cycle that does not cross any edges (including its own), then it has at least


 Myles Andrews
 2 years ago
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1 W prov th following thorm: Thorm If a K n is drawn in th plan in such a way that it has a hamiltonian cycl that dos not cross any dgs (including its own, thn it has at last n ( 4 48 π + O(n crossings Th basic ida of th proof is as follows Lt th vrtics of th graph b labld,, n in ordr along th noncrossing hamiltonian cycl Each dg of th graph not on this cycl must b on ithr th insid or th outsid Considr a pair of such dgs: (a, c and (b, d If a and c sparat b from d along th circl, and if ths two dgs ar drawn on th sam sid of th cycl, thn thy must cross This rducs our problm to finding th solution to a crtain MAXCUT problm W mak th following dfinitions: Dfinition For a, b, c, d Z/n, w say that (a, c crosss (b, d if, a, b, c, d can b assignd rprsntativs a, b, c, d so that ithr a < b < c < d < a + n or a > b > c > d > a n Not that (a, c crosss (b, d if and only if (c, a crosss (b, d This is bcaus if (without loss of gnrality a > b > c > d > a n, thn c > d > a n > b n > c n Similarly, (a, c crosss (b, d if and only if (b, d crosss (a, c Dfinition For positiv intgr n, lt G n b th graph whos vrtics ar unordrd pairs of distinct lmnts of Z/n, and whos dgs connct pairs {a, c} and {b, d} whn (a, c crosss (b, d Lmma If a K n is drawn in th plan in such a way that it has a hamiltonian cycl that dos not cross any dgs (including its own, thn it has at last E(G n MAXCUT(G n crossings Proof For vry such drawing of a graph, labl th vrtics along th hamiltonian cycl by lmnts of Z/n in ordr Th dgs of our K n now corrspond to th vrtics of G n in th obvious way Lt S b th subst of th vrtics of G n corrsponding to dgs of th K n that li within th dsignatd cycl Not that any two vrtics in S or any two vrtics not in S connctd by an dg, corrspond to pairs of dgs in th K n that must cross Thus th numbr of crossings of our K n is at last E(S, S + E( S, S = E(G n E(S, S E(G n MAXCUT(G n W hav thus rducd our problm to bounding th siz of th solution of a crtain family of MAXCUT problms W do this ssntially by solving th GomansWilliamson rlaxation of a limiting vrsion of this family of problms To st things up, w nd a fw mor dfinitions Dfinition By S hr w will man R/Z Givn a, b, c, d S w say that (a, c crosss (b, d if a, b, c and d hav rprsntativs a, b, c, d R rspctivly, so that ithr a > b > c > d > a or a < b < c < d < a +
2 Dfin th indicator function C(a, b, c, d := { if (a, c crosss (c, d, 0 othrwis W now prsnt th continuous vrsion of our MAXCUT problm: Proposition Lt f : S S {±}, thn (S 4 f(w, yf(x, zc(w, x, y, zdwdxdydz π W prov this by instad proving th following strongr rsult: Proposition 4 Lt f : S S C satisfy f(x, y for all x, y, thn f(w, yf(x, zc(w, x, y, zdwdxdydz (S π 4 Furthrmor, for any L function f : S S C, w hav that f(w, yf(x, zc(w, x, y, zdwdxdydz (S π f(x, y sin (π(x ydxdy 4 (S Th proof of Proposition 4 will involv looking at th Fourir transforms of th functions involvd Bfor w can bgin with this w nd th following dfinition: Dfinition Dfin th function (x := πix W now xprss th Fourir transform of C Lmma 5 W hav that C(w, x, y, z is qual to: π + π + π + ((nw nx + my mz + (nw mx + my nz (( mx + (n + my nz + (nw + my (n + mz (((n + mw nx mz (nw (n + mx + my (
3 Proof For x R/Z lt [x] b th rprsntativ of x lying in [0, ] For any w, x, y, z R/Z, it is clar that [x w] + [y x] + [z y] + [w z] Z It is not hard to s that this numbr is odd if and only if (w, y crosss (x, z Thus, ( [x w] ( C(w,x,y,z [y x] [z y] [w z] = ( [x w] = ( [y x] ( [z y] ( [w z] In ordr to comput th Fourir transform, w comput th Fourir transform of ach individual trm Not that ( [x w] ( α ( nx mwdxdw = ( mw n(α + wdαdw = ( (m + nw (n /αdαdw = δ m, n πi(n / Thrfor, by standard Fourir analysis, w can say that ( [x w] = i (aw ax π a + / W hav similar formula for thm togthr, w find that ( C(w,x,y,z = π 4 a,b,c,d Z ( [y x], a Z ( [z y], and ( [w z] Multiplying ((a dw + (b ax + (c by + (d cz (a + /(b + /(c + /(d + / W now nd to collct lik trms In particular, for vry 4tupl of intgrs α, β, γ, δ, th cofficint of (αw + βx + γy + δz quals th sum ovr 4tupls of intgrs a, b, c, d with α = a d,β = b a,γ = c b,δ = d c of π 4 (a + /(b + /(c + /(d + / Clarly, thr ar no such a, b, c, d unlss α+β +γ +δ = 0 If this holds, thn all such 4tupls ar of th form n, n + β, n + β + γ, n + β + γ + δ for n an arbitrary intgr Thus, w nd to valuat π 4 (n + /(n + β + /(n + β + γ + /(n + β + γ + δ + / n Z Considr th complx analytic function g(z = π cot(πz (z + /(z + β + /(z + β + γ + /(z + β + γ + δ + /
4 Not that along th contour max( R(z, I(z = m+/ for m a larg intgr, g(z = O(m 4 Thus th limit ovr m of th intgral of g ovr this contour is 0 This implis that th sum of all rsidus of g is 0 Not that g has pols only whn ithr z in an intgr or whn (z + /(z + β + /(z + β + γ + /(z + β + γ + δ + / = 0 At z = n, g has rsidu Thus, n Z (n + /(n + β + /(n + β + γ + /(n + β + γ + δ + / (n + /(n + β + /(n + β + γ + /(n + β + γ + δ + / = ρ Z Thrfor, ( C(w,x,y,z quals π 4 α+β+γ+δ=0 (αw + βx + γy + δz ρ Z Rs ρ (f α,β,γ,δ Not that all othr such rsidus ar at half intgrs Not furthrmor that cot(πz is an odd function around half intgrs Thus, g has a rsidu at z Z only if z is a root of (z +/(z +β +/(z +β +γ +/(z +β +γ +δ +/ of vn ordr, and in particular ordr at last In othr words, w hav rsidus only whn som pair of lmnts of (0, β, β + γ, β + γ + δ ar th sam, but no thr of thm ar unlss all four ar 0 In particular, w gt rsidus in th following cass: Whn β = 0, lt α = n, γ = m Thn, for (α, β, γ, δ = (n, 0, m, (n+m, π w hav a rsidu at ρ = / of so long as n, m 0 Whn γ = 0, lt β = n, δ = m Thn, for (α, β, γ, δ = (n + m, n, 0, m, w hav a rsidu at ρ = n / of π so long as n, m 0 Whn δ = 0, lt α = n, γ = m Thn, for (α, β, γ, δ = (n, (n+m, m, 0, π w hav a rsidu at ρ = n / of so long as n, m 0 Whn α = 0, lt β = n, δ = m Thn, for (α, β, γ, δ = (0, n, n + π m, m, w hav a rsidu at ρ = / of so long as n, m 0 Whn α+β = 0, lt α = n, γ = m Thn for (α, β, γ, δ = (n, n, m, m, w hav a rsidu at ρ = n / of π so long as n, m 0 Whn β +γ = 0, lt α = n, γ = m Thn for (α, β, γ, δ = (n, m, m, n, w hav a rsidu at ρ = / of π so long as n, m 0 Whn α = β = γ = δ, w hav a rsidu at ρ = / Rs ρ (g 4
5 Thus w hav that ( C(w,x,y,z quals π π π +D ((nw nx + my mz + (nw mx + my nz (( mx + (n + my nz + (nw + my (n + mz (((n + mw nx mz (nw (n + mx + my For som constant D Noting that C(w, x, y, z = ( C(w,x,y,z, w hav that C(w, x, y, z quals π + π + π +D ((nw nx + my mz + (nw mx + my nz (( mx + (n + my nz + (nw + my (n + mz (((n + mw nx mz (nw (n + mx + my On th othr hand, D = C(w, x, y, z Not that givn any w, x, y, z (S 4 distinct that of th thr ways to partition {w, x, y, z} into two pairs, xactly on givs a st of crossing pairs Thus C(w, x, y, z + C(w, y, x, z + C(w, x, z, y quals xcpt on a st of masur 0 Thus, sinc th intgral of ach of ths is D, w hav that D =, or that D = / This complts th proof Proof of Proposition 4 Sinc f is L w may writ f(x, y = a (nx + my Z for complx numbrs a with a < Notic that rplacing f(x, y by f(x,y+f(y,x dos not ffct th lft hand sid of Equation (, and can only incras th right hand sid Thus w can assum that f(x, y = f(y, x, and thrfor that a = a m,n By Lmma 5, th lft hand sid of Equation ( is π + π a a + a a m,n + π a n+m,0 a + a a n+m,0 + a 0,0a 0,0 a 0,n+m a m,n + a a 0,n+m 5
6 Using a = a m,n, this simplifis to π = π = π a a a a n+m,0 a n+m,0 a a a n+m,0 a n+m,0 + a 0,0 ( a a n+m,0 + π a k,0 k Z = 0 This can b sn by consid W claim that for k 0 that n+m=k ring th rsidus of th analytic function g(z = π cot(πz z(k z n+m=k + a 0,0 mn + a 0,0 Not that along th contour max( R(z, I(z = m+/ for m a larg intgr, g(z = O(m Thus th limit ovr m of th intgral of g ovr this contour is 0 This implis that th sum of all rsidus of g is 0 It is clar that g has rsidus only at intgrs At z = n for n 0, k, it has rsidu n(k n If k = 0, it has rsidu 0 at 0 and k Thus, th sum of rsidus is xactly n+m=k Furthrmor, if k = 0, n+m=k = n Z\{0} n π = ζ( = Thrfor, th lft hand sid of Equation ( is a a n+m,0 π ( π π a 0,0 + a 0,0 = π a a n+m,0 Th right hand sid of Equation ( is ( a a a n+,m a n,m+ π = π a (a a n+,m = π a a n+,m = π (a a n+m,0 (a n+,m a n+m,0
7 W now lt b = a a n+m,0 Notic that b 0,k = b k,0 = 0 Equation ( is now quivalnt to 0 b + b b n+,m 0 W will in fact prov th strongr statmnt that >0 b (n+m>0 b b n+,m W not by symmtry that w can assum that n, m > 0 W also not that it suffics to prov for ach k > 0 that >0,n+m=k b n+,m>0,n+m=k b b n+,m ( For fixd k, lt c n = b n,k n b n,k n+ By th symmtry xhibitd by th b s, th right hand sid of Equation ( is Manwhil, th right hand sid is k/ n= c n k/ n= n i= c i (k / n(k n + n= n i= c i n(k n ( Thus, th right hand sid is givn by a quadratic form in th c,, c k/ with positiv cofficints Thrfor, th biggst ratio btwn th right and lft and sids is obtaind by th uniqu ignvctor of this quadratic form for which all c i ar positiv W claim that this happns whn c n = k + n For ths c s, th drivativ of th xprssion in Equation ( with rspct to c m is k/ n=m ( n i= c (k / i n(k n + n=m It is asy to vrify that for this choic of c i that n c i = n(k n Thus, th abov rducs to k/ n=m (k / + n=m i= ( n i= c i n(k n = ( k/ m + + ( (k / m + = (k m + = c m 7
8 Thus, ths c i giv th uniqu positiv ignvctor Hnc it suffics to chck Equation ( whn c m = k m +, or quivalntly whn b n,k n = n(k n In this cas, th lft hand sid of Equation ( is k k n(k n = (kn n n= n= For this choic, th right hand sid is k (k + n = n= = k (k (k k(k k(k (k + = = k k k k + k 4kn + 4n + 4n n= = k + k k (k + + k k(k + (k + k(k + + = k + k k k + k k k + 4k + k + k = k k Thus, th largst possibl ratio btwn th lft and right hand sids of Equation ( is This complts our proof W ar now prpard to prov our main thorm Proof of Thorm W will procd by way of Lmma W not that E(G n = n 4 /4 + O(n W hav only to bound th siz of th MAXCUT of G n Considr any subst S of th vrtics of G n dfining a cut W wish to bound th numbr of dgs that cross this cut Dfin th function f S : S S {±} as follows: { if ( nx, ny S f S (x, y = othrwis Considr (S 4 f S (w, yf S (x, zc(w, x, y, zdwdxdydz (4 In ordr to valuat this xprssion, w considr th intgral ovr th rgion R a,b,c,d = [a/n, (a + /n] [b/n, (b + /n] [c/n, (c + /n] [d/n, (d + /n] for som a, b, c, d Z/n W not that ovr this rgion that f S (w, yf S (x, z is constant In particular, it is if (a, c and (b, d ar ithr both in S or both 8
9 not in S, and othrwis It should also b notd that if a, b, c, d ar distinct thn C(w, x, y, z is also constant on this rgion, and in particular is if G n contains an dg btwn (a, c and (b, d Thus th xprssion in Equation (4 is f S (w, yf S (x, zc(w, x, y, zdwdxdydz R a,b,c,d a,b,c,d = a,b,c,d nondistinct R a,b,c,d O( + a,b,c,d {{a,c},{b,d}} E(G n f S (a/n, c/nf S (b/n, d/n n 4 =8n 4 ( Edgs not crossing th cut Edgs crossing th cut + O(n On th othr hand, by Proposition, this is at last π Thus Edgs crossing th cut Edgs not crossing th cut n4 8π + O(n Adding th numbr of dgs of G n and dividing by, w find that ( Edgs crossing th cut n 4 π + + O(n 48 This provids an uppr bound on th siz of MAXCUT(G n Thus by Lmma, th crossing numbr of K n is at last ( n 4 48 π + O(n This complts our proof 9
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