The Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs
aa r X i v : . [ c s . CC ] N ov The Double Exponential Runtime is Tight for2-Stage Stochastic ILPs ⋆ Klaus Jansen , Kim-Manuel Klein − − − , and AlexandraLassota − − − X ] Department of Computer Science, Kiel University, Kiel, Germany [email protected] Department of Computer Science, Kiel University, Kiel, Germany [email protected] Department of Computer Science, Kiel University, Kiel, Germany [email protected]
Abstract.
We consider fundamental algorithmic number theoretic prob-lems and their relation to a class of block structured Integer Linear Pro-grams (ILPs) called 2-stage stochastic. A 2-stage stochastic ILP is an inte-ger program of the form min { c T x | A x = b, ℓ ≤ x ≤ u, x ∈ Z r + ns } wherethe constraint matrix A ∈ Z nt × r + ns consists of n matrices A i ∈ Z t × r onthe vertical line and n matrices B i ∈ Z t × s on the diagonal line aside.First, we show a stronger hardness result for a number theoretic problemcalled Quadratic Congruences where the objective is to compute anumber z ≤ γ satisfying z ≡ α mod β for given α, β, γ ∈ Z . This prob-lem was proven to be NP-hard already in 1978 by Manders and Adleman.However, this hardness only applies for instances where the prime fac-torization of β admits large multiplicities of each prime number. Wecircumvent this necessity proving that the problem remains NP-hard,even if each prime number only occurs constantly often.Then, using this new hardness result for the Quadratic Congruences problem, we prove a lower bound of 2 δ ( s + t ) | I | O (1) for some δ > | I | is the encoding lengthof the instance. This result even holds if r , || b || ∞ , || c || ∞ , || ℓ || ∞ and thelargest absolute value ∆ in the constraint matrix A are constant. Thisshows that the state-of-the-art algorithms are nearly tight. Further, itproves the suspicion that these ILPs are indeed harder to solve than theclosely related n -fold ILPs where the contraint matrix is the transposeof A . Keywords:
One of the most fundamental problems in algorithm theory and optimization isthe
Integer Linear Programming problem. Many theoretical and practical ⋆ This work was supported by DFG project JA 612/20-1 K. Jansen et al. problems can be modeled as integer linear programs (ILPs) and thus they serveas a very general but powerful framework for tackling various questions. Formally,the
Integer Linear Programming problem is defined asmin { c ⊤ x | A x = b, ℓ ≤ x ≤ u, x ∈ Z n } for some matrix A ∈ Z m × n , a right-hand side b ∈ Z m , an objective function c ∈ Z n and some lower and upper bounds ℓ, u ∈ Z n . The goal is to find asolution x such that the value of the objective function c ⊤ x is minimized. Ingeneral, this problem is NP-hard. Thus, it is of great interest to find structuresto these ILPs which make them solvable more efficiently. In this work, we consider2-stage stochastic integer linear programs where the constraint matrix admits aspecific block structure. Namely, the constraint matrix A only contains non-zeroentries in the first few columns and block-wise along the the diagonal aside. Thisyields the following form: A = A B . . . A B . . . ...... ... . . . . . . 0 A n . . . B n . Thereby A , . . . , A n ∈ Z t × r and B , . . . , B n ∈ Z t × s are integer matrices them-selves. The complete constraint matrix A has size nt × r + ns . Let ∆ denote thelargest absolute entry in A .Such 2-stage stochastic ILPs are a common tool in stochastic programmingand they are often used in practice to model uncertainty of decision makingover time [1,8,19,24]. Due to the applicability a lot of research has been done inorder to solve these (mixed) ILPs efficiently in practice. Since we focus on thetheoretical aspects of 2-stage stochastic ILPs in this chapter, we only refer thereader to the surveys [10,22,26] and the references therein regarding the practicalmethods.The current state-of-the-art algorithm to solve 2-stage stochastic ILPs admitsa running time of 2 (2 ∆ ) r s + rs n log ( n ) · L where L is the encoding length ofthe largest number in the linear program [9]. This result improves upon theresult in [20] due to Klein where the dependence on n was quadratic and thedependencies on the block dimensions and L were similar. The first result inthat respect was by Hemmecke and Schulz [14] who provided an algorithm witha running time of f ( r, s, t, ∆ ) · poly( n ) for some computable function f . However,due to the use of an existential result from commutative algebra, no explicitbound could be stated for f .Let us turn our attention shortly to the n -fold ILPs. These ILPs admit aconstraint matrix which is the transpose of the 2-stage stochastic constraintmatrix. Despite being so closely related, n -fold ILPs can be solved in time nearlinear in the blocks and only single exponentially in the block-dimensions of A Ti , B Ti [6,18]. he Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs 3 Thus, it is an intrinsic questions whether we can solve 2-stage stochastic ILPsmore efficient or – as the latest algorithms for them suggest – whether 2-stagestochastic ILPs are indeed harder to solve than the closely related n -fold ILPs.We answer this question by showing a double-exponential lower bound in therunning time for any algorithm solving the 2 -stage stochastic integer lin-ear programming (2 -stage ILP ) problem. Here, the 2 -stage ILP problem isthe corresponding decision variant which asks whether the ILP admits a feasiblesolution.To prove this hardness, we reduce from the Quadratic Congruences prob-lem. This problem asks whether there exists a z ≤ γ such that z ≡ α mod β for some γ, α, β ∈ N . This problem was proven to be NP-hard by Manders andAdleman [25] already in 1978 by showing a reduction from 3- SAT . This hardnesseven persists if the prime factorization of β is given [25]. By this result, Mandersand Adleman prove that it is NP-complete to compute the solutions of diophan-tine equations of degree 2. However, their reduction yields large parameters. Indetail, the occurrences of each prime factor in the prime factorization of β is toolarge to obtain the desired lower bound for the 2 -stage ILP problem. In fact,the occurrence of each prime factor is at least linear in the number of variablesand clauses of the underlying 3- SAT problem.We give a new reduction yielding a stronger statement: The
QuadraticCongruences problem is NP-hard even if the prime factorization of β is givenand each prime factor occurs at most once (except 2 which occurs four times).Beside being useful to prove the lower bounds for solving the 2-stage stochasticILPs, we think this results is of independent interest. We obtain a neat struc-ture which may be helpful in various related problems or may yield strongerstatements of past results which used the Quadratic Congruences problem.In order to achieve the desired lower bounds on the running time we makeuse of the Exponential Time Hypothesis (ETH) – a widely believed conjecturestating that the 3-
SAT problem cannot be solved in subexponentially time withrespect to the number of variables:
Conjecture 1 (ETH [15])
The - SAT problem cannot be solved in time lessthan O (2 δ n ) for some constant δ > where n is the number of variables inthe instance. Note that we use the index 3 for all variables of the 3-
SAT problem.Using the ETH, plenty lower bounds for various problems are shown, for anoverview on the techniques and results see e.g. [7]. So far, the best algorithmruns in time O (2 . n ), i. e., it follows that δ ≤ .
387 [7].In the following, we also need the Chinese Remainder Theorem (CRT) forsome of the proofs, which states the following:
Proposition 1 (CRT [17]).
Let n , . . . , n k be pairwise co-prime. Further, let i , . . . , i k be some integers. Then there exists integers x satisfying x ≡ i j mod n j for all j . Further, any two solutions x , x are congruent modulo Q kj =1 n j . K. Jansen et al.
Summary of Results – We give a new reduction from the 3-
SAT problem to the
Quadratic Con-gruences problem which proves a stronger NP-hardness result: The
QuadraticCongruences problem remains NP-hard, even if the prime factorization of β is given and each prime number greater than 2 occurs at most once andthe prime number 2 occurs four times. This does not follow from the originalproof. In contrast, the original proof generates each prime factor at least O ( n + m ) times, where m is the number of clauses in the formula. Our re-duction circumvents this necessity, yet neither introduces noteworthily morenor larger prime factors. The proof is based on the original one. We believethis result is of independent interest. – Based on this new reduction, we show strong NP-hardness for the so-called
Non-Unique Remainder problem. In this algorithmic number theoreticproblem we are given x , . . . , x n NR , y , . . . , y n NR , ζ ∈ N and pairwise coprimenumbers q , . . . , q n NR . The question is to decide whether there exists a num-ber z ∈ Z > with z ≤ ζ satisfying the following congruences: z mod q ∈ { x , y } z mod q ∈ { x , y } ... z mod q n NR ∈ { x n NR , y n NR } . In other words, either the residue x i or y i should be met for each equation.This problem is a natural generalization of the Chinese Remainder problemwhere x i = y i for all i . In that case, however, the problem can be solvedusing the Extended Euclidean algorithm. To the best of our knowledge the Non-Unique Remainder problem has not been considered in the literatureso far. – Finally, we show that the
Non-Unique Remainder problem can be mod-eled by a 2-stage stochastic ILP. Assuming the ETH, we can then concludea doubly exponential lower bound of 2 δ ( s + t ) | I | O (1) on the running time forany algorithm solving 2-stage stochastic ILPs. The double exponential lowerbound even holds if r = 1 and ∆, || b || ∞ , || c || ∞ ∈ O (1). This proves thesuspicion that 2-stage stochastic ILPs are significantly harder to solve than n -fold ILPs with respect to the dimensions of the block matrices and ∆ . Fur-thermore, it implies that the current state-of-the-art algorithms for solving2-stage stochastic ILPs is indeed (nearly) optimal. Further Related Work
In recent years there was significant progress in the devel-opment of algorithms for n -fold ILPs and lower bounds on the other hand. As-sume the parameters as of the transpose of the 2-stage stochastic constraint ma-trix, i. e., the blocks A Ti in the first few rows have dimension r × t and the blocks B Ti along the diagonal beneath admit a dimension of s × t . The best known algo-rithms to solve these ILPs have a running time of 2 O ( rs ) ( rs∆ ) O ( r s + s ) ( nt ) o (1) [6] he Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs 5 or respectively a running time of ( rs∆ ) r s + s L ( nt ) o (1) [18] where L denotethe encoding length of the largest number in the input. The best known lowerbound is ∆ δ n -fold ( r + s ) for some δ n -fold > n -fold ILPs. Yet, no superexponential lower bound for therunning time of any algorithm solving the 2 -stage ILP problem was shown.There is a lower bound for a more general class of ILPs in [9] that contain2-stage stochastic ILPs showing that the running time is double-exponentialparameterized by the topological height of the treedepth decomposition of theprimal or dual graph. However, the topological height of 2-stage stochastic ILPsis constant and thus no strong lower bound can be derived for this case.If we relax the necessity of an integral solution, the 2-stage stochastic LPproblem becomes solvable in time 2 ∆ O ( t n log ( n ) log( || u − ℓ || ∞ ) log( || c || ∞ ) [3].For the case of mixed integer linear programs there exists an algorithm solving2-stage stochastic MILPs in time 2 ∆ ∆tO ( t n log ( n ) log( || u − ℓ || ∞ ) log( || c || ∞ ) [3].Both results rely on the fractionality of a solution, whose size is only dependenton the parameters. This allows us to scale the problem such that it becomes anILP (as the solution has to be integral) and thus state-of-the-art algorithms for2-stage stochastic ILPs can be applied.There are also studies for a more general case called 4-Block ILPs where theconstraint matrix consists of non-zero entries in the first few columns, the firstfew rows and block-wise along the diagonal. This may be seen as the combinationof n -fold and 2-stage stochastic ILPs. Only little is known about them: They arein XP [13]. Further, a lower and upper bound on the Graver Basis elements(inclusion-wise minimal kernel elements) of O ( n r f ( k, ∆ )) was shown recently [4],where r is the number of rows in the submatrix appearing repeatedly in the firstfew rows and k denotes the sum of the remaining block dimensions. Structure of this Chapter
Section 2 presents the stronger hardness result forthe
Quadratic Congruences problem we derive by giving a new reductionfrom the 3-
SAT problem. Then we show that the
Quadratic Congruences problem can be modeled as a 2-stage stochastic ILP in Section 3. To do so,we introduce a new problem called the
Non-Unique Remainder problem asan intermediate step during the reduction. Finally, in Section 4 we bring thereductions together to prove the desired lower bound. This involves a construc-tion which lowers the absolute value of ∆ at the cost of slightly larger blockdimensions. This section proves that every instance of the 3-
SAT problem can be transformedinto an equivalent instance of the
Quadratic Congruences problem in poly-nomial time. Recall that the
Quadratic Congruences problem asks whetherthere exists a number z ≤ γ such that z ≡ α mod β holds. This problem was K. Jansen et al. proven to be NP-hard by Manders and Adleman [25] showing a reduction from 3-SAT. This hardness even persists when the prime factorization of β is given [25].However, we aim for an even stronger statement: The Quadratic Congru-ences problem remains NP-hard even if the prime factorization of β is givenand each prime number greater than 2 occurs at most once and the prime num-ber 2 occurs four times. This does not follow from the original hardness proof.In contrast, if n is the number of variables and m the number of clauses in the3-SAT formula then β admits a prime factorization with O ( n + m ) differentprime numbers each with a multiplicity of at least O ( n + m ). Even thoughour new reduction lowers the occurrence of each prime factor greatly, we neitherintroduces noteworthily more nor larger prime factors.While the structure of our proof resembles that of the original one from [25],adapting it to our needs requires various new observations concerning the be-haviour of the newly generated prime factors and the functions we introduce.The original proof heavily depends on the numbers being high powers of theprime factors whereas we employ careful combinations of (new) prime factors.This requires us to introduce other number theoretical results into the arguments.Before we present the reduction and show its correctness formally, we wantto give an idea of the hardness proof. The reduction may seem unintuitionallyat first as it only shows the final result of equivalent transformations betweenvarious problems until we reach the Quadratic Congruences one. In thefollowing, we list all these problems in order of their appearance whose strongNP-hardness is shown implicitly along the way. Afterwards, we give short ideasof their respective equivalence, which is then proved formally in separate claimsin the next theorem. Note that not all variables are declared at this point, butalso not necessary to follow the proof sketch. – (3- SAT ) Is there a truth assignment η that satisfies all clauses σ k of the3-SAT formula Φ simultaneously? – (P2) Are there values y k ∈ { , , , } and a truth assignment η such that0 = R k = y k − P x i ∈ σ k η ( x i ) − P ¯ x i ∈ σ k (1 − η ( x i )) + 1 for all k ? – (P3) Are there values α j ∈ {− , +1 } such that P νj =0 θ j α j ≡ τ mod 2 · p ∗ Q m ′ i =1 p i ? – (P5) Is there an x ∈ Z satisfying0 ≤ | x | ≤ H (P5.1) x ≡ τ mod 2 · p ∗ m ′ Y i =1 p i (P5.2)( H + x )( H − x ) ≡ K ? (P5.3) he Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs 7 – (P6) Is there an x ∈ Z satisfying0 ≤ | x | ≤ H (P6.1)( τ − x )( τ + x ) ≡ · p ∗ m ′ Y i =1 p i (P6.2)( H + x )( H − x ) ≡ K ? (P6.3) – ( Quadratic Congruences ) Is there a number x ≤ H such that (2 · p ∗ · Q m ′ i =1 p i + K ) x ≡ Kτ + 2 · p ∗ · Q m ′ i =1 p i H mod 2 · p ∗ · Q m ′ i =1 p i · K ?The 3- SAT problem is transformed to Problem (P2) by using the straight-forward interpretation of truth values as numbers 0 and 1 and the satisfiabilityof a clause as the sum of its literals being larger zero. Introducing slack variables y k yields the above form.Multiplying each equation of (P2) with exponentially growing factors andthen forming their sum preserves the equivalence of these systems. Introducingsome modulo consisting of unique prime factors larger than the outcome ofthe largest possible sum obviously does not influence the system. Replacingthe variables η ( x i ) and y k by variables α j with domain {− , +1 } , re-arrangingthe term and defining parts of the formula as the variables θ j and τ yieldsProblem (P3).We then introduce some Problem (P4) to integrate the condition x ≤ H . Theproblem asks whether there exists some x ∈ Z such that0 ≤ | x | ≤ H (P4.1)( H + x )( H − x ) ≡ K ? (P4.2)By showing that each solution to the system (P4) is of form P νj =0 θ j α j we cancombine (P3) and (P4) yielding (P5).Using some observations about the form of solutions for the second constraintof Problem (P5) we can re-formulate it as Problem (P6).Next, we use the fact that p ∗ Q m ′ i =1 p i and K are co-prime per definition andthus we can combine (P6.2) and (P6.3) to one equivalent equation. To do so, wetake each left-hand side of (P6.2) and (P6.3) and multiply the modulo of therespective other equation and form their overall sum. Using a little re-arrangingthis finally yields the desired Quadratic Congruences problem.
Theorem 1.
The
Quadratic Congruences problem is NP-hard even if theprime factorization of β is given and each prime factor greater than occurs atmost once and the prime factor occurs times.Proof. We show a reduction from the well-known NP-hard problem 3-
SAT wherewe are given a 3-SAT formula Φ with n variables and m clauses. Transformation:
First, eliminate duplicate clauses from Φ and those where somevariable x i and its negation ¯ x i appear together. Call the resulting formula Φ ′ , K. Jansen et al. the number of occurring variables n ′ and denote by m ′ the number of appearingclauses respectively. Let Σ = ( σ , . . . , σ m ′ ) be some enumeration of the clauses.Denote by p , . . . , p m ′ the first 2 m ′ + 1 prime numbers. Compute τ Φ ′ = − m ′ X i =1 i Y j =1 p j . Further, compute for each i ∈ , , . . . , n ′ : f + i = X x i ∈ σ j j Y k =1 p k and f − i = X ¯ x i ∈ σ j j Y k =1 p k . Set ν = 2 m ′ + n ′ . Compute the coefficients c j for all j = 0 , , . . . , ν as follows:Set c = 0. For j = 1 , . . . , m ′ set c j = − j Y i =1 p i for j = 2 k − c j = − j Y i =1 p i for j = 2 k. Compute the remaining ones for j = 1 , . . . n ′ as c m ′ + j = 1 / · ( f + j − f − j ).Further, set τ = τ ′ Φ + P νj =0 c j + P n ′ i =1 f − i .Denote by q , . . . , q ν +2 ν +1 the first ν +2 ν +1 prime numbers. Let p , , p , , . . . ,p ,ν , p , , . . . , p ν,ν be the first ( ν + 1) = ν + 2 ν + 1 prime numbers greaterthan (4( ν + 1)2 Q ν +2 ν +1 i =1 q i ) / ( ν +2 ν +1) and greater than p m ′ . Define p ∗ as the( ν + 2 ν + 2 m ′ + 13)th prime number.Determine the parameters θ j for j = 0 , , . . . , ν as the least θ j satisfying: θ j ≡ c j mod 2 · p ∗ m ′ Y i =1 p i ,θ j ≡ ν Y i =0 ,i = j ν Y k =0 p i,k ,θ j p j, . Set the following parameters: H = ν X j =0 θ j and K = ν Y i =0 ν Y k =0 p i,j . he Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs 9 Finally, set α = (2 · p ∗ m ′ Y i =1 p i + K ) − · ( Kτ + 2 · p ∗ m ′ Y i =1 p i · H ) ,β = 2 · p ∗ m ′ Y i =1 p i · K,γ = H. where (2 · p ∗ Q m ′ i =1 p i + K ) − is the inverse of (2 · p ∗ Q m ′ i =1 p i + K ) mod 2 · p ∗ Q m ′ i =1 p i · K . Correctness:
We show that the satisfiability of the formula Φ is equivalent to aline of (systems of) equations, i. e., the formula has a satisfying truth assignmenton the variables if and only if the (systems of) equations admit a solution. Bythis, we prove the hardness for various problems along the way. These are listedabove with their respective equivalence sketched. In the following, we separateeach of these steps by claims.However, before we start with the transformations of the formula, we firstobserve some properties about the generated prime factors. These come in handyfor the estimations later on. In particular we want to show that choosing p ∗ as the( ν +2 ν +2 m ′ +13)th prime factor satisfies p ∗ > p ν,ν : Suppose p m ′ ≥ (4( ν +1)2 · Q ν +2 ν +1 i =1 q i ) / ( ν +2 ν +1) . Then p ν,ν is the ( ν + 2 ν + 1 + 2 m ′ + 1)th prime numberand thus p ∗ > p ν,ν . Otherwise, if p m ′ < (4( ν + 1)2 Q ν +2 ν +1 i =1 q i ) / ( ν +2 ν +1) , webound the function values as follows:(4( ν + 1)2 ν +2 ν +1 Y i =1 q i ) / ( ν +2 ν +1) = 4 / ( ν +2 ν +1) ( ν + 1) / ( ν +2 ν +1) (2 ) / ( ν +2 ν +1) · ( ν +2 ν +1 Y i =1 q i ) / ( ν +2 ν +1) ≤ · · · (4 ν +2 ν +1 ) / ( ν +2 ν +1) = 2 · · · . The second transformation holds as the product of the first k prime numbers isbounded by 4 k [12]. There are 11 prime numbers in the interval [1 , p ν,ν is at most the (11 + ν + 2 ν + 1)th prime number and thus p ∗ > p ν,ν .Further, note that (4( ν + 1)2 · p ∗ m ′ Y i =1 p i ) / ( ν +2 ν +1) ≤ (4( ν + 1)2 ν +2 ν +1 Y i =1 q i ) / ( ν +2 ν +1) , i. e., p ∗ ≤ Q ν +2 ν +1 i = m ′ +1 q i : We can bound the value of the product from beneathas Q ν +2 ν +1 i = m ′ +1 q i ≥ q ν + νm ′ +1 . Estimating the value for p ∗ , we use that the valueof the next prime number after a number ρ is at most 2 ρ [2]. Thus, as thereare ν + 2 ν + m ′ + 11 prime numbers between p m ′ +1 and p ∗ , we get p ∗ ≤ q m ′ +1 · ν +2 ν + m ′ +11 . Setting q m ′ +1 = 5 to the smallest reasonable value for m ′ = 2 (if we only have one clause in the 3-SAT formula the problem becomeseasy), we get ν +2 ν +1 Y i = m ′ +1 q i ≥ q ν + νm ′ +1 ≥ ν + ν ≥ · ν +2 ν + m ′ +11 ≥ p ∗ . This follows as 5 ν + ν ≥ ν ν = 2 ν +2 ν is greater than 5 · ν +2 ν + m ′ +11 ≤ · ν +2 ν + m ′ +11 = 2 ν +2 ν + m ′ +14 ≤ ν +3 ν +11 for all reasonable values of ν ,i. e., ν ≥ m ′ = 2 clauses and n ′ = 3 variables). Obviously, thisrelation also holds for larger values of m ′ .Let us now focus on the transformations of the formula Φ yielding the firstclaim: Claim.
The 3-
SAT problem asking whether there is a truth assignment η thatsatisfies all clauses σ k of the 3-SAT formula Φ simultaneously is a yes-instance ifand only if Problem (P2) asking whether there are values y k ∈ { , , , } and atruth assignment η such that 0 = R k = y k − P x i ∈ σ k η ( x i ) − P ¯ x i ∈ σ k (1 − η ( x i ))+1for all k is a yes-instance. Proof.
Obviously, the reduced formula Φ ′ is satisfiable if and only if Φ is. The for-mula Φ ′ is satisfiable if there exists a truth assignment η : { x , . . . , x n ′ } → { , } assigning a logical value to each variable x , . . . , x n ′ which satisfies all clauses σ , . . . , σ m ′ simultaneously. This can be re-written to the following equation foreach clause σ k ∈ Φ k interpreting the truth values as numbers:0 = R k = y k − X x i ∈ σ k η ( x i ) − X ¯ x i ∈ σ k (1 − η ( x i )) + 1, y k ∈ { , , , } . For a clause σ k , this equation is only satisfiable if at least one variable x i ∈ σ k has value η ( x i ) = 1 or one variable occurring in its negation ¯ x i ∈ σ k has value η ( x i ) = 0. Otherwise, we have to set y k = − ⊓⊔ Note that we never have to set y k = 3 to satisfy the formula. However, weallow this value as it will come in handy later on when transforming the equation.Further, set 0 = R = α + 1 for α ∈ {− , +1 } for later convenience. Clearly,the new equation is satisfiable. Claim.
The Problem (P2) asking whether there are values y k ∈ { , , , } and atruth assignment η such that 0 = R k = y k − P x i ∈ σ k η ( x i ) − P ¯ x i ∈ σ k (1 − η ( x i ))+1for all k is a yes-instance if and only if Problem (P3) asking whether there arevalues α j ∈ {− , +1 } such that P νj =0 θ j α j ≡ τ mod 2 · p ∗ Q m ′ i =1 p i is a yes-instance. he Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs 11 Proof.
We can bound the values of R k for k ∈ { , , . . . , m ′ } by − ≤ R k ≤ y k = 0, all x i ∈ σ k have value η ( x i ) = 1 and all ¯ x i ∈ σ k have value η ( x i ) = 0. For the upper bound we set y k = 3, all x i ∈ σ k to η ( x i ) = 0 and ¯ x i ∈ σ k to η ( x i ) = 1. For R obviously0 ≤ R k ≤ R k = 0, ∀ k ∈ { , , . . . , m ′ } ⇔ m ′ X k =0 R k k Y i =0 p i = 0as the sum is zero if all R k = 0. For the opposite direction, if the sum is zero,then no R k = 0 as the product of the prime numbers grows too fast. Thus,the other summands cannot compensate for some R k = 0. We can bound theexpression further by | m ′ X k =0 R k k Y i =0 p i | ≤ m ′ X k =0 k Y i =0 p i ≤ m ′ + 1) m ′ Y i =0 p i < · p ∗ m ′ Y i =1 p i as p ∗ > p n,n > p m ′ > m ′ + 1. This yields R k = 0, ∀ k ∈ { , , . . . , m ′ } ⇔ m ′ X k =0 R k k Y i =0 p i ≡ · p ∗ m ′ Y i =1 p i (I)as the modulo has no impact on the satisfiability of the equation.Next, we aim to re-write R k by replacing the variables y k and η ( x i ) with newvariables admitting a domain of {− , } : y k = 1 / · [(1 − α k − ) + 2 · (1 − α k )], k ∈ { , . . . , m ′ } ,η ( x i ) = 1 / · (1 − α m ′ + i ), i ∈ { , . . . , n ′ } . Obviously the value domains of y k and η ( x i ) are preserved. Substituting thevariables and re-arranging the equation (I) yields ν X j =0 c j α j ≡ τ mod 2 · p ∗ m ′ Y i =1 p i , α j ∈ {− , +1 } . By definition of θ j this is equivalent to ν X j =0 θ j α j ≡ τ mod 2 · p ∗ m ′ Y i =1 p i , α j ∈ {− , +1 } proving the claim. ⊓⊔ Let H = P νj =0 θ j and K = Q νi =0 Q νj =0 p i,j be defined as before. Considerthe following system asking whether there is a x ∈ Z such that:0 ≤ | x | ≤ H (P4.1)( H + x )( H − x ) ≡ K (P4.2) We use this system to integrate the condition x ≤ H into the transformations.In the following, we prove that each solution of this system is of form x = P νj =0 α j θ j and thus Problem (P4) can be combined with Problem (P3) yieldingProblem (P5). Claim.
The Problem (P3) asking whether there are values α j ∈ {− , +1 } suchthat P νj =0 θ j α j ≡ τ mod 2 · p ∗ Q m ′ i =1 p i is a yes-instance if and only if the Prob-lem (P5) is a yes-instance. Proof.
The unique solutions x to the given system (P4) are of form x = ν X j =0 α j θ j , α ∈ {− , +1 } , j = 0 , , . . . , ν. Let us first verify that an x of such form solves the system. First | x | = | ν X j =0 α j θ j | ≤ ν X j =0 θ j = H satisfies (P4.1). Further, we have that each summand in the expanded for-mula ( H + x )( H − x ) has to contain all prime factors p i,j for i = 0 , , . . . , ν and j = 0 , , . . . , ν in its prime factorization to satisfy (P4.2). For ( H + x ) =( P nj =0 θ j + P nj =0 θ j α j ) it holds that each θ j where α j = +1 occurs twice whileeach θ j where α j = − H . The other way round holds for( H − x ). Thus, expanding the brackets yields that each summand is a productof some θ j and θ k where α j = +1 and α k = −
1. This implies that j = k . Aseach θ j contains all prime factors of K except p j, , . . . , p j,ν , the product of twodifferent θ j and θ k contains each prime factor occurring in K satisfying (P4.2).Regarding the uniqueness, observe that( H + x )( H − x ) ≡ ν Y j =0 p i,j , ∀ i = 0 , , . . . , ν. Assume there exists some number ˜ p = Q νj =0 p i,j for some i ∈ { , , . . . , ν } whichdivides ( H + x ) and ( H − x ) (without remainder). Thus, ( H + x ) + ( H − x ) ≡ p ⇔ H ≡ p . As ˜ p is a product of prime numbers greater than 2 isfollows that H ≡ p ⇔ P νj =0 θ j ≡ p . However, from the definition of θ j (third condition) it follows that for each j there exist different prime numbersnot present in the prime factorization of θ j contradicting the assumption. Thus,˜ p divides either ( H + x ) or ( H − x ) (without remainder). Define α i = ( +1 if ( H − x ) ≡ Q νj =0 p i,j − H + x ) ≡ Q νj =0 p i,j x ′ = ν X i =0 α i θ i . he Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs 13 In the following, we show that x ′ ≡ x mod Q νj =0 p i,j holds: x ′ ≡ x mod ν Y j =0 p i,j ⇔ ν X i =0 α i θ i ≡ x mod ν Y j =0 p i,j ⇔ α i θ i ≡ x mod ν Y j =0 p i,j ⇔ ν X k =0 α i θ k ≡ x mod ν Y j =0 p i,j ⇔ α i ν X k =0 θ k ≡ x mod ν Y j =0 p i,j ⇔ α i H ≡ x mod ν Y j =0 p i,j for all i ∈ { , , . . . , ν } . The first transformation simply inserts the definition of x ′ .Due to the definition of the θ k , only the summand θ i remains after calculating themodulo. Thus, we can sum up all θ k with arbitrary sign as they equal zero aftercalculating the modulo. In the last step we insert the definition of H . Now weeither have α j = +1. Then H ≡ x mod Q νj =0 p i,j , i. e., H − x ≡ Q νj =0 p i,j ,which is true by definition of α j = +1. Otherwise, α j = −
1. Then − H ≡ x mod Q νj =0 p i,j , i. e., H + x ≡ Q νj =0 p i,j , which is again true by thedefinition of α j . Thus, the initial statement is correct.So, as α j ∈ {− , +1 } for all j ∈ { , , . . . , ν } , it holds that − H ≤ x ≤ H . Since the same holds for x ′ it follows that | x − x ′ | ≤ H . Let us boundthe value of λ j = θ j / ( Q νi =0 ,i = j Q νk =0 p i,k ). It holds that θ j ≤ · p ∗ Q m ′ i =1 p i · Q νi =0 ,i = j Q νk =0 p i,k [27]. Further, for each i ∈ { , , . . . , ν } and j ∈ { , , . . . , ν } it holds that p i,j > (4( ν + 1)2 · p ∗ Q m ′ i =1 p i ) / ( ν +2 ν +1) as we estimated before.Thus, λ j = θ j Q νi =0 ,i = j Q νk =0 p i,k < · p ∗ Q m ′ i =1 p i · Q νi =0 ,i = j Q νk =0 p i,k ) Q νi =0 ,i = j Q νk =0 p i,k = 2 · p ∗ Q m ′ i =1 p i · K/ ( Q νk =0 p j,k ) Q νi =0 ,i = j Q νk =0 p i,k = 2 · p ∗ Q m ′ i =1 p i · K Q νi =0 Q νk =0 p i,k < · p ∗ Q m ′ i =1 p i · K Q νi =0 Q νk =0 p , < · p ∗ Q m ′ i =1 p i · Kp ν +2 ν +10 , ≤ · p ∗ Q m ′ i =1 p i · K (4( ν + 1)2 · p ∗ Q m ′ i =1 p i ) / ( ν +2 ν +1) ) ν +2 ν +1 = 2 · p ∗ Q m ′ i =1 p i · K ν + 1)2 · p ∗ Q m ′ i =1 p i = K ν + 1) . This term bounds each summand of H as it considers their largest value tosatisfies the constraints as well as the modulo when calculating the values (seedefinition of θ k ). It follows that 2 H = 2 P νj =0 θ j < · ( ν + 1) · K/ (2( ν + 1)) = K .Thus, x = x ′ and we conclude that solution of the form x = P νj =0 θ j α j are theunique solutions to the system (P4.1) and (P4.2).Thus, we can re-write ν X j =0 θ j α j ≡ τ mod 2 · p ∗ m ′ Y i =1 p i , α j ∈ {− , +1 } using the system (P4.1) and (P4.2) to the following one:0 ≤ | x | ≤ H , x ∈ Z (P5.1) x ≡ τ mod 2 · p ∗ m ′ Y i =1 p i (P5.2)( H + x )( H − x ) ≡ K (P5.3)proving their equivalence. ⊓⊔ Next, we re-write the system (P5) to:0 ≤ | x | ≤ H , x ∈ Z (P6.1)( τ − x )( τ + x ) ≡ · p ∗ m ′ Y i =1 p i (P6.2)( H + x )( H − x ) ≡ K. (P6.3) Claim.
The Problem (P5) is a yes-instance if and only if the Problem (P6) is ayes-instance.
Proof.
As only the second conditions differ, we focus on their equivalence in thefollowing. First, we prove that if (P5.2) holds, i. e., x ≡ τ mod 2 · p ∗ Q m ′ i =1 p i , he Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs 15 then (P6.2) holds, i. e., ( τ − x )( τ + x ) ≡ · p ∗ Q m ′ i =1 p i . We can re-write(P5.2) to x = λ · p ∗ Q m ′ i =1 p i + τ for some λ ∈ Z . Inserting this in (P6.2) yields:( τ + λ · p ∗ m ′ Y i =1 p i + τ )( τ − λ · p ∗ m ′ Y i =1 p i − τ )= (2 τ + λ · p ∗ m ′ Y i =1 p i )( λ · p ∗ m ′ Y i =1 p i ) ≡ · p ∗ m ′ Y i =1 p i as each factor is multiplied with λ · p ∗ Q m ′ i =1 p i .Next, we prove the opposite direction. First, observe that if ( τ − x )( τ + x ) ≡ · p ∗ Q m ′ i =1 p i then either ( τ − x ) ≡ or ( τ + x ) ≡ :As (P5.2) holds, ( τ + x ) = λ i · i and ( τ − x ) = λ j · j for some i, j ∈ Z and λ i , λ j τ + x ) + ( τ − x ) = λ i · i + λ j · j ⇔ τ = λ i · i + λ j · j ⇔ τ = λ i · i − + λ j · j − . As τ is odd per definition, either i or j has to be 1 and thus the other parameterhas to be 3. Using this, we know that if x satisfies ( P . τ − x ) ≡ or ( τ + x ) ≡ . In the first case, x directly corresponds to a solution of(P6.2) as x − τ is a multiple of 2 and thus x is a multiple of 2 with a residueof τ . Otherwise − x satisfies the condition using the same argument. Obviouslythe other conditions are also satisfied in both systems. ⊓⊔ Lastly, we re-write the system one final time to:0 ≤ x ≤ H , x ∈ Z (QC.1)2 · p ∗ · m ′ Y i =1 p i ( H − x ) + K ( τ − x ) ≡ · p ∗ · m ′ Y i =1 p i · K. (QC.2) Claim.
The Problem (P6) is a yes-instance if and only if the
Quadratic Con-gruences problem is a yes-instance.
Proof.
First, as we only consider x , we can suppose x ≥ · p ∗ · Q m ′ i =1 p i and K are co-prime. The first summand obviously alwayscontains the factor 2 · p ∗ · Q m ′ i =1 p i , thus we have to find an x such that ( H − x ) ≡ K which corresponds to (P6.3). The second summand clearly is a multiple of K , thus we have to assure that ( τ − x ) ≡ · p ∗ · Q m ′ i =1 p i . This matches(P5.2).Dissolving the brackets and rearranging the term (QC.2) we get(2 · p ∗ · m ′ Y i =1 p i + K ) x ≡ Kτ + 2 · p ∗ · m ′ Y i =1 p i H mod 2 · p ∗ · m ′ Y i =1 p i · K. As 2 · p ∗ · Q m ′ i =1 p i + K is relatively prime to 2 · p ∗ · Q m ′ i =1 p i · K it has an inversemodulo 2 · p ∗ · Q m ′ i =1 p i · K [23]. Thus, multiplying by the inverse we get thevalues for α, β and γ as in the transformation above. ⊓⊔ Overall, this proves that satisfying the formula Φ is equivalent to an instanceof the Quadratic Congruences problem admitting a feasible solution.
Running time:
All steps, numbers and their computation can be bounded in apolynomial dependent of n and m . First, we eliminate unnecessary clausesfrom the formula. Thus, we have to go through all clauses once. The first 2 m ′ + 1prime numbers have a value of at most O ( m ′ log( m ′ )) and can thus be foundin polynomial time via sieving. The function (4( ν + 1)2 Q ν +2 ν +1 i =1 q i ) / ( ν +2 ν +1) is at most 32 as shown before. Thus, we can also bound the value of the next ν + 2 ν + 1 prime numbers larger than 32 and p m ′ by a polynomial in n and m and we can compute them efficiently by sieving. All other numbers calculatedin the transformation are a product or sum over these prime numbers (eachoccurring at most once in the calculation) and thus their values are also inpoly( n , m ). We can compute the inverse (2 · p ∗ Q m ′ i =1 p i + K ) − in polynomialtime [23]. ⊓⊔ Now we have proved that the
Quadratic Congruences problem is NP-hard even in the restricted case where all prime factors in β only appear atmost once (except 2). To apply the ETH, however, we also have to estimate thedimensions of the generated instance. Denote by B = b β , . . . , b β n QC n QC the primefactorization of β where b , . . . , b n QC denotes the different prime factors of β and β i the occurrence of b i . The above reduction yields the following parameters: Theorem 2.
An instance of the - SAT problem with n variables and m clausesis reducible to an instance of the Quadratic Congruences problem in polyno-mial time with the properties that α, β, γ ∈ O (( n + m ) ) , n QC ∈ O (( n + m ) ) , max i { b i } ∈ O (( n + m ) log( n + m )) , and each prime factor occurs at mostonce except the prime factor which occurs four times.Proof. In Theorem 1 we already showed and proved a reduction from the 3-
SAT problem to the
Quadratic Congruences problem and argued the running he Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs 17 time. It remains to bound the parameters. To do so, we bound the numbersoccurring in the reduction above in order of their appearance.After eliminating the trivial clauses it obviously holds that m ′ ≤ m and n ′ ≤ n . Next we calculate τ Φ ′ . Its absolute value can be bounded as | τ Φ ′ | = | − m ′ X i =1 i Y j =1 p j | = m ′ X i =1 i Y j =1 p j ≤ m m Y j =1 p j ≤ m m ≤ O ( m ) since the product of the first k prime numbers is bounded by 4 k [12]. Simi-larly, we bound max i {| f + i | , | f − i |} ≤ P x i ∈ σ j Q jk =1 + P ¯ x i ∈ σ j Q jk =1 ≤ m · m ≤ O ( m ) and max j { c j } = max j { Q ji =1 p i , f + j + f − j } ≤ O ( m ) . Per definition, ν = 2 m ′ + n ′ = O ( n + m ). The largest prime number max i { b i } we generate inthe reduction is p ∗ , which is the ( ν + 2 ν + 2 m ′ + 13)th prime number. Thus, itsvalue is bounded by p ∗ ≤ O ( ν log( ν )) = O (( n + m ) log( n + m )) [11]. Dueto the modulo we can bound max j { Θ j } asmax j { Θ j } ≤ · p ∗ m ′ Y i =1 p i · ν Y i =0 ,i = j ν Y k =0 p i,k ≤ O (( n + m ) ) = 4 O (( n + m ) ) . Thus, we estimate H = P νj =0 Θ j ≤ ν · O (( n + m ) ) = 4 O (( n + m ) ) and K = Q νi =0 Q νk =0 p i,k ≤ O (( n + m ) ) . Finally, we can bound the main parameters. As α is bounded by the modulo of β is follows that α ≤ β . Further, β = 2 · p ∗ Q m ′ i =1 p i · K ≤ O (( n + m ) ) . Per definition γ = H and thus γ ≤ O (( n + m ) ) ,which finalizes the estimation of the numbers. ⊓⊔ This sections presents the reduction from the
Quadratic Congruences prob-lem to the 2 -stage ILP problem. First, we present a transformation of aninstance of the
Quadratic Congruences problem to an instance of the
Non-Unique Remainder problem. This problem was not considered so far and servesas an intermediate step in this chapter. However, it might be of independent in-terest as it generalizes the prominent Chinese Remainder theorem. Secondly, weshow how an instance of the
Non-Unique Remainder problem can be modeledas a 2-stage stochastic ILP. Recall that in the
Non-Unique Remainder prob-lem, we are given numbers x , . . . , x n NR , y , . . . , y n NR , q , . . . , q n NR , ζ ∈ N wherethe q i s are pairwise co-prime. The question is to decide whether there exists anatural number z satisfying the following integer linear program and which issmaller or equal to ζ : z mod q ∈ { x , y } z mod q ∈ { x , y } ... z mod q n NR ∈ { x n NR , y n NR } . In other words, we either should met the residue x i or y i . Thus, we can re-writethe equation as z ≡ x i mod q i or z ≡ y i mod q i for all i . Indeed, this problembecomes easy if x i = y i for all i , i. e., we know the remainder we want to satisfyfor each equation [27]: First, compute s i and r i with r i · q i + s i · Q n NR j =1 ,j = i q j = 1 forall i using the Extended Euclidean algorithm. Now it holds that s i · Q n NR j =1 ,j = i q j ≡ q i as q i and Q n NR j =1 ,j = i q j are coprime, and s i · Q n NR j =1 ,j = i q j ≡ q j for j = i . Thus, the smallest solution corresponds to z = P n NR i =1 x i · s i · Q n NR j =1 ,j = i q j due to the Chinese Remainder theorem [27]. Comparing z to the bound ζ finallyyields the answer. Also note that if n NR is constant we can solve the problemby testing all possible vectors ( v , . . . , v n NR ) with v i ∈ { x i , y i } and then use theChinese Remainder theorem as explained above. Theorem 3.
The
Quadratic Congruences problem is reducible to the
Non-Unique Remainder problem in polynomial time with the properties that n NR ∈ O ( n QC ) , max i ∈{ ,...,n NR } { q i , x i , y i } = O (max j ∈{ ,...,n QC } { b β j j } , and ζ ∈ O ( γ ) .Proof. Transformation: Set q = b β , . . . , q n NR = b β QC n QC and ζ = γ where β i denotes the occurrence of the prime factor b i in the prime factorization of β .Compute α i ≡ α mod q i . Set x i = α i if there exists such an x i ∈ Z q i . Further,compute y i = − x i + q i . If there is no such number x i and thus y i , produce atrivial no-instance. Instance size:
The numbers we generate in the reduction equal the prime numbersof the
Quadratic Congruences problem including their occurrence. Hence,it holds that max i ∈{ ,...,n NR } { q i } = O (max j ∈{ ,...,n QC } { b β j j } . Due to the modulo,this value also bounds x i and y i . The upper bound on a solution equals the onesfrom the instance of the Quadratic Congruences problem, i. e., ζ ∈ O ( γ ),and n NR = n QC holds. Correctness:
First, let us verify that producing a trivial no-instance is correct ifwe cannot find some x i . Indeed, this can be traced back to the Chinese Remain-der theorem: If and only if there is an x with x ≡ α mod β and q , . . . , q n NR (i. e., the equivalences to b β i i ) is the prime factorization of β , then x ≡ α i mod q i , α i ∈ Z q i for all i . In other words, it is has to be dividable by all b β i i yielding thesame remainder α (modulo b β i i ). Hence, if there does not exists a square root of α in one of the systems then x ≡ α mod β has no solution. he Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs 19 But if there exists x i and y i , these values are in Z q i as x i ≤ α i < q i perdefinition of x i and α i . Further, both values solve the problem x i , y i ≡ α mod q i as x i ≡ α i mod q i ≡ α i + λ · q i mod q i ≡ α mod q i for some λ ∈ N . Moreover, y i ≡ ( − x i + q i ) mod q i = q i − x i q i + x i mod q i ≡ x i mod q i ≡ α mod q i . The third equation holds as each summand except the last one is a multiple of q i . The last transformation is true due to the computation above.Note that for all prime numbers greater than 2 it holds that x i = y i . Thiscan easily be seen as we already argued that x i and y i are in Z p i . Let us supposeboth values are equal, i. e., x i = y i ⇔ α i = ( − x i + q i ) ⇔ α i = q i − q i x i + x i ⇔ α i = q i − q i x i + α i ⇔ q i x i = q i ⇔ x i = q i . The factor q i is a product of some prime number greater than 2 by the assump-tion above. Thus, there is no x i satisfying the formula.Let us now prove the equivalence of the reduction. ⇒ Let the instance of the
Quadratic Congruences problem be a yes-instance. Then there exists a z satisfying z ≡ α mod β with 0 < z ≤ γ . Thissolution directly corresponds to a solution of the generated instance of the Non-Unique Remainder problem. First, z ≤ γ = ζ . Secondly, z satisfies all equa-tions as it holds that z ≡ α mod β ≡ α mod n NR Y i =1 b β i i ≡ α mod b β i i for all i. The first equivalence holds as the b β i i s are the prime factorization of β . The secondequivalence is true as we can decompose the solution as follows: z = λ · Q ni =1 b β i i + α for some λ ∈ N . Thus, the first summand is not only divided without remainderby Q n NR i =1 b β i i but also by all primes along with their occurrences alone, leavingonly the second summand α as the remainder. Further, since x i , y i ≡ α mod q i as shown before, it holds that z ≡ α mod b β i i ≡ α mod q i ≡ x i ≡ y i for all i. Hence, this satisfies all equations of the generated instance of the
Non-UniqueRemainder problem making it a yes-instance. ⇐ Let the instance of the
Non-Unique Remainder problem be a yes-instance. Hence, we could verify that there exists a solution to the given equationssmaller than ζ . Let this solution be denoted as z ∗ . It holds that z ∗ ≡ x i mod q i or z ≡ y i mod q i . Let v i correspond to the residue that was satisfied, i. e., v i = x i or v i = y i . The solution z ∗ also solves the Quadratic Congruences problem.First, z ∗ ≤ ζ = γ . Further, it holds per definition of the numbers that( z ∗ ) ≡ ( v i ) ≡ α mod q i for all i. As it satisfies all equations simultaneously and the b i are pairwise co-prime, itfollows from the Chinese Remainder theorem that( z ∗ ) ≡ ( v i ) ≡ α mod q i for all i ≡ ( z ∗ ) ≡ α mod n NR Y i =1 q i ≡ α mod n QC Y i =1 b β i i ≡ α mod β as the b β i i s are the prime factorization of β . Running time:
Setting the variables accordingly can be done in time polynomialin n QC . Further, computing each x i , y i can be done in poly-logarithmic timeregarding the largest absolute number for each i ∈ { , . . . , n NR } [5]. ⊓⊔ Finally, we reduce the
Non-Unique Remainder problem to the 2 -stageILP problem. Note that the considered 2 -stage ILP problem is a decisionproblem. In other words, we only seek to determine whether there exists a feasiblesolution. We neither optimize a solution vector nor are we interested in thesolution vector itself.
Theorem 4.
The
Non-Unique Remainder problem is reducible to the -stageILP problem in polynomial time with the properties that n ∈ O ( n NR ) , r, s, t, || c || ∞ , || b || ∞ , || ℓ || ∞ ∈ O (1) , || u || ∞ ∈ O ( ζ ) , and ∆ ∈ O (max i { q i } ) .Proof. Transformation: Having the instance for the
Non-Unique Remainder problem at hand we construct our ILP as follows with n = n NR : A · x = − q x y . . . . . .
00 0 1 1 0 . . . . . . − . . . . . . q n x n y n . . . . . . · x = b = . All variables get a lower bound of 0 and an upper bound of ζ . We can set the ob-jective function arbitrarily as we are just searching for a feasible solution, hencewe set it to c = (0 , , . . . , ⊤ . he Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs 21 Instance size:
Due to our construction, it holds that t = 2 , r = 1 , s = 3. The num-ber n of repeated blocks equals the number n NR of equations in the instance ofthe Non-Unique Remainder problem. The largest entry ∆ can be bounded bymax i { q i } . The lower and upper bounds are at most || u || ∞ = O ( ζ ), || ℓ || ∞ = O (1).The objective function c is set to zero and is thus of constant size. The largestvalue in the right-hand side is || b || ∞ = 1. Correctness: ⇒ Let the given instance o the
Non-Unique Remainder problembe a yes-instance. Thus, there exists a solution z ∗ < ζ satisfying all equations. Asbefore, let v i correspond to the remainder that was satisfied in each equation i ,i. e., v i = x i or v i = y i . A solution to our integer linear program now looks asfollows: Set the first variable to z ∗ . Let the columns corresponding to x i and y i be set as follows for each i : If v i = x i then set this variable occurrence inthe solution vector to 1. Set the occurrence to the corresponding variable of y i to zero. Otherwise set the variables the other way round. Finally, the variablecorresponding to the columns of the q i are computed as ( z ∗ − v i ) /q i . It is easyto see that this solution is feasible and satisfies the bounds on the variable sizes. ⇐ Let the given instance of the 2 -stage ILP problem be a yes-instance. Bydefinition of the constraint matrix we have for every 1 ≤ i ≤ n that there existsa multiple λ i ≥ z = x i + λ i q i or z = y i + λ i q i . Hence z ≡ x i mod q i or z ≡ y i mod q i for every 1 ≤ i ≤ n . Further, z ≤ u . Thus, the solution z is asolution of the Non-Unique Remainder problem.
Running time:
Mapping the variables and computing the values for the q i s canall be done in polynomial time regarding the largest occurring number and n . ⊓⊔ This sections presents the proof that the double exponential running time inthe current state-of-the-art algorithms is nearly tight assuming the ExponentialTime Hypothesis (ETH). To do so, we make use of the reductions above showingthat we can transform an instance of the 3-
SAT problem to an instance of the2 -stage ILP problem.
Corollary 1.
The -stage ILP problem cannot be solved in time less than δ √ n for some δ > assuming ETH.Proof. Suppose the opposite. That is, there is an algorithm solving the 2 -stageILP problem in time less than 2 δ √ n . Let an instance of the 3- SAT problem with n variables and m clauses be given. Due to the Sparsification lemma we mayassume that m ∈ O ( n ) [16]. We can reduce such an instance to an instance ofthe Quadratic Congruences problem in polynomial time regarding n such that n QC ∈ O ( n ), max i { b i } ∈ O ( n log( n )), α, β, γ = 4 O ( n ) , see Theorems 1and 2.Next, we reduce this instance to an instance of the Non-Unique Remain-der problem. Using Theorem 3, this yields the parameter sizes n NR ∈ O ( n ),max i ∈{ ,...,n NR } { q i , x i , y i } = O ( n log( n )), and finally ζ ∈ O ( n ) . Note that allprime numbers greater than 2 appear at most once in the prime factorization of β and 2 appears 4 times. Thus, the largest q i , which corresponds to max i { b β i i } equals the largest prime number in the Quadratic Congruences problem:The largest prime number is at least the ( ν + 2 ν + 2 m ′ + 13) ≥ = 16.Finally, we reduce that instance to an instance of the 2 -stage ILP problemwith parameters r, s, t, || c || ∞ , || b || ∞ , || ℓ || ∞ ∈ O (1), || u || ∞ ∈ O ( n ) , n ∈ O ( n ),and ∆ ∈ O ( n log( n )), see Theorem 4.Hence, if there is an algorithm solving the 2 -stage ILP problem in time lessthan 2 δ √ n this would result in the 3- SAT problem to be solved in time less than2 δ √ n = 2 δ √ C n = 2 δ ( C n )) for some constants C , C . Setting δ ≤ δ/C , thiswould violate the ETH. ⊓⊔ To prove our main result, we still have to reduce the size of the coefficientsin the constraint matrix. To do so, we encode large coefficients into submatri-ces. This reduces the size of the entries greatly while just extending the matrixdimensions slightly. A similar approach was used for example in [20] to prove alower bound for the size of inclusion minimal kern-elements of 2-stage stochasticILPs or in [21] to decrease the value of ∆ in the matrices. Theorem 5.
The -stage ILP problem cannot be solved in time less than δ ( s + t ) | I | O (1) for some constant δ > , even if r = 1 , ∆, || b || ∞ , || c || ∞ , || b || ∞ ∈ O (1) , assuming ETH. Here | I | denotes the encoding length of the total input.Proof. First, we show that we can alter the resulting integer linear programsuch that we reduce the size of ∆ to O (1). We do so by encoding large co-efficients with base 2, which comes at the cost of enlarged dimensions of theconstraint matrix. Let enc( x ) be the encoding of a number x with base 2. Fur-ther, let enc i ( x ) be the i th number of enc( x ). Finally, enc ( x ) denotes thelast significant number of the encoding. Hence, the encoding of a number x is enc( x ) = enc ( x )enc ( x ) . . . enc ⌊ log( ∆ ) ⌋ ( x ) and x can be reconstructed by x = P ⌊ log( ∆ ) ⌋ i =0 enc i ( x ) · i .Let a matrix E be defined as, E = − . . .
00 2 − . . . . . . − . We re-write the constraint matrix as follows: For each coefficient a > a ) and beneath we put the matrix E . Furthermore, we he Double Exponential Runtime is Tight for 2-Stage Stochastic ILPs 23 have to fix the dimensions for the first row in the constraint matrix, the columnswithout great coefficients and the right-hand side b by filling the matrix at thecorresponding positions with zeros. In detail, the altered integer linear program A · x = b looks as follows: Note that the ones beneath the sub-matrices enc( x i ) − q ) enc( x ) enc( y ) 0 . . . . . . E . . . . . . . . . . . . . . . E . . . . . . . . .
00 0 . . . . . . E . . . . . .
00 0 . . . . . . . . . . . . . . . − . . . . . . q n ) enc( x n ) enc( y n )0 0 . . . . . . E . . . . . . . . . . . . . . . E . . .
00 0 . . . . . . . . . . . . E . . . . . . . . . . . . . . . · x = . and enc( y i ) correspond to enc ( x i ) and enc ( y i ). The independent blocks con-sisting of enc( a ) and the matrix E beneath correctly encodes the number a > x a be the number in the solution corre-sponding to the column with entry a of the original instance. The solution forthe altered column (i. e., the sub-matrix) is ( x a · , x a · , . . . , x a · ⌊ log( ∆ ) ⌋ ). Theadditional factor of 2 for each subsequent entry is due to the diagonal of E . Itis easy to see that a · x a = P ⌊ log( ∆ ) ⌋ i =0 enc i ( a ) · x a · i as we can extract x a on theright-hand side and solely the encoding of a remains. Thus, the solutions of theoriginal matrix and the altered one directly transfer to each other. Hence, thesolution space is preserved.Regarding the dimensions, each coefficient a > O (log( ∆ )) × O (log( ∆ ))) matrix. Thus, the dimension expand to t ′ = t + O (log( ∆ )) = O (log( ∆ )), s ′ = s · O (log( ∆ )) = O (log( ∆ )), while r and n stay the same. Further, we haveto adjust the bounds. The lower bound for all new variables is also zero. For theupper bounds we allow an additional factor of 2 i for the i th value of the encod-ing. Thus, || u ′ || ∞ = 2 ⌊ log( ∆ ) ⌋ || u || ∞ . Further, we get that the largest coefficientis bounded by ∆ ′ = O (1). The right-hand side b enlarges to a vector b ′ with O ( n log( ∆ )) entries.Now suppose there is an algorithm solving the 2 -stage ILP problem in timeless than 2 δ ( s + t ) | I | O (1) . The proof of Theorem 1 shows that we can transform aninstance of the 3- SAT problem with n variables and m clauses to an 2-stagestochastic ILP with parameters r, s, t, || c || ∞ , || b || ∞ , || ℓ || ∞ ∈ O (1), || u || ∞ ∈ O ( n ) , n ∈ O ( n ), and ∆ ∈ O ( n log( n )). Further, we explained above that we can transform this ILP to an equivalent one where t ′ = O (log( ∆ )) = O (log( n log( n ))) = O (log( n )) ,s ′ = t · O (log( ∆ )) = O (log( n log( n ))) = O (log( n )) ,∆ ′ = O (1) ,b ′ ∈ Z O ( n log( n )) , || u ′ || ∞ = 2 ⌊ log( ∆ ) ⌋ || u || ∞ = 2 ⌊ log( n log( n )) ⌋ O ( n ) = 4 O ( n ) , while r , and n stay the same. The encoding length | I | is then given by | I | = ( nt ′ ( r ′ + ns ′ )) log( ∆ ′ ) + ( r ′ + ns ′ ) log( || ℓ || ∞ )+( r ′ + ns ′ ) log( || u ′ || ∞ ) + nt ′ log( || b ′ || ∞ ) + ( r ′ + ns ′ ) log( || c || ∞ )= 2 O ( n ) . Hence, if there is an algorithm solving the 2 -stage ILP problem in time lessthan 2 δ ( s + t ) | I | O (1) this would result in the 3- SAT problem to be solved in timeless than2 δ ( s + t ) | I | O (1) = 2 δ ( C n C n n O (1)3 = 2 δC n n O (1)3 = 2 n δ · C n O (1)3 = 2 n δ · C for some constants C , C , C , C . Setting δ = δ ′ /C we get 2 n δC = 2 n δ ′ . As itholds for sufficient large x and ǫ < x ǫ < ǫx it follows that 2 n δ ′ < δ ′ n . Thisviolates the ETH. Note that this result even holds if r = 1, ∆, || c || ∞ , || b || ∞ , || ℓ || ∞ ∈ O (1) as constructed by our reductions. ⊓⊔ References
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