Bribery's Influence on Ranked Aggregation

Brib ery’s Influence on Rank ed Aggregation P allavi Jain Indian Institute of T ec hnology Jodhpur pallavi@iitj.ac.in Ansh ul Thakur Indian Institute of T ec hnology Jodhpur thakur.17@iitj.ac.com Abstract Kemen y Consensus is a w ell-known rank aggregation metho d in so cial choice theory . In this metho d, giv en a set of rankings, the goal is to find a ranking Π that minimizes the total Kendall tau distance to the input rankings. Computing a Kemen y consensus is NP -hard, and ev en v erifying whether a given ranking is a Kemeny consensus is coNP -complete. Fitzsimmons and Hemaspaandra [IJCAI 2021] established the computational in tractability of achieving a desired consensus through manipulative actions. Kemen y Consensus is an optimisation problem related to Kemeny’s rule. In this pap er, we consider a decision problem related to Kemeny’s rule, known as Kemeny Sc or e , in which the goal is to decide whether there exists a ranking Π whose total Kendall tau distance from the giv en rankings is at most k . Computation of Kemen y score is known to be NP -complete. In this pap er, we in vestigate the impact of sev eral manipulation actions on the Kemen y Score problem, in which giv en a set of rankings, an in teger k , and a ranking X , the question is to decide whether it is p ossible to manipulate the given rankings so that the total Kendall tau distance of X from the manipulated rankings is at most k . W e sho w that this problem can b e solved in p olynomial time for v arious manipulation actions. In terestingly , these same manipulation actions are known to b e computationally hard for Kemeny consensus. 1 In tro duction Rank aggregation, the task of com bining m ultiple rankings into a single consensus ranking, is a ubiquitous problem with wide-ranging applications. It is a fundamental problem in computational so cial c hoice theory and plays a crucial role in v arious domains suc h as meta-search engines [22], recommendation systems, mac hine learning [2], marketing and advertisemen t, bioinformatics [41], etc. One of the rank aggregation tec hniques, widely accepted in so cial c hoice theory , is Kemeny’s rule defined b y Kemen y in 1959 [36]. In the Kemeny Score problem, given a set of n rankings, sa y R , and an in teger k , the goal is to determine if there exists a ranking X such that the total Kendal l tau distanc e (defined later) b et ween the set of rankings in R and X is at most k [9, 34]. Kemen y’s rule is interesting in so cial choice theory as it satisfies Condorcet extension criterion [45] and is also maxim um likelihoo d estimator for a simple probabilistic model in which individual preferences are noisy estimates of an underlying “true” ranking [9, page 87]. Kemeny Score is w ell-studied computationally from theoretical computer science viewp oint [34, 18, 4, 5, 6, 7, 37, 31], machine learning [33], as well as heuristically [43, 44, 17]. Interestingly , Kemeny Score is p olynomial-time solv able for t wo rankings and NP-hard for four rankings [7], while the case of three rankings is a long-standing op en problem. One of the imp ortant directions of researc h in computational social c hoice theory is to study the com- putational complexit y of changing the outcome via manipulation, bribery , and v arious con trol actions, such as adding or deleting candidates or rankings [9, 1, 20, 40, 39, 27, 25]. T o the b est of our knowledge, de Haan [19] initiated the study of manipulation for Kemney in the context of judgement aggregation. Re- cen tly , Fitzsimmons and Hemaspaandra [29] studied the computational complexity of manipulation schemes for the Kemeny Consensus problem, where given the set of n rankings, R , and a ranking X , the goal is to manipulate the rankings R using a manipulative action so that X attains the minim um total Kendall tau distance to the profile among all rankings. The Kendall tau distance b et ween tw o rankings π and π ′ o ver a candidate set C is the n um b er of pairs of candidates whose relative ordering differs b etw een π and π ′ . Mathematically , 1 KT ( π , π ′ ) = |{{ x, y } ⊆ C : ( x ≺ π y ∧ y ≺ π ′ x ) ∨ ( y ≺ π x ∧ x ≺ π ′ y ) }| The notation x ≺ π y denotes that x is b efore y in the ranking π . The Kendall tau distance of a ranking π from the set of rankings R is the summation of distance from π to eac h ranking in R , i.e., P π ′ ∈R KT ( π , π ′ ). In this pap er, w e explore the computational complexity of v arious manipulation sc hemes for Kemeny Score , whic h has not b een studied so far to the b est of our knowledge. Kemen y Consensus vs Kemeny Score Kemeny Score is a decision problem of Kemen y’s rule and Kemeny Consensus is an optimisation problem of Kemen y’s rule [9, page 88]. Ho w ever, in terestingly , when manipulation is considered, there is a significan t difference b et ween the tw o problems. In the manipulation problems for Kemeny Score , the Kendall tau distance of X should be at most k from R . At the same time, in manipulation problems for Kemeny Consensus it is p ossible that the distance of X from R do es not change at all, rather the distance of other rankings increases so that X is the closest one. W e explain it using the following example, which is the same as Example 9 in [29]. Example 1. Consider thr e e c andidates a, b, c . Supp ose ther e ar e thr e e r ankings R 1 = R 2 = R 3 = a ≻ b ≻ c , two r ankings R 4 = R 5 = a ≻ c ≻ b , and one r anking R 6 = c ≻ b ≻ a . L et X = a ≻ c ≻ b . L et k = 3 . Note that if we delete r anking R 6 , then the Kendal l tau distanc e of X fr om R = { R 1 , . . . , R 5 } is 3 ; while X is not a Kemeny c onsensus. The Kemeny c onsensus r anking is a ≻ b ≻ c . However, if we delete R 1 , then X is the Kemeny c onsensus of R = { R 2 , . . . , R 6 } while the Kendal l tau distanc e of X fr om R is 4 . While man y problems, including the Kemeny Consensus and the Kemeny Winner (the top can- didate in the Kemen y consensus is the winner), are not susceptible to v arious brib ery-based manipulation sc hemes, we ha ve dev elop ed p olynomial-time algorithms for the Kemeny Score under v arious brib ery-based manipulativ e actions. F ormally , we study the follo wing problem. B -Kemen y Score Input: a set of candidates C , a set of rankings R ov er C , a ranking X ov er C , a cost function cost B (dep ending on the bribery action B ), a budget bgt ∈ N , and an in teger k . Question: Do es there exist a set of rankings R ′ that can b e obtained b y manipulating rankings in R using brib ery action B within the budget bgt , and the total Kendall tau distance of X from the rankings in R ′ is at most k ? W e consider the following brib ery actions. The cost of brib ery is additive for all brib ery actions. $ Brib ery . In this manipulation action, a ranking R ∈ R can b e changed to any arbitrary ranking. Note that for our problem, changing them to X is a better c hoice rather than anything else, as this makes the distance 0. The cost function is defined as follows: cost $ : R → N . Let S ⊆ R b e the set of rankings that are changed to X . Then, the cost of manipulation is P R ∈ S cost $ ( R ). Sw ap Brib ery . A swap ( c, c ′ ) R in the ranking R is called admissible if c and c ′ are consecutive in R . In this manipulation action, we can change a ranking b y admissible swaps, and the cost dep ends on the ranking and the num b er of swaps. F ormally , the cost function is defined as follows: cost SB : R → N . Note that cost SB ( R ) denotes the cost of one swap in the ranking R . If l 1 sw aps are done on ranking R 1 and l 2 sw aps are done on ranking R 2 then the total cost would b e l 1 · cost SB ( R 1 ) + l 2 · cost SB ( R 2 ) Ranking Deletion. In this manipulation action, w e are allo wed to delete rankings, and there is a cost asso ciated with the deletion. The cost function is defined as follo ws: cost RD : R → N . Candidate Deletion. In this manipulation action, w e are allo wed to delete candidates. Note that if we delete a candidate c ∈ C , then it is deleted from all the rankings in R as well as from X , and it do es not c hange the ordering of other candidates. The cost function is defined as follo ws: cost CD : C → N . In the te chnic al se ctions, we omit the subscript b elow cost sinc e e ach se ction is de dic ate d to a sp e cific brib ery action. Thus, it wil l b e cle ar fr om the c ontext. 2 Kemeny Score Kemeny Consensus Kemeny Winner $ Brib ery P [Theorem 8] coNP -complete [29, Observ ation 6] OPEN Sw ap Brib ery P [Theorem 12] OPEN OPEN Ranking Deletion P [Theorem 9] Σ p 2 -complete [29, Conjecture on page 199] OPEN Candidate Deletion k = 0: P [Theorem 16] Σ p 2 -complete [29, Theorem 8] Σ p 2 -complete [30, Theorem 5] |R| = 1: P [Theorem 19] OPEN (in general) P ossible Kemen y Score P [Theorem 6] OPEN OPEN T able 1: Summary of computational complexit y of v arious manipulation schemes for Kemeny Score , Kemeny Consensus , and Kemeny Winner . The computational complexit y of the cases mark ed with “-” are not known to the b est of our kno wledge and also mentioned in [39]. Candidate deletion and ranking deletion are usually considered as con trol actions. Since w e are consid- ering the cost, we consider these manipulativ e actions also as brib ery actions for brevit y . Note that the addition of rankings or candidates is not a meaningful manipulation action for us. This is b ecause adding rankings does not decrease the distance of X from the existing rankings. Additionally , since X is giv en to us, the addition of candidates is not a meaningful op eration for us. W e also study the problem when the rankings in R are not complete, i.e., all the rankings in R are partial; how ev er, X is a ranking ov er C . The ob jectiv e is to complete the rankings in R so that X is at most k distance aw a y from the new complete rankings o ver C . F ormally , w e study the following problem. P ossible Kemeny Score Input: a set of candidates C , a set of partial rankings R , a ranking X ov er C , and an integer k . Question: Can we extend rankings in R using the remaining candidates so that the total Kendall tau distance of X from the extended rankings in R is at most k ? By extending a ranking R , w e mean that the candidates of C \ C ′ are added to R , where R is ov er C ′ ; ho wev er, the relativ e ordering of the candidates of the set C ′ in the ranking R do es not change. T able 1 summarizes the complexity of v arious manipulation schemes for Kemeny Score , Kemeny Consensus , and Kemeny Winner . All the results pertaining to Kemeny Score are build in this pap er. Related W ork. Here, we discuss some more related work. In 1992, Bartholdi et al. [3] initiated the study of manipulating the election by adding/deleting the candidates or v oters so that a fa vorite candidate is the winner. Afterw ards, there is a lot of study on the manipulation actions for the several voting rules [9, 12, 26, 13, 16, 28, 35, 1, 20, 10, 23, 21], with several results pertaining to NP -hardness as well as algorithms. Manipulation is closely linked to the robustness of v oting rules, a connection iden tified a few years ago [11]. This insigh t has since inspired research into manipulation from a new p erspective [32, 8, 42, 24, 14]. This relation also highlights the imp ortance of algorithmic results for these problems. Road Map. W e begin our tec hnical sections from Section 3, whic h is dedicated to a p olynomial time algorithm for Possible Kemeny Score . Then, in Section 4, w e presen t results for $ -Briber y-Kemeny Score and Ranking Deletion-Kemeny Score . Section 5 and Section 6 are dedicated to Sw ap Briber y- Kemeny Score and Candida te Deletion-Kemeny Score , resp ectiv ely . Section 7 concludes the paper with some op en questions. 2 Preliminaries The notation [ n ] denotes the set of natural num b ers { 1 , . . . , n } . The Kendall tau distance betw een t wo rankings π, π ′ is denoted as KT ( π , π ′ ). F urthermore, we abuse the notations and use KT ( π , R ) to denote 3 the Kendall tau distance of ranking π from the set of rankings R . F or a ranking π , let S ( π ) denote the set of elements that are ranked in π , i.e., π is a ranking ov er S ( π ). W e also use KT ( π , π ′ ) to denote the distance b et ween π and π ′ when either S ( π ) ⊆ S ( π ′ ) or S ( π ′ ) ⊆ S ( π ). The distance is computed only for the elemen ts in S ( π ) ∩ S ( π ′ ). It will be clear from the con text. When w e sa y that a ranking π is k distance a wa y from another ranking π ′ (or set of rankings R ), we mean that KT ( π, π ′ ) ≤ k (or KT ( π, R ) ≤ k ). A pair of candidates c, c ′ disagree in π and π ′ if their relativ e ordering differs in π and π ′ . Let π , π ′ b e tw o rankings and S ⊆ S ( π ) ∩ S ( π ′ ). W e say that π and π ′ agr e es o ver S if for every { c, c ′ } ⊆ S , the relative ordering of c and c ′ is same in b oth π and π ′ , i.e., either c is b efore c ′ in π and π ′ or either c ′ is b efore c in b oth the rankings. Let π b e a ranking. Let Y ⊆ S ( π ). Then, π | Y denotes the ordering π r estricte d to Y , i.e., it is an ordering ov er Y and the relative ordering of every pair of elements in Y is the same as that in π . W e denote the i th elemen t of π as π ( i ) and the num b er of elements in π as | π | . The notation x ≺ π y denotes that x is b efore y in the ordering π . The num b er of disagr e ements with resp ect to x in π and π ′ is defined as follo ws # disagree ( π , π ′ , x ) = |{ ( x, y ) : ( x ≻ π y ∧ y ≻ π ′ x ) ∨ ( y ≻ π x ∧ x ≻ π ′ y ) }| . W e treat the set of rankings R as an ordered tuple ( R 1 , . . . , R n ) for conv enience in the technical sections. F or a function f : P → Q and a subset S ⊆ P , f ( S ) = ∪ s ∈ S f ( s ). F or a graph G = ( V , E ), V ( G ) denotes the vertex set of G and E ( G ) denotes the edge set of G . F or a subset S ⊆ V ( G ), G − S denotes the graph on the vertex set V ( G ) \ S and the edge set { uv ∈ E ( G ) : { u, v } ⊆ V ( G ) \ S } . 3 P ossible Kemen y Score In this section, we design a p olynomial-time algorithm for Possible Kemeny Score problem. Recall that in Possible Kemeny Score w e need to complete the partial ranking set R so that the Kendall tau distance b et w een the given ranking X and the extended ranking set of R , sa y R ′ , is at most k . Note that there is no cost associated with the extension of rankings. Thus, we first design an algorithm to extend a ranking R o ver C ′ ⊆ C to R ′ using the candidates in C ′′ = C \ C ′ suc h that KT ( X , R ′ ) is least among all the extensions of R . W e call R ′ as an optimal extension of R . W e will argue that w e can extend all the rankings in R using this algorithm and return Yes if and only if we are given a yes-instance. The algorithm proceeds as follows. W e hav e a ranking R ov er C ′ ⊆ C as an input. Let C ′′ = C \ C ′ and Π = X | C ′′ . Let R ′ = R . F or i ∈ [ |C ′′ | ], we insert Π( i ) after the candidates Π(1) , . . . , Π( i − 1) in R ′ at the leftmost p osition that minimizes the Kendall tau distance of R ′ from X after inserting Π( i ). Algorithm 1 describ es the pro cedure formally . By inserting a candidate c in the ranking R ′ at position j , we mean that c is inserted betw een R ′ | C ′ ( j ) and R ′ | C ′ ( j + 1) and after all the candidates of C ′′ b et w een R ′ | C ′ ( j ) and R ′ | C ′ ( j + 1) in R ′ . W e denote this op eration by insert ( R ′ , c, j ). F or example, let C ′ = { 1 , 2 , . . . , 100 } , C ′′ = { a, b, . . . , z } , and R = 1 ≺ 2 ≺ 3 ≺ a ≺ b ≺ c ≺ 4 ≺ . . . ≺ 100. Then insert ( R, d, 3) will result in ranking R ′ = 1 ≺ 2 ≺ 3 ≺ a ≺ b ≺ c ≺ d ≺ 4 ≺ . . . ≺ 100. Note that insert ( R ′ , c, 0) means that c is inserted b efore R ′ | C ′ (1) and insert ( R ′ , c, | C ′ | ) means that c is inserted after the last candidate in R ′ | C ′ in R ′ . Algorithm 1: P ossible-Kemeny-Score Input: Ranking X , P artial ranking R o ver C ′ ⊆ C Output: a ranking R ′ o ver C whic h is an extension of R 1 Initialize set C ′′ = C \ C ′ , ranking Π = X | C ′′ , and R ′ 0 = R 2 l = 0 3 for i ∈ [ |C ′′ | ] do 4 S = { l, . . . , | C ′ |} 5 q = min(arg min j ∈ S KT ( X, insert ( R ′ , Π( i ) , j ))) 6 R ′ i = insert ( R ′ i − 1 , Π( i ) , q ) 7 l = q 8 return R ′ | C ′′ | Before proving that R ′ is an optimal exte nsion of R , we first pro ve the following structural prop ert y of all optimal extensions, which is crucial for our algorithm. 4 Figure 1: Π( i ) , t i , and t f are as discussed in ab o v e paragraph. Circles represen t the candidates of C ′ and squares denote the candidates of C ′′ . Big squares denote a ch unk of C ′′ that are consecutiv e in the ranking. Lemma 1. L et ˆ R b e an optimal extension of R . Then ˆ R and X agr e e over C ′′ . Pr o of. Let ˆ R b e an optimal extension of R . Supp ose that { c, c ′ } ⊆ C ′′ suc h that c ≺ X c ′ . T o wards the con tradiction, suppose that c ′ ≺ ˆ R c . Let R ⋆ b e an ordering ov er C obtained b y swapping c ′ and c in ˆ R . Clearly , KT ( X , ˆ R ) ≤ KT ( X , R ⋆ ) as ˆ R is an optimal extension. Note that if there is no candidate b et ween c and c ′ in ˆ R , then KT ( X, R ⋆ ) < KT ( X , ˆ R ), whic h is a contradiction. Next we consider the case when there exist candidates c 1 , c 2 , . . . , c n b et w een c and c ′ in ˆ R , i.e., c ′ ≺ ˆ R . . . ≺ ˆ R c i ≺ ˆ R . . . ≺ ˆ R c . Recall that { c, c ′ } con tributes to KT ( X , ˆ R ), but not to KT ( X , R ⋆ ). Let c i ∈ { c 1 , c 2 , . . . , c n } . If c i is b et ween c and c ′ in X , i.e., c ≺ X c i ≺ X c ′ , then note that { c, c i } and { c ′ , c i } contribute to KT ( X, ˆ R ), but not to KT ( X, R ⋆ ). If c i is after c ′ in X , i.e., c ≺ X c ′ ≺ X c i , then { c, c i } con tributes to KT ( X, ˆ R ) and not to KT ( X, R ⋆ ), while { c ′ , c i } con tributes to KT ( X , R ⋆ ) and not to KT ( X, ˆ R ). Similarly , if c i is b efore c in X , i.e., c i ≺ X c ≺ X c ′ , then { c ′ , c i } contributes to KT ( X , ˆ R ) and not to KT ( X, R ⋆ ), while { c, c i } con tributes to KT ( X , R ⋆ ) and not to KT ( X , ˆ R ). Since w e do not c hange the position of an y other candidates, their relativ e ordering does not c hange in R ⋆ . Th us, KT ( X, R ⋆ ) < KT ( X , ˆ R ), whic h is a contradiction. Th us, every optimal extension of R agrees with X o ver C ′′ . Lemma 1 provides the intuition for placing Π( i ) after Π(1) , . . . , Π( i − 1) in Algorithm 1. Let A b e a ranking o ver C , suc h that A and X agree ov er C ′′ . Let A i = A | C ′ ∪{ Π(1) ,..., Π( i ) } . Before analysing the correctness of Algorithm 1, let us consider the follo wing scenario. Supp ose the candidate Π( i ) is p ositioned at t i in the ranking A i , and there exists a p osition t f ≥ t i suc h that placing Π( i ) at p osition t f minimises the num b er of disagreements b etw een Π( i ) and the candidates of C ′ as measured b et ween A and X compared to any p osition after t i . If we mo ve Π( i ) along with all candidates ℓ > i, Π( ℓ ) ∈ C ′′ p ositioned b et ween t i and t f in A to t f , this op eration will not increase Kendall tau distance b et w een A and X as pro ved in Lemma 2. Note that we conserve the relative order b etw een candidates of C ′′ in the aforemen tioned op eration. W e illustrate this operation and the imp ortance of Lemma 2 using Figure 1. Note that in the figure, it is clear that moving Π( i ) from t i to t f reduces the num b er of disagreemen ts inv olving Π( i ). How ever, it is not obvious why moving other candidates of C ′′ in red squares do es not increase the total Kendall tau distance. Lemma 2. Moving Π( i ) , along with every c andidate Π( ℓ ) ∈ C ′′ for ℓ > i that is p ositione d b etwe en t i and t f in A , to p osition t f do es not incr e ase the Kendal l tau distanc e b etwe en A and X . Pr o of. T o streamline our arguments, we begin by defining a sequential operation that mov es Π( i ) and all the candidates of C ′′ after Π( i ) that are p ositioned b et ween t i and t f to t f in A . The pro cedure is as follows, first w e mo ve all the candidates of C ′′ after Π( i ) positioned at t i to t i + 1, then we mo ve all the candidates of C ′′ p ositioned at t i + 1 to t i + 2, note that in the second op eration, the candidates of C ′′ p ositioned at t i + 1 include all the candidates mov ed to t i + 1 after the first op eration. W e contin ue to do this operation un til we mov e all the candidates Π( ℓ ) ∈ C ′′ , ℓ ≥ i , positioned b et ween t i and t f to t f . W e denote the c hange in the num b er of disagreements of c ∈ C ′′ with the candidates in C ′ when we mov e a candidate c from its p osition x to y as ∆ x,y c , i.e., ∆ x,y c is the num b er of disagreements when c is at p osition y min us the num b er of disagreemen ts when c is at position x . Claim 3. F or every j ∈ { i, . . . , f − 1 } , ∆ t j ,t j +1 Π( ℓ ) ≤ ∆ t j ,t j +1 Π( i ) , wher e Π( i ) and Π( ℓ ) ar e at p osition j , ℓ > i . Pr o of. Note that mo ving Π( i ) from t j to t j +1 , either increases or decreases the num b er of disagreemen ts of Π( i ) with the candidates in C ′ b y 1. W e consider these t wo cases separately . 5 Case 1. If moving Π( i ) to t j + 1 increases the n umber of disagreements by 1 (i.e., ∆ t j ,t j +1 Π( i ) = +1), this implies that, Π( i ) ≺ X A | C ′ ( t j + 1). In this case, we cannot definitively predict the effect of moving the candidate Π( ℓ ) forward to t j + 1, as Π( i ) ≺ X Π( ℓ ). Therefore, ∆ t j ,t j +1 Π( ℓ ) can b e either p ositiv e or negativ e 1 when mo v ed. Hence ∆ t j ,t j +1 Π( ℓ ) ≤ ∆ t j ,t j +1 Π( i ) Case 2. If moving Π( i ) to t j + 1 decreases the n umber of disagreements by 1 (i.e., ∆ t j ,t j +1 Π( i ) = − 1), this implies that, A | C ′ ( t j + 1) ≺ X Π( i ). Since A | C ′ ( t j + 1) ≺ X Π( i ) ≺ X Π( ℓ ), mo ving Π( ℓ ) up b y one p osition in A will also decrease the n umber of disagreemen ts b y 1 (i.e., ∆ t j ,t j +1 Π( ℓ ) = − 1). Hence ∆ t j ,t j +1 Π( ℓ ) ≤ ∆ t j ,t j +1 Π( i ). Due to our sequential procedure and Claim 3, we kno w that for every Π( ℓ ), where ℓ > i , at t i + p , where p ∈ { 0 , . . . , f − 1 } X 0 ≤ j i that is p ositioned betw een t i and t f in A , to p osition t f do es not increase the Kendall tau distance betw een A and X . Finally , w e are ready to argue that the ranking returned b y Algorithm 1 is an optimal extension of R . Lemma 4. L et R ′ b e the r anking r eturne d by Algorithm 1. Then, R ′ is an optimal extension of R . Pr o of. W e first claim that R ′ is an extension of R . Note that R ′ | C ′ = R | C ′ as we do not change the relativ e ordering of any pair of candidates of C ′ in the algorithm, i.e., for all { c, c ′ } ⊆ C ′ , if c ≺ R c ′ , then c ≺ R ′ c ′ . Let Π i = S i q =1 Π( q ) and R ′ i b e the ranking after i th iteration of Algorithm 1. Next, we argue the optimality in the following claim. Claim 5. F or al l i ∈ [ | C ′′ | ] , ther e exists an optimal extension R ⋆ of R such that and R ⋆ | C ′ ∪ Π i = R ′ i Pr o of. W e prov e this claim using induction on i . F or i = 1, consider the case where R ⋆ | C ′ ∪ Π 1  = R ′ 1 , which means that t 1  = t ′ 1 where t 1 is the position of Π(1) in R ⋆ and t ′ 1 is the position where Π(1) was inserted in R ′ 1 . If we mov e Π(1) to the p osition t ′ 1 in R ⋆ , we note that the num b er of disagreemen ts b et ween R ⋆ and X do not increase as t ′ 1 is the p osition returned b y Algorithm 1 such that insertion of Π(1) at that p osition leads to the least num b er of disagreemen ts b et ween our ranking and X . Let this new ranking b e ˜ R . W e can see that ˜ R is an optimal extension of R such that ˜ R | C ′ ∪ Π 1 = R ′ 1 . Supp ose that Claim 5 holds for i ≤ ℓ . Then there exists an optimal extension R ⋆ of R suc h that R ⋆ | C ′ ∪ Π l = R ′ l . Next w e argue for i = ℓ + 1. Consider the case where R ⋆ | C ′ ∪ Π ℓ +1  = R ′ ℓ +1 , which means that t ℓ +1  = t ′ ℓ +1 where t ℓ +1 is the p osition of Π( ℓ + 1) in R ⋆ and t ′ ℓ +1 is the p osition where Π( ℓ + 1) w as inserted in R ′ ℓ +1 . So let us assume the case where t ′ ℓ +1 < t ℓ +1 . Note that t ℓ = t ′ ℓ ≤ t ′ ℓ +1 as R ⋆ | C ′ ∪ Π l = R ′ ℓ . W e can alw ays mov e Π( ℓ + 1) to t ′ ℓ +1 in R ⋆ and it do es not change the relativ e ordering of candidates of C ′′ . If we mov e Π( ℓ + 1) to the position t ′ ℓ +1 in R ⋆ , w e note that the n umber of disagreements b et ween R ⋆ and X inv olving R ⋆ do es not increase as t ′ ℓ +1 is the p osition returned b y Algorithm 1 suc h that insertion of Π( ℓ + 1) at that p osition leads to the least n um b er of disagreements b etw een our ranking and X , note that w e are only moving Π( ℓ + 1) in R ⋆ , so the c hanges in the Kendall tau distance will only b e caused due to Π( ℓ + 1). Hence, the Kendall tau distance does not increase. Let this new ranking b e ˜ R , w e can see that ˜ R is an optimal extension of R suc h that ˜ R | C ′ ∪ Π ℓ +1 = R ′ ℓ +1 . F or the second case where t ℓ +1 < t ′ ℓ +1 , w e can say that mo ving Π( ℓ + 1) and every candidate Π( x ) , x > ℓ + 1, p ositioned betw een t ℓ +1 and t ′ ℓ +1 in R ⋆ to t ′ ℓ +1 in R ⋆ do es not increase the n umber of disagreements b et w een R ⋆ and X as a direct consequence of Lemma 2. Note that t ′ ℓ +1 is p osition returned by Algorithm 1 such that insertion of Π( ℓ + 1) at that p osition leads to the least n umber of disagreemen ts b etw een our ranking and X . Let this new ranking b e ˜ R , we can see that ˜ R is an optimal extension of R such that ˜ R | C ′ ∪ Π ℓ +1 = R ′ ℓ +1 . 6 As Claim 5 holds for all i ∈ [ | C ′′ | ], w e can say that there exists an optimal extension R ⋆ of R such that R ⋆ | C ′ ∪ C ′′ = C = R ′ . This prov es that R ′ returned b y Algorithm 1 is an optimal extension to R . Lemma 4 giv es an optimal extension of a ranking in R . W e complete the algorithm in the follo wing theorem b y extending all the rankings using the same pro cedure and argue the running time. Theorem 6. Possible Kemeny Score c an b e solve d in p olynomial time. Pr o of. Let I = ( C , R , X, k ) b e an instance of Possible Kemeny Score . Recall that rankings in R are o ver C ′ ⊆ C . Let R = ( R 1 , R 2 , . . . , R n ). F or every R i ∈ R , w e find an optimal extension R ′ i using Algorithm 1. Let R ′ = ( R ′ 1 , R ′ 2 , . . . , R ′ n ) b e the set of optimal extensions obtained by applying Algorithm 1 to eac h ranking in R . If KT ( X, R ) ≤ k , w e return Yes ; otherwise No . Next, w e pro ve the correctness of our algorithm. Clearly , if w e return Yes , then I is a yes-instance of Possible Kemeny Score . W e next argue that if I is a yes-instance of Possible Kemeny Score we return an extension R ′ suc h that KT ( X, R ′ ) ≤ k . Let us supp ose that there exists a set of extended rankings R ⋆ = ( R ⋆ 1 , R ⋆ 2 , . . . , R ⋆ n ), such that KT ( X , R ⋆ ) ≤ k . Due to Lemma 4, w e can sa y that for all i ∈ [ n ], KT ( X, R ′ i ) ≤ KT ( X , R ⋆ i ), which implies that, KT ( X, R ′ ) = P i KT ( X, R ′ i ) ≤ P i KT ( X, R ⋆ i ) = KT ( X, R ⋆ ) ≤ k . Next, we argue the running time of the algorithm. Note that Algorithm 1 runs in time O ( m 3 ) and w e call Algorithm 1 n times. Th us, the algorithm runs in time O ( n · m 3 ). Note that in Algorithm 1 at line n umber 5, the Kendall tau distance calculation runs in time O ( m ) because we only need to compute the n umber of disagreemen ts b et w een Π( i ) and the candidates of C ′ . This completes the pro of. In our algorithm, w e assume that all the rankings in R are ov er the subset C ′ ⊆ C . How ev er, this is only for ease of explanation, and the algorithm works even when rankings in R are not o ver the same subset. This is due to the reason that we call Algorithm 1 indep enden tly for eac h partial ranking in R . 4 $ -Brib ery-Kemen y Score and Ranking Deletion Kemeny Score W e begin with designing a p olynomial-time algorithm for $ -Briber y-Kemeny Score . Recall that in $ - Briber y-Kemeny Score we need to c hange some rankings to X so that X is at most k distance aw a y from the new set of rankings. The problem inherently seems to b e closer to the Knapsa ck problem, in which giv en a set of items, A = { a 1 , . . . , a n } , a weigh t function w : A → N , a v alue function v : A → N , and tw o natural num bers T , W ; the goal is to decide if there exists a subset S ⊆ A such that P a ∈ S w ( a ) ≤ W and P a ∈ S v ( a ) ≥ T . W e formalize this in tuition below by giving a p olynomial time reduction to the Knapsa ck problem. Lemma 7. Ther e is a p olynomial time r e duction fr om the $ -Briber y-Kemeny Score pr oblem to the Knapsack pr oblem, wher e the values ar e p olynomial ly b ounde d in the input size. Pr o of. Giv e an instance I = ( C , R , X , cost , bgt , k ) of the $ -Briber y-Kemeny Score problem, we create an instance J = ( A, w , v , T , W ) of the Knapsack problem as follows. Let R = ( R 1 , . . . , R n ). The set of items is A = { a 1 , . . . , a n } . The weigh t function w : A → N is defined as follows: w ( a i ) = cost ( R i ) and the v alue function v : A → N is defined as follows: v ( a i ) = KT ( X, R i ). F urthermore, W = bgt and T = D − k , where D = KT ( X, R ). Note that the v alue of every item is b ounded by |C | 2 and T is at most n |C | 2 . Next, w e prov e the correctness. In the forw ard direction, supp ose that I is a y es-instance of $ -Briber y-Kemeny Score . Let S = R \ R ′ where R ′ is the new set of rankings, i.e., S is the set of rankings in R that are manipulated. Note that KT ( X, R ′ ) ≤ k and P R ∈ S cost ( R ) ≤ bgt . W e construct a set S ′ as follows: S ′ = { a i ∈ A : R i ∈ S } . W e claim that S ′ is a solution to J . Due to the construction, we know that P a ∈ S ′ w ( a ) ≤ bgt . As discussed in the definition of $ -Briber y-Kemeny Score , w e can assume that every ranking in S is the same as X . Th us, KT ( X , R ′ ) = KT ( X , R \ S ). Hence, KT ( X, S ) + KT ( X , R ′ ) = D . Since KT ( X, R ′ ) ≤ k , it follows that KT ( X, S ) ≥ D − k . Thus, P a ∈ S ′ v ( a ) ≥ D − k . In the reverse direction, supp ose that J is a yes-instance of the Knapsack problem. Let S ⊆ A b e a solution to J . W e construct a set of rankings S ′ as follows: S ′ = { R i ∈ R : a i ∈ S } . W e claim that S ′ is a solution to I . Let R ′ b e the rankings obtained by changing S ′ to X and the remaining rankings are same in 7 R and R ′ . Clearly , P R ∈ S ′ cost ( R ) ≤ bgt . Since, P a ∈ S v ( a ) ≥ D − k , using the same arguments as earlier, KT ( X, R ′ ) ≤ k . This completes the proof. Since there is a pseudo-p olynomial time algorithm for Knapsack that solves the problem in p olynomial time if either weigh ts or v alues are p olynomially bounded [38] and the ab ov e reduction is p olynomial time, w e hav e the following result. Theorem 8. $ -Briber y-Kemeny Score c an b e solve d in p olynomial time. Note that c hanging a ranking to X is equiv alent to deleting this ranking as the contribution of this ranking to Kendall tau distance b ecomes 0 due to b oth the op erations. Thus, we ha ve the following result due to the same algorithm, with the difference that instead of changing it to X , we delete the ranking. Theorem 9. Ranking Deletion-Kemeny Score c an b e solve d in p olynomial time. 5 Sw ap Brib ery-Kemen y Score In this section, we design a p olynomial-time algorithm for Sw ap Briber y-Kemeny Score problem. Recall that in Sw ap Briber y-Kemeny Score we need to p erform some admissible sw aps on the set of rankings R , so that X is at most k distance aw ay from the new set of rankings R ′ . Let A be a ranking ov er C . Consider a pair of candidates c, c ′ suc h that c ≺ X c ′ and c ′ ≺ A c . If c, c ′ are consecutive in A , then w e sa y that { c, c ′ } is an admissible disagreement betw een A and X , otherwise, it is a non-admissible disagreement. In our algorithm, w e sw ap candidates only when they are in disagreemen t. Before presenting the algorithm, w e first establish an imp ortan t prop ert y regarding admissible disagreements. Sp ecifically , if there exists a non-admissible disagreemen t { a, b } b et ween rankings A and X , then it implies the existence of an admissible disagreemen t b et w een A and X . This prop erty is crucial b ecause it implies that for any ranking A , either there exists an admissible disagreement or there is no disagreement b et ween A and X (i.e., KT ( X , A ) = 0). Consequently , this ensures that in our algorithm, we can alwa ys p erform an admissible swap action if KT ( X , A ) > 0. W e formalise all these prop erties b elow. Throughout this section, when w e refer to p erforming a swap action b etwe en c andidates , w e implicitly mean that the candidates are in disagreement. Throughout the section, after p erforming a swap action on a ranking R , we reuse the notation R for the new ranking. W e b egin with the follo wing observ ation. Observ ation 1. One swap op er ation in A of an admissible disagr e ement b etwe en A and X r e duc es the Kendal l tau distanc e b etwe en them by exactly one. Lemma 10. The pr esenc e of a non-admissible disagr e ement b etwe en r ankings A and X implies the existenc e of an admissible disagr e ement b etwe en A and X . Pr o of. Supp ose that there exists a non-admissible disagreement { a, b } b et ween A and X . T o wards the con tradiction, supp ose that there are no admissible disagreements betw een A and X . Since { a, b } is a disagreemen t, let a ≺ A b and b ≺ X a . Note that { a, b } is a non-admissible disagreement, so there must b e some candidates c 1 , c 2 , . . . , c p rank ed b etw een a and b in A . Let A = . . . a ≺ c 1 ≺ . . . ≺ c p ≺ b ≺ . . . . Since there is no admissible disagreement b et w een A and X , we know that a ≺ X c 1 , c i ≺ X c i +1 , for all i ∈ [ p − 1], and c p ≺ X b . This implies a ≺ X b due to transitivity in the ranking X , whic h is a contradiction. Due to Lemma 10, w e hav e the following useful prop ert y . Corollary 11. We c an always p erform an admissible swap action in the r anking A if KT ( X , A ) > 0 . By p erforming s swaps in the ranking A , we mean executing s sw ap op erations sequentially , each applied to the up dated ranking resulting from the previous sw ap. It is important to note that a sw ap which is initially non-admissible may b ecome admissible after performing some admissible swaps, allowing us to p erform that swap at a later stage. Due to Observ ation 1 and Corollary 11, w e can alw ays p erform a sequence of 0 ≤ s ≤ KT ( X , A ) admissible swap actions in A and that res ults in a drop of s in the Kendall tau distance b etw een A and X . W e finally mov e to wards designing our algorithm. 8 Theorem 12. Sw ap Briber y-Kemeny Score c an b e solve d in p olynomial time. Pr o of. Let I = ( C , R , X , cost , bgt , k ) b e a given instance of the Sw ap Briber y-Kemeny Score problem. W e design a dynamic programming algorithm to solv e this problem. Let R = ( R 1 , R 2 , . . . , R n ). W e define the dynamic programming table as follows: T [ i, s 1 , s 2 , j ] = minim um cost incurred in performing admissible sw ap actions on the first i rankings such that s 1 sw aps are done to R i , s 2 sw aps are done to the set of rankings ( R 1 , R 2 , . . . , R i − 1 ), and the Kendall tau distance b et ween the set of rankings ( R 1 , R 2 , . . . , R i ) and X is at most j . F or i = 1, there are no previous rankings to consider; ho wev er, we write T [1 , s 1 , s 2 , j ] for brevity , and for s 2 > 0, it is an inv alid entry . F or 1 < i ≤ |R| , if either s 1 > KT ( X , R i ) or s 2 > KT ( X , ( R 1 , R 2 , . . . , R i − 1 )), then also the entry T [1 , s 1 , s 2 , j ] is inv alid. W e store ∞ for the inv alid en tries or if the allo wed num b er of sw aps is insufficien t to ac hieve the desired Kendall tau distance. W e compute the remaining table entries as follows. Base Case: F or i = 1, 0 ≤ s 1 ≤ KT ( X , R 1 ), s 2 = 0 and 0 ≤ j ≤ k , we compute the table as follows. T [1 , s 1 , 0 , j ] = ( s 1 · cost ( R 1 ) If KT ( X , R 1 ) − s 1 ≤ j ∞ otherwise . (1) Recursiv e Step: F or all 1 < i ≤ |R| , 0 ≤ s 1 ≤ KT ( X, R i ), 0 ≤ s 2 ≤ KT ( X, ( R 1 , R 2 , . . . , R i − 1 )), and 0 ≤ j ≤ k , w e compute the table entries as follo ws. T [ i, s 1 , s 2 , j ] = min s ′ 2 + s ′′ 2 = s 2 T [ i − 1 , s ′ 2 , s ′′ 2 , j − KT ( X , R i ) + s 1 ] + s 1 · cost ( R i ) (2) Next, w e pro ve the correctness of this dynamic programming algorithm in the follo wing lemma. Lemma 13. F or every i ∈ [ |R| ] , 0 ≤ s 1 ≤ KT ( X, R i ) , 0 ≤ s 2 ≤ KT ( X, ( R 1 , R 2 , . . . , R i − 1 )) , and 0 ≤ j ≤ k , Equations 1 and 2 c orr e ctly c ompute T [ i, s 1 , s 2 , j ] . Pr o of. W e will pro ve the correctness of the dynamic programming solution through induction on i . F or the base case, when i = 1, if KT ( X , R 1 ) − s 1 ≤ j , then b y p erforming s 1 sw aps, the distance b et ween the resultan t ranking and X is at most j , otherwise s 1 sw aps do es not decrease the distance to j . Hence, T [1 , s 1 , 0 , j ] is computed correctly by Equation 1. F or the inductive step, let us suppose that the table en tries are computed correctly for all i ≤ l , 0 ≤ s 1 ≤ KT ( X, R i ), 0 ≤ s 2 ≤ KT ( X, ( R 1 , R 2 , . . . , R i )) and 0 ≤ j ≤ k . W e denote the v alue of T [ l + 1 , s 1 , s 2 , j ] computed by Equation 2 as al g . Let OPT b e a set of sw ap op erations corresp onding to the minimum cost incurred in p erforming admissible sw ap actions on the first l + 1 rankings suc h that s 1 sw aps are done to R l +1 , s 2 sw aps are done to the set of rankings ( R 1 , R 2 , . . . , R l ), and the Kendall tau distance b et w een the set of rankings ( R 1 , R 2 , . . . , R l +1 ) and X is at most j . Let OPT cost b e the cost of p erforming sw aps in OPT . Clearly , OPT cost ≤ alg . Let us supp ose that s l admissible swaps are p erformed on R l in OPT , this implies that s 2 − s l admissible swaps are p erformed on ( R 1 , R 2 , . . . , R l − 1 ). W e claim that T [ l , s l , s 2 − s l , j − KT ( X , R l +1 ) + s 1 ] + s 1 · cost ( R l +1 ) = OPT cost . Due to Equation 2, alg ≤ T [ l, s l , s 2 − s l , j − KT ( X , R l +1 ) + s 1 ] + s 1 · cost ( R l +1 ). Since alg ≥ OPT cost , T [ l , s l , s 2 − s l , j − KT ( X , R l +1 ) + s 1 ] + s 1 · cost ( R l +1 ) ≥ OPT cost . Since OPT cost − s 1 · cost ( R l +1 ) is the cost of performing s 2 admissible sw aps on the set of rankings ( R 1 , R 2 , . . . , R l ) where s l admissible sw aps are p erformed on R l and s 2 − s l sw aps are p erformed on the set of rankings ( R 1 , R 2 , . . . , R l − 1 ) such that the Kendall tau distance is at most j − ( KT ( X , R l +1 ) − s 1 ) = j − KT ( X , R l +1 ) + s 1 from X , due to induction h yp othesis, T [ l, s l , s 2 − s l , j − KT ( X , R l +1 ) + s 1 ] ≤ OPT cost − s 1 · cost ( R l +1 ). Hence, T [ l, s l , s 2 − s l , j − KT ( X, R l +1 ) + s 1 ] + s 1 · cost ( R l +1 ) = OPT cost . This implies that al g ≤ OPT cost . Hence al g = OPT cost . Thus Equations 1 and 2 correctly compute the dynamic programming table. The minim um cost incurred in performing admissible swap actions on R , suc h that the Kendall tau distance betw een R and X is at most j is min s 1 ,s 2 T [ |R| , s 1 , s 2 , k ]. So w e return Yes if min s 1 ,s 2 T [ |R| , s 1 , s 2 , k ] ≤ bgt , otherwise No . The correctness follo ws due to Lemma 13 W e compute at most |R| 2 m 2 k en tries and each en try can b e computed in time O ( m 2 ). Thus, the algorithm runs in p olynomial time. 9 6 Candidate Deletion Kemen y Score In this section, w e b egin with the case when k = 0, and design a polynomial time algorithm by reducing the problem to the Hea viest Common Increasing Subsequence ( HCIS ) on permutations problem. In the HCIS problem, given p sequences of n umbers, a weigh t function w on the n umbers, and an integer W , the goal is to decide whether there exists a sequence of num b ers with a total w eight of at least W that is an increasing subsequence of each of the given p sequences. In our reduction, these sequences are p erm utations. Lemma 14. Ther e exists a p olynomial time r e duction fr om Candida te Deletion-Kemeny Score to HCIS on p ermutations when k = 0 . Pr o of. Let I = ( C , R , X , cost , bgt , k ) b e an instance of Candida te Deletion-Kemeny Score where k = 0. Let R = ( R 1 , . . . , R n ). W e define a function f : C → [ m ] as follows: f ( X ( i )) = i . Recall that X is a p erm utation ov er C and C has m candidates. W e construct a set of p ermutations J 1 , . . . , J n as follows. Each J i is an m -length sequence, where for j ∈ [ m ], J i ( j ) = f ( R i ( j )). W e can observe that J i ( j )  = J i ( k ) when j  = k as R i ( j )  = R i ( k ) when j  = k and f is injective, whic h implies that the elements of J i are unique for all i and due to the construction the maxim um v alue of f and the length of J i are b oth m . Hence J i is a p erm utation for all i . The weigh t of num b er i is w ( i ) = cost ( f − 1 ( i )). Let W = P c ∈C cost ( c ) − bgt . Let J = ( J 1 , . . . , J n , w , W ) b e an instance of HCIS on p erm utations. Next, we pro ve the equiv alence b et w een the t wo instances. In the forw ard direction, let Z b e a solution to I . Let ˜ Z = f ( C \ Z ) W e claim that J 1 | ˜ Z is a solution to J . Since P c ∈ Z cost ( c ) ≤ bgt , P c ∈C \ Z cost ( c ) ≥ W . Th us P i ∈ ˜ Z w ( i ) ≥ W . Supp ose that J 1 | ˜ Z is not an increasing subsequence of the p erm utation J i , i ∈ [ n ]. Then, there exists { ℓ, ℓ ′ } ⊆ ˜ Z such that ℓ < ℓ ′ , and ℓ ′ is b efore ℓ in J i . Since ℓ < ℓ ′ , it follows that f − 1 ( ℓ ) ≺ X f − 1 ( ℓ ′ ). Since ℓ ′ is b efore ℓ in J i it follows that f − 1 ( ℓ ′ ) ≺ R i f − 1 ( ℓ ′ ). F urthermore, since { f − 1 ( ℓ ) , f − 1 ( ℓ ′ ) } ⊆ C \ Z , it con tradicts that Z is a solution to I . In the reverse direction, let Π b e a solution to J . Let S b e the set of n umbers that are in Π and ˜ S b e the set of n umbers that are not in Π. Let Z = { f − 1 ( s ) : s ∈ ˜ S } . W e claim that Z is a solution to I . Since P i ∈ S w ( i ) ≥ W it follo ws that P c ∈ Z cost ( c ) ≤ bgt . Supp ose Z is not a solution to I , then there exists a pair of candidates { c, c ′ } ⊆ C \ Z and a ranking R i suc h that c ≺ X c ′ and c ′ ≺ R i c . Supp ose that c = X ( j ) and c ′ = X ( j ′ ). Thus j < j ′ and j ′ is b efore j in the sequence J i . Since { c, c ′ } ⊆ C \ Z it follows that { j, j ′ } ⊆ S . This contradicts the fact that Π is an increasing common subsequence of J 1 , . . . , J n . Chan et al. [15] designed an algorithm for HCIS with unit weigh ts. The same algorithm can be used for non-unit weigh ts b y storing the maximum w eight of the common increasing subsequence as the rank of a matc h instead of the longest length. W e present the complete algorithm for completeness. Lemma 15. HCIS c an b e solve d in p olynomial time when the se quenc es ar e p ermutations. Pr o of. Let I = ( J 1 , J 2 , . . . , J m , w , W ) b e the giv en instance of the problem, where for all i ∈ [ m ], J i is a p erm utation ov er [ n ] and for all x ∈ [ n ], w ( x ) ≥ 0. Let J − 1 i ( x ) denote the p osition of x in J i and let J k i denote the prefix of J i with k elements. W e define the notations match , dominating match , and r ank as follo ws. A match is a m -tuple, ( i 1 , . . . , i m ) where J 1 ( i 1 ) = J 2 ( i 2 ) = . . . = J m ( i m ), the v alue of this match is J 1 ( i 1 ). A matc h ( δ 1 , . . . , δ m ) is called a dominating matc h of another match ( θ 1 , . . . , θ m ) if J 1 ( δ 1 ) < J 1 ( θ 1 ) and δ j < θ j for all 1 ≤ j ≤ m . Note that ∆ j denotes the match with the v alue of j and ∆ j is a unique match b ecause our sequences are p erm utations. The rank of a match ∆ = ( δ 1 , . . . , δ m ), denoted by R (∆), is the weigh t of the HCIS of J δ 1 1 , J δ 2 2 , . . . , J δ m m suc h that J 1 ( δ 1 ) is the last element of the HCIS . F or ∆ 1 , . . . , ∆ n , R (∆ j ) is computed in increasing order of j as follows, R (∆ j ) =      w (∆ j ) if no match dominates ∆ j . w (∆ j ) + max { R (∆ k | ∆ k dominates ∆ j } otherwise . (3) It is easy to see that HCIS w ould be max ∆ j R (∆ j ). F or the decision version, we accept iff max ∆ j R (∆ j ) ≥ W . W e will now pro ve the correctness of the dynamic programming solution through induction on j . Note that b ecause the sequences are p ermutations, there are n matc hes with v alues 1, 2, . . . , n . W e can compute 10 all the matc hes in time O ( m · n ) and w e pro cess them in the increasing order of their v alue i.e., for a match with v alue j , the ranks of all the matches with v alue 1 , . . . , j − 1 should hav e b een computed first as a match with v alue j can only be dominated by a matc h whic h has a v alue less than j . F or the base case, R (∆ 1 ) = w (∆ 1 ) = w (1) as there is no match that can dominate ∆ 1 and hence the HCIS ending with v alue 1 only has one elem en t with w eigh t w (1). Let us suppose that for all y < j , R (∆ y ) is computed correctly . F or the first case assume w OP T as the optimal rank of match ∆ j that is the w eight of the sequence O P T = j i.e. the HCIS ending at match ∆ j con tains only the element j . It is obvious in this case that the rank computed b y Equation 3 = w OP T . F or the second case assume w OP T as the optimal rank of the matc h ∆ j that is w eight of the sequence OP T = . . . , x, j . W e kno w that the rank of match ∆ j computed by Equation 3 ≤ w OP T as w OP T is the optimal rank of the match ∆ j . Since x is b efore j in O P T , x has to b e before j in every sequence J i and x < j , hence ∆ x dominates ∆ j whic h means that the rank of matc h ∆ j computed b y Equation 3 ≥ w (∆ j ) + R (∆ x ) ≥ w OP T . Hence the rank of match ∆ j computed b y Equation 3 = w OP T . Hence Equation 3 correctly computes the rank of all matches. W e precompute all the n matches in time O ( m · n ), then for each match w e c heck all the n matches for domination in time O ( m ), hence the algorithm runs in time O ( m · n 2 ) Theorem 16. Candid a te Deletion-Kemeny Score c an b e solve d in p olynomial time when k = 0 . Next, we sho w that for one ranking (i.e., |R| = 1), the problem is equiv alen t to the P ar tial Ver tex Cover ( WPVC ) problem for p erm utation graphs. In WPVC , giv en a v ertex-weigh ted graph G with w eight function w : V ( G ) → N , and in tegers W , t , the goal is to find a subset S ⊆ V ( G ) of weigh t at most W that co vers at least t edges, i.e., for at least t edges, at least one of the endp oin ts is in S . A p ermutation gr aph is a graph whose vertices represen t the elements of a p ermutation, and whose edges represent pairs of elements that are reversed by the permutation, and w e say that such a permutation realises the giv en p ermutation graph. The complexity of WPV C for p erm utation graph is not known. How ever, if WPVC for p erm utation graph is NP-hard , then Candida te Deletion-Kemeny Score is also NP-hard ; and if it is p olynomial-time solv able, then Candida te Deletion-Kemeny Score is polynomial time solv able for one ranking. Lemma 17. Ther e is a p olynomial time r e duction fr om WPVC on p ermutation gr aph to Candida te Deletion-Kemeny Score . Pr o of. Let I = ( G, w , W, t ) b e an instance of WPVC where G is a p erm utation graph. Let Π b e a p erm u- tation realising the p erm utation graph G . Let n = | V ( G ) | . Let C = V ( G ), X = 1 ≻ 2 ≻ . . . ≻ n , R = Π, cost = w , bgt = W , and k = | E ( G ) | − t . Let J = ( C , R , X , cost , bgt , k ) b e an instance of Candida te Deletion-Kemeny Score . Next, we pro ve the equiv alence b et ween the tw o instances. In the forward direction, let Z b e a solution to I . W e claim that Z is also a solution to J . Note that the n umber of edges in E ( G − Z ) is at most | E ( G ) | − t , thus due to the definition of p erm utation graph, the num b er of pairs of v ertices such that i < j and Π( i ) > Π( j ), where { Π( i ) , Π( j ) } ⊆ G − Z , is at most | E ( G ) | − t . Thus, KT ( X | C \ Z , R | C \ Z ) ≤ | E ( G ) | − t . Clearly , cost ( Z ) ≤ bgt . In the reverse direction, let Z be a solution to J . Then, w e claim that Z is also a solution to I . Clearly , w ( Z ) ≤ W . Note that KT ( X , R ) = | E ( G ) | b y the definition of p ermutation graph. Since KT ( X | C \ Z , R | C \ Z ) ≤ | E ( G ) | − t , for at least t disagreemen ts, one of the candidate is in Z . Thus, Z cov ers at least t edges in G . Due to Lemma 17, if WPVC on a p ermutation graph is NP-hard , then Candida te Deletion-Kemeny Score is also NP-ha rd . Obtaining a p olynomial-time algorithm for WPVC on a p erm utation graph is also b eneficial due to the following re sult. Lemma 18. Ther e is a p olynomial time r e duction fr om Candida te Deletion-Kemeny Score when |R| = 1 to WPVC on p ermutation gr aph. Pr o of. Let I = ( C , R , X, cost , bgt , k ) b e an instance of Candida te Deletion-Kemeny Score where R has only one ranking, sa y R . W e construct an instance J = ( G, w , W, t ) of the WPV C problem as follo ws. Let f : C → [ m ] b e as function defined as follows: for eac h i ∈ [ m ], f ( X ( i )) = i . Let Π = f ( R (1)) , . . . , f ( R ( m )) b e a p erm utation and G Π b e a p erm utation graph corresp onding to the permutation Π. Let w = cost , W = bgt , and t = | E ( G ) | − k . Let J = ( G Π , w , W , t ) b e an instance of WPV C where G Π is a p ermutation graph. Next, we show the equiv alence betw een the t wo instances. 11 In the forw ard direction, let Z b e a solution to I . W e claim that Z is also a solution to J . Clearly , w ( Z ) ≤ W . Since | E ( G ) | = KT ( X , R ) and KT ( X | C \ Z , R | C \ Z ) ≤ k , for at least KT ( X, R ) − k edges one of the endp oin t is in Z . In the rev erse direction, supp ose that Z is a solution to J . Then, G − Z has at most | E ( G ) | − k edges. Note that if there exists a pair of candidates x, y that disagree b et w een R and X , then f ( x ) and f ( y ) are reversed and there is an edge betw een then in G Π . Th us, there can b e at most k = | E ( G ) | − t disagreemen ts b et ween R and X . This completes the pro of. Lemma 17 gives a p olynomial-time reduction from WPVC on p erm utation graphs to Candida te Deletion- Kemeny Score , and the fact that our construction fixes | R | = 1 do es not w eaken the result, an y instance of WPVC is still mapped to an equiv alent instance of the Candida te Deletion-Kemeny Score . Con- v ersely , Lemma 18 reduces that same | R | = 1 v ariant bac k to WPVC. Because both reductions op erate on the iden tical restricted case, they establish a full p olynomial-time equiv alence, which not only yields Theorem 19 but also transfers hardness: if WPVC on permutation graphs were shown NP-hard, the same would hold for Candida te Deletion-Kemeny Score with | R | = 1 and hence for Candida te Deletion-Kemeny Score in general Th us, due to Lemma 17, if WPVC on a p ermutation graph is solv able in p olynomial time, then Can- dida te Deletion-Kemeny Score is also solv able in p olynomial time for one ranking. Thus, due to Lemma 17 and 18, w e hav e the following result. Theorem 19. Candida te Deletion-Kemeny Score c an b e solve d in p olynomial time for one r anking if and only if WPVC c an b e solve d in p olynomial time for p ermutation gr aphs unless P = NP . 7 Conclusion In this pap er, w e studied the Kemeny Score problem under some w ell-studied brib ery schemes. 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