The Complexity Landscape of Two-Stage Robust Selection Problems with Budgeted Uncertainty

The Complexit y Landscap e of T w o-Stage Robust Selection Problems with Budgeted Uncertaint y Marc Go e rigk 1 , Dorothee Henk e 1 , and Lasse W ulf 2 1 Business Decisions and Data Science, Univ ersit y of P assau, German y . {marc.goerigk,dorothee.henke}@uni-passau.de 2 Theoretical Computer Science, IT Univ ersit y of Copenhagen, Denmark. lasw@itu.dk A standard t ype of uncertaint y set in robust optimization is budgeted un certain t y , where an in terv al of p ossible v alues for eac h parameter is giv en and the total deviation from their lo w er b ounds is b ounded. In the t w o-stage setting, discrete and contin uous budgeted uncertain ty ha ve to b e distinguished. The complexity of suc h problems is largely unexplored, in particular if the underlying nominal optimization problem is simple, such as for selection problems. In this pap er, we giv e a comprehensive answ er to long-standing op en complexit y questions for three t ypes of selection problems and three types of budgeted uncertain t y sets. In particular, w e demonstrate that the t wo-stage selection problem with contin uous budgeted uncertain t y is NP-hard, while the corresp onding t w o-stage represen tativ e selection problem is solv able in p olynomial time. Our hardness result implies that also the tw o-stage assignmen t problem with contin uous budgeted uncertaint y is NP-hard. Keyw o rds robust optimization, tw o-stage robustness, budgeted uncertain ty , selec- tion problem, assignment problem, computational complexit y Mathematics Subject Classiﬁcation 90C17, 90C27, 68Q25 1 Intro duction In com binatorial optimization, w e often study problems of the t yp e min x ∈ X n X i =1 c i x i , (Nom) where X ⊆ { 0 , 1 } n denotes the discrete set of feasible solutions, and c ∈ R n ≥ 0 is a cost vector that represen ts the costs of each item that w e can choose. In practice, problem data is often uncertain, whic h has led to the p opular framew ork of robust com binatorial optimization, where the uncertain t y in the cost v ector c is included b y a worst-case p erspective ov er a set of p ossible cost scenarios. The focus of this paper will be on t w o-stage robustness. Before we discuss this framew ork, we introduce the more fundamental setting of single-stage robustness. The robust coun terpart of ( Nom ) then is to solv e min x ∈ X max c ∈ U n X i =1 c i x i (MinMax) 1 for an uncertain t y set U ⊆ R n ≥ 0 that con tains all scenarios against which w e wish to protect the solution x . F or a general ov erview on robust com binatorial optimization, we refer to the recen t b o ok [ GH24 ]. Unfortunately , robust problems ( MinMax ) ma y b ecome harder to solv e than their nominal coun terparts ( Nom ) , ev en for simple uncertaint y sets. Most nominal problems that can b e solved in polynomial time b ecome NP-hard in their robust version, ev en for the case that there are tw o p ossible cost scenarios, i.e., U = { c 1 , c 2 } ; see [ KZ16 ]. As a consequence, there has b een in tense researc h on sp eciﬁc sp ecial cases of X and U that allow better tractability and establish the b oundary cases of complexity . Regarding the set X , v ariants of so-called selection problems ha ve b een studied. In the most general case, we deﬁne the nominal multi-r epr esentative sele ction pr oblem as follo ws. Let a set of n items and a partition T 1 ∪ T 2 ∪ · · · ∪ T m of this set b e given, along with in tegers p j for eac h j ∈ [ m ] = { 1 , . . . , m } suc h that p j ∈ { 1 , . . . , | T j |} . F or a cost v ector c ∈ R n ≥ 0 , the task is to select p j items from each set T j suc h that the total costs are minimized, i.e., to solve min x ∈ X P n i =1 c i x i with X = { x ∈ { 0 , 1 } n : X i ∈ T j x i = p j ∀ j ∈ [ m ] } . T wo sp ecial cases of this problem are the r epr esentative sele ction pr oblem , where p j = 1 for all j ∈ [ m ] , and the sele ction pr oblem , where m = 1 . Robust v ariants of the representativ e selection problem ha ve b een studied, e.g., in [ Büs11 ] and [ DK12 ], while the selection problem has b een studied, e.g., in [ Do e13 ] and [ LL W21 ]. The analysis of types of selection problems in robust optimization pla ys a cen tral role, as they represent one of the easiest nontrivial com binatorial problems that are p ossible. Regarding the set U , a seminal breakthrough has b een achiev ed with the in tro duction of budgeted uncertaint y sets in [ BS03 ; BS04 ], whic h allo w some degree of mo deling ﬂexibilit y , while the complexit y class of nominal combinatorial optimization problems is preserved for their robust coun terpart. Such sets are deﬁned b y an interv al [ c i , c i + d i ] for eac h item with uncertain costs, and an additional b ound Γ on the total cost deviation. Given c i and d i , w e deﬁne c i := c i + d i . In the current literature, three v arian ts of this concept ha v e b een established. The c ontinuous budgete d unc ertainty set is deﬁned as U C = { c ∈ R n ≥ 0 : ∃ δ ∈ [ 0 , 1] n s.t. c i = c i + d i δ i ∀ i ∈ [ n ] , n X i =1 δ i ≤ Γ } , while the discr ete budgete d unc ertainty set is given by U D = { c ∈ R n ≥ 0 : ∃ δ ∈ { 0 , 1 } n s.t. c i = c i + d i δ i ∀ i ∈ [ n ] , n X i =1 δ i ≤ Γ } . Observ e that, in the discrete case, item costs are either at their respective low er b ound or at their upper b ound (in which case w e also sa y that an item is attack ed). In the con tin uous case, it is p ossible to distribute the attac k budget Γ contin uously o v er items. If Γ is an integer, then min-max problems with discrete or contin uous uncertaint y sets result in the same worst-case scenario and th us b oth problems are equiv alent. The third v ariant of budgeted uncertain t y that has been studied in the literature, whic h we call alternative c ontinuous budgete d unc ertainty , is deﬁned as U AC = { c ∈ R n ≥ 0 : ∃ δ ∈ n Y i =1 [0 , d i ] s.t. c i = c i + δ i ∀ i ∈ [ n ] , n X i =1 δ i ≤ Γ } . In this setting, δ do es not represen t the (relative) degree by whic h costs are increased from c to c , but rather the (absolute) amount of cost increase itself. Accordingly , w e can assume Γ ≤ n for the ﬁrst t w o types of uncertain t y , but Γ ≤ P n i =1 d i for the last t yp e of set. In Figure 1 , w e 2 0 c 1 0 c 2 1 2 3 4 1 2 3 4 (a) Con tin uous. 0 c 1 0 c 2 1 2 3 4 1 2 3 4 (b) Discrete. 0 c 1 0 c 2 1 2 3 4 1 2 3 4 (c) Alternativ e con tin uous. Figure 1: Example budgeted uncertain t y sets. visualize the diﬀerences betw een these three sets. Figure 1a shows a con tin uous budgeted set with c = (2 , 1) , d = (2 , 3) , and Γ = 1 . 5 . The discrete budgeted set with the same parameters is presen ted in Figure 1b . Finally , we show the case with Γ = 4 for the alternativ e con tinuous budgeted set in Figure 1c , where the sum of additional costs is b ounded rather than the sum of relativ e deviations. Min-max problems as deﬁned in ( MinMax ) are static in the sense that once a decision is made, it is imp ossible to react to further developmen ts. In tw o-stage robust optimization, as ﬁrst in tro duced in [ Ben+04 ], it is p ossible to decide some v ariables after the uncertaint y has been rev ealed; see also the surv ey [ YGd19 ]. In the con text of combinatorial optimization, we deﬁne the set Y ( x ) as all second-stage solution v ectors that can b e obtained for a given ﬁrst-stage solution x , i.e., Y ( x ) = { y ∈ { 0 , 1 } n : x + y ∈ X } . The set of feasible ﬁrst-stage solutions is deﬁned as those v ectors x where feasible second-stage solutions exist, i.e., X ′ = { x ∈ { 0 , 1 } n : Y ( x )  = ∅} . In the con text of multi-represen tative selection problems, w e ha ve X ′ = { x ∈ { 0 , 1 } n : X i ∈ T j x i ≤ p j ∀ j ∈ [ m ] } and Y ( x ) = { y ∈ { 0 , 1 } n : X i ∈ T j ( x i + y i ) = p j ∀ j ∈ [ m ] , x i + y i ≤ 1 ∀ i ∈ [ n ] } . Then, the tw o-stage robust problem is to solve min x ∈ X ′ n X i =1 C i x i + max c ∈ U min y ∈ Y ( x ) n X i =1 c i y i ! (T woStage) for some given ﬁrst-stage cost v ector C ∈ R n ≥ 0 . T w o-stage selection problems remain some of the few kno wn cases where it is p ossible to ac hiev e polynomial-time solv ability . In [ Cha+18 ], it was shown that the tw o-stage selection problem with alternativ e conti n uous budgeted uncertain ty can be solv ed in p olynomial time. On the other hand, in [ GL W22 ], a pro of has b een given that the same problem with discrete budgeted uncertain t y is NP-hard. In [ GKZ22 ], hardness w as sho wn for general p olyhedral uncertain ty sets in outer description, while the case of inner descriptions w as discuss ed in [ GKZ20 ]. A p olynomial-time algorithm w as presen ted for tw o-stage representativ e selection with alternative con tin uous budgeted uncertaint y in [ GKZ22 ]. T o the b est of our knowledge, no other positive or negative complexity results for tw o-stage selection problems with budgeted uncertain t y are kno wn. Other decision criteria, such as min-max regret, hav e b een studied in the context of selection problems as w ell; see [ AL04 ] for positive and [ A v e01 ] for negative results. T wo-stage selection problems with other t yp es of 3 Discrete (D) Con tin uous (C) Alt. Cont. (A C) Repr. Sel. (2RS) NP-hard (Thm. 19 ) P (Thm. 5 ) P [ GKZ22 ] Sel. (2S) NP-hard [ GL W22 ] NP-hard (Thm. 8 ) P [ Cha+18 ] Multi-Repr. Sel. (2MRS) NP-hard [ GL W22 ] NP-hard (Thm. 8 ) P (Thm. 18 ) T able 1: Ov erview of kno wn and new complexit y results for t wo-stage robust selection problems with budgeted uncertaint y . uncertain t y sets hav e b een studied in [ KZ17 ], and the so-called min-max-min selection problem with con tinuous budgeted uncertaint y was studied in [ BGK24 ]. Single-stage robust v ariants of the selection problem with budgeted uncertaint y , which are motiv ated b y the t wo-stage setting, ha v e b een introduced and in v estigated in [ Lho+25 ]. In this paper, we give a comprehensive answer to all op en complexity questions for tw o-stage robust selection problems with budgeted uncertaint y . T able 1 presen ts an o v erview on known and new complexity results for these problems. Observ e that hardness results for sp ecial cases (selection, represen tative selection) extend to hardness results for the more general case (multi- represen tativ e selection), while p olynomial-time solv ability extends from the general case tow ards the special cases. In particular, we prov e that the complexit y of the selection and represen tativ e selection problems ma y diﬀer, and so do es the complexit y b etw een the tw o v ariants of con tinuous budgeted uncertain ty sets. These are the ﬁrst results of this kind, as far as w e are aw are. Our results also provide a partial answer to the op en problem num b er 7 in [ GH24 ], whic h is to determine the complexity of tw o-stage and reco v erable robust selection problems with contin uous budgeted uncertain t y . Our reduction requires a deep er understanding of ho w the optimal v alue function b eha ves dep ending on the choice of a dual v ariable, which may explain why these complexit y questions hav e remained op en for quite some time (the related recov erable case w as ﬁrst noted as an open problem in [ Büs11 ]). Finally , our main hardness result implies that also the tw o-stage assignmen t problem with con tin uous budgeted uncertain t y is NP-hard (see Corollary 14 ), which gives a partial answer to the open problem n um ber 15 in [ GH24 ]. Throughout the pap er, w e use the follo wing shorthands to denote problems w e consider. W e write “2RS” for t wo-stage represen tative selection, “2S” for tw o-stage selection, and “2MRS” for t w o-stage m ulti-representativ e selection. F or the uncertain t y s ets, w e write “D” for discrete, “C” for con tinuous, and “AC” for alternativ e contin uous budgeted uncertain t y . These tw o problem name comp onen ts are connected with a dash, e.g., w e write (2RS-C) for the t w o-stage robust represen tativ e selection problem with con tin uous budgeted uncertain t y . The remainder of this pap er is structured as follo ws. In Section 2 , we discuss ho w to reformulate the t wo-stage problem ( T woStage ) , whic h is cen tral for the further discussion. In the follo wing sections, we deriv e the complexit y results of T able 1 . W e provide p olynomial-time algorithms for (2RS-C) in Section 3 and discuss the hardness of (2S-C) in Section 4 (which implies hardness of (2MRS-C) as well). W e then sho w that (2MRS-AC) can b e solv ed in p olynomial time in Section 5 , b efore showing hardness of (2RS-D) in Section 6 . W e conclude our pap er in Section 7 . 2 Refo rmulation as a Mixed-Integer Linea r Program In this section, w e derive mixed-in teger linear programming reformulations of the problems (2MRS-C) and (2MRS-AC), following a standard dualization pro cedure. The reform ulations will b e used as the basis for the results in the later sections. As b efore, we denote by n the n um ber of items, which are partitioned into the sets [ n ] = T 1 ∪ · · · ∪ T m . Let ﬁrst-stage costs C ∈ R n ≥ 0 and an uncertaint y set U ⊆ R n ≥ 0 b e giv en. Recall 4 that the tw o-stage problem we consider is of the form min x ∈ X ′ n X i =1 C i x i + max c ∈ U min y ∈ Y ( x ) n X i =1 c i y i ! , where X ′ = { x ∈ { 0 , 1 } n : P i ∈ T j x i ≤ p j ∀ j ∈ [ m ] } and Y ( x ) = { y ∈ { 0 , 1 } n : P i ∈ T j ( x i + y i ) = p j ∀ j ∈ [ m ] , x i + y i ≤ 1 ∀ i ∈ [ n ] } for x ∈ X ′ . Observ e that the second-stage problem min y ∈ Y ( x ) P n i =1 c i y i , for an y ﬁxed x ∈ X ′ and c ∈ U , is equiv alen t to its linear programming relaxation, i.e., the integralit y of y can b e relaxed, b ecause the matrix in the description of Y ( x ) is totally unimo dular. The dual of this linear program can b e written as follows: max m X i =1 α j ( p j − X i ∈ T j x i ) + n X i =1 β i (1 − x i ) s . t . α j + β i ≤ c i ∀ j ∈ [ m ] , i ∈ T j α ∈ R m , β ∈ R n ≤ 0 F or ease of presen tation, let us begin with contin uous budgeted uncertaint y sets. The adv ersarial problem max c ∈ U min y ∈ Y ( x ) P n i =1 c i y i , for an y ﬁxed x ∈ X ′ , is then equiv alent to the follo wing linear program: max m X i =1 α j ( p j − X i ∈ T j x i ) + n X i =1 β i (1 − x i ) s . t . α j + β i ≤ c i + δ i d i ∀ j ∈ [ m ] , i ∈ T j n X i =1 δ i ≤ Γ α ∈ R m , β ∈ R n ≤ 0 , δ ∈ [ 0 , 1] n The dual of this linear program can b e written as follo ws: min n X i =1 c i y i + Γ π + n X i =1 ρ i s . t . X i ∈ T j ( x i + y i ) = p j ∀ j ∈ [ m ] x i + y i ≤ 1 ∀ i ∈ [ n ] π + ρ i ≥ d i y i ∀ i ∈ [ n ] y ∈ [0 , 1] n , π ∈ R ≥ 0 , ρ ∈ R n ≥ 0 Com bining this formulation with the ﬁrst-stage decision ﬁnally results in the following mixed- in teger programming formulation of (2MRS-C): min n X i =1 C i x i + n X i =1 c i y i + Γ π + n X i =1 ρ i (2MRS-C) s . t . X i ∈ T j ( x i + y i ) = p j ∀ j ∈ [ m ] x i + y i ≤ 1 ∀ i ∈ [ n ] π + ρ i ≥ d i y i ∀ i ∈ [ n ] x ∈ { 0 , 1 } n , y ∈ [0 , 1] n , π ∈ R ≥ 0 , ρ ∈ R n ≥ 0 5 F or (2MRS-AC), the same dualization procedure gives the follo wing problem formulation: min n X i =1 C i x i + n X i =1 c i y i + Γ π + n X i =1 d i ρ i (2MRS-A C) s . t . X i ∈ T j ( x i + y i ) = p j ∀ j ∈ [ m ] x i + y i ≤ 1 ∀ i ∈ [ n ] π + ρ i ≥ y i ∀ i ∈ [ n ] x ∈ { 0 , 1 } n , y ∈ [0 , 1] n , π ∈ R ≥ 0 , ρ ∈ R n ≥ 0 Note that the position of the d i co eﬃcien ts has c hanged. F or discrete budgeted uncertain t y sets, compact reform ulations are p ossible (see [ Cha+18 ]), but not necessary for this pap er. 3 Rep resentative Selection with Continuous Budgeted Sets In this section, w e consider problem (2RS-C), i.e., problem ( 2MRS-C ) with p j = 1 for all j ∈ [ m ] . Note that the constraints x i + y i ≤ 1 in ( 2MRS-C ) are redundan t in this case. W e sho w that (2RS-C) can be solved in time O ( n log n ) b y a combinatorial algorithm. As input, w e are giv en cost co eﬃcients C i , c i , d i ∈ R ≥ 0 for i ∈ [ n ] , together with a budget parameter Γ ∈ R ≥ 0 and a partition [ n ] = T 1 ∪ · · · ∪ T m . In order to reduce the technical details necessary to present our result, throughout this section, w e make the simplifying assumption that d i  = 0 for all i ∈ [ n ] . Hence, division b y d i is alw a ys deﬁned. W e explain in the end of the section ho w this assumption can be remov ed. Observ e that, if π ∈ R ≥ 0 is ﬁxed, the problem decomposes into m indep enden t problems, one for eac h j ∈ [ m ] : min X i ∈ T j C i x i + X i ∈ T j c i y i + X i ∈ T j ρ i s . t . X i ∈ T j ( x i + y i ) = 1 π + ρ i ≥ d i y i ∀ i ∈ T j x ∈ { 0 , 1 } T j , y ∈ [0 , 1] T j , ρ ∈ R T j ≥ 0 F or an optimal solution to this problem, there are the following t w o options (whichev er is c heap er): • In the ﬁrst case, one item i ∈ T j is selected in the ﬁrst stage, i.e., x i = 1 and all other x and y v ariables are set to 0 . Then i m ust be an item with minimal ﬁrst-stage cost C i = min k ∈ T j C k . Moreo ver, ρ only has to satisfy the constrain ts ρ i ≥ − π for all i ∈ T j and can therefore b e set to 0 . • In the second case, no item is selected in the ﬁrst stage, i.e., x i = 0 for all i ∈ T j . W e then get an optimal solution of the following problem ( ∗ ) . Denote its optimal v alue, dep ending on π , b y the function f j : R ≥ 0 → R ≥ 0 . f j ( π ) := min X i ∈ T j c i y i + X i ∈ T j ρ i ( ∗ ) s . t . X i ∈ T j y i = 1 π + ρ i ≥ d i y i ∀ i ∈ T j y ∈ [0 , 1] T j , ρ ∈ R T j ≥ 0 6 By the ab o ve observ ations, the minimal ac hiev able v alue of (2RS-C), for any ﬁxed π , can b e describ ed b y the function f : R ≥ 0 → R ≥ 0 with f ( π ) := Γ π + m X j =1 min { min k ∈ T j C k , f j ( π ) } . ( ∗∗ ) The optimal v alue of (2RS-C) is th us given by min π ∈ R ≥ 0 f ( π ) . In the following, we will study the functions f j and sho w that they are piecewise linear and con tin uous. Hence, also the function f is piecewise linear and contin uous, and its minim um can b e found b y iterating o v er its breakp oints. By describing how to compute the sets of the breakp oin ts of the functions f j and then of the function f , w e sho w ho w this can be done in p olynomial time. W e ﬁrst reformulate ( ∗ ) in to the following equiv alent problem: min X i ∈ T j c i y i + X i ∈ T j c i y i ( ∗ ’) s . t . X i ∈ T j ( y i + y i ) = 1 y i ∈ [0 , π i ] ∀ i ∈ T j y i ∈ [0 , 1 − π i ] ∀ i ∈ T j , where π i := min { 1 , π /d i } for all i ∈ T j . Recall that c = c + d . Lemma 1. The pr oblems ( ∗ ) and ( ∗ ’ ) ar e e quivalent. Pr o of. First, let y, ρ b e an optimal solution to ( ∗ ). Deﬁne y i and y i as follo ws, for all i ∈ T j : • If y i ≤ π i , then set y i := y i and y i := 0 . • If y i > π i , then set y i := π i and y i := y i − π i . Note that this alwa ys ensures y = y + y . Therefore, y and y form a feasible solution to ( ∗ ’ ). The abov e tw o cases are closely related to the constrain ts π + ρ i ≥ d i y i in ( ∗ ): • If y i ≤ π i , then y i ≤ π /d i and it is optimal to set ρ i = 0 . • If y i > π i , whic h is only p ossible if π i = π /d i < 1 , then it is optimal to set ρ i = d i y i − π = d i ( y i − π i ) . Hence, w e ma y assume that ρ from the given optimal solution has these prop erties. This implies that the ob jectiv e v alue of the solution y , y in ( ∗ ’ ) is X i ∈ T j c i y i + X i ∈ T j c i y i = X i ∈ T j c i y i + X i ∈ T j d i y i = X i ∈ T j c i y i + X i ∈ T j ,y i >π i d i ( y i − π i ) = X i ∈ T j c i y i + X i ∈ T j ρ i , i.e., the same v alue as the objective v alue of y , ρ in ( ∗ ). F or the other direction, let y and y b e an optimal solution to ( ∗ ’ ) . Set y := y + y , and deﬁne ρ i as follo ws, for all i ∈ T j : • If y i + y i ≤ π i , then set ρ i := 0 . • If y i + y i > π i , then set ρ i := d i y i − π . This is a feasible solution to ( ∗ ). Observ e that, as y and y form an optimal solution of ( ∗ ’ ) and c i ≤ c i , w e ma y assume that the follo wing holds for all i ∈ T j : • If y i + y i ≤ π i , then y i = 0 . 7 • If y i + y i > π i , then y i = π i and π i = π /d i < 1 . Therefore, the ob jectiv e v alue of the solution y , ρ in ( ∗ ) can b e written as X i ∈ T j c i y i + X i ∈ T j ρ i = X i ∈ T j c i ( y i + y i ) + X i ∈ T j ,y i + y i >π i ( d i ( y i + y i ) − π ) = X i ∈ T j c i y i + X i ∈ T j c i y i + X i ∈ T j ,y i + y i >π i ( d i y i − π ) + X i ∈ T j d i y i = X i ∈ T j c i y i + X i ∈ T j c i y i . Th us, b oth problems ( ∗ ) and ( ∗ ’ ) ha v e the same optimal v alue, and any optimal solution to ( ∗ ) can easily b e transformed into an optimal solution to ( ∗ ’ ), and vice v ersa. Scaling the ranges of all v ariables in ( ∗ ’ ) to [0 , 1] leads to another equiv alen t problem formu- lation, which now has the structure of a contin uous knapsack problem (in the minimization v ersion): min X i ∈ T j c i π i z i + X i ∈ T j c i (1 − π i ) z i ( ∗ ”) s . t . X i ∈ T j ( π i z i + (1 − π i ) z i ) = 1 z i , z i ∈ [0 , 1] ∀ i ∈ T j Lemma 2. The pr oblems ( ∗ ’ ) and ( ∗ ” ) ar e e quivalent. Pr o of. First observ e that b oth problems are equiv alent for π = 0 , so let us assume π > 0 . Given an optimal solution y , y to ( ∗ ’ ) , setting z i := y i /π i and z i := y i / (1 − π i ) (or z i := 0 in case π i = 1 ) leads to a feasible solution to ( ∗ ” ) of the same ob jective v alue. Analogously , giv en an optimal solution z , z to ( ∗ ” ) , setting y i := π i z i and y i := (1 − π i ) z i leads to a feasible solution to ( ∗ ’ ) of the same objective v alue. The structure of the problem ( ∗ ” ) is no w simple enough that it can be solved in a greedy fashion, analogously to Dantzig’s algorithm for the con tin uous knapsac k problem [ Dan57 ]: Sort the 2 | T j | items corresp onding to the v ariables z i and z i b y their ratios of cost (i.e., the v ariable’s co eﬃcien t in the ob jective function) and s ize (i.e., the v ariable’s co eﬃcien t in the constrain t), in a nondecreasing order. F ormally , consider the set Z := S i ∈ T j { z i , z i } . F or i ∈ T j , let ratio ( z i ) := c i (whic h is equal to c i π i /π i for π i  = 0 ) and ratio ( z i ) := c i (whic h is equal to c i (1 − π i ) / (1 − π i ) for π i  = 1 ). The algorithm computes an order O Z of Z that sorts the items in Z nondecreasingly b y their ratio. Then it selects items in this order until the total size of 1 in the constraint is reac hed, possibly selecting the last item only fractionally . Observ e that ratio ( z ) is actually independent of π for all z ∈ Z . Hence, when the problem ( ∗ ” ) is solv ed for diﬀerent v alues of π , we may assume that the order O Z of the items do es not c hange. Only the second part of the algorithm b eha v es diﬀeren tly , as the items’ sizes change and therefore the total size of 1 migh t b e reached earlier or later when iterating through the ordered items. Figure 2 sho wcases an example. Based on these insights, w e will now inv estigate in more detail ho w the optimal v alue of ( ∗ ” ) c hanges when π c hanges. Because of Lemmas 1 and 2 , this tells us ho w the function f j b eha v es. F or ease of notation, w e assume T j = [ n j ] for n j := | T j | in the following. Since c i ≤ c i for all i ∈ T j , we may assume, without loss of generality , that the order of the ratios corresp onding to O Z starts with c 1 ≤ c 2 ≤ · · · ≤ c r ≤ c s , for some r ∈ T j = [ n j ] and some s ∈ [ r ] . It will turn out that the algorithm nev er pac ks anything that app ears later in the order, for an y π . W e start by considering large v alues of π and, step by step, study the algorithm’s behavior when decreasing π . (Figure 2 sho ws an example of this evolution and Figure 3a depicts the 8 c 1 c 2 π 1 = 1 3 z 1 = 1 π 2 = 2 3 z 2 = 1 π = 4 3 c 2 c 1 1 − π 2 = 1 3 z 2 = 0 1 − π 1 = 2 3 z 1 = 0 c 2 c 1 π 1 = π 2 = 0 z 1 = z 2 = 1 1 − π 2 = 1 z 2 = 1 1 − π 1 = 1 z 1 = 0 π = 0 c 1 c 2 1 − π 2 = 1 − π 1 = 0 z 2 = z 1 = 0 π 1 = 1 z 1 = 1 π 2 = 1 z 2 = 0 π = 4 c 1 c 2 π 1 = 1 4 z 1 = 1 π 2 = 1 2 z 2 = 1 π = 1 c 2 c 1 1 − π 2 = 1 2 z 2 = 1 2 1 − π 1 = 3 4 z 1 = 0 1 1 1 1 i = 1 i = 2 T 1 c i d i c i 1 +4 5 2 +2 4 Figure 2: Problem ( ∗ ” ) can b e understo o d as a con tin uous min-knapsack problem dependent on parameter π . W e w an t to pack 1 unit size into the knapsack while minimizing the cost. Index i is asso ciated with t w o items of size π i = min { 1 , π /d i } and 1 − π i , resp ectiv ely . W e depict an example instance with m = 1 and | T 1 | = 2 for diﬀeren t v alues of π . The order of the four items is giv en by their ratios c 1 = 1 < c 2 = 2 < c 2 = 4 < c 1 = 5 . corresp onding function f j ( π ) for this example.) If π ≥ d 1 , then π 1 = 1 and the algorithm only pac ks the ﬁrst item, i.e., sets z 1 = 1 and all other z v ariables to 0 . The cost of this solution is c 1 . Hence, for π ≥ d 1 , the function f j is constan t with f j ( π ) = c 1 . When decreasing π , i.e., when setting π = d 1 − ε for some (small) ε > 0 , then π 1 = π /d 1 b ecomes smaller than 1 . Therefore, the algorithm starts to pac k the second item in order to reac h the required total size of 1 . Since b oth the ﬁrst and the second item b ecome smaller when decreasing π , solutions consisting of the ﬁrst item and some (increasing) fraction of the second item are selected un til π is so small that the tw o items hav e a total size of 1. This happ ens when π 1 + π 2 = π /d 1 + π /d 2 = 1 , i.e., at π = 1 / (1 /d 1 + 1 /d 2 ) . F or π ∈ [1 / (1 /d 1 + 1 /d 2 ) , d 1 ] , w e can describ e the solutions and their costs more precisely as follo ws: The fraction of the second item that is needed to ﬁll up to a total size of 1 is z 2 = (1 − π 1 ) /π 2 = (1 − π /d 1 ) /π 2 . The total cost of the solution is therefore f j ( π ) = c 1 π 1 + c 2 π 2 z 2 = c 1 π /d 1 + c 2 (1 − π /d 1 ) = c 2 + π ( c 1 − c 2 ) /d 1 . This sho ws that the function f j is linear with a slop e of ( c 1 − c 2 ) /d 1 ≤ 0 in the range π ∈ [1 / (1 /d 1 + 1 /d 2 ) , d 1 ] . Note that the fraction z 2 that is selected of the second item changes nonlinearly , while the size and the cost of the fractional item change linearly . When decreasing π further, starting from π = 1 / (1 /d 1 + 1 /d 2 ) , the third item starts to be pack ed in a similar fashion. Generalizing the observ ations from the previous paragraphs, w e obtain the follo wing Lemma 3 . Giv en j ∈ [ m ] , let us deﬁne the order O Z = O ( j ) Z as abov e. Set n j := | T j | and let d ( j ) 1 , . . . , d ( j ) n j b e a reordering of the co eﬃcien ts ( d i ) i ∈ T j according to the app earance of the corresponding v ariables ( z i ) i ∈ T j in the order O ( j ) Z . Moreo ver, deﬁne the indices r ( j ) , s ( j ) ∈ [ n j ] as ab ov e, 9 corresp onding to the b eginning of the order O ( j ) Z . Now deﬁne the v alues b ( j ) l := l X l ′ =1 1 d ( j ) l ′ ! − 1 for all l ∈ [ r ( j ) ] , b ( j ) r ( j ) +1 := 0 , and the set B j := n b ( j ) l : l ∈ [ r ( j ) + 1] o . Then B j has size O ( n j ) and we can show the following result. Lemma 3. F or al l j ∈ [ m ] , the function f j is c ontinuous, c onvex, and pie c ewise line ar with br e akp oints in the set B j . The function values in the br e akp oints ar e given by f j ( b ( j ) l ) = b ( j ) l l X l ′ =1 c l ′ d ( j ) l ′ for al l l ∈ [ r ( j ) ] , f j ( b ( j ) r ( j ) +1 ) = c s ( j ) . Pr o of. First, by a general argumen t, eac h function f j is contin uous, con v ex, and piecewise linear. Indeed, b y deﬁnition, its v alues equal the optimal v alues of the linear program ( ∗ ) for v arying v alues of π . F or each ﬁxed π , ( ∗ ) is b ounded, and π app ears in its righ t-hand side vector. More precisely , the right-hand side vector is an aﬃne function with resp ect to π . Therefore, the linear program’s optimal v alue, as a function of π , is piecewise linear and con vex (see, e.g., [ BT97 , Section 5.2]). The rest of this pro of will be concerned with inv estigating the piecewise linear structure in more detail, in order to describ e the breakpoints. Fix some j ∈ [ m ] , and for ease of notation, assume T j = [ n j ] and let r := r ( j ) , s := s ( j ) , and d l ′ := d ( j ) l ′ for l ′ ∈ [ n j ] , i.e., the order O ( j ) Z starts with c 1 ≤ c 2 ≤ · · · ≤ c r ≤ c s . Moreov er, write b l := b ( j ) l for l ∈ [ r + 1] . Note that the v alues b l , as deﬁned ab ov e, are monotonically decreasing in l , i.e., the indices giv e an ordering of the dedicated breakpoints from right to left. Denote the actual breakp oints of the piecewise linear function f j (in the most general case) by ˜ b 1 , ˜ b 2 , . . . , also from righ t to left. W e will argue that these breakp oin ts ˜ b l agree with the v alues b l . Recall that π l ′ = min { 1 , π /d l ′ } for all l ′ ∈ [ n ] . F or π ≥ d 1 , w e ha ve π 1 = 1 . Therefore, in this range, only the ﬁrst item is pac k ed and the function f j ( π ) is constant equal to c 1 . This gives rise to the rightmost breakp oint ˜ b 1 = d 1 = b 1 with f j ( b 1 ) = c 1 . F or l ∈ { 2 , . . . , r } , a solution consisting of the ﬁrst l − 1 items and a fraction of the l -th item is selected in the range π ∈ [ ˜ b l , ˜ b l − 1 ] . The breakp oin t ˜ b l corresp onds to the solution where the ﬁrst l items are pac k ed. It is therefore characterized by the equation P l l ′ =1 π l ′ = P l l ′ =1 π /d l ′ = 1 . This equation is clearly solv ed b y π = 1 / ( P l l ′ =1 1 /d l ′ ) = b l . Hence, ˜ b l = b l for all l ∈ [ r ] . The cost of the solution where the ﬁrst l ∈ [ r ] items are pack ed, i.e., the v alue of f j at π = b l , is giv en b y f j ( b l ) = l X l ′ =1 c l ′ π l ′ = l X l ′ =1 c l ′ π d l ′ = l X l ′ =1 c l ′ b l d l ′ = b l l X l ′ =1 c l ′ d l ′ . Finally , when decreasing π b elo w b r , w e start pac king the item z s . Note that no item after z s is pac k ed: The total size of all items in the order up to z s is alw a ys at least 1 b ecause the items z s and z s ha v e a total size of π s + (1 − π s ) = 1 already , for an y π . When π reac hes 0 , the size 1 − π s of the item z s b ecomes 1 and all previous items v anish (they ha ve size and cost 0 now). This leads to the left end of the function f j : The leftmost linear piece, corresp onding to the item z s , has the range π ∈ [0 , b r ] . Hence, the leftmost breakp oin t is given by ˜ b r +1 = 0 = b r +1 and f j ( b r +1 ) = c s . Recall that the functions f j are a part of the description of the function f ; see ( ∗∗ ) . W e can no w deriv e that also the function f is contin uous and piecewise linear: 10 1 4 1 5 f j ( π ) π (a) F unction f j is piecewise lin- ear and con v ex. 1 4 1 5 π min { min k ∈ T j C k , f j ( π ) } (b) T aking the point wise mini- m um of f j and a constant in tro duces at most one ad- ditional breakp oin t. 1 4 1 5 π f ( π ) = Γ π + P m j =1 min { min k ∈ T j C k , f j ( π ) } (c) F unction f is piecewise lin- ear as a sum of the piecewise linear functions from (b) for all j ∈ [ m ] and the linear function Γ π . Figure 3: Illustration of the piecewise linear functions f j and f . F or every j ∈ [ m ] , the term min k ∈ T j C k is a constan t, therefore min { min k ∈ T j C k , f j ( π ) } arises from f j ( π ) b y replacing its left part that is larger than this constant (if it exists) b y a constant function. Note that the resulting function is not con v ex anymore and a new breakpoint b ⋆ j migh t arise at the intersection of f j and the constant function (see Figure 3b ). In the next step, several functions of this type, one for eac h j ∈ [ m ] , are (p oint wise) added. This leads to a con tinuous piecewise linear function again. Each breakp oint of one of the functions can lead to a breakpoint in their sum. Finally , the linear function Γ π is added, which also preserves con tinuit y and the piecewise linear structure (see Figure 3c ). Thus, we hav e: Lemma 4. The function f is c ontinuous and pie c ewise line ar. Its O ( n ) br e akp oints ar e c ontaine d in the set S j ∈ [ m ] ( B j ∪ { b ⋆ j } ) . In order to solve the problem (2RS-C), w e need to determine the minim um of the function f o v er all π ≥ 0 . This can be done b y iterating through all its breakpoints. Theorem 5. The pr oblem (2RS-C) c an b e solve d in O ( n log n ) time. Pr o of. F or each j ∈ [ m ] , w e can use a sorting algorithm to compute the order O ( j ) Z , and therefore compute the set B j of breakpoints of f j in O ( n j log n j ) time. Note that, since B j is deﬁned in terms of partial sums, we can compute the whole set B j in only O ( n j ) time once the order is giv en. F or eac h breakp oint b ∈ B j , w e compute and store also the corresp onding v alue f j ( b ) . By Lemma 3 , also these v alues can b e expressed using partial sums. Hence, they can be computed in a total time of O ( n j ) as well. By iterating o v er the breakp oints B j in O ( n j ) time, w e can ﬁnd the linear piece that intersects with the constan t function min k ∈ T j C k and compute the additional breakp oint b ⋆ j (or decide that b ⋆ j do es not exist). This deﬁnes the function g j ( π ) := min { min k ∈ T j C k , f j ( π ) } with breakpoints B ′ j := { 0 , b ⋆ j } ∪ { b ∈ B j : b > b ⋆ j } . W e then obtain the set B := S j ∈ [ m ] B ′ j of breakp oints of f and sort it in O ( n log n ) time. Finally , we need to ev aluate f ( π ) for all π ∈ B in order to ﬁnd the minim um π ⋆ . F or this, w e ﬁrst compute the v alue f (0) = P m j =1 g j (0) and then iterate o v er all O ( n ) breakpoints in B in increasing order, determining the v alue of f at eac h of them. In order to p erform eac h ev aluation in constant time, w e keep trac k of the current slop e of f . The slop e of a linear piece of f is the sum of Γ and the sum of the slop es of the linear pieces of all g j in the corresponding range. Eac h slope of a linear piece of g j can be obtained from the tw o adjacent breakp oints and their function v alues in O (1) time. In each iteration corresp onding to a breakp oin t in B , w e adjust the curren t slop e of f according to the slop e c hange of the function g j whose breakpoint is curren tly considered. This enables a total running time of O ( n ) for the whole loop. 11 In order to compute the corresp onding ﬁrst-stage selection x ∈ { 0 , 1 } n that attains the same minim um for (2RS-C), we need to keep trac k of the set { j ∈ [ m ] : b ⋆ j ≥ π } during the search for π ⋆ . This set for π = π ⋆ describ es for which T j the ﬁrst-stage solution x should select a c heap est item (with costs min k ∈ T j C k ), while all other x v ariables should be set to 0 . Hence, x can b e determined in O ( n ) in the end of the algorithm. Discussion of the case d i = 0 . Finally , we talk ab out the case where our simplifying assumption do es not hold and we ha ve d i = 0 for at least one i ∈ [ n ] . In this case, π /d i is undeﬁned. Ho w ev er, it can b e sho wn that all our arguments are still v alid. T o this end, we deﬁne π i := min { 1 , π /d i } if d i  = 0 , and π i := 1 if d i = 0 . Then it can b e sho wn that Lemma 1 is still true. Informally , this is due to the fact that π ≥ 0 and so π /d i can be in terpreted as p ositive inﬁnity . Lemma 2 remains true since it describ es only a linear scaling. Lemma 3 remains true for the following reason. Consider j ∈ [ m ] suc h that for some i ∈ T j w e hav e d i = 0 . Observ e that ratio ( z i ) = c i = c i = ratio ( z i ) are well-deﬁned, even if d i = 0 . Hence, the order O ( j ) Z is also w ell-deﬁned and we ma y assume that z i , z i are consecutiv e items in O ( j ) Z . Analogously to before, without loss of generalit y , assume the indexing of the items in T j to b e such that c 1 ≤ c 2 ≤ · · · ≤ c r ≤ c s where c s is the ﬁrst o ccurrence of some item z s in the order ( s ≤ r ). Then w e ha v e r ≤ i since the items z i , z i are consecutiv e. W e can then consider the deﬁnition of the breakp oin ts b ( j ) l . F or all l < r , there is no problem since we do not divide b y 0. F or the last case of l = r , w e can encoun ter a division b y 0. Then it can b e sho wn that this v alue can b e treated as a breakp oint at “ 0 = ∞ − 1 ”, which is already included in the set B j (as b r +1 = 0 ). Lemma 4 and Theorem 5 do not dep end on d i  = 0 . In total, w e ha ve sho wn ho w to solv e (2RS-C) in O ( n log n ) time, ev en if d i = 0 for some i ∈ [ n ] . 4 Selection with Continuous Budgeted Sets In this section, w e consider the problem (2S-C), i.e., problem ( 2MRS-C ) with m = 1 , and pro v e its NP-hardness. Let us ﬁrst recall the problem setting. Giv en are a set of n items, denoted b y [ n ] = { 1 , . . . , n } , a num b er p ∈ [ n ] , ﬁrst-stage costs C ∈ R n ≥ 0 , and an uncertain t y set U ⊆ R n ≥ 0 of possible second-stage costs, giv en as U = { c ∈ R n ≥ 0 : ∃ δ ∈ [ 0 , 1] n s.t. c i = c i + d i δ i ∀ i ∈ [ n ] , n X i =1 δ i ≤ Γ } with c, d ∈ R n ≥ 0 and Γ ∈ R ≥ 0 . As b efore, let c = c + d . The task is to select p items from [ n ] , either in the ﬁrst stage, where item i has a cost of C i , or in the second stage, where item i has a cost of c i , after the (worst-case) second-stage costs c ∈ U ha v e b een realized. As a sp ecial case of the formulation ( 2MRS-C ) derived in Section 2 , the problem can b e written as: min n X i =1 C i x i + n X i =1 c i y i + Γ π + n X i =1 ρ i (2S-C) s . t . n X i =1 ( x i + y i ) = p x i + y i ≤ 1 ∀ i ∈ [ n ] π + ρ i ≥ d i y i ∀ i ∈ [ n ] x ∈ { 0 , 1 } n , y ∈ [0 , 1] n , π ∈ R ≥ 0 , ρ ∈ R n ≥ 0 In order to understand this problem, w e will study it for ﬁxed x and ﬁxed π . F or all ﬁrst-stage solutions x ∈ { 0 , 1 } n that are feasible for ( 2S-C ) , i.e., with P n i =1 x i ≤ p , we deﬁne the set 12 R x := { i ∈ [ n ] : x i = 0 } of items not selected in the ﬁrst stage and the remaining budget p x := p − P n i =1 x i . Observe that, for an item i ∈ [ n ] with x i = 1 , the constrain t x i + y i ≤ 1 forces the corresponding v ariable y i to b e 0 , and the corresp onding v ariable ρ i is set to 0 in any optimal solution as w ell. Therefore, when x is ﬁxed, we may omit these v ariables and fo cus on the indices in R x . W e no w introduce a function g that describ es the optimal v alues of the remaining problem when x and π are ﬁxed. Note that the parts of the ob jective of ( 2S-C ) that are constant for ﬁxed x and π , namely P n i =1 C i x i + Γ π , are omitted in this deﬁnition. F or all x ∈ { 0 , 1 } n with P n i =1 x i ≤ p and all π ∈ R ≥ 0 , w e set: g ( x, π ) := min X i ∈ R x c i y i + X i ∈ R x ρ i ( △ ) s . t . X i ∈ R x y i = p x π + ρ i ≥ d i y i y ∈ [0 , 1] R x , ρ ∈ R n ≥ 0 Then the following problem formulation is equiv alent to problem ( 2S-C ): min n X i =1 C i x i + Γ π + g ( x, π ) (2S-C’) s . t . n X i =1 x i ≤ p x ∈ { 0 , 1 } n , π ∈ R ≥ 0 Similarly to Section 3 , w e can rewrite ( △ ) as follo ws. First observe that the optimal v alue for ρ satisﬁes ρ i = max { 0 , d i y i − π } for all i ∈ [ n ] . Splitting each v ariable y i in to separate v ariables y i and y i , representing the tw o cases where d i y i − π ≤ 0 and d i y i − π > 0 , results in the following form ulation (see also ( ∗ ’ ) and Lemma 1 ): g ( x, π ) = min X i ∈ R x c i y i + X i ∈ R x c i y i s . t . X i ∈ R x ( y i + y i ) = p x y i ∈ [0 , π i ] ∀ i ∈ R x y i ∈ [0 , 1 − π i ] ∀ i ∈ R x Here, we again set π i := min { 1 , π /d i } for all i ∈ [ n ] . In the construction that w e use later, we ensure that d i > 0 for all i ∈ [ n ] , whic h means that the term is alwa ys well-deﬁned. In the next step, we scale the ranges of the v ariables y and y , lik e in ( ∗ ” ) and Lemma 2 , and get a contin uous min-knapsack problem: g ( x, π ) = min X i ∈ R x c i π i z i + X i ∈ R x c i (1 − π i ) z i ( △ ’) s . t . X i ∈ R x ( π i z i + (1 − π i ) z i ) = p x z i , z i ∈ [0 , 1] ∀ i ∈ R x Our aim is to prov e that the problem ( 2S-C’ ) is NP-hard, b oth in general and for a ﬁxed v alue of π . W e will use the sp ecial structure of the con tinuous min-knapsack problem ( △ ’ ) in order to understand how its optimal solution c hanges when selecting diﬀeren t ﬁrst-stage solutions x ∈ { 0 , 1 } n and diﬀeren t v alues π ∈ R ≥ 0 . Lik e in Section 3 , an optimal solution to such a 13 con tin uous min-knapsack problem can b e found b y sorting the items by their ratios of cost and size, whic h are again giv en by ratio ( z i ) := c i π i /π i = c i and ratio ( z i ) := c i (1 − π i ) / (1 − π i ) = c i here, and then selecting items in this order until the capacit y p x is reached. Note that this order of all 2 n items (corresp onding to z i and z i for all i ∈ [ n ] ) is indep endent of x and of π . Diﬀerent ﬁrst-stage solutions x lead to diﬀeren t subsets of these 2 n items b eing considered (corresponding to R x ) and diﬀerent capacities p x . Diﬀerent v alues of π lead to diﬀerent sizes π i and 1 − π i of the items, and corresp onding costs c i π i and c i (1 − π i ) . The following result describes the b ehavior of the contin uous min-knapsack problem ( △ ’ ) for v arying v alues of π , i.e., the b eha vior of the function g ( x, · ) : R ≥ 0 → R ≥ 0 in π . It is a generalization of Lemma 3 . Note that the only diﬀerence b etw een the contin uous min-knapsac k problems ( ∗ ” ) and ( △ ’ ) concerns the knapsac k’s capacity . Lemma 6. F or every x ∈ { 0 , 1 } n with P n i =1 x i ≤ p , the function g ( x, · ) is c ontinuous, c onvex, and pie c ewise line ar. Pr o of. Consider the deﬁnition of the function g ( x, · ) in ( △ ) . Analogously to the pro of of Lemma 3 , this problem is a b ounded linear program for eac h ﬁxed π , where π app ears in the righ t-hand side vector. Therefore, by general linear programming arguments (see, e.g., [ BT97 , Section 5.2]), g ( x, · ) is a piecewise linear and conv ex function in π . The function g ( x, · ) can be easily transformed in to a function h x : R ≥ 0 → R ≥ 0 that describ es the objective v alues of ( 2S-C’ ) for the ﬁxed x and v arying v alues of π . Indeed, we only need to add the term P n i =1 C i x i , whic h is constan t for ﬁxed x , and the linear function Γ π , i.e., we set, for all π ∈ R ≥ 0 : h x ( π ) := n X i =1 C i x i + Γ π + g ( x, π ) Clearly , the structural prop erties of the function g ( x, · ) are preserv ed b y these op erations. Thus, w e ha ve: Lemma 7. F or every x ∈ { 0 , 1 } n with P n i =1 x i ≤ p , the function h x is c ontinuous, c onvex, and pie c ewise line ar. In the pro of b elow, w e will mak e use of these prop erties when iden tifying the minim um of suc h a function h x . More sp eciﬁcally , we will show that the minim um is attained for some sp eciﬁc π ⋆ b y analyzing the slop es of the linear pieces left and righ t of π ⋆ . If the slop e on the left is negativ e and the slop e on the right is p ositive, then we hav e found the unique minimum due to the conv exity . Iden tifying the minima of the functions h x is important b ecause the optimal v alue of ( 2S-C’ ) correspon ds to the minimum, ov er all feasible x , of these minima. Based on these general structural observ ations, we now present our hardness result: Theorem 8. The pr oblem (2S-C) is NP-har d. In order to prov e Theorem 8 , w e give a reduction from the NP-hard (binary) knapsack problem (see [ Kar72 ]) to the equiv alent reform ulation ( 2S-C’ ) of the problem (2S-C). Let an instance a, v ∈ N m , b ∈ N of the knapsack problem be given. The goal is to select a subset S ⊆ [ m ] of the items such that their total size ﬁts in the knapsack, i.e., P i ∈ S a i ≤ b , and their total v alue P i ∈ S v i is maximized. W e can assume without loss of generality that a i ≤ b − 1 for all i ∈ [ m ] . Moreo v er, we assume that P m i =1 a i is divisible by b (whic h is without loss of generality , as w e can alw ays add one dummy item with v alue 0 for this purp ose). W e scale the sizes of the items and of the knapsack b y the factor 1 /b and deﬁne a ′ i := a i /b for all i ∈ [ m ] . Then the original constraint P i ∈ S a i ≤ b is equiv alent to P i ∈ S a ′ i ≤ 1 . Our assumptions imply that a ′ i ≤ 1 − 1 /b for all i ∈ [ m ] and that A ′ := P m i =1 a ′ i < m is an in teger. Moreo v er, set V := P m i =1 v i . 14 item i 1 2 3 4 P i ∈ S a i P i ∈ S v i feasible? robust v alue a i 2 3 4 5 a ′ i 2 / 7 3 / 7 4 / 7 5 / 7 v i 3 6 7 9 S 1 0 0 0 0 0 0 ✓ 8425.00 S 2 0 0 0 1 5 9 ✓ 8416.00 S 3 0 0 1 0 4 7 ✓ 8418.00 S 4 0 0 1 1 9 16 8667.97 S 5 0 1 0 0 3 6 ✓ 8419.00 S 6 0 1 0 1 8 15 8534.72 S 7 0 1 1 0 7 13 ✓ 8412.00 S 8 0 1 1 1 12 22 8948.00 S 9 1 0 0 0 2 3 ✓ 8422.00 S 10 1 0 0 1 7 12 ✓ 8413.00 S 11 1 0 1 0 6 10 ✓ 8415.00 S 12 1 0 1 1 11 19 8817.75 S 13 1 1 0 0 5 9 ✓ 8416.00 S 14 1 1 0 1 10 18 8696.65 S 15 1 1 1 0 9 16 8588.40 S 16 1 1 1 1 14 25 8925.00 T able 2: Example knapsac k instance and solution candidates. W e now deﬁne an instance of the problem ( 2S-C’ ) as follo ws. It consists of n := 2 m − A ′ + 3 items: one for each knapsack item, one additional “exp ensiv e” item, and m − A ′ + 2 additional “c heap” items. Let M := 3 mbV b e a large constant. The items’ costs are deﬁned b y C i := M a ′ i , c i := v i / (1 − a ′ i ) , d i := M / (1 − a ′ i ) , for i ∈ { 1 , . . . , m } , C i := ( m + 3) M , c i := M , d i := M , for i = m + 1 , C i := ( m + 3) M , c i := 0 , d i := M , for i ∈ { m + 2 , . . . , n } . Finally , set the n um ber of items to select as p := 2 m − 2 A ′ + 3 and the uncertaint y budget as Γ := m − A ′ + 1 . T o ease the follo wing discussion, we illustrate this construction with some sp eciﬁc v alues. Example 9. W e consider a knapsac k instance with m = 4 items, for which the item sizes and v alues are given in T able 2 . W e use b = 7 . A dditionally , we sho w all p ossible solution candidates S 1 , . . . , S 16 for this knapsack instance, along with their sizes and v alues. Only some of these candidates are indeed feasible, whic h is marked in the column “feasible?” . In T able 3 , w e sho w the tw o-stage selection instance that is constructed in our reduction. By our deﬁnitions, we obtain A ′ = 2 , V = 25 , n = 9 , p = 7 , Γ = 3 , and M = 2100 . Observ e that selecting any of the last ﬁv e items in the ﬁrst stage immediately results in high costs. F or eac h of the solution candidates that only pick ﬁrst-stage items amongst the ﬁrst four items, there is a corresp onding knapsack solution candidate. In T able 2 , the ﬁnal column lab eled “robust v alue” shows the ob jective v alue of this solution for the t w o-stage selection problem. W e can observe that knapsack solution candidates that are infeasible alwa ys result in higher ob jective v alues than solution candidates that are feasible. F urthermore, we note that the best knapsack solution ( S 7 , with v alue 13) corresp onds to the best tw o-stage selection 15 knapsac k items exp ensiv e item c heap items i 1 2 3 4 5 6 7 8 9 C i 600 900 1200 1500 14700 14700 14700 14700 14700 c i 4 . 2 10 . 5 16 . 3 31 . 5 2100 0 0 0 0 d i 2940 3675 4900 7350 2100 2100 2100 2100 2100 T able 3: Example t w o-stage selection instance constructed in our reduction. solution (with v alue 8412). Indeed, w e show b elow that, for feasible knapsack solutions x , the ob jectiv e v alue of the corresponding solution of the constructed tw o-stage selection instance is (Γ + 1) M + V − P m i =1 v i x i = 8425 − P m i =1 v i x i . With this example in mind, we no w return to the pro of of Theorem 8 . W e kno w from Lemma 7 that, for ev ery ﬁxed feasible x , the ob jectiv e v alue of ( 2S-C’ ) is a con tin uous, con v ex, and piecewise linear function h x in π . The optimal v alue of ( 2S-C’ ) can b e written as the minimum of the minima of all these functions, i.e., as min x ∈{ 0 , 1 } n : P n i =1 x i ≤ p min π ∈ R ≥ 0 h x ( π ) . In the following, we will analyze the functions h x , ﬁrst for x that corresp ond to feasible knapsack solutions (Lemma 10 ) and second for x that do not correspond to feasible knapsack solutions (Lemma 11 ). In the former case, we will show that the minimal v alue of h x is negativ ely related to the knapsack solution’s v alue. In the latter case, we will sho w that the minimal v alue of h x is alw a ys larger than the minimal v alues o ccurring in the ﬁrst case. T ogether, this prov es that the optim um of ( 2S-C’ ) corresp onds to an optimal knapsack solution, ﬁnishing our reduction. Lemma 10. L et x ∈ { 0 , 1 } n b e a ve ctor such that x 1 , . . . , x m enc o des a fe asible solution of the given knapsack instanc e, i.e., such that P m i =1 a ′ i x i ≤ 1 , and x i = 0 for al l i ∈ { m + 1 , . . . , n } . Then, the fol lowing holds: (a) x is fe asible for pr oblem ( 2S-C’ ) , i.e., P m i =1 x i ≤ p . (b) When ﬁxing x in ( 2S-C’ ) , the unique optimal value for π is π = M , i.e., argmin π ∈ R ≥ 0 h x ( π ) = { M } . (c) When ﬁxing x in ( 2S-C’ ) , the optimal obje ctive value is given by min π ∈ R ≥ 0 h x ( π ) = (Γ + 1) M + V − P m i =1 v i x i . Pr o of. W e ﬁrst show (a) . Due to P m i =1 a ′ i x i ≤ 1 and the assumption that a ′ i < 1 holds for all i ∈ [ m ] , we hav e m X i =1 x i + A ′ = m X i =1 x i + m X i =1 a ′ i x i + m X i =1 a ′ i (1 − x i ) ≤ m X i =1 x i + 1 + m X i =1 (1 − x i ) = m + 1 , and therefore P m i =1 x i ≤ m − A ′ + 1 < p . Hence, x is feasible for problem ( 2S-C’ ). Next, w e prov e that h x ( M ) = (Γ + 1) M + V − P m i =1 v i x i . After that, we will argue that this is in fact the unique minim um of h x , whic h then implies (b) and (c) . In order to determine h x ( M ) , w e study the b ehavior of the con tin uous min-knapsack prob- lem ( △ ’ ) for our ﬁxed x and the ﬁxed v alue π = M . As explained abov e, an optimal solution of this problem can b e describ ed by sorting its items by the v alues c and c . Note that, if the sorting is not unique due to sev eral of the v alues b eing equal, any order of them works. Here, the items corresp onding to c i = 0 for i ∈ { m + 2 , . . . , n } come ﬁrst in this order, follo w ed b y the items corresp onding to c i for i ∈ [ m ] ∩ R x (and we are not in terested in their precise order). Note that, 16 for all i ∈ [ m ] , we hav e 0 ≤ c i = v i / (1 − a ′ i ) ≤ bv i < M . Next are the items corresponding to c m +1 = M and c i = M for i ∈ { m + 2 , . . . , n } . Finally , the items corresp onding to c i > M for i ∈ [ m + 1] ∩ R x follo w (again, w e are not in terested in their precise order). W e will see that the items corresponding to c i are nev er selected. Regarding the sizes of the items in ( △ ’ ) , note that w e ha v e π ≤ d i for all i ∈ [ n ] , and therefore π i = π /d i = M / ( M / (1 − a ′ i )) = 1 − a ′ i for all i ∈ [ m ] and π i = M / M = 1 for all i ∈ { m + 1 , . . . , n } . W e now inv estigate when the capacity p x is reac hed when adding the sizes ( π i or (1 − π i ) ) of the items in the given order. The items corresp onding to c i for i ∈ R x \ { m + 1 } hav e a total size of X i ∈ R x \{ m +1 } π i = ( m − A ′ + 2) + m X i =1 π i (1 − x i ) = ( m − A ′ + 2) + m X i =1 (1 − a ′ i )(1 − x i ) = ( m − A ′ + 2) + m − m X i =1 a ′ i − m X i =1 x i + m X i =1 a ′ i x i = 2 m − 2 A ′ + 2 − m X i =1 x i + m X i =1 a ′ i x i = p − 1 − m X i =1 x i + m X i =1 a ′ i x i = p x + m X i =1 a ′ i x i − 1 ≤ p x . Hence, all of these items are pac ked, i.e., z i = 1 for all i ∈ R x \ { m + 1 } . It remains to select a fraction of 1 − P m i =1 a ′ i x i ≥ 0 of the next item in the order, which corresp onds to c m +1 and has size π m +1 = 1 . Hence, w e set z m +1 = 1 − P m i =1 a ′ i x i . Therefore, the optimal v alue of ( △ ’ ) can b e written as g ( x, M ) = X i ∈ R x c i π i z i = m X i =1 c i π i (1 − x i ) + c m +1 π m +1 z m +1 = m X i =1 v i (1 − x i ) + M (1 − m X i =1 a ′ i x i ) . T ogether with the ﬁrst-stage costs P m i =1 C i x i = M P m i =1 a ′ i x i , the total ob jectiv e v alue of the solution consisting of x and π = M in ( 2S-C’ ) is h x ( M ) = m X i =1 C i x i + Γ M + g ( x, M ) = (Γ + 1) M + V − m X i =1 v i x i . It remains to show that this is the unique minimum of the function h x . F or this, recall from Lemma 7 that h x is con tin uous, con v ex, and piecewise linear. W e will show that π = M is a breakp oin t of h x , the linear piece left of π = M has negativ e slope, and the linear piece righ t of π = M has p ositive slope. As h x is con vex, this implies that π = M is the unique minim um. Recall that changing π c hanges the item sizes and costs in the con tin uous min-knapsac k problem ( △ ’ ) , while the ordering of the items remains the same. Let us ﬁrst consider the situation where π is sligh tly smaller than M . The intuition is that all of the items that are pac k ed in the solution described ab ov e, corresp onding to c i for i ∈ R x , become slightly smaller as w ell, whic h leads to a sligh tly larger fraction z m +1 of the exp ensive fractional item b eing pac k ed, thus increasing the total objective v alue. More precisely , when decreasing π , the size π i = π /d i of eac h item corresp onding to some c i decreases linearly at a rate of 1 /d i . The total size of the fully pac k ed items, corresp onding to c i for all i ∈ R x \ { m + 1 } , th us decreases at a rate of P i ∈ R x \{ m +1 } 1 /d i . Accordingly , the size of the additionally pack ed fraction of the item corresp onding to c m +1 = M increases at the same rate. The total cost of the solution then 17 c hanges linearly as well, at a rate of − X i ∈ R x \{ m +1 } c i d i + c m +1 X i ∈ R x \{ m +1 } 1 d i = X i ∈ R x \{ m +1 } c m +1 − c i d i = ( m − A ′ + 2) M − 0 M + m X i =1 (1 − x i ) M − v i / (1 − a ′ i ) M / (1 − a ′ i ) = Γ + 1 + m X i =1 (1 − x i )(1 − a ′ i ) − m X i =1 (1 − x i ) v i M ≥ Γ + 1 − V M > Γ . Hence, the slop e of the linear piece left of π = M is less than − Γ for the function g ( x, · ) and negativ e for the function h x , whic h includes the additional linear function Γ π . Similarly , consider the situation where π is slightly larger than M . Here, the items cor- resp onding to c i for i ∈ [ m ] ∩ R x b ecome slightly larger, while the size of the cheap items i ∈ { m + 2 , . . . , n } is constant π i = min { 1 , π / M } = 1 now. This leads to a slightly smaller fraction z m +1 of the expensive fractional item being pack ed or, if z m +1 = 0 holds already for π = M (i.e., if P m i =1 a ′ i x i = 1 ), to some small fraction of the previous item in the order being unpac k ed. This previous item corresp onds to max i ∈ [ m ] ∩ R x c i < M . Similarly as ab ov e, the c hange in the total cost of the solution is linear again and happ ens at a rate of at least X i ∈ [ m ] ∩ R x c i d i − M X i ∈ [ m ] ∩ R x 1 d i = m X i =1 (1 − x i ) c i − M d i = m X i =1 (1 − x i ) v i / (1 − a ′ i ) − M M / (1 − a ′ i ) = m X i =1 (1 − x i ) v i M − m X i =1 (1 − x i )(1 − a ′ i ) = m X i =1 (1 − x i ) v i M − m + m X i =1 x i + m X i =1 a ′ i − m X i =1 a ′ i x i > − m + A ′ − 1 = − Γ . Hence, the slop e of the linear piece right of π = M is greater than − Γ for the function g ( x, · ) and positive for the function h x . This shows that the con v ex function h x indeed attains its unique minim um in π = M and th us ﬁnishes the pro of of (b) and (c) . Lemma 11. L et x ∈ { 0 , 1 } n b e such that it is fe asible for pr oblem ( 2S-C’ ) , i.e., P n i =1 x i ≤ p , and do es not me et the c onditions of L emma 10 , i.e., P m i =1 a ′ i x i > 1 or x i = 1 for some i ∈ { m + 1 , . . . , n } . Then, when ﬁxing x in ( 2S-C’ ) , the optimal obje ctive value satisﬁes min π ∈ R ≥ 0 h x ( π ) > (Γ + 1) M + V . Pr o of. First consider the case where x i = 1 for some i ∈ { m + 1 , . . . , n } . Then the ﬁrst-stage costs of the solution x are v ery high and we hav e min π ∈ R ≥ 0 h x ( π ) ≥ P n i =1 C i x i ≥ ( m + 3) M > (Γ + 2) M > (Γ + 1) M + V , which prov es the claim in this case. F or the rest of the pro of, assume that P m i =1 a ′ i x i > 1 and x i = 0 for all i ∈ { m + 1 , . . . , n } . Deﬁne π ⋆ = M p x / ( p x + P m i =1 a ′ i x i − 1) . Similarly to the pro of of Lemma 10 , we will ﬁrst determine the v alue h x ( π ⋆ ) and sho w that h x ( π ⋆ ) > (Γ + 1) M + V , and then argue that π = π ⋆ is the unique minimum of h x b y in vestigating the slop es left and righ t of this point. As in the pro of of Lemma 10 , h x ( π ⋆ ) can b e determined b y studying the b ehavior of the con tin uous min-knapsack problem ( △ ’ ) for x and π ⋆ . In fact, its optimal solution is given by z i = 1 for all i ∈ R x \ { m + 1 } b ecause these are again the ﬁrst items in the greedy order and 18 their total size precisely ﬁlls the capacit y p x : X i ∈ R x \{ m +1 } π i = ( m − A ′ + 2) π ⋆ M + m X i =1 (1 − x i ) π ⋆ d i = p x p x + P m i =1 a ′ i x i − 1 m − A ′ + 2 + m X i =1 (1 − x i )(1 − a ′ i ) ! = p x p x + P m i =1 a ′ i x i − 1 p x + m X i =1 a ′ i x i − 1 ! = p x The optimal v alue of ( △ ’ ) is therefore g ( x, π ⋆ ) = X i ∈ R x c i π i z i = m X i =1 (1 − x i ) c i π ⋆ d i = m X i =1 v i (1 − x i ) π ⋆ M = V − m X i =1 v i x i ! π ⋆ M . The total ob jectiv e v alue of ( x, π ⋆ ) in ( 2S-C’ ) can then b e written as h x ( π ⋆ ) = m X i =1 C i x i + Γ π ⋆ + g ( x, π ⋆ ) = M m X i =1 a ′ i x i + Γ π ⋆ + V − m X i =1 v i x i ! π ⋆ M = M + M m X i =1 a ′ i x i − 1 ! + Γ M − Γ M  1 − π ⋆ M  + V − V  1 − π ⋆ M  − m X i =1 v i x i π ⋆ M = (Γ + 1) M + V + ε, where ε = M ( P m i =1 a ′ i x i − 1) − Γ M (1 − π ⋆ / M ) − V (1 − π ⋆ / M ) − P m i =1 v i x i π ⋆ / M . It remains to sho w that ε > 0 b ecause this implies the claimed assertion h x ( π ⋆ ) > (Γ + 1) M + V . W e can deriv e ε = M m X i =1 a ′ i x i − 1 − Γ  1 − π ⋆ M  ! − V + m X i =1 v i (1 − x i ) π ⋆ M ≥ M m X i =1 a ′ i x i − 1 − Γ P m i =1 a ′ i x i − 1 p x + P m i =1 a ′ i x i − 1 ! − V = M m X i =1 a ′ i x i − 1 ! δ − V , where δ := 1 − Γ / ( p x + P m i =1 a ′ i x i − 1) . Note that we hav e δ > 1 / (3 m ) because p x + P m i =1 a ′ i x i − 1 < p + A ′ = 2 m − A ′ + 3 ≤ 3 m and p x + m X i =1 a ′ i x i − 1 − Γ = m − A ′ + 1 − m X i =1 x i + m X i =1 a ′ i x i = m X i =1 (1 − a ′ i )(1 − x i ) + 1 ≥ 1 . T ogether with P m i =1 a ′ i x i − 1 ≥ 1 /b (whic h follo ws from P m i =1 a ′ i x i > 1 due to the integralit y of a and b ) and the deﬁnition of M , this ﬁnally gives ε > M / (3 mb ) − V = 0 . Next, w e pro ve that the function h x actually attains its minim um at π = π ⋆ . The slop e of h x left of π = π ⋆ is exactly the same as in the pro of of Lemma 10 and therefore again negativ e. In the analysis of the slop e right of π = π ⋆ , there are tw o diﬀerences to the pro of of Lemma 10 . First, since π ⋆ < M , not only the sizes of the items corresponding to c i for i ∈ [ m ] ∩ R x increase, but also the sizes π i = π / M of the cheap items corresponding to c i for i ∈ { m + 2 , . . . , n } . Second, w e ha ve z m +1 = 0 already at π = π ⋆ , so the item w e unpack a fraction of is alwa ys the previous 19 one in the order, corresp onding to c ⋆ := max i ∈ [ m ] ∩ R x c i . Note that c i = v i / (1 − a ′ i ) ≤ bV for all i ∈ [ m ] and therefore c ⋆ ≤ bV . The total cost of the solution of ( △ ’ ) , when sligh tly increasing π starting from π = π ⋆ , no w changes linearly at a rate of X i ∈ [ m ] ∩ R x c i d i − c ⋆ X i ∈ R x \{ m +1 } 1 d i ≥ − c ⋆ m X i =1 (1 − x i ) 1 d i + ( m − A ′ + 2) 1 M ! = − c ⋆ M m X i =1 (1 − x i )(1 − a ′ i ) + m − A ′ + 2 ! = − c ⋆ M 2Γ − m X i =1 x i (1 − a ′ i ) ! ≥ − Γ 2 c ⋆ M ≥ − Γ 2 bV M > − Γ Hence, like in the pro of of Lemma 10 , the slop e of the linear piece righ t of π = π ⋆ is greater than − Γ for the function g ( x, · ) and p ositiv e for the function h x . This sho ws that π ⋆ indeed represen ts the unique minimum of h x and th us ﬁnishes the pro of. As outlined ab o ve, we ha ve now presented all ingredients for the pro of of Theorem 8 : Pr o of of The or em 8 . W e ha v e described a reduction from the NP-hard knapsack problem to the problem ( 2S-C’ ) . In Lemmas 10 and 11 , w e hav e analyzed t w o types of feasible solutions x of the constructed instance of ( 2S-C’ ) . The cost of the solutions considered in Lemma 11 is alw a ys larger than the cost of the solutions considered in Lemma 10 . The solutions considered in Lemma 10 directly corresp ond to the feasible solutions of the given knpasack instance. Moreo v er, according to Lemma 10(c) , the ob jective v alue achiev ed by such a solution x , together with the b est corresp onding c hoice of π , is minimal when the v alue of the kn apsac k solution is maximal. Th us, an optimal solution of problem ( 2S-C’ ) directly giv es an optimal solution of the knapsac k problem. Our reduction can also be used to conclude that problem ( 2S-C’ ) is still NP-hard when the v alue of π is ﬁxed: Theorem 12. The pr oblem ( 2S-C’ ) with a ﬁxe d value of π is NP-har d. Pr o of. A reduction from the NP-hard knapsac k problem can be obtained using the same construction as for Theorem 8 and additionally setting the ﬁxed v alue of π as π := M . Due to Lemma 10(b) , the solutions x of the constructed instance of ( 2S-C’ ) that corresp ond to feasible knapsac k solutions b ehav e exactly as b efore. Lemma 11 implies that, for all other feasible solutions x of ( 2S-C’ ) , the cost h x ( M ) ≥ min π ∈ R ≥ 0 h x ( π ) > (Γ + 1) M + V is again worse than the one of any solution of the ﬁrst kind. Th us, also when ﬁxing π = M , an optimal solution of problem ( 2S-C’ ) giv es an optimal solution of the knapsack problem. Remark 13. While Theorem 8 implies NP-hardness of (2S- C), it is a priori not clear whether the decision v ariant of (2S-C) is contained in NP . How ever, the con tainment in NP follows from the argumen ts presented in [ GL W24 , Corollary 4]. Therefore, the decision v ariant of (2S-C) is NP-complete. Finally , we demonstrate that the NP-hardness result of Theorem 8 can b e extended from the t w o-stage robust selection problem to tw o-stage versions of other nominal problems, in particular the assignmen t problem. This partially answ ers the open problem num b er 15 in [ GH24 ]. In the assignmen t problem, a bipartite graph with edge costs is given, and the task is to ﬁnd a p erfect matc hing of minimum cost. 20 Corollary 14. The two-stage r obust assignment pr oblem with c ontinuous budgete d unc ertainty is NP-har d. Pr o of. The assignmen t problem can b e considered to b e a sp ecial case of the selection problem; see, e.g., [ KKZ13 ]. Suc h a construction can easily b e transferred to the t w o-stage v ersions of the problems: Let an instance ( n, p, C, c, d, Γ) of the t wo-stage selection problem with budgeted uncertain t y b e giv en. Deﬁne a bipartite graph G = ( V ∪ W, E ) with | V | = | W | = 2 n − p , and denote its vertices by V = { v 1 , . . . , v 2 n − p } and W = { w 1 , . . . , w 2 n − p } . The edge set is given by E = {{ v i , w i } : i ∈ [ n ] } ∪ {{ v i , w j } : i ∈ [ n ] , j ∈ { n + 1 , . . . , 2 n − p }} ∪ {{ v i , w j } : i ∈ { n + 1 , . . . , 2 n − p } , j ∈ [ n ] } . Deﬁne the t w o-stage assignment problem’s ﬁrst-stage costs C ′ ∈ R E ≥ 0 and the parameters c ′ ∈ R E ≥ 0 and d ′ ∈ R E ≥ 0 of the uncertaint y set as follo ws: F or all i ∈ [ n ] , set C ′ { v i ,w i } := C i , c ′ { v i ,w i } := c i , and d ′ { v i ,w i } := d i . F or all other edges e ∈ E , set C ′ e = c ′ e = d ′ e := 0 . Finally , set Γ ′ := Γ . It can b e easily chec ked that every p erfect matc hing in G con tains exactly p edges from the set {{ v i , w i } : i ∈ [ n ] } . Hence, there is a direct corresp ondence b etw een the feasible item selections and the p erfect matchings. Finally , note that we may assume that a ﬁrst-stage solution of the t w o-stage assignment instance only consists of edges in {{ v i , w i } : i ∈ [ n ] } , since all other edges can still b e selected at cost 0 in the second stage. 5 Multi-Rep resentative Selection with Alternative Continuous Budgeted Sets W e show that the problem ( 2MRS-A C ) can be solved in p olynomial time, using similar arguments as in [ Cha+18 ]. First, we substitute the v ariables y i . Similarly to the reform ulations in Section 3 , w e replace their lo w-cost part where y i ∈ [0 , π ] using v ariables z i , and their high-cost part where y i ∈ [ π , 1] using v ariables z i , to obtain the follo wing reform ulation: min n X i =1 C i x i + n X i =1 c i π z i + n X i =1 c i (1 − π ) z i + Γ π s . t . X i ∈ T j ( x i + π z i + (1 − π ) z i ) = p j ∀ j ∈ [ m ] x i + z i ≤ 1 ∀ i ∈ [ n ] x i + z i ≤ 1 ∀ i ∈ [ n ] x ∈ { 0 , 1 } n , z , z ∈ [0 , 1] n , π ∈ [0 , 1] F or each j ∈ [ m ] , we deﬁne the function f j : [0 , 1] → R ≥ 0 as follo ws: f j ( π ) = min X i ∈ T j C i x i + X i ∈ T j c i π z i + X i ∈ T j c i (1 − π ) z i ( ◦ ) s . t . X i ∈ T j ( x i + π z i + (1 − π ) z i ) = p j x i + z i ≤ 1 ∀ i ∈ T j x i + z i ≤ 1 ∀ i ∈ T j x ∈ { 0 , 1 } T j , z , z ∈ [0 , 1] T j Notice that the optimal v alue of ( 2MRS-A C ) is given b y min π ∈ [0 , 1]  P m j =1 f j ( π ) + Γ π  . The follo wing result is a direct adaptation of [ GH24 , Lemma 6.22]. 21 Lemma 15. F or al l j ∈ [ m ] , the function f j is pie c ewise line ar and has br e akp oints within the set Π j = [0 , 1] ∩  p j − a − b c − b : a, b ∈ { 0 , . . . , p j } , c ∈ { 0 , . . . , | T j |} , a + b ≤ p j , c > b  . Observ e that, if eac h f j is piecewise linear with breakp oin ts in Π j , then P m j =1 f j ( π ) + Γ π is also piecewise linear with breakp oints in Π = S j ∈ [ m ] Π j . The solution strategy to ( 2MRS-A C ) is to calculate eac h v alue f j ( π ) for π ∈ Π , and to choose the best of these solutions. By Lemma 15 , w e th us ensure that w e hav e found an optimal solution. So let π ∈ Π and j ∈ [ m ] b e ﬁxed. Lemma 16. F or every j ∈ [ m ] and π ∈ [0 , 1] , ther e is an optimal solution to the pr oblem ( ◦ ) wher e at most one of the variables S i ∈ T j { z i , z i } is fr actional. Pr o of. F or ev ery ﬁxed j ∈ [ m ] , π ∈ [0 , 1] , and x ∈ { 0 , 1 } n , the problem ( ◦ ) is a linear program that has the structure of a con tin uous min-knapsac k problem, similarly to ( ∗ ” ) in Section 3 and ( △ ’ ) in Section 4 . Such a problem can b e solv ed b y a greedy algorithm [ Dan57 ], whic h giv es an optimal solution with at most one fractional v ariable. In particular, this holds when ﬁxing x to the v alues these v ariables attain in an optimal solution of ( ◦ ) , whic h concludes the pro of. Using the existence of a single fractional item, we can no w solv e problem ( ◦ ): Lemma 17. F or every j ∈ [ m ] and π ∈ [0 , 1] , the pr oblem ( ◦ ) c an b e solve d in p olynomial time. Pr o of. Using Lemma 16 , we ﬁrst guess one v ariable z i or z i that is allo w ed to be fractional. Using a dynamic program, w e construct the set of P areto solutions among all binary v ariables (i.e., all remaining z and z v ariables and all x v ariables) with respect to minimum costs and minim um size. As all p ossible item sizes are 1, π , or 1 − π , this set cannot con tain more than O ( | T j | 3 ) elemen ts. Among these solutions, we then consider all solutions with total size in [ p j − π , p j ] or [ p j − (1 − π ) , p j ] (if the fractional v ariable is of type z i or of type z i , respectively), and extend them to a feasible solution using the fractional v ariable. A solution constructed in this w ay with minimum costs is then an optimal solution to problem ( ◦ ). Theorem 18. The pr oblem (2MRS-A C) c an b e solve d in p olynomial time. Pr o of. W e solve the problem min π ∈ [0 , 1]  P m j =1 f j ( π ) + Γ π  b y enumerating a p olynomial num ber of breakp oints using Lemma 15 . According to Lemma 17 , each v alue f j ( π ) can then be calculated in p olynomial time, whic h means that the optimal v alue is found in p olynomial time. Note that w e also construct a corresp onding feasible solution in the pro cess. 6 Rep resentative Selection with Discrete Budgeted Sets Finally , we study the complexit y of problem (2RS-D), which is the only problem with discrete budgeted uncertain ty we consider. Indeed, tw o-stage problems with discrete budgeted uncertain t y are often kno wn to b e hard. In this con text, the following is an easy adaptation from [ GL W22 ], whic h w e present for the sak e of completeness. Theorem 19. The pr oblem (2RS-D) is NP-har d. Pr o of. Let an instance of the NP-hard partition problem (see [ Kar72 ]) b e given, consisting of (distinct) v alues a 1 , . . . , a n ∈ N . Let Q := 1 / 2 P n i =1 a i b e half of their total sum. The task is to decide if there is a set I ⊆ [ n ] suc h that P i ∈ I a i = Q . W e construct an instance of the t wo-stage robust represen tative selection problem with discrete budgeted uncertain t y . Let M > 2 Q b e a large constan t, and let W b e an even larger constant deﬁned later. The instance has 2 n items, n + 1 buc k ets, and Γ = n . The cost of all items is describ ed in T able 4 . 22 T 1 T 2 . . . T n +1 i 1 . . . n n + 1 . . . 2 n C i W . . . W a 1 . . . a n c i M . . . M 0 . . . 0 c i M + 2 Q . . . M + 2 Q 2 a 1 . . . 2 a n T able 4: Instance constructed in the reduction for the pro of of Theorem 19 . The buck et T 1 is the largest and contains n items, eac h ha ving C i = W , c i = M , and c i = M + 2 Q . W e will choose W as suc h a large v alue that pac king an item from T 1 in the ﬁrst stage immediately disqualiﬁes a solution from being optimal. The follo wing analysis sho ws that W > M + 4 Q suﬃces. The items in the remaining buc k ets T 2 , . . . , T n +1 eac h corresp ond to one of the given v alues a i . More precisely , for all i ∈ { 2 , . . . , n + 1 } , the buck et T i con tains a single item n + i − 1 with C n + i − 1 = a i − 1 , c n + i − 1 = 0 , and c n + i − 1 = 2 a i − 1 . Note that, for each of these buc kets, the optimal solution m ust trivially pac k its single item. How ever, there is still the non trivial choice whether this happ ens in the ﬁrst or in the second stage. Note that an optimal ﬁrst-stage solution does not select an item in T 1 . Therefore, the item from T 1 is alw a ys selected in the second stage. The adv ersary can therefore choose to in v est all their budget in to T 1 , causing a cost increase of 2 Q . How ever, if not all n items of T 1 are attac k ed sim ultaneously , the attac k has no eﬀect. Hence, the adv ersary has eﬀectively tw o v alid strategies: Either the costs of all items in T 1 are increased, or the costs of all items in T 2 ∪ · · · ∪ T n +1 are increased. In b oth cases, a second-stage solution will select the remaining items in T 2 ∪ · · · ∪ T n +1 and one item in T 1 . In the ﬁrst strategy , the cost of the item from T 1 is M + 2 Q , and in the second case, it is M . Let X b e the total ﬁrst-stage cost of the items in T 2 ∪ · · · ∪ T n +1 that are pack ed by some solution. The total cost of this solution is X + max { M + 2 Q, M + 4 Q − 2 X } = M + 2 Q + max { X , 2 Q − X } . This is less than or equal to M + 3 Q if and only if X = Q , whic h can b e achiev ed if and only if the giv en instance of the partition problem is a y es-instance. Remark 20. While Theorem 19 implies NP-hardness of (2RS-D), it is a priori not clear whether the decision v ariant of (2RS-D) is contained in NP . How ever, the containmen t in NP follo ws from the following easy argument: Note that it suﬃces to sho w that the adv ersarial problem b elonging to (2RS-D) can be solved in p olynomial time. Consider a ﬁxed ﬁrst-stage selection x . F or eac h buck et T j , for j ∈ [ m ] , we kno w if there is still an item missing. If this is not the case, the adversary do es not increase the cost of any item of this buc ket. If an item still needs to be selected in the second stage, it will b e an item with minimal cost. This means that the adv ersary will increase the costs of the items b y ordering them from smallest to largest costs c i . F or j ∈ [ m ] , assume that the adversary in vests Γ j ∈ { 0 , . . . , Γ } of their budget into making items of T j more exp ensiv e. Let α j (Γ j ) denote the optimal cost that the adversary can cause in buc k et T j assuming they in vest Γ j of their budget. The v alue α j (Γ j ) can therefore be computed in p olynomial time. Given all v alues α j (Γ j ) , the adv ersary then solves the problem of maximizing P m j =1 α j (Γ j ) sub ject to P m j =1 Γ j = Γ . This can b e done with a simple dynamic program. In summary , we conclude that the decision v ariant of (2RS-D) is NP-complete. 23 7 Conclusions Selection problems are fundamen tal to robust com binatorial optimization due to their simplicity . In many cases, p olynomial-time solv ability results w ere ﬁrst kno wn for problems of this t ype, and sometimes remain the only such results to this day . Despite having b een studied p ossibly more than any other type of problem in the con text of robust combinatorial optimization, our kno wledge of the complexit y landscape under diﬀeren t types of uncertain ty sets has remained fragmen tal. This paper giv es a systematic and thorough understanding of the tw o-stage problem complexit y for the three commonly studied v ariants of selection problems, in com bination with the three commonly studied t yp es of budgeted uncertain t y sets. In particular, w e settle the long- standing op en problem regarding the complexit y of tw o-stage selection with contin uous budgeted uncertain t y , by showing that this problem is NP-hard. Moreo ver, w e show that the same is true for the t w o-stage assignment problem with contin uous budgeted uncertaint y . T o the b est of our kno wledge, these are the ﬁrst suc h hardness results f or contin uous budgeted uncertain t y sets, whic h leads us to exp ect that similar hardness results for more complex underlying problems can be found in the future. Our hardness result is based on a reduction from the knapsac k problem, so it does not sho w strong NP-hardness. It remains an op en question whether the t w o-stage selection problem with con tin uous budgeted uncertaint y can b e solv ed in pseudop olynomial time. There remain special cases to our setting that are yet unexplored and an in teresting c hallenge for further researc h. If we consider the multi-represen tative selection problem with the restriction that eac h p j is in O (1) , w e obtain a generalization of the representativ e selection problem that ma y b e easier to solv e than the m ulti-representativ e selection problem itself. W e conjecture that, in case of con tin uous budgeted uncertain t y , this problem can b e solv ed in p olynomial time by extending the argumen ts from Section 3 . F urthermore, our hardness results require the budget Γ to b e arbitrary , so they do not imply hardness for sp eciﬁc b ounds, such as Γ = 1 . Finally , it remains open to explore recov erable problem v ariants in the same wa y as t wo-stage problems ha v e b een studied in this paper. A ckno wledgements. Lasse W ulf w as supp orted b y Ev a Rotenberg’s Carlsb erg F oundation Y oung Researcher F ellowship CF21-0302 “Graph Algorithms with Geometric Applications” . 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The Complexity Landscape of Two-Stage Robust Selection Problems with Budgeted Uncertainty

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