Further Parameterizations

3 Collectively Accepted Ballots

toQ⁰. The essential idea for deciding whether a ballotQ⁰is crucial is to simply guess and search for a certificate showing that the guess was correct.

More precisely,Pdoes the following nondeterministic preprocessing. For each subset Q⁰⊆Q, program P guesses whetherQ⁰ is crucial. ThenP guesses

(a) the position of the table entry z[Q⁰] in the registers if Q⁰ is guessed crucial or

(b) the position of the table entryz[Q⁰⁰] for some pseudo-crucial subsetQ⁰⁰⊆ Q⁰ifQ⁰is guessed non-crucial.

For Case (a), programPchecks whether the augmented entry corresponds to ballotQ⁰ andz[Q⁰]>0: If this is the case, then the guess was correct and P stores the position of the table entry for later access; otherwise, Preturns ‘no’. For Case (b), programPchecks whether the augmented entry corresponds to some pseudo-crucial subset ofQ⁰⁰⊆Q⁰(includingQ⁰itself) and whetherz[Q⁰⁰]=0: If this is the case, thenQ⁰⁰is pseudo-crucial which means that Q⁰ is non-crucial and that the guess was correct; otherwise, Preturns ‘no’.

It is easy to see that if we have a yes-instance and an appropriate ballotQ, then there exist correct guesses such that for at least one set of guesses the machine will answer ‘yes’, as needed. If we have a no-instance andP returns ‘yes’, then obviously one of the guesses was wrong. However, this is not possible sincePvalidates the correctness of each single guess or returns

‘no’.

3.2 Theoretical Results

parameterizations are often good candidates for fixed-parameter tractabil-ity [GHN04]. However, both problems becomeNP-complete when this mini-mum size is one less than the guaranteed(m+1)/2e, even without hidden agenda. This implies that there is no hope for fixed-parameter tractability with respect to the “below guarantee parameter”b_gap.

Theorem 3.6. Every instance of UNANIMOUSLYACCEPTED BALLOTor MAJORITYWISEACCEPTEDBALLOTwhere each voter i satisfies|B_i| >m/2 is a yes-instance.UNANIMOUSLYACCEPTEDBALLOTandMAJORITYWISE ACCEPTEDBALLOTareNP-complete even without hidden agenda and when each voter i satisfies|B_i| >m/2−1.

Proof. As for the first statement, choosingQ=P makes every voter happy.

To show the second statement, we many-one reduce from theNP-complete VERTEXCOVERproblem.

VERTEXCOVER(VC)

Input: An undirected graphG=(U,E) and an integerk.

Question: Is there avertex coverof at mostkvertices, that is, a setU⁰⊆U with|U⁰| ≤ksuch that for eache∈Eit holdse∩U⁰6= ;? Let I=(G=(U,E),k) with vertex setU={u₁, . . . ,u_r}and edge setE= {e₁, . . . ,e_s}be a VC instance where we assume without loss of generality thatr≥k+2. We first reduce from it to an instanceI⁰for UNAAB and then extend this reduced instanceI⁰to an instanceI⁰⁰for MAJAB.

Both instancesI⁰andI⁰⁰have the same proposal setP. It consists of one special proposalα, of all vertices inU, ofkdummy proposalsβj (1≤j≤k), and ofr−kadditional dummy proposalsγj⁰(1≤j⁰≤r−k). Thus,|P| =2r+1.

InstanceI⁰contains four types of voters: one voterv₀, one voterv₀,s edge voters, and r−k vertex haters. Voterv₀supports proposalαand all the rdummy proposals. Voterv₀also supports proposalα, and all the vertices inU. For 1≤i≤s, theith edge voter’s favorite ballotAiconsists of the two vertices inei, of all thekdummy proposalsβj, and ofr−k−2 arbitrarily chosen dummy proposals from {γ1, . . . ,γ_r−k}. For 1≤i⁰≤r−k, the favorite ballotB_i0 of vertex hateri⁰consists ofαand of all dummy proposals butγi⁰. In total, the number of voters inI⁰iss+r−k+2, with each voter supporting at least r= b|P|/2cproposals. SetQ₊:= ;and q₊:=0. Obviously, this reduction can be computed in polynomial time; it is illustrated with an example inFigure 3.5.

3 Collectively Accepted Ballots

u₁

u2

u4

u₅ u₃

k=2

u1 u2 u₃ u4 u₅ α β1 β2 γ1 γ2 γ3

v₀ - - - + + + + + +

v₀ + + + + + + - - - -

-{u₁,u₄} + - - + - - + + + -

-{u₁,u₅} + - - - + - + + + -

-{u₂,u₄} - + - + - - + + + -

-{u2,u5} - + - - + - + + + -

-{u₃,u₅} - - - + - + + + + -

-{u₄,u₅} - - - + + - + + + -

-1⁰ - - - + + + - + +

2⁰ - - - + + + + - +

3⁰ - - - + + + + +

-special voter

edge voter

vertex hater

Figure 3.5: Illustration of the reduction from VERTEXCOVER to UNANI -MOUSLYACCEPTEDBALLOT. Top: The graph of the VERTEX COVERinstance. The verticesu₄andu₅cover all edges. Bottom:

The corresponding UNANIMOUSLYACCEPTEDBALLOTinstance.

All voters are happy with the ballot {u₄,u₅,α,β1,β2}(marked in gray).

3.2 Theoretical Results

To show the reduction’s correctness, we have to show thatIhas a vertex cover of size at mostkif and only if there is a ballotQ⊆P that all the voters inIare happy with.

For the “only if ” part, suppose thatU⁰⊆Uwith|U⁰| ≤kis a vertex cover.

We show that every voter is happy withQ={α}∪{βj|1≤j≤ |U⁰|}∪U⁰. First, the size ofQ is 2|U⁰| +1. To make a voter happy, at least|U⁰| +1 of his favorite proposals must be also inQ. Obviously, votersv₀,v₀and all vertex haters are happy withQ. For eachi∈{1, . . . ,s},Q∩A_icontains all dummy proposalsβjwith 1≤j≤ |U⁰|and at least one vertex proposalv_j0 withv_j0∈ e_i∩U⁰sinceU⁰is a vertex cover. This sums up to at least|U⁰| +1 proposals.

Hence, every edge voter is also happy withQ.

For the “if ” part, by applyingObservation 3.1(ii) to the ballots of votersv₀ andv₀, ballotQ must containα, and furthermore,Q contains an equal numberxof vertex proposals and dummy proposals. For eachi⁰∈{1, . . . ,r−k}, ballotQcannot contain dummy proposalγi⁰ since otherwise|B_i0∩Q| =x<

b|Q|/2c +1. Thus, vertex hater i⁰ would not be happy. Therefore, the x dummy proposals must come from {β1, . . . ,βk}andx≤k. To make theith edge voter happy, ballotQmust satisfy the condition|Q∩Ai| ≥x+1. But since no edge voter supports proposalα, ballotQmust contain at least one proposalu_j∈A_i. By definition ofA_i, the corresponding vertexu_jis incident to edgee_i. This implies that thexvertices inQform a vertex cover forG.

Next, we extend instance I⁰ to instance I⁰⁰ for MAJAB by adding r− k vertex lovers who have the same favorite ballot U, and s edge-inverse voters such that for 1≤i≤s, edge-inverse voter i’s favorite ballot C_i= (U∪{γ1, . . . ,γr−k}) \A_i. Thus,C_iandA_iare disjoint for all 1≤i≤s. In total, I⁰⁰has 2(s+r−k)+2 voters. Since each of the newly added voters supports exactlyrproposals, the constraint that each voter’s proposal set has at least r= b|P|/2cholds. This extension can also be computed in polynomial time.

Now we show the correctness of the extended reduction, that is,Ihas a vertex cover of size at mostkif and only if there is a ballotQ⊆P which more than half of the voters inI⁰⁰are happy with.

For the “only if ” part, the ballotQas constructed in the “only if ” part above makes all voters inI⁰happy. This sums up tos+r−k+2. SinceI⁰⁰ contains all the voters fromI⁰and has 2(s+r−k)+2 voters, this also means that more than half of the voters inI⁰⁰are happy withQ.

For the “if ” part, for 1≤i≤s, theith edge voter and theith edge-inverse voter do not share a common favorite proposal. Furthermore, no vertex hater’s favorite ballot intersects any vertex lover’s favorite ballot. Hence,

3 Collectively Accepted Ballots

by applyingObservation 3.1(i), any ballot can make at mostsvoters from the edge voters and the edge-inverse voters happy, and can make at most r−kvoters from the vertex haters and the newly constructed vertex lovers happy. But I⁰⁰ has 2(s+r−k)+2 voters. This means that in order to be a solution ballot forI⁰⁰,Qmust make bothv₀andv₀happy. By applying Observation 3.1(ii),Qmust then containα, and, furthermore,Qcontains the same numberxof vertex proposals and dummy proposals. The ballotQ cannot make any vertex lover happy since his favorite ballot andQhave an intersection of sizexwhich is smaller thanb|Q|/2c +1. Thus,Qneeds to make all vertex haters happy. Then, Q cannot contain any dummy proposalγi⁰since otherwise the vertex hateri⁰is not happy due to|B_i0∩Q| = x< b|Q|/2c +1. Hence,Qcontainsxdummy proposals from {β1, . . . ,βk}with x≤k. Then, no edge-inverse voter is happy withQsince at mostxproposals from his favorite ballot are inQ. This means that all edge voters must be happy with Q. To make the ith edge voter happy,Q must intersect withAiin at least one vertexuj∈Ai. By definition ofAi, the corresponding vertexujis incident to edgeei. Thus, thexvertices inQform a vertex cover forG.

We conclude this subsection with a brief discussion on the relation be-tween the parameters “maximum sizeb_maxof the favorite ballots” and “the sizeh_maxof the maximum symmetric difference between any two favorite ballots”. As the following proposition shows, for the cases without hidden agenda the two parametersh_maxandb_maxare “equivalent” in terms of pa-rameterized complexity theory: The fact that for two parametersxandy one hasx=Θ(y) implies that the parameterization byxand the parame-terization by yare in the same level of the W-hierarchy and yield the same parameterized hardness results.

Proposition 3.1. For any instance of UNANIMOUSLYACCEPTEDBALLOT or MAJORITYWISE ACCEPTEDBALLOTit holds that hmax≤2bmax, where hmax denotes the size of the maximum symmetric difference between two favorite ballots and bmaxdenotes the maximum size of the given favorite ballots. Instances of UNANIMOUSLYACCEPTEDBALLOTorMAJORITYWISE ACCEPTEDBALLOTare yes-instances if h_max<b_max/2and Q₊= ;.

Proof. The first statement follows ash_maxequals max_i,j∈V(|B_i\B_j| + |B_j\ B_i|), which is bounded by 2b_max. For the second statement, note thath_max<

b_max/2 implies that every voter is happy with the favorite ballotB_` of a