8.2 Number of votes
8.2.2 Borda
After considering Possible Winner for k-approval and a constant number of votes, we now investigate the same setting for the Borda rule. We first show that Possible Winner for Borda is NP-complete for a profile in which the number of partial votes is at least three making use of an unbounded number of linear votes. After this, we discuss how to obtain NP-hardness on profiles for a constant number of linear votes for specific constant numbers of partial votes. The results of this section are based on a many-one reduction from a special case of the 3-Partition problem.
3-Partition
Given: A multi-set A ={a1, . . . , an} of positive integers and B := (3/n) · P
ai∈Aai.
Question: Is there a partition of A into size-3 subsets A1, . . . , An/3 such that P
ai∈Ajai= B for each j∈ {1, . . . , n/3}?
Note that in this subsection n does not denote the number of votes (which is constant) but, following the literature, the number of integers from the 3-Partition instance. The 3-Partition problem is strongly NP-complete [118]. This implies that the NP-hardness still holds when the integers of A are polynomially bounded in n.
We denote the special case that each integer ai ∈ A must be a multiple of n as 3-n-Partition. The 3-n-Partition problem is strongly NP-hard since every 3-Partition instance can be easily reduced to a 3-n-Partition instance by multiplying all input integers with n. Now, we provide the main result of this section.
Theorem 8.3. For Borda, Possible Winner is NP-complete for an instance with partial profile Vl∪ Vp if|Vp| ≥ 3 with Vl being the set of linear and Vp being the set of partial votes.
8.2 Number of votes 133 Proof. Let A = {a1, . . . , an} denote a 3-n-Partition instance with B := (3/n) · P
ai∈Aai. To ease the presentation, we first give a construction for the case that ai < ai+1 for i = 1, . . . , n− 1 and a1 = n and after this discuss the general case. We construct a partial profile P = Vp∪ Vl with |Vp| = 3 over a set C of candidates in which the distinguished candidate c∈ C is a possible unique winner if and only if A is a yes-instance for 3-n-Partition. The set of candidates is
C :={c} ⊎ E ⊎ T ⊎ D,
with one candidate for every member of A, that is, E := {ei | ai ∈ A}, candi-dates representing the subsets resulting from the partition into 3-sets, that is, T :=
{t1, . . . , tn/3}, and a set of dummy candidates D := Un
i=1Di only needed to “fill”
positions and further specified in the following.
The set of partial votes Vp consists of three identical partial votes v1, v2, and v3. Each of them is defined as follows:
c≻ T and c ≻ D1≻ e1≻ D2≻ · · · ≻ Di ≻ ei≻ · · · ≻ Dn≻ en
with |D1| = a1− 1 and |Di| = ai− ai−1− 1 for i ∈ {2, . . . , n}. This definition fixes the number of dummy candidates; more precisely,
|D| = |D1| + |D2| + · · · + |Dn| = a1− 1 + Xn i=2
(ai− ai−1− 1) = an− n.
Thus, the total number of candidates is
m = 1 +|E| + |T | + |D| = 1 + n + n/3 + an− n = an+ n/3 + 1.
Since 3-n-Partition is strongly NP-complete, we can assume that an and, thus, m is polynomial in n. Having a closer look at the partial votes, for every candidate ei
corresponding to an integer ai from A, the following holds within every partial vote:
|{s ∈ C \ T : s ≻ ei}| = a1− 1 + Xi j=2
(aj− aj−1− 1) + i − 1 + 1 = ai. (8.8)
Furthermore, the position and thus the total score of the distinguished candidate c is already fixed. In contrast, every subset candidate tj ∈ T can be “inserted” at any position behind c in the three partial votes. The basic idea of this construction is that the “choice” of the positions for tjin the three partial orders corresponds to the choice of three numbers from A into the corresponding subset Aj. For example, inserting tj
directly before the candidate ei in one of the partial votes means that ai ∈ Aj. To this end, our goal is to ensure the following two properties for every winning extension of P :
• Every number of A is selected exactly once, that is, for every candidate ei ∈ E\{e1} there is exactly one candidate tj ∈ T with “ei−1 >· · · > tj>· · · > ei” within one of the three extended votes from Vp, and one candidate tj∈ T with
“tj>· · · > e1”.
• For every tj ∈ T , the sum corresponding to the three “number candidates”
from E selected by tj is B.
As argued in the remainder of the proof, these two points can be realized by set-ting the linear orders Vl such that the following maximum partial scores hold (us-ing Lemma 7.1).
• smaxp (ei) = 3(m− 1) − 3ai− i for all ei∈ E,
• smaxp (tj) = 3(m− 1) − B for all tj∈ T , and
• smaxp (d)≥ 3(m − 1) for all d ∈ D.
This implies that a candidate d ∈ D cannot beat c in any extension. For the other candidates, we interpret the maximum partial scores as follows. Since the maximum amount of points a candidate can make within an extension of the partial votes is 3(m− 1), the maximum partial scores also provide the number of points a candidate must loose to be beaten by c. More specifically, every ei ∈ E must loose at least 3ai+ i points and every tj ∈ T must loose at least B points. Herein, a candidate looses one point for every other candidate that is placed before it in an extension of one of the partial votes. Now, we show the following.
Claim: There is a solution for 3-n-Partition if and only if there is an extension of P such that c wins.
“⇒”: Let {A1, . . . , An/3} with Aj={aj1, aj2, aj3} denote a solution of 3-n-Partition for A. Then, extend v1 such that “Dj1 > tj > ej1” for every j ∈ {1, . . . , n/3}, v2 such that “Dj2 > tj > ej2” for every j ∈ {1, . . . , n/3}, and v3 such that “Dj3 >
tj > ej3” for every j ∈ {1, . . . , n/3}; that is, every candidate tj is inserted directly before the three candidates corresponding to the integers from Aj. Since all integers from A are pairwisely distinct, this extension is unambiguous. In every partial vote, for every ei∈ E, there are exactly aicandidates s∈ C\T with s ≻ ei(see Equation (8.8)).
Thus, without inserting any tj ∈ T before ei, ei looses 3ai points. For q ∈ {1, 2, 3}, let τi,q denote the number of candidates from T that are inserted before ei in the considered extension of vq. Then, for every ei, we have τi,1+ τi,2+ τi,3= i since for each z ∈ {1, . . . , i} a candidate from T is inserted directly before ez in one of the three partial votes. Thus, ei looses 3ai+ i points in this extension and c beats ei. It remains to show that c beats tjfor each tj∈ T . Since aj1+ aj2+ aj3= B, analogously to Equation (8.8), one can compute that the number of candidates that are “better”
than tj in the three partial votes is at least
|{s ∈ C \ T | s ≻ ej1 in v1}| +
|{s ∈ C \ T | s ≻ ej2 in v2}| +
|{s ∈ C \ T | s ≻ ej3 in v3}| = B.
It follows that within the considered extension tjmakes at most 3(m−1)−B = smaxp (tj) points and thus is beaten by c.
“⇐”: Consider a winning extension of the partial votes. As explained before without
“counting” any ti∈ T before ei, ei “looses” 3aipoints (see Equation (8.8)). Hence, in every winning extension at least i times a candidate from T must be inserted before ei, that is,
τi1+ τi2+ τi3≥ i.
We denote this as selection property (which will be crucial in the following argumenta-tion). For every winning extension, we show the following two properties (also stated above):
8.2 Number of votes 135 1. For every candidate tj ∈ T selecting ej1, ej2, ej3, one must haveP3
q=1ajq = B.
2. Every candidate from E, that is, every number of A, is selected exactly once.
To show the first property, we devise a proof by contradiction showing that one can neither have P3
q=1ajq < B nor P3
q=1ajq > B. Assume that in a winning extension there is a tj selecting three candidates ej1, ej2, ej3 such that P3
q=1ajq < B. Then, the minimum number of points that tj makes in this extension is 3m− 3 minus the number of candidates which are placed before tj in this extension (summing over all three votes). In Vq, q∈ {1, 2, 3}, due to Equation (8.8) at most ajq candidates of C\ T can be placed before tjsince otherwise tj would not have selected ejq (but a candidate from E which is right from ejq for at least one q). Since|T \ {tj}| = n/3 − 1 and every candidate from T \ {tj} can be inserted at most once before tj in every partial vote, tj must make at least points in the partial votes, and tj thus beats c, a contradiction.
Now, assume that there is a tjwithP3
q=1ajq > B in a winning extension. Consider the amount of points all remaining candidates from T\{tj} together can loose by candidates from C\ T . Due to the selection property, at least i candidates from T must be inserted before every ei∈ E. Clearly, inserting every candidate from T \ {tj} as far right as possible (that is, directly before the selected candidate) maximizes the amount of points the candidates from T\{tj} can loose. Using Equation (8.8) and further using the assumption thatP3
q=1ajq > B , this amount is
This amount can be “contributed” to the candidates only in multiples of n since ai
differs from aj at least by n. Hence, at least one candidate t of the n/3− 1 candidates from T\{tj} must loose less than B points and, thus, can loose at most B −n points by candidates from C\ T . Again, t can loose at most n − 3 additional points by inserting other candidates from T before it. Thus, the minimum score that t will make is
3(m− 1) − B + n − n + 3 > smaxp (t) and t will beat c, a contradiction.
To see the second property, it remains to show that every number in A is selected exactly once. We give a proof by induction: The “last” number candidate en cannot be selected twice (or more times) since this would imply that at most n− 2 times a
candidate from T could be inserted before en−1 and this would violate the selection property. Now, for any i∈ {1, . . . , n − 1} consider ei and assume that every ej, j > i, has been selected exactly once. Then, selecting ei twice (or more times) implies that at most n− 2 − (n − i) = i − 2 times a candidate from T can be inserted before ei−1, again violating the selection property.
Summarizing, we have shown that in every winning extension, every candidate from E is selected exactly once and every candidate from T selects three candidates whose corresponding numbers sum up to B. Hence, the subsets Aj :={aj1, aj2, aj3 | tj selects ej1, ej2, ej3} for j ∈ {1, . . . , n/3} form a solution of the 3-n-Partition in-stance. This finishes the proof of the Claim.
Now, we briefly discuss how to modify the construction if the numbers from A are not pairwisely different. Then, the candidate subset E consists of one candidate for every different number in A and we want to have that a candidate ei∈ E representing s equal numbers is selected s times. This is achieved by adapting the maximum partial score of ei as follows:
smaxp (ei) = 3(m− 1) − 3ai− |{aj∈ A | aj≥ ai}|.
Then, the proof for this case works in complete analogy to the given proof for the special case.
Finally, for every fixed number s > 3 of partial votes, one can apply an analogous reduction from s-n-Partition (or allow to fix all but three partial votes and adapt the maximum partial scores appropriately).
The reduction used for the proof of Theorem 8.3 can be adapted to the co-winner case by decreasing the maximum partial score of every non-distinguished candidate by one.
Number of linear votes. Theorem 8.3 makes use of Lemma 7.1 to construct a multiset of linear votes leading to the required maximum partial scores. We briefly discuss how to find an equivalent constant-size set of linear votes for some cases. To this end, it is not hard to construct three linear votes that “realize” the maximum partial scores for the candidates from A and E as defined within the proof (for exactly three partial votes). For example, one can set c such that it makes exactly one point in total and, then, set eito position ai in two of the votes and to position ai+ i in the third vote. It remains to place the candidates from D such that they are beaten by c in every extension. This can be done without adding further candidates by rearranging them in a sophisticated way (see [133]). Another possibility is to append a set D′ of further “dummy” candidates with |D′| = |D| which is used to fill the remaining positions “between” the candidates from{c} ∪ A ∪ E in the three linear votes whereas the candidates from D are appended at the end of the linear votes. Clearly, appendig the candidates from D, changes the total scores of the other candidates in the partial votes but does not affect their relative differences. Then, modifying the partial votes by appending the candidates from D′at the end (in arbitrary order) yields a construction for three linear and three partial orders as needed for the NP-hardness proof.
Corollary 8.2. For Borda, Possible Winner is NP-complete for a partial profile consisting of three partial and three linear votes.
8.3 Measures of incompleteness 137