A O(1/eps^2)^n Time Sieving Algorithm for Approximate Integer Programming

A O (1 /ǫ 2 ) n -time Sie ving Algorithm for Approxi mate Inte ger Programm ing Daniel Dadush ∗ November 27, 2024 Abstract The Integer Programming Pro blem (IP) for a polytop e P ⊆ R n is to find an integer point in P or decide that P is integer free. W e giv e a rando mized algorithm for an app roxima te version of this prob lem, which correctly decid es whether P contain s an integer point or whether a (1 + ǫ ) scaling of P around its barycen ter is integer free in time O (1 /ǫ 2 ) n with overwhelming p robab ility . W e reduce this ap proxim ate IP question to an appr oximate Closest V ector Problem (CVP) in a “near-symmetric” semi-n orm, wh ich we solve via a rando mized sie vin g tech nique first developed by Ajtai, Kumar, and Siv ak umar (STOC 2001) . Our main technical co ntribution is an extension of the AKS sieving techniqu e which works for any near-symmetric semi-no rm. Ou r results also extend to general con vex bodies and lattices. Ke ywords. Intege r Programming, Shortest V ecto r Problem, Closes t V ector Proble m . 1 Introd uction The Inte ger Programming (IP) P roblem, i.e. the pr oblem of decid ing wheth er a polyt ope co ntains an in- teg er point, is a classic probl em in Operation s Research and C omputer S cience . Algorithms for IP were first de veloped in the 1950s when Gomory [Gom58] gav e a finite cuttin g plane algorit hm to solv e genera l (Mixe d)-Intege r Programs. H o w e ver , the first algorit hms with complexity guarantee s (i.e. better than finite- ness) came much late r . The first such algorith m was the break through result of Lentra [Len83], which gav e the first fixed dimension polynomial time algorithm for IP . Lenstr a’ s approach rev olv ed on fi nding “flat” inte ger di rections o f a po lytope, and achiev ed a leading complexi ty term of 2 O ( n 3 ) where n is the num- ber of va riables. Lenstra’ s approach was generaliz ed and sub stantially impro ved upon by Kannan [Kan87], who decreased the complexit y to O ( n 2 . 5 ) n . Recently , D adush et al [DPV11] improv ed this comple xity to ˜ O ( n 4 3 ) n by using a solv er for th e Short est V ector Problem (SVP) in ge neral no rms. Follo wing the works of Lenstra and Kannan , fixed dimensio n polyn omial algorit hms were disco vered for counting the number of intege r points in a rational polyhe dron [Bar94], parametr ic inte ger programming [Kan90, ES08], and inte ger opti mization ov er quasi-con v ex polynomia ls [Hei05, H K10]. Howe ver , over the last twenty years the kno wn algori thmic co mplexity of IP has only modestly decreased. A centr al open problem in the area therefo re remains the follo wing: ∗ H. Milton Stewa rt Schoo l of Industrial and Systems Engineering, Geo rgia In sti tute of T echn ology , 765 F erst Driv e NW , Atlanta, GA 30332-020 5, USA dndadush@gat ech.edu 1 Question: Does there exi st a 2 O ( n ) -time algorith m for Integer Progra mm ing? In this pape r , w e sho w that the answer to this ques tion is affirmati ve as long as we are willing to acce pt an approx imate notion of containmen t. More precise ly , we gi ve a randomized algo rithm which c an corr ectly distin guish whether a polytop e P contain s an inte ger point or if a small “blo wup” of P contains no integer points in O (1 /ǫ 2 ) n time with overwhel ming p robability . Our resul ts naturally ex tend to setting of genera l con v ex bodies and lattices, w here the IP problem in this context is to decide for a con v ex body K and lattice L in R n whether K ∩ L = ∅ . T o obtain the approxima te IP result, we reduce the problem to a (1 + ǫ ) -appro ximate Closest V ector Problem (CVP) under a “near -symmetric” semi-norm. Giv en a lattice L ⊆ R n (inte ger combinat ions of a basis b 1 , . . . , b n ∈ R n ) the SVP is to find min y ∈ L \{ 0 } k y k , and gi ven x ∈ R n the CVP is to find min y ∈ L k y − x k , where k · k is a gi ven (semi-)n orm. A semi-norm k · k satisfies all norm proper ties excep t symmetry , i.e. w e allo w k x k 6 = k − x k . Our method s in this setting are based on a ra ndomized sie ving t echnique first dev eloped by Ajtai, K umar and Siv akumar [AKS01, AKS02] for solving the S hortest (SVP) and C losest V ector Problem (CVP ). In [AKS01], they gi ve a 2 O ( n ) sie ving algorithm for S VP in the ℓ 2 norm, exten ding this in [A KS02] to gi ve a 2 O ( 1 ǫ ) -time algorithm for (1 + ǫ ) -CVP in the ℓ 2 norm. I n [BN07], a sie ve based 2 O ( n ) -time algorithm for SVP and O (1 /ǫ 2 ) n -time algo rithm for (1 + ǫ ) -CVP in any ℓ p norm is giv en. In [AJ08 ], the previ ous results are exte nded to giv e a 2 O ( n ) -time S VP algorithm in any norm (though not semi-n orm). In [EHN11], a techni que to boost any 2 -app roximation alg orithm for ℓ ∞ CVP is gi ven which yields a O (ln( 1 ǫ )) n algori thm for (1 + ǫ ) -CVP und er ℓ ∞ . Our main tec hnical contrib ution is a n exten sion of t he AKS sie ving tech nique to gi ve a O (1 /ǫ 2 ) n algori thm for CVP under any near-s ymmetric semi-norm . 1.1 Definitions In what follo ws, K ⊆ R n will denote a con ve x body (a full dimens ional compac t con vex set) and L ⊆ R n will denote an n -dimension al lattice (all integer combination s of a basis of R n ). K w ill be prese nted by a members hip oracle in the stand ard way (see sect ion 2), and L will be pres ented by a gen erating basis b 1 , . . . , b n ∈ R n . W e define the barycen ter (or centr oid) of K as b ( K ) = 1 v ol ( K ) R K xdx . For sets A, B ⊆ R n and scalars s , t ∈ R define the Minko wski S um s A + tB = { sa + tb : a ∈ A, b ∈ B } . in t ( A ) denotes the in terior of the set A . Let C ⊆ R n be a con ve x b ody w here 0 ∈ in t( C ) . Define the semi-no rm induced by C (or gauge function of C ) as k x k C = inf { s ≥ 0 : x ∈ sC } for x ∈ R n . k · k C satisfies all no rm proper ties excep t symmetry , i.e. k x k C 6 = k − x k C is allo wed. k · k C (or C ) is γ -symmetric, for 0 < γ ≤ 1 , if vol( C ∩ − C ) ≥ γ n v ol( C ) . Note C is 1 -symmetric if f C = − C . For a lattice L and semi-norm k · k C , define the fi rst minimum of L under k · k C as λ 1 ( C, L ) = inf z ∈ L \{ 0 } k z k C (lengt h of shortes t non-zero vector). For a tar get x , lattice L , and semi-nor m k · k C , define the distanc e from x to L under k · k C as d C ( L, x ) = inf z ∈ L k z − x k C . 1.2 Results W e stat e our m ain result in terms of general co n vex bodies and lattices. W e recov er the standard int eger progra mm ing setting by setting L = Z n , the standard integer lattice, and K = { x ∈ R n : Ax ≤ b } , a general polyto pe. Fo r simp licity , we of ten omit standa rd polynomial f actors from the ru ntimes of our algorith ms (i.e. p olylog terms asso ciated with boun ds on K or the bit length of the basis for L ). Our main result is the follo wing: 2 Theor em 1.1 (Approximate IP Feasibi lity) . F or 0 < ǫ ≤ 1 2 , ther e e xists a O (1 /ǫ 2 ) n time algorithm which with pr obability at least 1 − 2 − n either outputs a point y ∈ (1 + ǫ ) K − ǫb ( K ) ∩ L or decides that K ∩ L = ∅ . Furthemor e, if 1 1 + ǫ K + ǫ 1 + ǫ b ( K ) ∩ L 6 = ∅ , the algorit hm r eturns a point z ∈ K ∩ L with pr obabi lity at least 1 − 2 − n . The abo ve theorem substanti ally improve s the comple xity of IP in the case w here K con tains a “deep” lattice point (i.e. within a slight sc aling of K aroun d its barycenter ). Compared to exact algorithms, our methods are competiti ve or fast er as long as 1 1 + n − 1 / 2 K + n − 1 / 2 1 + n − 1 / 2 b ( K ) ∩ L 6 = ∅ , where we ac hiev e compl exity O ( n ) n (which is th e conjectured complexity of th e IP algorithm in [DPV11]). Hence to impro ve the co m plex ity of IP belo w O ( n δ ) n , for any 0 < δ < 1 , one may assume that al l the inte ger point s lie close to the bound ary , i.e. that 1 1 + n − 1 2 δ K + n − 1 2 δ 1 + n − 1 2 δ b ( K ) ∩ L = ∅ . The abo ve stat ement lends creden ce to the intuition that exact IP is hard because of lattice points lying very near the bound ary . Starting with the above algorithm, w e can use a binary search proc edure to go from approximate feasi- bility to appro ximate optimizati on. This yields the follo wing theo rem: Theor em 1.2 (Approximate Integer Optimizatio n) . F or v ∈ R n , 0 < ǫ ≤ 1 2 , δ > 0 , ther e exist s a O (1 /ǫ 2 ) n p olylog( 1 δ , k v k 2 ) algorit hm which with pr obability at least 1 − 2 − n either outputs a point y ∈ K + ǫ ( K − K ) ∩ L suc h that sup z ∈ K ∩ L h v , z i ≤ h v , y i + δ or corr ectly decides that K ∩ L = ∅ . The abo ve theorem state s that if we wish to opti m ize ov er K ∩ L , we can find a la ttice point in a slig ht blo wup of K whose object ive v alue is essentially as goo d as any point in K ∩ L . W e re mark that the blo wup is worse than in Theorem 1.1, since (1 + ǫ ) K − ǫx ⊆ K + ǫ ( K − K ) for any x ∈ K . This stems from the ne ed to call the feasi bility algorit hm on multip le rest rictions of K . T o giv e a clearer unde rstanding of this notion, the ne w constraints of the “blowup ” body can be underst ood from the follo wing formula: sup x ∈ K + ǫ ( K − K ) h v , x i = sup x ∈ K h v , x i + ǫ  sup x ∈ K h v , x i − in f x ∈ K h v , x i  . Hence each v alid constraint h v , x i ≤ c for K , is relaxed by an ǫ -fraction of its v ’ s v ariation over K . 3 1.3 Main T ool W e now describe the main tool used to deri ve both of the abov e algorithms. At the heart of Theore m 1.1, is the follo wing algorithm: Theor em 1.3. L et k · k C denote a γ -symmetri c semi-nor m . F or x ∈ R n , 0 < ǫ ≤ 1 2 , ther e e xists an O ( 1 γ 4 ǫ 2 ) n time algori thm which compute s a point y ∈ L satisfying k y − x k C ≤ (1 + ǫ ) d C ( L, x ) with pr obab ility at least 1 − 2 − n . Furthermor e, if d C ( L, x ) ≤ tλ 1 ( C, L ) , for t ≥ 2 , then an e xact close st vector can be found in time O ( t 2 γ 4 ) n with pr obability at least 1 − 2 − n . The above algori thm ada pts the AKS siev e to work for general semi-no rms. As mention ed pr eviou sly [BN07] ga ve the abov e result for ℓ p norms, and [AJ08] gav e a 2 O ( n ) -time ex act S VP solver for all norms (also implied by the abov e since SVP ≤ CVP , see [GMS S99]). In [DPV 11], a Las V egas algorith m (where only the runtime is pro babilistic, not the corre ctness) is giv en for the exact v ersions of the abov e results (i.e. w here an exact closest / short est vector is found) with similar asymptotic complexi ty usin g completely dif ferent technique s, Hence compared with pre vious results, the no velty of the abo ve algorithm is the exte nsion of the AKS sie ving techniq ue for (1 + ǫ ) -CVP in gen eral semi-norms. As seen from Theorems 1.1 and 1.2, the signif- icance of this exte nsion is in its direct application s to IP . Furthermore , we belie ve our results illustr ate th e ver satility of the AKS sie ving paradigm. From a high le vel, o ur algor ithm uses t he same frame work as [BN07, AJ08]. W e first s how that the AKS sie ve can be used to solve the S ubspac e A voidi ng Prob lem (SAP), which was first defined in [BN07], and use a reduct ion from CVP to SAP to get the final r esult. The techni cal cha llenge we o vercome , is finding the correc t genera lizations of the each of the steps perfor m ed in previo us algorit hms to the asymmetric setting . W e discuss this further in section 3.2. 1.4 Organization In secti on 2, we gi ve some general backgro und in con ve x geometr y and lattices . In section 3.1, we describ e the reduction s from Appr oximate Integ er Programming to Approx imate CV P as well as Approximate In- teg er Optimization to Approx imate In teger P rogramming. In sectio n 3.2, we present the algorithm for the Subspace A voidi ng Prob lem, and in sectio n 3.3 we gi ve the reduction from CVP to SAP . In section 4, we presen t our conclusi ons and open pro blems. 2 Pr eliminaries Computation Model: A con vex body K ⊆ R n if ( a 0 , r , R ) -centered if a 0 + r B n 2 ⊆ K ⊆ a 0 + R B n 2 , where B n 2 is the unit euclide an ball. All the con ve x bodies in this pape r w ill be ( a 0 , r , R ) -centered unless otherwis e specified . T o interact with K , algor ithms are gi ven access to a membership oracle for K , i.e. an oracle O K such that O K ( x ) = 1 if x ∈ K and 0 . In some situat ions, an e xact membership orac le is d ifficult to implemen t (e.g. decidin g whether a m atrix A has operator norm ≤ 1 ), in w hich situation w e settle for a “weak”-memb ership o racle, w hich only guarantee s its answer for points that are either ǫ -deep inside K or ǫ -far from K (the error toler ance ǫ is provi ded as an input to the oracle). 4 For a (0 , r, R ) -centered K the gaug e function k · k K is a semi-n orm. T o interac t w ith a semi- norm, algori thms are giv en a distance oracle, i.e. a funct ion w hich on inp ut x return s k x k K . It is not hard to check that giv en a members hip oracle for K , one can compute k x k K to within an y desired accurac y using binary search. Also we remembe r that k x k K ≤ 1 ⇔ x ∈ K , hence a distance oracle can easily implement a membership oracle . All the algorith ms in this paper can be made to work with weak-ora cles, but for simplicit y in presen tation, we assume that o ur oracles a re all exac t and tha t the con vers ion between dif ferent types of oracles occurs automaticall y . W e note that when K is a polytop e, all the necessa ry oracles can be implemente d exac tly and without dif ficulty . In the oracle model of computa tion, complexity is measured by the number of oracles calls and arith- metic operati ons. Pro b ability: For ran dom v ariables X , Y ∈ Ω , we define the total v ariation distance between X and Y as d T V ( X, Y ) = sup A ⊆ Ω | Pr( X ∈ A ) − Pr( Y ∈ A ) | The follo wing lemma is a sta ndard fact in pr obability theory : Lemma 2.1. Let ( X 1 , . . . , X m ) ∈ Ω m and ( Y 1 , . . . , Y m ) ∈ Ω m denote indep endent rand om variable s variab les satisfyin g d T V ( X i , Y i ) ≤ ǫ for i ∈ [ m ] . Then d T V (( X 1 , . . . , X m ) , ( Y 1 , . . . , Y m )) ≤ mǫ Algorithms on Con vex Bodies: For the purpose s of our sie ving algor ithm, we will need an algorith m to sample unifo rm points from K . The follo wing result of [DFK89] provid es the result: Theor em 2.2 (Uniform Sampler) . Given η > 0 , ther e e xists an a lgorithm whic h ou tputs a rand om point X ∈ K whose dis tribu tion has total vari ation distance at most η fr om the uni form distrib ution on K , usi ng at most p oly ( n, ln( 1 η ) , ln ( R r )) calls to the oracle and arithmetic operat ions. W e call a random v ector X ∈ K η -uniform if the total v ariatio n distance betwee n X and a uniform vec tor on K is at most η . Our main IP algorithm will prov ide a gua rantee with res pect to the baryce nter of K . The follo wing lemma allo ws us to ap proximate a poin t near b ( K ) w ith ov erwhelming probability : Lemma 2.3 (A pprox. Barycenter) . F or ǫ > 0 , let b = 1 N P N i =1 X i , N = cn 2 ǫ 2 , c > 0 an absolute constan t, and wher e X 1 , . . . , X N ar e iid 4 − n -unifo rm samples on K ⊆ R n . Then Pr[ k ± ( b − b ( K )) k K − b ( K ) > ǫ ] ≤ 2 − n Lattices: An n -dimensio nal latti ce L ⊆ R n is formed by integral combinat ions of linea rly indepen dent vec tors b 1 , . . . , b n ∈ R n . Letting B = ( b 1 , . . . , b n ) , for a poin t x ∈ R n we define the modulus operator as x mo d B = B ( B − 1 x − ⌊ B − 1 x ⌋ ) where for y ∈ R n , ⌊ y ⌋ = ( ⌊ y 1 ⌋ , . . . , ⌊ y n ⌋ ) . W e note that x mo d B ∈ B [0 , 1) n , i.e. the funda m ental paralle lipiped of B and that x − ( x mod B ) ∈ L , hence x mo d B is the uniqu e representat ive of the cose t x + L in B [0 , 1) n . 5 Con vex Geometry: The f ollowing lemma p rovides us some simple e stimates on th e eff ects of re centering the semi-norm assoc iated with a con vex body . Lemma 2.4. T ake x, y ∈ K satisfyin g k ± ( x − y ) k K − y ≤ α < 1 . Then for z ∈ R n we have that 1. z ∈ τ K + (1 − τ ) y ⇔ k z − y k K − y ≤ τ 2. k z − y k K − y ≤ k y − x k K − x + α | 1 − k z − x k K − x | 3. k z − x k K − x ≤ k z − y k K − y + α 1 − α | 1 − k z − y k K − y | The follo wing theorem of Milman and Pajor , tells us that K − b ( K ) is 1 2 -symmetric . Theor em 2.5 ([MP00]) . Assume b ( K ) = 0 . T hen v ol( K ∩ − K ) ≥ 1 2 n v ol( K ) . Using the abo ve theorem, we gi ve a simple exten sion which sho ws tha t near -symmetry is a stable prop- erty . Cor ollary 2.6. A ssume b ( K ) = 0 . Then for x ∈ K we have that K − x is 1 2 (1 − k x k K ) -symmetric . 3 Algorithms 3.1 Integer Pr ogramming W e descri be th e basic reduction from Approx imate In teger Programmin g to Approximat e CVP , as well as the reducti on from Approximate Intege r Optimization to Approximate Integ er Programming. Pr oof of Theor em 1.1 (Appr oximate Int e ger P r ogr amming). W e are giv en 0 < ǫ ≤ 1 2 , and we wish to find a lattice point in (1 + ǫ ) K − ǫb ( K ) ∩ L or decide that K ∩ L = ∅ . The algorithm, which we denote by ApproxIP ( K, L, ǫ ) , will be the follo wing: Algorithm: 1. Compute b ∈ K , sati sfying k ± ( b − b ( K )) k K − b ( K ) ≤ 1 3 , using Lemma 2.3 (see detail s belo w). 2. Compute y ∈ L such that y is 1 + 2 ǫ 5 approx imate closes t lattice vect or to b under the semi-no rm k · k K − b using Approx-CVP (Theorem 3.7). 3. Return y if y ∈ k y − b k K − b ≤ 1 + 3 ǫ 4 , and other w ise retu rn “EMPTY” (i.e. K ∩ L = ∅ ). Corr ectness: Assuming tha t steps (1) and (2) return corr ect output s (which occurs with o verwhelming probab ility), we sho w that the final outp ut is correct. First note that if k y − b k K − b ≤ 1 + 3 ǫ 4 , then by Lemma 2.4 we ha ve that k y − b ( K ) k K − b ( K ) ≤ k y − b k K − b + 1 3 | 1 − k y − b k K − b | ≤ 1 + 3 ǫ 4 + 1 3 3 ǫ 4 = 1 + ǫ as required. Now assume that K ∩ L 6 = ∅ . Then we can tak e z ∈ L suc h that k z − b k K − b ≤ 1 . S ince y is a 1 + 2 ǫ 5 closes t vector , we m ust ha ve that k y − b k K − b ≤ 1 + 2 ǫ 5 . Hence by the reaso ning in the pre vious paragr aph, we hav e that k y − b ( K ) k K − b ( K ) ≤ 1 + ǫ as neede d. 6 For the fu rthermore, we as sume that 1 1+ ǫ K + ǫ 1+ ǫ b ( K ) ∩ L 6 = ∅ . So we may pick z ∈ L su ch that k z − b ( K ) k K − b ( K ) ≤ 1 1+ ǫ . By Lemma 2.4, we ha ve that k z − b k K − b ≤ k z − b ( K ) k K − b ( K ) + 1 3 1 − 1 3   1 − k z − b ( K ) k K − b ( K )   ≤ 1 1 + ǫ + 1 2 ǫ 1 + ǫ = 1 + ǫ 2 1 + ǫ Next by the assumptions on y , we hav e th at k y − b k K − b ≤ 1+ ǫ 2 1+ ǫ (1 + 2 ǫ 5 ) ≤ 1 since 0 < ǫ ≤ 1 2 . Hence y ∈ K ∩ L as needed. Runtime: For step (1), by Lemma 2.3 we can comput e b ∈ K , satis fying k ± ( b − b ( K )) k K − b ( K ) ≤ 1 3 , with probabi lity at least 1 − 2 − n , by lettin g b be the a verage of O ( n 2 ) 4 − n -unifo rm sa mples ove r K . By Theorem 2.2, each of these samples can be computed in p oly( n, ln( R r )) time. For step (2), we fi rst note that by C orollary 2.6, K − b is (1 − 1 3 ) 1 2 = 1 3 - symmetric. Therefore , the call to the Approximat e CVP algo rithm, with error parameter 2 ǫ 5 return s a v alid approximatio n vector with probability at least 1 − 2 − n in time O (3( 5 2 ǫ ) 2 ) n = O (1 /ǫ 2 ) n . H ence the entire algorithm tak es time O (1 /ǫ 2 ) n and outpu ts a correct answer with probabil ity at least 1 − 2 − n +1 as need ed. Pr oof of Theor em 1.2 (Appr oximate Int e ger O ptimizati on). W e are giv en v ∈ R n , 0 < ǫ ≤ 1 2 , and δ > 0 where we wish to find a lattice point in K + ǫ ( K − K ) ∩ L whose objecti ve value is w ithin an add itiv e δ of the best poin t in K ∩ L . W e remember that K is ( a 0 , r , R ) -centered . Since we lose nothing by makin g δ small er , we shall ass ume that δ ≤ k v k 2 r . W e will sho w that Algorith m 1 c orrectly solves th e optimizati on proble m. Algorithm 1 Algorit hm ApproxOPT ( K, L, v , ǫ, δ ) Input: ( a 0 , r , R ) -centered con vex b ody K ⊆ R n presen ted by membership o racle, lattice L ⊆ R n gi ven b y a basis, objec tiv e v ∈ R n , toleranc e parameters 0 < ǫ ≤ 1 2 and δ > 0 Output: “EMP TY” if K ∩ L = ∅ or z ∈ K + ǫ ( K − K ) ∩ L satisfying sup y ∈ K ∩ L h v , y i ≤ h v , z i + δ 1: z ← ApproxIP ( K, L, ǫ ) 2: if z = “EMP TY” then 3: r eturn “EMPTY” 4: Compute x l , x u ∈ K using the ellipsoid algorithm satis fying inf x ∈ K h v , x i ≥ h v , x l i − δ 12 and sup x ∈ K h v , x i ≤ h v , x u i + δ 12 5: Set l ← h v, z i and u ← h v , x u i + δ 12 6: wh ile u − l > δ do 7: m ← 1 2 ( u + l ) 8: y ← ApproxIP ( K ∩ { x ∈ R n : m ≤ h v, x i ≤ u } , L, ǫ ) 9: if y = “EMPTY ” t h en 10: u ← m 11: y ← ApproxI P ( K ∩ { x ∈ R n : l ≤ h v , x i ≤ m } , L, ǫ ) 12: if y = “EMPTY ” t h en 13: Set u ← l and y ← z 14: if h v , z i < h v , y i then 15: Set z ← y and l ← h v , z i 16: retur n z 7 Corr ectness: Assuming that all t he calls to the Approx IP solv er output a correct re sult (which occurs with ov erwhelming probab ility), we sho w that Algorithm 1 is correct . As can be seen, the algorithm performs a standa rd binary search ov er the objec tiv e va lue. During iterat ion of the while loop, the valu e u repr esents the curren t best upper b ound on sup y ∈ K ∩ L h v , y i , where th is bound is achie ved first by bound ing s u p x ∈ K h v , x i (line 5), or by showing the lattice infeasibi lity of appriopria te res trictions of K (line 10 and 13) . S imilarly , the v alue l represe nts the objecti ve v alue of th e best lattice po int found thus f ar , which is deno ted by z . Now as long as the v alue of z is not null, we claim that z ∈ K + ǫ ( K − K ) . T o see this note tha t z is the output of some call to A pprox IP , on K a,b = K ∩ { x ∈ R n : a ≤ h v , x i ≤ b } for some a < b , the lattice L , with toleran ce parameter ǫ . Hence if z is none null, we are guarantee d that z ∈ (1 + ǫ ) K a,b − ǫ b ( K a,b ) = K a,b + ǫ ( K a,b − b ( K a,b )) ⊆ K a,b + ǫ ( K a,b − K a,b ) ⊆ K + ǫ ( K − K ) (3.1) since b ( K a,b ) ⊆ K a,b ⊆ K . Therefore z ∈ K + ǫ ( K − K ) as requ ired. Now , the alg orithm returns if “EMPTY” if K ∩ L = ∅ (line 3), or z if u − l < δ (line 17 ). Hence the algorit hms outp ut is v alid as requi red. Runtime: Assuming that each call to Approx IP returns a correc t result, we first bound the numbe r of of iterati ons of the whil e loop. After this, usin g a unio n boun d ov er the fa ilure probability of ApproxIP , we get a bound on the probabi lity that the algorith m does not perform as described by the analysis. First, we sho w that gap u − l decreases by a fac tor of at l east 3 4 after eac h iteration of the loo p. N ote th at by constru ction of m , if K m,u is decla red “EMPTY ” in line 8, then clearly u − l decreas es by 1 2 in the next step (since u becomes m ). Next, if a lattice point y is returned in line 8, we know by Equation (3.1) that y ∈ K m,u − ǫ ( K m,u − K m,u ) . Therefo re h v , y i ≥ inf x ∈ K m,u h v , y i − ǫ sup x ∈ K m,u h v , y i − in f x ∈ K m,u h v , y i ! ≥ m − ǫ ( u − m ) (3.2) Since m = 1 2 ( l + u ) , and ǫ ≤ 1 2 , we see that u − h v , y i ≤ ( u − m ) + ǫ ( u − m ) ≤ 1 2 ( u − l ) + 1 4 ( u − l ) = 3 4 ( u − l ) as needed . From here , we claim that we p erform at mos t ⌈ ln( 4 R k v k 2 ) ln( δ ) / ln ( 4 3 ) ⌉ iterations of the for loop. N o w since K ⊆ a 0 + RB n 2 , note that the vari ation of v ov er K (max minus m in) is at most 2 R k v k 2 . Therefore, using Equation (3.2), the initia l val ue of u − l (line 6) is at mos t 2 R k v k 2 + ǫ (2 R k v k 2 ) + δ 12 ≤ 2 R k v k 2 + 1 2 (2 R k v k 2 ) + k v k r 12 ≤ 4 R k v k 2 Since u − l decreas es by a facto r at least 3 4 at each iteratio n, it takes at most ⌈ ln( 4 R k v k 2 δ ) / ln 4 3 ⌉ iterations before u − l ≤ δ . Since we call Approx IP at most twice at each iteration, the probabil ity that any one of these calls fa ils (wheru pon the above ana lysis does not hol d) is at most 2 ⌈ ln ( 4 R k v k 2 δ ) / ln 4 3 ⌉ F , where F is the failur e probab ility of a call to ApproxIP . For the purposes of this algorithm, we claim that the error probab ility of a call to Approx IP ca n be made arbitrari ly small by repetition . T o see this, no te an y lattice vec tor returned by ApproxIP ( K a,b , L, ǫ ) is alway s a succes s for our purposes, since by the algorithm’ s design any retur ned vector is always in K a,b + ǫ ( K a,b − K a,b ) ∩ L (which is suf fi cient for us). Hence the only failure possibility is that Appro xIP returns that K a,b ∩ L = ∅ w hen this is not the case . By the 8 guaran tees on ApproxIP , the probab ility that this occurs ov er k independent repetitio ns is at most 2 − nk . Hence by repeating each call to ApproxIP O (1 + 1 n ln ln R k v k δ ) times, the total er ror probabilit y over all calls can be reduced to 2 − n as requi red. Hence with probabilit y at least 1 − 2 − n , the algori thm correc tly terminate s in at most ⌈ ln( 4 R k v k 2 δ ) / ln 4 3 ⌉ iteratio ns of the while loop. Lastly , w e mus t check that each call t o ApproxIP is done o ver a well c entered K a,b , i.e. we must be able to pro vide to Appr oxIP a cen ter a ′ 0 ∈ R n and radii r ′ , R ′ such that a ′ 0 + r ′ B n 2 ⊆ K a,b ⊆ a ′ 0 + R ′ B n 2 , where each of a ′ 0 , r ′ , R ′ ha ve size poly nomial in the input parameter s. Here we can show that appropria te con ve x combina tions of the poin ts x l , x u (line 4) and a 0 ∈ K allow us to get well-c entered points insid e each K m,u (line 8) and K l,m (line 11). For s implicity in the pres entation, we delay thi s discu ssion until the full versio n of the paper . Giv en the abo ve, since we call ApproxI P at most twice in each iterat ion (o ver a well-cen tered con vex body) , with probabili ty at leas t 1 − 2 − n the total runni ng time is O ( 1 ǫ 2 ) n p olylog( R , r , δ, k v k 2 ) as requir ed. 3.2 Subspace A voiding Problem In the follo wing two section s, C ⊆ R n will denote be a (0 , r , R ) -centered γ -symmetric con vex body , and L ⊆ R n will denote an n -dimension al lattice. In th is section, we intro duce the Subspa ce A void ing Problem of [BN07], and outli ne how the AKS sie ve can b e adapted to solv e it unde r general semi-norms. W e defe r most of the analy sis to the ful l version o n the paper . Let M ⊆ R n be a linear subspa ce where d im( M ) = k ≤ n − 1 . Let λ ( C, L, M ) = inf x ∈ L \ M k x k C . Note that under this definitio n, we hav e the identity λ 1 ( C, L ) = λ ( C, L, { 0 } ) . Definition 3.1. The ( 1 + ǫ ) -Approximat e Subspace A void ing Problem with respect C , L and M is to find a lattice vec tor y ∈ L \ M such that k y k C = (1 + ǫ ) λ ( C, L, M ) . For x ∈ R n , let k x k ∗ C = min {k x k C , k x k − C } . For a point x ∈ R n , define s ( x ) = 1 if k x k C ≤ k x k − C and s ( x ) = − 1 if k x k C > k x k − C . From the not ation, we ha ve that k x k ∗ C = k x k s ( x ) C = k s ( x ) x k C . W e begin with an ex tension of the AKS sie ving lemma to the a symmetric setting. The foll owing lemma will pro vide the centr al tool for the SAP algorit hm. Lemma 3.2 (Basic Sie ve) . Let ( x 1 , y 1 ) , ( x 2 , y 2 ) , . . . , ( x N , y N ) ∈ R n × R n denote a list of pair s satisfy ing y i − x i ∈ L , k x i k ∗ C ≤ β and k y i k ∗ C ≤ D ∀ i ∈ [ N ] . Then a clustering , c : { 1 , . . . , N } → J , J ⊆ [ N ] , satisfy ing: 1 . | J | ≤ 2  5 γ  n 2 . k y i − y c ( i ) + x c ( i ) k ∗ C ≤ 1 2 D + β 3 . y i − y c ( i ) + x c ( i ) − x i ∈ L for all i ∈ [ N ] \ J , can be computed in deterministi c O ( N  5 γ  n ) -time . Pr oof. Algorithm: W e bu ild the set J and cluste ring c itera tiv ely , starting from J = ∅ , in the follo w ing manner . For eac h i ∈ [ N ] , check if there exists j ∈ J such that k y i − y j k s ( x j ) C ≤ D 2 . If such a j exist s, set c ( i ) = j . Otherwise, appen d i to the set J and set c ( i ) = i . R epeat. 9 Analysis: W e first note, t hat for any i, j ∈ [ N ] , we ha ve th at y i − y j + x j − x i = ( y i − x i ) − ( y j − x j ) ∈ L since by assumpt ion both y i − x i , y j − x j ∈ L . Hence, property (3) is trivia lly satisfied by the cluste ring c . W e no w check that the clustering satisfies propert y (2). For i ∈ [ N ] \ J , note that by constructio n we ha ve that k y i − y c ( i ) k sC ≤ D 2 where s = s ( x c ( i ) ) . Therefo re by the triangle inequa lity , we hav e that k y i − y c ( i ) + x c ( i ) k ∗ C ≤ k y i − y c ( i ) + x c ( i ) k sC ≤ k y i − y c ( i ) k sC + k x c ( i ) k sC = k y i − y c ( i ) k sC + k x c ( i ) k ∗ C ≤ D 2 + β as requi red. W e no w sho w that J satisfies prop erty (1) . By con struction of J , we kno w that for i, j ∈ J , i < j that k y j − y i k s ( x i ) C > D 2 . Therefore we ha ve that k y j − y i k s ( x i ) C > D 2 ⇒ k y j − y i k C ∩− C = k y i − y j k C ∩− C > D 2 ( by symmetry of C ∩ − C ) From here, we claim that y i + D 4 ( C ∩ − C ) ∩ y j + D 4 ( C ∩ − C ) = ∅ . (3.3) Assume not, then we may pick z in the intersect ion above . Then by d efinition, we ha ve that k y j − y i k C ∩− C = k ( y j − z ) + ( z − y i ) k C ∩− C ≤ k y j − z k C ∩− C + k z − y i k C ∩− C = k z − y j k C ∩− C + k z − y i k C ∩− C ≤ D 4 + D 4 = D 2 a clear contr adiction. For eac h i ∈ [ N ] , we hav e by assumpt ion that k y i k ∗ C ≤ D ⇔ y i ∈ D ( C ∪ − C ) . Therefor e, we see that y i + D 4 ( C ∩ − C ) ⊆ D ( C ∪ − C ) + D 4 ( C ∩ − C ) = D (( C + 1 4 ( C ∩ − C )) ∪ ( − C + 1 4 ( C ∩ − C ))) ⊆ D (( C + 1 4 C ) ∪ ( − C + 1 4 ( − C ))) = 5 4 D ( C ∪ − C ) (3.4) From (3.3), (3.4), and since J ⊆ [ N ] , we hav e that | J | = v ol( { y i : i ∈ J } + D 4 ( C ∩ − C )) v ol( D 4 ( C ∩ − C )) ≤ v ol ( 5 4 D ( C ∪ − C )) v ol( D 4 ( C ∩ − C )) ≤  5 4  n (v ol( D C ) + vo l ( − D C ))  γ 4  n v ol( D C ) = 2  5 γ  n as need ed. T o boun d the runni ng time of the cluster ing algorithm is straightfo rward. For each element of [ N ] , we iterate o nce throu gh the parti ally construc ted set J . Since | J | ≤ 2  5 γ  n throug hout the entire algorit hm, w e ha ve that the entire runtime is bounded by O ( N  5 γ  n ) as required . 10 Definition 3.3 (Sie ving P rocedu re) . For a list of pairs ( x 1 , y 1 ) , . . . , ( x N , y N ) as in Lemma 3.2, we call an applic ation of the Sie ving Pr ocedur e the process of computin g the cluste ring c : [ N ] → J , and outputt ing the list of pairs ( x i , y i − y c ( i ) + x c ( i ) ) for all i ∈ [ N ] \ J . Note that the Sieving Pr ocedu re deletes the set of pairs associated with the cluster cente rs J , and com- bines the remaini ng pairs with their associate d centers. W e remark some diff erences with the stan dard AKS sie ve. Here the Sie ving Procedure does not guar an- tee that k y i k C decrea ses after each iterat ion. In stead it shows that at least one of k y i k C or k − y i k C decrea ses approp riately at e ach step. Hence the re gion we must contr ol is in fact D ( C ∪ − C ) , which we note is g ener- ally non-c on vex . Additional ly , our ana lysis sho ws that ho w well we can use k · k C to sie ve only depen ds on v ol( C ∩ − C ) / vol( C ) , which is a ve ry flex ible globa l quantity . For e xample, if C = [ − 1 , 1] n − 1 × [ − 1 , 2 n ] (i.e. a cube with one highly sk ewed coor dinate) then C is still 1 2 -symmetric , and henc e the sie ve bar ely notice s the asymmetry . The algorithm for approxi m ate SA P we describe present ly will constru ct a list of lar ge pairs as abov e, and use repeate d applicatio ns of the Sie ving P r ocedur e to creat e shorter and shorter vect ors. The nex t lemma allo ws us to get a crude estimate on the value of λ ( C, L, M ) . Lemma 3.4. Let C ⊆ R n a (0 , r, R ) -center ed con vex body , L ⊆ R n be an n -dimensional lattice, and M ⊆ R n , dim( M ) ≤ n − 1 , be a lin ear subsp ace. T hen a number ν > 0 satisfying ν ≤ λ ( C , L, M ) ≤ 2 n R r ν can be computed in polynomial time. The abo ve l emm a foll ows direc tly from Lemma 4.1 of [BN09 ]. They prov e it for ℓ p balls, b ut it is eas ily adapte d to the abov e setti ng using the relation ship 1 r k x k 2 ≤ k x k C ≤ 1 R k x k 2 (since C is (0 , r, R ) -cente red). The follo wing technical lemma will be needed in the a nalysis of the SAP algori thm. Lemma 3.5. T ake v ∈ R n wher e β ≤ k v k C ≤ 3 2 β . D efine C + v = β C ∩ ( v − β C ) and C − v = ( β C − v ) ∩ − β C . Then v ol( C + v ) v ol( β C ) = v ol( C − v ) v ol ( β C ) ≥  γ 4  n Furthermor e, in t( C + v ) ∩ in t( C − v ) = ∅ . The follo wing is the co re subrou tine for the SAP solv er . W e relat e some important details about th e the SA P algorith m. Our algorithm for SAP follo w s a standa rd proced ure. W e fi rst guess a v alue β satisf ying β ≤ λ ( C, L, M ) ≤ 3 2 β , and then run ShortV ector s on inputs C, L, M , β and ǫ . W e sho w tha t for this value of β , ShortV ectors outpu ts a (1 + ǫ ) ap proximate so lution with ov erwhelming probabil ity . As we can be seen above , the main task of the ShortV ectors algorithm, is to generate a larg e qua ntity of random vector s, and sie ve them until they are all of relati vely small size (i.e. 3 β ≤ 3 λ ( C, L, M ) ). ShortV ectors the n examin es all the dif ferences betwee n the sie ved v ectors in the ho pes of finding o ne of s ize (1 + ǫ ) λ ( C , L, M ) in L \ M . ShortV ectors, in fact, needs to balan ce certain tradeof fs. O n the one hand, it must sie ve enou gh times to guaran tee that the ve ctor differ ences ha ve small size. O n the other , it must use “lar ge” perturbation s sampled from β ( C ∪ − C ) , to guara ntee that these dif ferences do no t all lie in M . W e note that the main algori thmic diffe rences with res pect to [BN07, AJ08] is the use of a modified sie ving procedure as well as a diffe rent samplin g distrib ution for the pertur bation vecto rs (i.e. ov er β ( C ∪ − C ) instead o f jus t β C ). These d ifferen ces also make the a lgorithm’ s analysis more techn ically challen ging. 11 Algorithm 2 ShortV ectors( C ,L,M, β , ǫ ) Input: (0 , r , R ) -centered γ -symmetric con vex body C ⊆ R n , basis B ∈ Q n × n for L , linear subspa ce M ⊆ R n , scaling parameter β > 0 , tolerance parameter 0 < ǫ ≤ 1 2 1: D ← n max 1 ≤ i ≤ n k B i k C 2: N 0 ← 4 ⌈ 6 ln  D β  ⌉  20 γ 2  n + 8  36 γ 2 ǫ  n , η ← 2 − ( n +1) N 0 3: Create pairs ( x 0 1 , y 0 1 ) , ( x 0 2 , y 0 2 ) , . . . , ( x 0 N 0 , y 0 N 0 ) as follo ws: fo r each i ∈ [ N 0 ] , compute X an η -uniform sample o ver β C (using T heorem 2.2) and a uni form s in {− 1 , 1 } , and set x 0 i ← sX and y 0 i ← x 0 i mo d B . 4: t ← 0 5: wh ile D ≥ 3 β d o 6: Apply Sie ving Pr ocedu re to ( x t 1 , y t 1 ) , . . . , ( x t N t , y t N t ) yieldin g ( x t +1 1 , y t +1 1 ) , . . . , ( x t +1 N t +1 , y t +1 N t +1 ) 7: D ← D 2 + β and t ← t + 1 8: retur n { y t i − x t i − ( y t j − x t j ) : i, j ∈ [ N t ] } \ M Theor em 3.6 (Approximate-SAP) . F or 0 < ǫ ≤ 1 2 lattice vector y ∈ L \ M such that k y k C ≤ (1 + ǫ ) λ ( C, L, M ) can be computed in time O ( 1 γ 4 ǫ 2 ) n with pr obabili ty at least 1 − 2 − n . Furthermor e, if λ ( C, L, M ) ≤ tλ 1 ( C, L ) , t ≥ 2 , a vector y ∈ L \ M satisfyin g k y k C = λ ( C, L, M ) , can be with computed in time O  1 γ 4 t 2  n with pr obability at least 1 − 2 − n . Pr oof. Algorithm: The algor ithm for (1 + ǫ ) -SAP is as follo ws: 1. Using Lemma 3.4 compute a val ue ν satis fying ν ≤ λ ( C , L, M ) ≤ 2 n R r ν . 2. For each i ∈ 0 , 1 , . . . , ⌈ ln(2 n R/r ) / ln(3 / 2) ⌉ , let β = (3 / 2) i ν and run ShortV ectors ( C , L, β , ǫ ) . 3. Return the shortes t vector found with respect to k · k C in the abo ve runs of ShortV ectors. Pre liminary Analysis: In w ords, th e algor ithm first gues ses a good approx imation of λ ( C, L , M ) (among polyn omially many ch oices) and runs the Short V ectors algorithm on thes e gues ses. By des ign, there will be one iteration of the algor ithm where β sa tisfies β ≤ λ ( C , L, M ) ≤ 3 2 β . W e prov e that for this setting of β the algorit hm returns a (1 + ǫ ) -approxi mate solu tion to the SAP probl em with probability at least 1 − 2 − n . T ake v ∈ L \ M denote an optimal solution to the SAP proble m, i.e. v satisfies k v k C = λ ( C, L, M ) . W e will show that w ith prob ability at least 1 − 2 − n , a small pertubation o f v (Claim 4) w ill be in the set return ed by ShortV ectors when ru n on inputs C , L , M , β , and ǫ . W ithin the ShortV ectors algori thm, we w ill assume that the samples generat ed ov er β C (line 3) are exa ctly uniform. B y doing this, we claim the probability that ShortV ectors returns a (1 + ǫ ) -appro ximate soluti on to the SAP problem by changes by at most 2 − ( n +1) . T o see this, note that we gene rate exactly N 0 such samples, all of which a re η -uniform. Therefor e by Lemma 2.1, we hav e that th e total v ariatio n distance between th e vector of appro ximately unifo rm sampl es and truly unif orm samples is at most N 0 η = 2 − ( n +1) . Lastly , t he e vent that S hortV ectors returns (1+ ǫ ) -approxi m ate so lution is a random function of these samples, and hence when switchi ng uniform samples for η -unifor m ones, the probabilit y of this eve nt change s by at most 2 − ( n +1) . Therefo re to prove the theorem, it suffice s to show that the fa ilure proba bility under truly unifor m samples is at most 2 − ( n +1) . 12 In th e proof, we adopt all the names of para meters and va riables defined in the ex ecution of ShortV ector . W e denote the pairs at stage t as ( x t 1 , y t 1 ) , . . . , ( x t N t , y t N t ) . W e also let C + v , C − v be as in L emma 3.2. For any stag e t ≥ 0 , we define the pai r ( x t i , y t i ) , i ∈ [ N t ] , as good if x t i ∈ in t( C + v ) ∪ int ( C − v ) . Claim 1: Let G denote the eve nt th at the re are at least 1 2  γ 4  n N 0 good pa irs at stage 0 . Then G occurs with proba bility least 1 − e − 1 48 γ n N 0 . Let G i = I [ x 0 i ∈ in t ( C + v ) ∪ in t( C − v )] for i ∈ [ N 0 ] d enote the indicator random variab les denoting whether ( x 0 i , y 0 i ) is good or not. Let s i , i ∈ [ N 0 ] , denote the {− 1 , 1 } random variab le indicating whether x 0 i is sampled unifor mly from β C or − β C . Since β ≤ k v k C ≤ 3 2 β , by lemma 3.5 we ha ve that Pr[ G i = 1] ≥ Pr ( x 0 i ∈ in t( C + v ) | s i = 1) Pr( s i = 1) + Pr( x 0 i ∈ in t( C − v ) | s i = − 1) Pr ( s i = − 1) = 1 2 v ol( β C ∩ ( v − β C )) v ol( β C ) + 1 2 v ol (( β C − v ) ∩ ( − β C )) v ol( − β C ) ≥  γ 4  n From the abov e we see that E [ P N i =1 G i ] ≥  γ 4  n N 0 . Since the G i ’ s are iid Bernoul lis, by the Chernof f bound we get that Pr[ G ] = Pr[ P N i =1 G i < 1 2  γ 4  n N 0 ] ≤ e − 1 48 γ n N 0 , as needed . Claim 2: L et T deno te the last s tage of the sie ve (i.e. va lue o f t at end of the whi le lo op). Then cond itioned on G , the number of good pairs at stage T is at least N G = 4  9 γ ǫ  n . Examine ( x 0 i , y 0 i ) for i ∈ [ N 0 ] . W e first claim that k y 0 i k ∗ C ≤ D . T o see this note that y 0 i = B z where z = B − 1 x 0 i − ⌊ B − 1 x 0 i ⌋ ∈ [0 , 1) n . Hence k y 0 i k ∗ C ≤ k y 0 i k C = k n X i =1 B i z i k C ≤ n X i =1 z i k B i k C ≤ n max 1 ≤ i ≤ n k B i k C = D as need ed. L et D t = max {k y t i k ∗ C : i ∈ [ N t ] } , where we note that abo ve shows that D 0 ≤ D . By Lemma 3.2, we kno w that N t ≥ N t − 1 − 2  5 γ  n and that D t ≤ 1 2 D t − 1 + β for t ≥ 1 . For D t ≥ 3 β , we see that 1 2 D t + β ≤ 5 6 D t . Gi ven the pre vious bounds, an easy computation rev eals that T ≤ ⌈ ln( D 3 β ) ln( 6 5 ) ⌉ ≤ ⌈ 6 ln( D β ) ⌉ . From the abo ve, we see that during the entire sievin g phase we remov e at most T (2)  5 γ  n ≤ 2 ⌈ 6 ln ( D β ) ⌉  5 γ  n pairs. Since we ne ver modify the x t i ’ s d uring the sie ving oper ation, any pair that starts o ff as go od stays good as long a s it survi ves through the last st age. Since we start with at least 1 2  γ 4  n N 0 good pairs in stage 0 , we are left with at least 1 2  γ 4  n N 0 − 2 ⌈ 6 ln  D β  ⌉  5 γ  n ≥ 4  9 γ ǫ  n = N G good pair s at stage T as required . Modifying the outp ut: Here we will analyz e a way of modi fying the outp ut the ShortV ectors, which will maintain the output distrib ution b ut make the output analysis far simpler . 13 Let w ( x ) = I [ x ∈ β C ] + I [ x ∈ − β C ] . L etting X be uniform in β C and s be uniform in {− 1 , 1 } (i.e. the distrib ution in lin e 3), for x ∈ β ( C ∪ − C ) w e ha ve that d Pr[ sX = x ] = d Pr[ X = x ] Pr [ s = 1] + d Pr[ X = − x ] Pr[ s = − 1] = 1 2  I [ x ∈ C ] v ol( β C ) + I [ x ∈ − C ] v ol( β C )  = w ( x ) 2 vol( β ( C )) (3.5) Examine the functio n f v : β ( C ∪ − C ) → β ( C ∪ − C ) defined by f v ( x ) =      x − v : x ∈ in t( C + v ) x + v : x ∈ in t( C − v ) x : otherwis e Since in t ( C + v ) ∩ int( C − v ) = ∅ , it is easy to see that f v is a well-defined bijectio n on β ( C ∪ − C ) satisfy ing f v ( f v ( x )) = x . Furthemore by con struction, we see that f v (in t( C + v )) = int( C − v ) and f v (in t( C − v )) = in t ( C + v ) . Lastly , note that for any x ∈ β ( C ∪ − C ) , th at f v ( x ) ≡ x mo d B since f v ( x ) is jus t a lattice vec tor shi ft of x . Let F v denote the random funct ion where F v ( x ) = ( x : with probabili ty w ( x ) w ( x )+ w ( f v ( x )) f v ( x ) : with prob ability w ( f v ( x )) w ( x )+ w ( f v ( x )) Here, we inte nd that dif ferent applicati ons of the function F v all occur w ith indepe ndent randomness. Next we define the funct ion c v as c v ( x, y ) = ( f v ( x ) : k y − f v ( x ) k ∗ C < k y − x k ∗ C x : otherwise For an y stage t ≥ 0 , define ¯ x t i = c v ( x t i , y t i ) . Claim 3: For any stage t ≥ 0 , the pairs ( x t 1 , y t 1 ) , . . . , ( x t N t , y t N t ) , and ( F v ( ¯ x t 1 ) , y t 1 ) , . . . , ( F v ( ¯ x t N t ) , y t N t ) are identic ally distrib uted. Furth ermore, this remains true after cond itioning on the ev ent G . T o prov e th e claim, it suf fi ces to sho w that the pairs in (1) ( x t 1 , y t 1 ) , . . . , ( x t N t , y t N t ) , and (2) ( F v ( ¯ x t 1 ) , y t 1 ) , ( x t 2 , y t 2 ) , . . . , ( x t N t , y t N t ) are identi cally distrib uted (both before and after conditioni ng on G ). Our analys is will be indepen dent of the ind ex anal yzed, hence the claim will follow by app lying the proof inducti vel y on each remaining pair in the second list. The pairs in (1) cor respond exactly to the induced distrib ution of the algorith m on the stage t variab les. W e think o f the pai rs in (2) as the in duced distrib ution of a modi fi ed algor ithm on these v ariabl es, where the modified algorithm just runs the normal algorith m and replaces ( x t 1 , y t 1 ) by ( F v ( c v ( x t 1 , y t 1 )) , y t 1 ) in stage t . T o sho w the distrib utional equiv alenc e, we sho w a probability preser ving correspond ance between runs of the normal and modified algori thm ha ving the same stage t v ariables. For 0 ≤ k ≤ t , let the pairs ( x k i , y k i ) , i ∈ [ N k ] , deno te a v alid run of the normal algorithm through stage t . W e label this as run A . Let us denote the sequen ce of ancestors of ( x t 1 , y t 1 ) in the normal algorithm by ( x k a k , y k a k ) for 0 ≤ k ≤ t − 1 . By definition of this sequenc e, we ha ve that x 0 a 0 = x 1 a 1 = · · · = x t 1 . Since 14 the ShortV ectors algorithm is deterministic giv en the initial samples, the probability density of this ru n is simply d Pr  ∩ i ∈ [ N 0 ] { x 0 i = x 0 i }  = d Pr  x 0 a 0 = x t 1  Y i ∈ [ N 0 ] ,i 6 = a 0 d Pr  x 0 i = x 0 i  (3.6) by the indepe ndence of the sample s and since x 0 a 0 = x t 1 . Notice if we condition on the e vent G , assuming the run A belon gs to G (i.e. that there ar e enough good pairs at stag e 0 ), the abov e pro bability density is simply di vided by Pr [ G ] . If x t 1 / ∈ in t( C + v ) ∪ in t ( C − v ) , note that F v ( c v ( x t 1 , y t 1 )) = F v ( x t 1 ) = x t 1 , i.e. the action of F v and c v are tri vial. In this case, we associa te run A with the identic al run for the m odified algorithm, which is clearly v alid and has t he same pro bability . Now assume that x t 1 ∈ C + v ∪ C − v . In this case, we asso ciate run A to tw o runs of the modified algorithm: ˜ A . identica l to run A , C . run A with ( x k a k , y k a k ) rep laced by ( f v ( x k a k ) , y k a k ) for 0 ≤ k ≤ t − 1 . Note th at both of the associat ed runs ha ve the same stage t v ariables as run A by constr uction. W e must check that both runs are indee d val id for the modified algorithm. T o see this, note that up till stage t , the modified algorithm jus t runs the normal algorithm. Run ˜ A inherents v alidity for these stag es from the fact that run A is v alid for the normal alg orithm. T o see that run C is v alid, w e first no te that f v ( x 0 a 0 ) ≡ x 0 a 0 ≡ y 0 a 0 mo d B ( B is the lattice basis) , which giv es v alidity for stage 0 . By design of the normal sie ving algorithm, note that during run A , th e algorithm ne ver inspects the conte nts of x k a k for 0 ≤ k < t . Therefor e, if ( x k a k , y k a k ) denotes a va lid ancest or sequ ence in run A , then so doe s ( f v ( x a k ) , y k a k ) in run B for 0 ≤ k < t . For sta ge t , not e that the normal algorith m , gi ven the sta ge 0 inputs of run ˜ A would outpu t ( x t 1 , y t 1 ) , . . . , ( x N t , y N t ) for the stage t vari ables, and that gi ven the stage 0 inputs of run C would outpu t ( f v ( x t 1 ) , y t 1 ) , ( x t 2 , y t 2 ) , . . . , ( x t N t , y t N t ) . Hence in run ˜ A , the modified algorithm retains the normal algori thms output, and in run C , it swiches it from f v ( x t 1 ) back to x t 1 . Therefo re, both runs are indeed v alid for the modi fi ed algorithm. Furt hermore, note th at if run A is in G , then both the stage 0 v ariable s of ˜ A and C inde x a good run for t he norma l algori thm since the pair ( x 0 a 0 , y a 0 ) is good if f ( f v ( x 0 a 0 ) , y a 0 ) is good . Hence we see that correspon dance des cribed is v alid both before and after conditi oning on G . Lastly , it is clear th at for an y run o f the modified algo rithm, the abo ve cor respondanc e yields a u nique run of the normal algori thm. It remains to sho w that the corres pondance is probabi lity preservin g. W e must therefore compute th e probab ility density assoc iated with the u nion of run ˜ A and C for the mod ifi ed algorith m. Using the a nalysis from the pre vious paragraph and the computation in (3.6), we see tha t this proba bility density is  d Pr[ x 0 a 0 = x t 1 ] Pr[ F v ( c v ( x t 1 , y t 1 )) = x t 1 ] + d Pr[ x 0 a 0 = f v ( x t 1 )] P r[ F v ( c v ( f v ( x t 1 ) , y t 1 )) = x t 1 ]  Y i ∈ [ N 0 ] ,i 6 = a 0 d Pr  x 0 i = x 0 i  (3.7) On the first line abo ve, the first term cor responds to run ˜ A which samples x t 1 in stage 0 and then choo ses to kee p ( x t 1 , y t 1 ) in stag e t , and the second term corresponds to C which samples f v ( x t 1 ) in stag e 0 and ch ooses to flip ( f v ( x t 1 ) , y t 1 ) to ( x t 1 , y t 1 ) in stage t . Now by definition of F v and c v , we ha ve that Pr[ F v ( c v ( f v ( x t 1 ) , y t 1 )) = x t 1 ] = Pr[ F v ( c v ( x t 1 , y t 1 )) = x t 1 ] = w ( x t 1 ) w ( x t 1 ) + w ( f v ( x t 1 )) 15 Therefore , using the abov e and Equatio n (3.5), we ha ve that the first line of (3.7) is equal to  d Pr[ x 0 a 0 = x t 1 ] + d Pr[ x 0 a 0 = f v ( x t 1 )]  w ( x t 1 ) w ( x t 1 ) + w ( f v ( x t 1 )) =  w ( x t 1 ) 2 vol( β C ) + w ( f v ( x t 1 )) 2 vol( β C )  w ( x t 1 ) w ( x t 1 ) + w ( f v ( x t 1 )) = w ( x t 1 ) 2 vol( β C ) = d Pr[ x 0 a 0 = x t 1 ] Hence the probab ilities in Equa tions (3.6) an d (3.7) are equal as needed . Lastl y , note that when con ditioning on G , both of the correspon ding prob abilities are di vided by Pr[ G ] , and henc e equal ity is maintained . Output Analysis: Claim 4: Let T denote the last stage of the sie ve. Then conditione d on the eve nt G , with probabilit y at least 1 −  2 3  1 2 N G there exists a lattice vector w ∈ { y T i − x T i − ( y T j − x T j ) : i, j ∈ [ N t ] } satisfying ( † ) w ∈ L \ M , w − v ∈ M ∩ L and k w − v k C < ǫβ . Furthermor e, any latt ice vec tor satisfying ( † ) is a (1 + ǫ ) -appr oximate solution to SAP . Let ( x T 1 , y T 1 ) , . . . , ( x T N t , y T N T ) den ote any va lid instantia tion of the stag e T varia bles correspond ing to a good run of th e alg orithm (i.e. one belonging to G ). L et ( ¯ x T i , y T i ) = ( c v ( x T i , y T i ) , y T i ) for i ∈ [ N T ] . By claim 3, it suf fi ces to prov e the claim for the pairs ( F v ( ¯ x T 1 ) , y T 1 ) , . . . , ( F v ( ¯ x T N T ) , y T N T ) . This follo ws since the probab ility of “success ” (i.e. the existe nce of the desired vect or) conditioned on G , is simply an av erage ov er all instantiat ions abo ve of the conditio nal probability of “success”. Since our instan tiation correspo nds to a goo d run , by Claim 2 we hav e at least N G good pairs in stag e T . Since c v preser ves good pairs, the same holds true for ( ¯ x T i , y T i ) , i ∈ [ N t ] . For notational con venie nce, let us as sume that the pairs ( ¯ x T i , y T i ) , i ∈ [ N G ] are all good . W e no te then that f v ( ¯ x T i ) = ¯ x T i ± v and f v ( f v ( ¯ x T i )) = ¯ x T i for i ∈ [ N G ] . First, since T is last the stage, we know that k y T i k ∗ C ≤ 3 β for i ∈ [ N G ] . Next, f or i ∈ [ N G ] , b y defini tion of c v we ha ve that k y T i − ¯ x T i k ∗ C = min {k y T i − x T i k ∗ C , k y T i − f v ( x T i ) k ∗ C } Let s = s ( y T i ) , i.e. k y T i k ∗ C = k y T i k sC . Since ( x T i , y T i ) is goo d at leas t one of − x T i , − f v ( x T i ) ∈ β sC . W ithout loss of gener ality , we as sume − x T i ∈ β sC . T herefor e, we get that k y T i − ¯ x T i k ∗ C ≤ k y T i − x T i k ∗ C ≤ k y T i − x T i k sC ≤ k y T i k sC + k − x T i k sC ≤ 3 β + β = 4 β (3.8) Let S denote the set { y T i − ¯ x T i : i ∈ [ N G ] } . Since ¯ x T i ≡ y T i mo d B , we note that S ⊆ L . Also by Equation (3.8) w e hav e that S ⊆ 4 β ( C ∪ − C ) ∩ L . L et Λ ⊆ S denote a maximal subset such that x + int( ǫ 2 β ( C ∩ − C )) ∩ y + int( ǫ 2 β ( C ∩ − C )) = ∅ for distinct x, y ∈ Λ . Since S ⊆ 4 β ( C ∪ − C ) , we se e that for x ∈ S x + ǫ 2 β ( C ∩ − C ) ⊆ 4 β ( C ∪ − C ) + ǫ 2 β ( C ∩ − C ) ⊆  4 + ǫ 2  β ( C ∪ − C ) Therefore we see that | Λ | ≤ v ol((4 + ǫ 2 ) β ( C ∪ − C )) v ol( ǫ 2 β ( C ∩ − C )) =  8 + ǫ ǫ  n v ol ( C ∪ − C ) v ol ( C ∩ − C ) ≤  8 + ǫ ǫ  n 2 vol( C ) γ n v ol( C ) ≤ 2  9 γ ǫ  n ≤ 1 2 N G 16 Since Λ is maximal, we note that for any x ∈ S , there e xists y ∈ Λ such that in t ( x + ǫ 2 β ( C ∩ − C )) ∩ int( y + ǫ 2 β ( C ∩ − C )) 6 = ∅ ⇔ k x − y k C ∩− C < ǫβ (3.9) Let c 1 , . . . , c | Λ | ∈ [ N G ] , denote indices such that Λ = { y T c i − ¯ x T c i : 1 ≤ i ≤ | Λ |} , and let C = { c j : 1 ≤ j ≤ | Λ |} . For j ∈ { 1 , . . . , | Λ |} , recursi vely define the sets I j = { i ∈ [ N G ] : k ( y T i − ¯ x T i ) − ( y T c j − ¯ x T c j ) k C ∩− C < ǫβ } \  C ∪ ( ∪ j − 1 k =1 I k )  (3.10) Giv en Equation 3.9, we hav e by constr uction that the sets C , I 1 , . . . , I | Λ | partiti on [ N G ] . For each j ∈ { 1 , . . . , | Λ |} , we examine the dif ference s S j = ±{ ( y T i − F v ( ¯ x T i )) − ( y T c j − F v ( ¯ x T c j )) : i ∈ I j } W e will show that S j fail s to contain a ve ctor satisf ying ( † ) with probabili ty at m ost  2 3  | I j | . First we note that S j ⊆ L since y T i ≡ ¯ x T i ≡ F v ( ¯ x T i ) mod B for i ∈ [ N G ] . W e first condit ion on the v alue of F v ( ¯ x T c j ) which is either ¯ x T c j , ¯ x T c j − v or ¯ x T c j + v . W e examine the case where F v ( ¯ x T c j ) = ¯ x T c j , the analysi s for the other two cas es is simila r . N o w , for i ∈ I j , we analyze the dif ference y T i − F v ( ¯ x T i ) − ( y T c j − ¯ x T c j ) (3.11) Let δ i = ( y T i − ¯ x T i ) − ( y T c j − ¯ x T c j ) . Dependin g on the output of F v ( ¯ x T i ) , note that the vector (3.11) is either (a) δ i or of the form (b) ± v + δ i (since f v ( ¯ x T i ) = ¯ x T i ± v ) . W e claim that a vector of form (b) satisfies ( † ) . T o see this, not e that after possib ly negatin g the v ector , it can be broug ht to the form v ± δ i ∈ L , where we ha ve that k ± δ i k C < k ± δ i k C ∩− C = k δ i k C ∩− C < ǫβ ≤ ǫλ ( C, L, M ) < λ ( C, L , M ) , (3.12) since i ∈ I j and ǫ ≤ 1 2 . Since δ i ∈ L and k δ i k C < λ ( C, L, M ) , we must hav e that ± δ i ∈ M ∩ L . Next, since v ∈ L \ M and ± δ i ∈ M , we hav e that v ± δ i ∈ L \ M . Lastly , note that k v ± δ i k C ≤ k v k C + k ± δ i k C < λ ( C, L, M ) + ǫβ ≤ (1 + ǫ ) λ ( C, L, M ) as requi red. No w the probability the v ector in (3.11) is of form (b) is Pr[ F v ( ¯ x T i ) = f v ( ¯ x T i )] = w ( f v ( ¯ x T i )) w ( ¯ x T i ) + w ( f v ( ¯ x T i )) ≥ 1 3 since for any x ∈ β ( C ∪ − C ) we ha ve that 1 ≤ w ( x ) ≤ 2 . S ince each i ∈ I j inde xes a vecto r in S j not satisfy ing ( † ) with probab ility at most 1 − 1 3 = 2 3 , the pro bability that S j contai ns no v ector satisfyin g ( † ) is at most  2 3  | I r | (by indepe ndence) as needed. Let F j , j ∈ { 1 , . . . , | Λ |} , denote the eve nt that S j does not contain a v ector sati sfying ( † ) . Note that F j only depend s on the pairs ( F v ( ¯ x T c j ) , y T c j ) and ( F v ( ¯ x T i ) , y T i ) for i ∈ I j . Since the sets I 1 , . . . , I | Λ | , C partition [ N G ] , these depende ncies are all disjoint, and hence the eve nts are independen t. Therefore the probability that none of S 1 , . . . , S | Λ | contai n a vecto r satisfyi ng ( † ) is at m ost Pr[ ∩ | Λ | j =1 F j ] ≤ | Λ | Y j =1  2 3  | I j | =  2 3  N G −| Λ | ≤  2 3  1 2 N G as need ed. 17 Runtime and Fa ilure Probability: W e first analyze the runtime. First, we m ake O ( n log R r ) guesses for the v alue of Λ( C, L, M ) . W e run the ShortV ectors algorithm once for each such guess β . During one iterati on of the siev ing algorithm, we first generate N 0 η -uniform samples from β C (line 3). By Theor em 2.2, this ta kes p oly( n, ln 1 η , ln β , ln R, ln r ) time per sample. W e al so mod each sample by the basis B for L , which take s p oly( | B | ) time ( | B | is the bit size of the basis). Ne xt, by the analysis of Claim 2 , w e apply the sie ving procedure at most ⌈ 6 ln D β ⌉ times (runs of while loop at line 5), where each iteration of the siev ing proced ure (line 6) takes at most O ( N 0  5 γ  n ) time by Lemma 3.2. L astly , we return the set of diffe rences (line 8) , which takes a t most O ( N 2 0 ) time. Now by s tandard argu ments, one h as that the v alues D and β (for each gues s) each ha ve size (bit desc ription lengt h) polynomia l in the input, i.e. polyno m ial in | B | (bit size of the basis of L ), n , ln R , ln r . Since N 0 = O (ln( D β )( 36 γ 2 ǫ ) n ) , we ha ve that the total running time is p oly( n , ln R, ln r, | B | , 1 / ǫ ) O  1 γ 4 ǫ 2  n as need ed. W e now analyze the succ ess probabi lity . Here we only exa m ine the guess β , where β ≤ λ ( C , L, M ) ≤ 3 2 β . Assuming perfe ctly uniform samples ov er β C , by the analysi s of Claim 4 , we ha ve that con ditioned on G , we fail to output a (1 + ǫ ) appro ximate solution to SAP with prob ability at m ost  2 3  1 2 N G . Hence , under the uniform samplin g assumption, the total probabi lity of f ailure is at mos t  2 3  1 2 N G + Pr[ G c ] ≤  2 3  1 2 N G + e − 1 48 γ n N 0 ≪ 2 − ( n +1) by Claim 2 . When switching to η -unifo rm samples, as ar gued in the preliminary analysis , this failure probab ility incre ases by at most the total v ariation distanc e, i.e. by at most η N 0 = 2 − ( n +1) . Therefor e, the algori thm succeed s with probabilit y at least 1 − 2 − ( n +1) − 2 − ( n +1) = 1 − 2 − n as needed. Exact SAP: Here w e are giv en the guarantee that λ ( C , L, M ) ≤ tλ 1 ( C, L ) , and we wish to use our SAP solv er to get an exa ct minimizer to the SAP . T o solv e this, we run the approx imate S AP so lver on C, L, M with parameter ǫ = 1 t , which tak es O ( t 2 /γ 4 ) n time. Let v ∈ L \ M be a lattice vec tor satisfying k v k C = λ ( C , L, M ) . By Claim 4 , w ith prob ability at least 1 − 2 − n we are guarant eed to output a lattice vec tor w ∈ L \ M , such that k w − v k C < ǫλ ( C, L, M ) ≤  1 t  tλ 1 ( C, L ) = λ 1 ( C, L ) Ho w e ver , since w − v ∈ L and k w − v k C < λ 1 ( C, L ) , we must ha ve that w − v = 0 . Therefore w = v and our SAP solv er return s an exact minimize r as needed . 3.3 Closest V ector Pr oblem In this secti on, we present a reduction from Approximate-CVP to Approximate-SAP for genera l semi-no rms. In [BN07], it is sho wn that ℓ p CVP redu ces to ℓ p SAP in one highe r dimension . B y relaxi ng the condition that the lifted SA P probl em remain in ℓ p , we giv e a very simple reduction which reduces CVP in any semi- norm to SAP in one high er dimensio n under a differ ent semi-nor m that is essentially as symmetric. Giv en the genera lity of our SAP solver , such a redu ction is suf fices. 18 Theor em 3.7 (Approximat e-CVP ) . T ake x ∈ R n . Then for any ǫ ∈ (0 , 1 3 ) , y ∈ L satisfying k y − x k C ≤ (1 + ǫ ) d C ( L, x ) can be computed in time O ( 1 γ 4 ǫ 2 ) n with pr obabil ity at least 1 − 2 − n . F urthermor e, if d C ( L, x ) ≤ tλ 1 ( C, L ) , t ≥ 2 , then a vector y ∈ L satisfying k y − x k C = d C ( L, x ) can be computed in time O ( t 2 γ 4 ) n with pr obabil ity at least 1 − 2 − n . Pr oof. T o sho w the the orem, we use a slightly modified v ersion of Kannan’ s lifting techn ique to reduce CVP to SAP . Let us define L ′ ⊆ R n +1 as the lattice generate d by ( L, 0) and ( − x, 1) . In the standard way , we first gu ess a v alue β > 0 satisfy ing β ≤ d C ( L, x ) ≤ 3 2 β . Now let C ′ = C × [ − 1 β , 1 2 β ] . For ( y, z ) , y ∈ R n , z ∈ R , we ha ve that k ( y , z ) k C ′ = max {k y k C , β z , − 2 β z } Also, no te that C ′ ∩ − C ′ = ( C ∩ − C ) × [ − 1 2 β , 1 2 β ] . No w we se e that v ol ( C ′ ∩ − C ′ ) = 1 β v ol( C ∩ − C ) and v ol( C ′ ) = 3 (2 β ) v ol( C ) . Therefore, we get that v ol( C ′ ∩ − C ′ ) = 1 β v ol( C ∩ − C ) ≥ 1 β γ n v ol ( C ) = 2 3 γ n v ol ( C ′ ) Hence C ′ is γ (1 − 1 /n ) -symmetric. L et M = { y ∈ R n +1 : y n +1 = 0 } . Define m : L → L ′ \ M by m ( y ) = ( y − x, 1) , where it is easy to see that m is well-defined and injecti ve. Define S = { y ∈ L : k y − x k C ≤ (1 + ǫ ) d C ( L, x ) } and S ′ = { y ∈ L ′ \ M : k y k ′ C ≤ (1 + ǫ ) λ ( C ′ , L ′ , M ) } . W e claim that m defines a norm preserving bijection between S and S ′ . T aking y ∈ L , we see that k m ( y ) k C ′ = k ( y − x, 1) k C ′ = max {k y − x k C , β , − 2 β } = k y − x k C since β ≤ d C ( L, x ) ≤ k y − x k C by co nstruction. So we hav e that k m ( y ) k C ′ = k y − x k C , and hence λ ( C ′ , L ′ , M ) ≤ inf y ∈ L k y − x k C = d C ( L, x ) . Nex t take ( y , z ) ∈ L ′ \ M , y ∈ R n , z ∈ R , such th at k ( y , z ) k C ′ ≤ (1 + ǫ ) λ ( C ′ , L ′ , M ) . W e claim that z = 1 . Assume not, then since ( y , z ) ∈ L ′ \ M , we must ha ve that either z ≥ 2 or z ≤ − 1 . In either case, we ha ve that k ( y , z ) k C ′ = max {k y k C , β z , − 2 β z } ≥ max { β z , − 2 β z } ≥ 2 β No w since β ≤ d C ( L, x ) ≤ 3 2 β , ǫ ∈ (0 , 1 3 ) , and that λ ( C ′ , L ′ , M ) ≤ d C ( L, x ) , we get that k ( y , z ) k C ′ ≥ 2 β = (1 + 1 3 )( 3 2 β ) ≥ (1 + 1 3 ) d C ( L, x ) > (1 + ǫ ) d C ( L, x ) ≥ (1 + ǫ ) λ ( C ′ , L ′ , M ) a clear contra diction to our initi al assumption. Since z = 1 , we may write y = w − x where w ∈ L . Therefore , we see that k ( y , z ) k C ′ = k ( w − x, 1) k C ′ = max {k w − x k C , β , − 2 β } = k w − x k C since k w − x k C ≥ d C ( L, x ) ≥ β . S o we hav e that (1 + ǫ ) λ ( C ′ , L ′ , M ) ≥ k ( y , z ) k C ′ = k w − x k C ≥ d C ( L, x ) . Since the pre vious statement still holds when choosing ǫ = 0 , we must ha ve that λ ( C ′ , L ′ , M ) ≥ d C ( L, x ) and hence λ ( C ′ , L ′ , M ) = d C ( L, x ) . From the abov e, for y ∈ S , we hav e that k m ( y ) k C ′ = k y − x k C ≤ (1 + ǫ ) d C ( L, x ) = (1 + ǫ ) λ ( C ′ , L ′ , M ) , and henc e m ( y ) ∈ S as need ed. Next if ( y , z ) ∈ S ′ , from the abov e we hav e that z = 1 , 19 and hence ( y , z ) = ( w − x, 1) where w ∈ L . Therefore ( y , z ) = m ( w ) , w here k w − x k C = k ( y , z ) k C ′ ≤ (1 + ǫ ) λ ( C ′ , L ′ , M ) = (1 + ǫ ) d C ( L, x ) , and hence w ∈ S . S ince the m ap m is injecti ve, we get that m defines a norm preser ving bijectio n between S and S ′ as claimed . Hence solvin g (1 + ǫ ) -CVP with respect to C , L, x is equi v alent to solving (1 + ǫ ) -SAP with respect to C ′ , L ′ , M . W e get the desired result by applying 1 + ǫ approx imation algorith m for S AP describ ed in theore m 3.6. For exact CVP , we are gi ven the g uarantee that d C ( L, x ) ≤ tλ 1 ( C, L ) . From a nalysis abo ve, we see that λ 1 ( C ′ , L ′ ) = min { λ 1 ( C ′ , L ′ , M ) , inf y ∈ L \{ 0 } k ( y , 0) k C ′ } = min { d C ( L, x ) , λ 1 ( C, L ) } Therefore λ ( C ′ , L ′ , M ) = d C ( L, x ) = min { d C ( L, x ) , tλ 1 ( C, L ) } ≤ t min { d C ( L, x ) , λ 1 ( C, L ) } = tλ 1 ( C ′ , L ′ ) Hence we may again use the SAP solve r in theor em 3.6 to solve the exact CVP problem in O ( t 2 γ 4 ) n time with proba bility at least 1 − 2 − n as required . 4 Conclusions and Open Proble m s In this pape r , we hav e sho w n that an approxi m ate v ersion of Intege r Programming (IP) can be solved via an e xtension of the A KS si eving tech niques to ge neral semi-norms. Furthermore, we giv e algorit hms to solv e both the (1 + ǫ ) -Approxima te Subspace A void ing Problem (SAP) and Closest V ector Problem (CVP ) in general semi-norms using these techniques . Due to the reliance on a probabi listic si eve , the algorith ms presen ted here onl y guarante e the correct ness of th eir outputs with high probabi lity . In [DP V11], it was sho wn that thi s shortcoming can so m etimes be a voide d, by gi ving a Las V egas algo rithm for SVP in gene ral norms (which is deterministi c for ℓ p norms) achie ving simil ar asymptotic runni ng times as the AKS siev e based methods. The follo wing question is still open: Pr o blem: D oes there e xists a Las V egas or det erministic algo rithm for (1 + ǫ ) -SAP or CVP in g eneral (semi-)no rms achie ving the same asympto tic running time as the AKS sie ve based methods? As for potentia l improv ements in the comple xity of (1 + ǫ ) -SAP / CVP , th e follo wing question is open: Pr o blem: C an the comple xity of the AKS sie ve based methods for (1 + ǫ ) -SAP / CVP in general (semi-)no rms be reduced to O ( 1 γ 2 ǫ ) n ? 5 Ackno wledgments I would lik e to than k my advis or Santosh V empala for useful dis cussions relatin g to this problem. Refer ences [AJ08] V . Arvind and P . S. Joglek ar . Some sievi ng algorithms for lattice prob lems. In FSTT CS , pages 25–36 . 2008. 20 [AKS01] M. Ajtai, R. Kumar , and D. Siv akumar . A s iev e algor ithm for the sh ortest latt ice vec tor proble m . In STOC , pag es 601–6 10. 2001. [AKS02] M. A jtai, R. Ku mar , and D. S iv akumar . Sampling short lattice v ectors and th e closes t lattice vec tor proble m. In IE EE C onfer ence on Computatio nal Complexit y , pages 53–57 . 2002. [Bar94] A. Barvino k. A polynomia l time algorith m for counting inte gral points in poly hedra when the dimensio n is fixed. Mathematic s of Operation s Resear ch , 19(4):76 9–779, 1994. [BN07] J. Bl ¨ omer and S. Nae we. Samplin g method s for sho rtest v ectors, closest vector s and succe ssiv e minima. In ICALP , pages 65–77. 2007. [BN09] J. Bl ¨ omer and S. Nae we. Samplin g method s for sho rtest v ectors, closest vector s and succe ssiv e minima. Theor etical Computer Sci ence , 110:1 648–1665, 2009. [DFK89] M. E. Dyer , A. M. F rieze, and R. Kannan. A random polynomia l time algorithm for appro xi- mating the vol ume of con vex bodies . In STOC , pages 375–38 1. 1989. [DPV11] D. Dadush, C. P eiker t, and S . V empala. Enumerati ve lattice algorithms in any no rm via m- ellipso id cov erings. In FOCS . 2011. [EHN11] F . Eisenbr and, N. H ¨ ahnle, and M. Niemei er . C ov ering cubes and the clos est vec tor pro blem. In Pr oceedi ngs of the 27th annual ACM symposium on Computationa l ge ometry , SoCG ’11 , pages 417–4 23. A CM , Ne w Y ork, NY , USA, 201 1. [ES08] F . Eisenb rand and G. Shmonin . Pa rametric in teger p rogramming in fixed dimension. M athemat- ics of Operation s Resear ch , 33(4) :839–850, 2008. [GMSS99] O. Goldre ich, D. Miccianci o, S. S afra, and J.-P . Seifert . Approximatin g shortest latt ice vector s is not harder than approxi m ating clos est lattice vect ors. Inf. Pr ocess. Lett. , 71(2):5 5–61, 1999. [Gom58] R. G omory . An outlin e of an algorithm for solvin g inte ger programs. Bulletin of the American Mathematic al Society , 64(5):275 –278, 1958 . [Hei05] S. H einz. Complex ity of inte ger quasic on vex polynomia l optimiza tion. J ournal of Comple xity , 21(4): 543 – 556, 2005. Festschrift for the 70th Birthday of A rnold Schonh age. [HK10] R. Hild ebrand and M . K ¨ op pe. A new lenstra-ty pe al gorithm for quasicon vex polyn o- mial integ er minimiz ation with compl exity 2 O ( n log n ) . Arxiv , Report 100 6.4661, 2010. http: //arx iv.o rg . [Kan87] R. Kannan. Mink owski’ s con ve x body theore m and integer programming . Mathematic s of oper ations r esear ch , 12(3) :415–440, August 1987. 1987. [Kan90] R. Kannan. T est sets for inte ger programs, ∀∃ sentences . In DIMACS Series in Discr ete Mathe- matics and Theor etical Computer Scie nce V olume 1 , pages 39–47. 1990. [KLS95] R. Kannan, L . Lov ´ asz, and M. Simonovits . Isoperimetr ic problems for con ve x bodies and a localiz ation lemma. Discr ete & Computational Geometry , 13:541–55 9, 1995. [Len83] H. W . Lenstra. Integer prog ramming with a fixed number of var iables. Mathematics of Oper a- tions Resear ch , 8(4):5 38–548, Nove mber 198 3. 21 [MP00] V . Milman and A. Pajor . Entrop y and asymptotic geometry of non-symmetric co n vex bodies. Advances in Mathematic s , 152(2):314 – 335, 2000. [Pao0 6] G. Paouris . Concentra tion of mass on isotr opic con ve x bodi es. Comptes Rendus Mathematiqu e , 342(3 ):179–182, 2006. A A ppendix Pr oof of Lemma 2.3 (Appr ox. Barycenter ). Let X 1 , . . . , X N denote iid uniform samples ov er K ⊆ R n , where N =  2 cn ǫ  2 . W e will sho w that for b = 1 N P n i =1 X i , that the follo wing hold s Pr[ k ± ( b − b ( K )) k K − b ( K ) > ǫ ] ≤ 4 − n (A.1) Since the abov e statement is in vari ant under affine transfo rmations, we may assume K is isotrop ic, i.e. b ( K ) = E[ X 1 ] = 0 , the origin , and E[ X 1 X t 1 ] = I n , the n × n iden tity . Since K is isotropi c, we ha ve that B n 2 ⊆ K (see [KL S95]). Therefore to sho w (A.1) it suf fices to pro ve that Pr[ k b k 2 > ǫ ] ≤ 4 − n . Since the X i ’ s are iid iso tropic rando m vector s, we see that E[ b ] = 1 N P N i =1 E[ X i ] = 0 and E[ bb t ] = 1 N 2 X i,j ∈ [ N ] E [ X i X t j ] = 1 N 2 n X i =1 E [ X i X t i ] = 1 N I n No w since the X i s are log-con cav e, we ha ve t hat b is also log-con cav e (sin ce its dis tribut ion is a con vol ution of log-c oncav e distrib utions ). Now , gi ven that b has cov arianc e matrix 1 N I n , by the c oncentratio n inequ ality of Paou ris [Pao0 6 ], we ha ve that Pr[ k b 2 k 2 > ǫ ] = Pr[ k b 2 k 2 > 2 c n √ N ] < e − 2 n < 4 − n as cl aimed. T o pr ove the t heorem, we note tha t when switching the X i ’ s from truly u niform to 4 − n unifor m , the abo ve prob ability chang es by at mo st cn 2 ǫ 2 4 − n by Lemma 2.1. Therefore the tot al erro r proba bility under 4 − n -unifo rm samples is at most 2 − n as need ed. Pr oof of Lemma 2.4 (Estimates for semi-norm rec entering). W e hav e z ∈ R n , x, y ∈ K satisfying ( † ) k ± ( x − y ) k K − y ≤ α < 1 . W e prov e the statements as fo llows: 1. k z − y k K − y ≤ τ ⇔ ( z − y ) ∈ τ ( K − y ) ⇔ z ∈ τ K + (1 − τ ) y as needed . 2. Let τ = k z − x k K − x . Then by (1) , we ha ve that z ∈ τ K + (1 − τ ) x . N o w note that (1 − τ )( x − y ) ⊆ | 1 − τ | α ( K − y ) by assumptio n ( † ) and (1) . Therefore z ∈ τ K + (1 − τ ) x = τ K + (1 − τ ) y + (1 − τ )( x − y ) ⊆ τ K + (1 − τ ) y + α | 1 − τ | ( K − y ) = ( τ + α | 1 − τ | ) K + (1 − τ − α | 1 − τ | ) y Hence by (1) , we ha ve that k z − y k K − y ≤ τ + α | 1 − τ | = k z − x k K − x + α | 1 − k z − x k K − x | as needed. 22 3. W e first sho w that ± ( y − x ) ∈ α 1 − α ( K − x ) By (1) and ( † ) we hav e that ( x − y ) ∈ α ( K − y ) ⇔ ( x − y ) − α ( x − y ) ∈ α ( K − y ) − α ( x − y ) ⇔ (1 − α )( x − y ) ∈ α ( K − x ) ⇔ ( x − y ) ∈ α 1 − α ( K − x ) as needed. Next s ince 0 ≤ α ≤ 1 , we hav e that | 1 − 2 α | ≤ 1 . Therefore by ( † ) we hav e that (1 − 2 α )( y − x ) ∈ | 1 − 2 α | α ( K − y ) ⊆ α ( K − y ) since 0 ∈ K − y . Now not e that (1 − 2 α )( y − x ) ∈ α ( K − y ) ⇔ (1 − 2 α )( y − x ) + α ( y − x ) ∈ α ( K − y ) + α ( y − x ) ⇔ (1 − α )( y − x ) ∈ α ( K − x ) ⇔ ( y − x ) ∈ α 1 − α ( K − x ) as needed. Let τ = k z − y k K − y . Then by (1) , we ha ve that z ∈ τ K + (1 − τ ) y . Now n ote tha t z ∈ τ K + (1 − τ ) y = τ K + (1 − τ ) x + (1 − τ )( y − x ) ⊆ τ K + (1 − τ ) x + α 1 − α | 1 − τ | ( K − x ) = ( τ + α 1 − α | 1 − τ | ) K + (1 − τ − α 1 − α | 1 − τ | ) x Hence by (1) , we ha ve that k z − x k K − x ≤ τ + α 1 − α | 1 − τ | = k z − y k K − y + α 1 − α | 1 − k z − y k K − y | as needed. Pr oof of Cor ollary 2.6 (Stability of symmetry). W e claim that (1 − k x k K )( K ∩ − K ) ⊆ K − x ∩ x − K . T ake z ∈ K ∩ − K , th en note that k x + (1 − k x k K ) z k K ≤ k x k K + (1 − k x k K ) k z k K ≤ k x k K + (1 − k x k K ) k z k K ∩− K ≤ k x k K + (1 − k x k K ) = 1 hence x + (1 − k x k K )( K ∩ − K ) ⊆ K ⇔ (1 − k x k K )( K ∩ − K ) ⊆ K − x . Next n ote that k − x + (1 − k x k K ) z k − K ≤ k − x k − K + (1 − k x k K ) k z k − K ≤ k x k K + (1 − k x k K ) k z k K ∩− K ≤ k x k K + (1 − k x k K ) = 1 hence − x + (1 − k x k K )( K ∩ − K ) ⊆ − K ⇔ (1 − k x k K )( K ∩ − K ) ⊆ x − K , as need ed. Now we see that v ol (( K − x ) ∩ ( x − K )) ≥ v ol((1 − k x k K )( K ∩ − K )) = (1 − k x k K ) n v ol( K ∩ − K ) and so the claim follo ws from T heorem 2.5. 23 Pr oof of Lemma 3.5 (Inters ection Lemma). S ince k v k C ≤ 3 2 β , we see that v 2 ∈ 3 / 4 C . Now we get that v 2 + 1 / 4 β ( C ∩ − C ) ⊆ 3 / 4 β C + 1 / 4 β C = β C Furthermor e since k v 2 − v k − C = k − v 2 k − C = 1 / 2 k v k C ≤ 3 / 4 β , w e also ha ve that v 2 ∈ v − 3 / 4 C . Therefore v 2 + 1 4 β ( C ∩ − C ) ⊆ ( v − 3 4 β C ) + 1 4 β ( − C ) = v − β C W e therefore conclude that v ol( β C ∩ ( v − β C )) v ol( β C ) ≥ v ol( v 2 + 1 / 4 β ( C ∩ − C )) v ol( C ) =  1 4  n v ol ( C ∩ − C ) v ol( C ) ≥  γ 4  n as need ed. For the further m ore, we remember that k x k C = inf { s ≥ 0 : x ∈ sC } = sup {h x, y i : y ∈ C ∗ } and that ( − C ) ∗ = − C ∗ . Now assume there e xists x ∈ C + v ∩ C − v . Then x = v − β k 1 = β k 2 − v where k 1 , k 2 ∈ C . Choose y ∈ C ∗ such that h y , v i = k v k C . Note that h y , v − β k 1 − ( β k 2 − v ) i = 2 h y , v i − β ( h y , k 1 i + h y , k 2 i ) = 2 k v k C − β ( h y , k 1 i + h y , k 2 i ) ≥ 2 k v k C − β ( k k 1 k C + k k 2 k C ) ≥ 2 β − 2 β = 0 Since v − β k 1 − ( β k 2 − v ) = x − x = 0 by c onstruction , all of th e above inequal ities must hold at e quality . In particu lar , we must ha ve that 1 = k k 1 k C = k k 2 k C = h y, k 1 i = h y , k 2 i . S ince − y ∈ ( − C ) ∗ , we kno w that v − β C ⊆ { x ∈ R n : h− y , x − v i ≤ β } and since h− y , ( v − β k 1 ) − v i = β h y , k 1 i = β , w e m ust hav e that v − β k 1 ∈ ∂ C + v . V ia a symmetric ar gument, we get that β k 2 − v ∈ ∂ C − v . Therefore C + v ∩ C − v ⊆ ∂ C + v ∩ ∂ C − v ⇔ in t( C + v ) ∩ int( C − v ) = ∅ , as needed . 24