) ad movg ethe to the east, wth pobablty p, j), o oth, wth pobablty 1 p, j), s, afte +s z 1 steps, at oe of the vetces the set of lattce pots o the le x+y = +s 1 that le o o above the le y = s. Smlaly, l s the pobablty that a sgle wale statg at z, 0) s, afte +s z 1 steps, the set of lattce pots o the le x+y = +s 1 that le o o to the ght of the le x =. If the tastoal pobabltes p, j) A ae a fucto of + j oly, say p, j) = p +j, the we ca exted p, j) so that ths emas tue. I ths case, l s also the pobablty that a sgle wale statg at 0, z) s, afte +s z 1 steps, the set of lattce pots o the le x+y = +s 1 that le o o below the pot z, s+z 1). The expesso u+l 1 thus smplfes to yeld the followg: Theoem 5 If the tastoal pobablty at, j) s a fucto of + j oly, the B, s, z) s equal to the pobablty that a sgle ucostaed wale statg at 0, z) ad walg fo + s z 1 steps wthout baes eds up at oe of the z pots { t, s + t 1) : 1 t z }. 11

12 Coollay 2 If all the tastoal pobabltes p j of movg east have the same value p, the ) z + s z 1 B, s, z) = p t 1 p) s+t z 1. t t=1 I patcula, f two wales stat at the og 0, 0) ad move as above, the the pobablty that they do ot meet aga utl the pot, s) s smply 2 +s 2) 1 p 1 p) s. Refeeces [1] W. N. Baley, Geealzed Hypegeometc Sees. New Yo: Hafe, 1972 ogally publshed 1935 by Cambdge Uvesty Pess). [2] N. Deshowtz ad S. Zas, Eumeato of odeed tees, Dscete Math ) [3] M.E. Fshe, Wals, walls, wettg, ad meltg, J. Statst. Phys ), [4] A. J. Guttma ad T. Pellbeg, Stacase polygos, ellptc tegals, Heu fuctos, ad lattce Gee fuctos, Phys. Rev. E ) R2233 R2236. [5] T. V. Naayaa, Su les tells fomés pa les pattos d u ete et leus applcatos à la théoes des pobabltés, C. R. Acad. Sc. Pas ) [6] T. V. Naayaa, A patal ode ad ts applcato to pobablty, Sahya ) [7] H. S. Wlf, geeatgfuctoology 2d ed.), Academc Pess, New Yo, [8] D. Zelbege, The method of ceatve telescopg, J. Symbolc Comput )