Integrality Gap for the Knapsack Problem

Transcription

1 Proof, beiefs, ad agorihms hrough he es of sum-of-squares 1 Iegraiy Gap for he Kapsack Probem Noe: These oes are si somewha rough. Suppose we have iems of ui size ad we wa o pack as may as possibe i a kapsack of size r. The, we ca pack a mos r umber of iems i he kapsack. I his ecure, we show ha his seemigy simpe reasoig ha oe ca pack a fracioa" iem is o capured by ow-degree SoS agorihm. Specificay, we prove Grigoriev s heorem (Grigoriev [2001]: 1. Theorem (SoS Hardess for Kapsack. For ay r /2, here is a pseudodisribuio of degree Ω(r suppored o pois x {0, 1} such ha x i = r. (1 i=1 Observe ha for o-iegra r he above yieds a iegraiy gap for he specia case of kapsack preseed above of degree Ω(r. Grigoriev s origia proof of his resu was a eega argume ha aayzed a aura, maximay symmeric pseudodisribuio ha we wi momeariy describe. This argume esseiay iveed severa basic resus abou he specrum of marices from he Johso Scheme sudied i agebraic combiaorics. We highy recommed readig Grigoriev s origia proof - i he ecure however, we wi rey o sadard resus from he heory of associaio schemes ad ge a shorer proof. This argume was firs preseed by Meka ad Wigderso (Meka ad Wigderso [2013] i a work ha aemped o show he firs SoS ower boud for he Paed Cique probem. The Pseudodisribuio Fix d = Θ(r o be chose aer. The idea for cosrucig degree d pseudodisribuio is very aura - he cosrai i x i = r is symmeric i r. Thus, if we had a arbirary degree 2d pseudodisribuio, we coud average i over a permuaios σ of [] ad obai aoher degree d pseudodisribuio over he hypercube ha is symmeric w.r.. permuaios of [] ad saisfies Eq. (1. Thus, for ay symmeric pseudodisribuio µ, here exiss a fucio f such ha for every muiiear moomia x S = Π i S x i, Ẽ[x S ] = f ( S. I urs ou ha here s o choice i f eiher.

2 Boaz Barak ad David Seurer 2 Firs, i=1 Ẽµ[x i ] = f (1. O he oher had, sice µ is suppored o x saisfyig Eq. (1, i=1 Ẽ[x i] = r. Thus, f (1 = r/ for every i. r f (1 = r Ẽ[x i ] = Ẽ[x i ( j=1 x j ] = Ẽ[x i ] + ( 1 Ẽ[x i x j ] = f (1 + ( 1 f (2, which impies ha f (2 = (r 1/( 1. Oe ca repea his argume o obai ha for every S such ha S 2d. (2 Ẽ[x S ] = f ( S = ( r S ( S, (3 I is easy o check ha Ẽ cosruced here saisfies he cosrai Eq. (1. Thus, as usua, we are ef wih showig he posiiviy of Ẽ. Reducio o Posiiviy over Squared Homogeous Poyomias Because of he iear equaiy cosrai Eq. (1, i urs ou ha i is eough o prove posiiviy of Ẽ[p 2 ] for ay homogeous degree d poyomia. 2. Lemma (Posiiviy over Squared Homogeous Poyomias is Eough. Suppose M be ay iear operaor o degree d poyomias ha is cosise wih he cosrais xi 2 = x i for every i [] ad i=1 x i r = 0. Suppose M(p 2 0 for every degree d homogeous poyomia p. The, M(q 2 0 for ay degree d poyomia q. Proof Skech. The idea is ha ay poyomia p of degree d ca be wrie as p 1 + i=1 (x 2 i x i p i + q( i=1 x i r (4 for a homogeous degree d poyomia p 1, degree a mos d 2 poyomias p i ad degree a mos d 1 poyomia q. This ca be show by poyomia divisio, for exampe. Oce we have he above represeaio, squarig he righ had side of Eq. (4 yieds p 2 1 pus erms ha have eiher (x2 i x i or ( i=1 x i r as a facor. Appyig M o he RHS, usig ieariy ad he fac ha M saisfies boh Booea ad he kapsack cosrais hus yieds ha M(p 2 = M(p We ca ow go o he marix view of higs i order o show posiiviy of Ẽ o homogeous degree d poyomias - we have see his argume severa imes i he ecures before.

3 Proof, beiefs, ad agorihms hrough he es of sum-of-squares 3 3. Exercise (Mome Marix. Le M R ( d ( d be defied by M(S, T = Ẽ[x S T ] = f (2d S T for ay S, T ( d. Show ha Ẽ[p 2 ] 0 for every homogeous degree d poyomia if ad oy if M 0. Noe ha M is he ( d ( d dimesioa pricipa sub-marix of he usua mome-marix of Ẽ. Johso Scheme The discussio here is based o (Meka ad Wigderso [2013] which i ur is based o (Godsi [1993]. Associaio schemes are wesudied objecs i Agebraic Combiaorics. For our purposes, we ca hik of hem as a commuaive agebra of square marices - i.e. addig or muipyig ay wo marices from he se yieds aoher marix i he se ad for ay wo marices A, B i he se, AB = BA. We are ieresed i oe such we-sudied scheme. 4. Defiiio (Johso Scheme. Le d < /2 N be parameers. The Johso Scheme J,d of order d o [] is he iear subspace of a marices J i R ( d ( d ha are se symmeric i.e. J(I, J = h( I J for ay I, J ( d. I oher words, ay ery of ay marix i he subspace depeds oy o he size of he iersecio of he row ad coum idex ses. We ow defie wo basis for J,d. 5. Defiiio (D Basis. For 0 d, e D R ( d ( d be he marix defied by 1 if I J = D (I, J = (5 0 oherwise.. D 0 is he he we-sudied Se Disjoiess marix from commuicaio compexiy. I is easy o check ha D for d spa J,d. Furher, i s aso easy o verify ha he D s commue wih each oher - ad hus every pair of marices i J,d commue esabishig ha J,d is ideed a commuaive agebra of marices. For he purposes of provig PSDess, aoher basis of J,d is very usefu - he P-Basis. 6. Defiiio (P Basis. For 0 d, e P R ( d ( d be defied by ( I J P (I, J =, (6

4 Boaz Barak ad David Seurer 4 where i s udersood ha if I J = 0, he P (I, J = 0. There s a equivae defiiio of P s ha s hepfu i cacuaios. 7. Exercise (Aerae Defiiio of P-Basis. Le R T be he rak 1 marix defied by R T (I, J = 1(I T1(J T. Show ha P (I, J = T [], T = R T. The above exercise ca be used o obai expici basis chage coefficies bewee he D ad he P basis. 8. Exercise (Chage bewee P ad D Basis. Show ha 1. For 0 d, we have P = d = ( D. 2. For 0 d, we have D = ( 1 ( P. The mai resus from he heory of associaio schemes of ieres o us is he characerizaio of eigespaces ad eigevaues of he marices i J,d. This is doe usig he fac ha here s a aura acio (he reabeig acio o eemes of [] of S ha commues wih every marix i J,d (because of se-symmery of J,d -marices. This impies ha marices i J,d mus share eigespaces wih hose correspodig o he acio of he symmeric group S. The aer are we udersood as represeaios of S ad oe ca use his udersadig o arrive a a expici descripio of eigevaues ad eigespaces of marices i J,d. For our purposes, we wi jus sae he resus required for us. 9. Lemma (Eigespaces of Johso Scheme. For P = P,,d defied as above, here exis pairwise orhogoa subspaces V 0, V 1, V 2,..., V d such ha 1. V 0, V 1, V 2,..., V d are eigespaces for P for every 0 r ad cosequey for every marix i J,d. 2. dim(v j = ( j ( j For ay marix J J,d, e λ i (J deoe he eigevaue of J o he subspace V i. The, ( i d λ i (P = (d i i if i 0 oherwise. (7 The above emma heps us esimae he eigevaues of ay marix ha ca be wrie as a iear combiaio of he marices P or D marices. I paricuar, we wi use he foowig esimae o he

5 Proof, beiefs, ad agorihms hrough he es of sum-of-squares 5 eigevaues of such marices i our aaysis of M from he previous secio. 10. Lemma (Eigevaue Esimaes for Johso Scheme. Le Q = α D J,d ad β = ( α for α 0. The, for 0 j r, ( j λ j (Q β d j ( d j j. (8 Proof. α D = α ( ( 1 ( P α = P ( ( 1 ( P ( α = β P (9 Usig ha Q ad P share eigespaces ad appyig Lemma Lemma 9, we obai he esimae caimed. PSDess of Ẽ 11. Lemma (M is PSD. M 0. Proof. We are goig o use Lemma Lemma 10. For ay o-egaive α 1, α 2,..., α d, oe ha: 0 α P = r =0 ( =0 ( α D (10 From he defiiio of M, we kow ha M = d =0 f (D 2d. We wi hus be doe from above if we ca fid o-egaive α such ha ( =0 α ( = f (2d for every. Observe ha f (2d = f (2d ( 2d+ ( r 2d+. Choose α = α = f (d ( d We ca ow verify: ( 2d + f (2d = f (2d = = ( r. ( r 2d+ 1 α =0 ( ( r r 2d + 1 ( ( r 2d + 1 r 2d + 1 =0 =0 α ( ( r 2d +. (11

6 Boaz Barak ad David Seurer 6 The emma ow foows ad i fac shows ha he miimum eigevaue of M is α d. Refereces Chris D. Godsi. Agebraic combiaorics. Chapma ad Ha mahemaics series. Chapma ad Ha, Dima Grigoriev. Compexiy of posiivseesaz proofs for he kapsack. Compuaioa Compexiy, 10(2: , Raghu Meka ad Avi Wigderso. Associaio schemes, ocommuaive poyomia coceraio, ad sum-of-squares ower bouds for paed cique. Eecroic Cooquium o Compuaioa Compexiy (ECCC, 20:105, 2013.