3.2 Theoretical Results
3.2.3 Small Ballots
3 Collectively Accepted Ballots
the “integer feasibility problem” (that is, an ILP without objective function) withρvariables andLinput bits can be solved inO(ρ2.5ρ+o(ρ)L) time [FT87;
Kan87;Len83], MAJAB is fixed-parameter tractable with respect to the numbernof voters.
If we delete Constraint (3.1) and the variableszi, and replace the right-hand sides of Constraint set (3.3) with 1, then we gain an ILP for UNAAB with at most 2·2n variables and 2·2n+n+1 constraints. Thus, UNAAB is also fixed-parameter tractable with respect to parametern.
UnlessNP⊆coNP/poly, even without hidden agenda, both problems do not have a polynomial kernel with respect to the parametern:Reduction 3.1 is a polynomial-time reduction from theNP-complete problem HITTINGSET; the numbernof voters in the reduced instance is linearly bounded by the numberrof sets in the instance one reduces from; andQ+= ;. A polynomial kernel of UNANIMOUSLYACCEPTEDBALLOTwithQ+= ;parameterized by n would yield a polynomial kernel for HITTING SET parameterized by r. However, this is not possible unless NP⊆coNP/poly(cf. [Her+13, Theorem 5]). Thus, even without hidden agenda, UNAAB does not admit a polynomial kernel. Neither does MAJAB admit a polynomial kernel even without hidden agenda due toObservation 3.2.
3.2 Theoretical Results
accepted ballot of size at most h for I if and only if there is a unanimously accepted ballot of size exactly h for I0.
Proof. For the “only if ” part, suppose that a ballotQwith|Q| ≤his accepted by all voters inI. We assume that|Q|andhare odd since every voter that is happy with a ballot of even size is still happy if one removes an arbitrary proposal. Then, adding (h− |Q|)/2 dummy proposals and (h− |Q|)/2 copy proposals toQresults in a ballot of sizehthat is accepted by all voters inI0. For the “if ” part, suppose that a ballotQ0with|Q0| =his accepted by all voters inI0. Then delete all dummy and copy proposals fromQ0to obtain a ballotQthat is accepted by all voters inI. By applyingObservation 3.1(i) to any original voter and its duplicate one knows thatQis non-empty. Assume towards a contradiction that there is a voter inIthat is not happy withQ.
Since the corresponding voter inI0is happy withQ0, ballotQ0contains more dummy proposals than copy proposals. Then, the corresponding duplicate voter is not happy withQ0, a contradiction.
Note thatIandI0have the same agenda (Q+,q+) (we initially copiedI to obtainI0) and adding or removing proposals that are not inQ+from a ballotQhas no influence on whether|Q+∩Q| ≥q+.
Due toLemma 3.2we can assume without loss of generality that given any upper boundh on the size of the solution ballot every unanimously accepted ballot has size exactlyh. We use this in theW[2]-membership proof for UNANIMOUSLYACCEPTEDBALLOTparameterized byhleading to the following theorem.
Theorem 3.4. Parameterized by the size h of the solution ballot,UNAN -IMOUSLY ACCEPTEDBALLOTisW[2]-complete andMAJORITYWISE AC -CEPTED BALLOT is W[2]-hard. Both results hold even without hidden agenda.
Proof. Due toLemma 3.1,Reduction 3.1is a parameterized reduction from theW[2]-hard HITTINGSETparameterized by the sizekof the hitting set to UNANIMOUSLYACCEPTEDBALLOTparameterized by the sizehof the solution ballot without hidden agenda. Hence, UNANIMOUSLYACCEPTED BALLOTparameterized byhisW[2]-hard. Because ofObservation 3.2, this impliesW[2]-hardness for MAJORITYWISEACCEPTEDBALLOT parameter-ized byheven without hidden agenda.
3 Collectively Accepted Ballots
p1 p2 p3 p4
B1 + + -
-B2 + - +
-B3 - + + +
h=3,Q+={p2,p3,p4},q+=2 [h]={1, 2, 3},H ={H1,H2,H3} H1={1, 2},H2={2, 3},H3={1, 3}
p11 p12 p13 p14
d1
p21 p22 p23 p24
d2
p31 p32 p33 p34
d3
layer 1 layer 2 layer 3
a1
a2
v11 v12 v13
v21 v22 v23
v31 v32 v33
Figure 3.3: Illustration of the reduction from UNANIMOUSLY ACCEPTED BALLOTto INDEPENDENTDOMINATINGSET(IDS). Top: UNAN -IMOUSLYACCEPTED BALLOTinstance with four proposals and three voters and the corresponding auxiliary set familyH. All entries in columns corresponding to proposals from the agenda are written in boldface. All voters are happy with the ballot {p1,p2,p3}. Bottom: The corresponding graph of the IDS in-stance. Vertices inside gray bars form cliques. The three vertices p31,p22, andp13form an independent dominating set.
3.2 Theoretical Results
It remains to show that UNANIMOUSLY ACCEPTED BALLOT is inW[2]. We describe a parameterized reduction from UNANIMOUSLY ACCEPTED BALLOTparameterized by the size h of the solution ballot to INDEPEN -DENTDOMINATINGSETparameterized by the solution sizek. (TheW[2] -membership of IDS follows by a straightforward reduction to WEIGHTED WEFT-tCIRCUITSATISFIABILITY[DF13].)
INDEPENDENTDOMINATINGSET(IDS)
Input: An undirected graphG=(U,E) and an integerk.
Question: Is there anindependent dominating setof kvertices, that is, a vertex setU0⊆U with|U0| =ksuch that no pair of vertices fromU0is adjacent and each vertex fromU\U0is adjacent to at least one vertex fromU0?
In the following, we use [z] to denote the set {1, . . . ,z}withz∈N. Let I:=(P,V, (Q+,q+),h) be an instance of UNANIMOUSLYACCEPTEDBALLOT withnvoters and mproposals where hdenotes an upper bound for the size of the solution ballot. LetB1, . . . ,Bndenote the favorite ballots of the voters. We construct an instanceI0=((U,E),k) of IDS as follows. The vertex set U consists of proposal vertices, voter vertices, agenda vertices, and dummy vertices. There arehlayers ofproposal verticeseach containing one vertex for each proposal. We say that the proposal vertexp`j corresponds to proposal jin layer`. We enumerate all subsets of [h] of sizedh/2e. In the following letHxdenote thexth of these subsets and setH :={H1, . . . ,Hh∗} withh∗=¡ h
dh/2e
¢. For each voterithere is onevoter vertexfor each element inH. We say that voter vertexvsi corresponds to voteriand subsetHs. There areq+agenda vertices a1, . . . ,aq+andh dummy vertices d1, . . . ,dh.
The edge set E is constructed as follows. Two proposal vertices are adjacent if they correspond to the same proposal or are in the same layer, that is, {p`j,p`j00}∈Eif j=j0or`=`0. Each dummy vertexd` is adjacent to all proposal vertices in layer `, that is, {d`,p`j}∈E for each j∈[m].
Proposal vertex p`j and voter vertex vsi are adjacent if voter i supports proposal jand`∈Hs. Proposal vertexp`jand agenda vertexa`are adjacent if proposal jis fromQ+.
We set the sizekof the independent dominating set toh. This completes the construction which is illustrated inFigure 3.3.
Now, we highlight two properties of the constructed instance which help to prove the correctness of the reduction. First, the proposal vertices from
3 Collectively Accepted Ballots
the same layer and the proposal vertices corresponding to the same proposal form a complete subgraph, respectively. Hence, an independent dominating set may contain at most one proposal vertex from each layer and no two proposal vertices corresponding to the same proposal. Second, taking a different proposal vertex from each layer into the dominating set is the only way to form an independent set of sizeksuch that all dummy vertices and all other proposal vertices are dominated.
Due toLemma 3.2it remains to show that every voter inIis happy with some ballotQwith|Q| =hand|Q+∩Q| ≥q+if and only if the constructed graph has an independent dominating set of sizek=h.
For the “only if ” part, assume that every voter is happy with the ballot Q:={j1, . . . ,jh}and|Q+∩Q| ≥q+. Herein, without loss of generality let {j1, . . . ,jq+}⊆Q+, that is, we fix an ordering of the proposals inQ such that the firstq+proposals are fromQ+. We show that the vertex setU0= {p1j
1, . . . ,phj
h}is an independent dominating set forI. Obviously,U0 is an independent set and all dummy vertices as well as proposal vertices are either in U0 or adjacent to a vertex inU0. So, suppose for the sake of contradiction that a voter vertexvsi is not dominated byU0. This means thatU0∩{p`j|j∈Bi∧`∈Hs}= ;. Hence, at mostbh/2clayers may contain proposal vertices p`j with j∈Bi that are in vertex setU0. SinceU0does not contain two proposal vertices from the same layer, there are at most bh/2cproposal vertices p`j inU0 with j∈Bi. Thus, voter iis not happy withQ, a contradiction. Analogously, suppose that agenda vertexa`is not dominated byU0. However, we already know that p`j
`∈Q+and there is an edge between p`j
`anda`, a contradiction.
For the “if ” part, due to the observations above we know that the vertices forming an independent dominating set must be proposal vertices (fromh different layers). Now, suppose thatU0:={p1j
1, . . . ,phj
h}is an independent dominating set of size h. First, we show that every voter in I is happy withQ:={j1, . . . ,jh}. The ballotQis of sizehbecauseU0is an independent set, and hence, there are no two proposals vertices inU0corresponding to the same proposal. Suppose for the sake of contradiction that voter iis not happy withQ. Then|Q∩Bi| ≤ bh/2cwhich means|Q\Bi| ≥ dh/2e. Let X={`∈[h]|p`j∈U0∧j∈Q\Bi}and letvsibe a voter vertex withHs⊆X. This vertex exists since|X| ≥ dh/2e. Letp`j be a proposal vertex inU0that dominates vsi. Then `∈Hs⊆X and j∈Bi, a contradiction. Second, we show that|Q∩Q+| ≥q+. Recall thatU0containshproposal vertices from
3.2 Theoretical Results
hdifferent layers corresponding tohdifferent proposals. In particular, this implies that the q+agenda vertices are adjacent to q+proposal vertices fromU0corresponding to pairwise different proposals. Hence,Qcontains q+proposals from the agenda setQ+.
The membership of MAJORITYWISEACCEPTEDBALLOTparameterized by the sizehof the solution ballot for the class W[2] remains open. Note that theW[2]-hardness reduction in the proof ofTheorem 3.4does not rely on (an upper bound for)h being given as part of the input. That is, the problem is computationally hard also for the cases where the size of ballotQ is not explicitly required to be bounded byh.
Except for the parameterhwhere we only know that MAJORITYWISE ACCEPTEDBALLOTisW[2]-hard while UNANIMOUSLYACCEPTEDBALLOT is even W[2]-complete, all results shown so far are the same for unanimous acceptance and majority acceptance. The following theorem shows that this changes when considering the parameterbmaxwhere UNANIMOUSLYAC -CEPTEDBALLOTremains fixed-parameter tractable but for MAJORITYWISE ACCEPTEDBALLOTwe showW[1]-completeness.
Theorem 3.5. Parameterized by the maximum size bmax of the favorite ballots,
1. UNANIMOUSLY ACCEPTEDBALLOTcan be solved in O(b2bmaxmax·nm) time implying fixed-parameter tractability; however, even without hid-den agenda it admits no polynomial kernel unlessNP⊆coNP/poly, and
2. MAJORITYWISEACCEPTEDBALLOTparameterized by bmaxisW[1] -complete.
In the remainder of this subsection we prove Theorem 3.5. The non-existence of a polynomial kernel for UNANIMOUSLY ACCEPTED BALLOT with respect to parametermshown inTheorem 3.2also holds for parame-terbmax, asbmax≤m. Although there is no hope for polynomial kernels, we at least show fixed-parameter tractability for UNANIMOUSLYACCEPTED BALLOTparameterized bybmaxby a depth-bounded search tree algorithm.
Lemma 3.3. UNANIMOUSLYACCEPTEDBALLOTcan be solved in O(b2bmaxmax· nm)time.
3 Collectively Accepted Ballots
Proof. Let I=(P,V, (Q+,q+)) denote an instance of UNANIMOUSLY AC -CEPTED BALLOT with|Bi| ≤bmax,i∈V. Observe that any ballot Q that makes every voter happy contains at most 2bmax−1 proposals, otherwise the intersection ofQ and any favorite ballot has at mostbmax proposals but bmax≤ |Q|/2. Using this observation, we describe a depth-bounded search tree algorithm solving the optimization version of UNANIMOUSLY ACCEPTEDBALLOT, that is, it computes a solution ballotQwith the largest intersectionQ∩Q+such that every voter is happy withQ, or it returns
‘no’ if there is no such ballot. The algorithm works as follows. Start with branching over the upper bound sizeh∈{1, . . . , 2bmax−1} of the solution and initializeQ← ;in each branch. Repeat the following until all voters are satisfiedor|Q| =h: Mark every voteriwith|Bi∩Q| > bh/2cas satisfied and branch into adding one proposal fromBj\QtoQfor an arbitrary unsatisfied voter j. Finally, if all voters are satisfied, then the computed ballotQmakes every voter happy since|Bi∩Q| > bh/2c ≥ |Q|/2 for each i∈V; otherwise discard this branch, because this path of the search tree can not lead to a solution as the size of ballotQreaches the upper bound hbut there is some voter jwith|Bj∩Q| ≤ bh/2c. Finally, if|Q| <h, then fill upQwith h− |Q|arbitrary proposals fromQ+. It is easy to verify that the ballotQ having the largest intersection withQ+among all ballots in the leaves of the search tree is an optimal solution, that is, among all possible ballots that make all voters happy, ballotQalso has the largest intersection with Q+.
The search tree has depth at most 2bmax, since the size ofQis increased in each branching step and|Q| ≤2bmax. Clearly, the number of branching possibilities in each step is at mostbmax. Altogether, the algorithm takes O(b2bmaxmax·nm) time, because each branching step needsO(nm) time.
In contrast to UNANIMOUSLYACCEPTEDBALLOT, MAJORITYWISEAC -CEPTEDBALLOTbecomesW[1]-complete for the parameterbmax. We first prove theW[1]-hardness by a reduction from MAJORITYVERTEXCOVER. Lemma 3.4. MAJORITYWISEACCEPTED BALLOTparameterized by the maximum size bmaxof the favorite ballots isW[1]-hard.
Proof. We describe a parameterized reduction from theW[1]-hard MAJOR -ITYVERTEXCOVERproblem [Fel+10] (cf. [GNW07]).
3.2 Theoretical Results
u7 u8 u1
u2 u3
u4
u5
u6
k=2
u1 u2 u3 u4 u5 u6 u7 u8 d1
{u1,u2} + + - - - - - - +
{u2,u3} - + + - - - - - +
{u1,u7} + - - - - - + - +
{u2,u7} - + - - - - + - +
{u3,u7} - - + - - - + - +
{u7,u8} - - - - - - + + +
{u4,u5} - - - + + - - - +
{u5,u6} - - - - + + - - +
{u4,u8} - - - + - - - + +
{u5,u8} - - - - + - - + +
{u6,u8} - - - - - + - + +
Figure 3.4: Illustration of the reduction from MAJORITYVERTEXCOVER (MVC) to MAJORITYWISEACCEPTEDBALLOT. Top: The graph of the MVC instance. Vertices u7 andu8 cover seven out of eleven edges. Bottom: The corresponding MAJORITYWISEAC -CEPTEDBALLOTinstance. The agenda is ({u1, . . . ,u8}, 2). Seven out of eleven voters (as highlighted) are happy with the ballot {u7,u8,d1}.
3 Collectively Accepted Ballots
MAJORITYVERTEXCOVER(MVC)
Input: An undirected graphG=(U,E) and an integerk.
Question: Is there a subset ofkvertices which covers a majority of the edges ofG, that is, is there a size-ksubsetU0⊆U such that
|{e∈E|e∩U06= ;}| > |E|/2?
Let (G,k) be an instance of MVC withU denoting the set of vertices and E={e1, . . . ,es}denoting the set of edges. Now, construct a MAJORITYWISE ACCEPTED BALLOTinstance as follows. The proposal set P is defined asU∪DwithDbeing a set ofk−1 dummy proposals. For each edgeej∈E there is one voter with the favorite ballotBj=ej∪D. Furthermore, the agenda consists ofQ+=U and q+=k. This completes the construction which clearly runs in polynomial time. It is illustrated by an example in Figure 3.4. Next, we show that (G,k) is a yes-instance for MVC if and only if the constructed instance is a yes-instance for MAJORITYWISEACCEPTED BALLOT.
For the “only if ” part, suppose that there is a subsetU0⊆U of k ver-tices covering more thans/2 edges. Then,Q=U0∪Dis a solution for our MAJORITYWISE ACCEPTED BALLOTinstance: Clearly, |Q| =2k−1 and
|Q+∩Q| =k. Since|Bj∩Q| = |D| + |U0∩ej| ≥kfor each covered edgeej, more thans/2 voters are happy.
For the “if ” part, we first assume thats/2≥¡k+2 2
¢. Otherwise, add to the graphGa tree consisting of a root with¡k+3
2
¢children each of which has a single leaf. This results in an equivalent instance ((U0,E0),k0=k+1) with
|E0|/2=s/2+¡k0+2 2
¢≥¡k0+2 2
¢.
Suppose that there is a ballotQ⊆P which a majority of voters is happy with, and|Q∩Q+| ≥k. Note that adding all dummy proposals toQalso results in a feasible solution. Thus, we assume thatQcontains allk−1 dummy proposals. Then, ballotQmust contain at least one vertex proposal in each of the happy voters’ favorite ballots. This implies that the edges corresponding to the happy voters are covered by the vertices corresponding to the voters inQ. As a majority of voters is happy, a majority of edges is covered. Finally, we show that the number of vertex proposals inQis exactly k. Observe that ballotQcan have at mostk+2 vertex proposals since otherwise no voter is happy: each voters supports at most two vertex proposals andk−1 dummy proposals, and hence, a solution can not have more thankunsupported vertex proposals. Furthermore, there are at most
3.2 Theoretical Results
¡k+2
2
¢happy voters such that both their vertex proposals are contained inQ.
But sinces/2≥¡k+2
2
¢, there must be at least one happy voter j such that only one vertex proposal in his favorite ballotBj is contained inQ. This implies thatQcan only be of size 2k−1, and hence, contains exactlykvertex proposals, since otherwise voter jis not happy.
As the most technical part of the proof ofTheorem 3.5we finally show that MAJORITYWISEACCEPTEDBALLOTis contained inW[1]forbmax. Lemma 3.5. MAJORITYWISE ACCEPTED BALLOTparameterized by the maximum size bmaxof the favorite ballots is inW[1].
Proof. We use a characterization of W[1][FG06, Theorem 6.22.] which states that a parameterized problemLwith parameterkis inW[1]if and only if there is atail-nondeterministic k-restricted nondeterministic random access machine (NRAM)program decidingLas follows: On each input (x,k), it
(a) performs at mostf(k)·poly(|x|) steps with only the lastf(k) steps being nondeterministic,
(b) uses only the firstf(k)·poly(|x|) registers, and
(c) has numbers of value at most f(k)·poly(|x|) in any register at any time where f is a function only depending on the parameter kand poly is a polynomial function. See Flum and Grohe [FG06, Chapter 6] for further information about the machine characterization of problems inW[1].
To show theW[1]containment, we describe a tail-nondeterministicbmax -restricted NRAM programPto decide MAJORITYWISEACCEPTEDBALLOT. Given a MAJORITYWISEACCEPTEDBALLOTinstanceI=(P,V, (Q+,q+)) with|P| =m, we say that a ballotQ0⊆P is crucialif it is a subset of the favorite ballotBj for some voter j∈V, that is, ∃j∈V :Q0⊆Bj. The deterministic phase ofPworks as follows. For each crucial ballotQ0 the programPcounts and stores the number of voters whose favorite ballots are supersets ofQ0; we denote this number asz[Q0] in the following. Since the maximum size of the favorite ballots is bmax, table z can be filled in f(bmax)·(nm)ctime where f is a function only depending onbmaxandcis a constant. Every number inzhas value at mostn. The programPuses an additional counterz0to store the number of happy voters.
3 Collectively Accepted Ballots
Algorithm 1Nondeterministic part of programP
Require: Table entryz[Q0] contains the number of voters whose favorite ballots are supersets ofQ0.
functionFINDBALLOT(P,n,z)
guessa ballotQ⊆P with|Q+∩Q| ≥q+and|Q| ≤2bmax
initializez0←0
fors=min(bmax,|Q|)downtod(|Q| +1)/2edo for eachcrucial subsetQ0⊆Qof sizesdo
z0←z0+z[Q0]
for eachcrucial subsetQ00⊆Q0do z[Q00]←z[Q00]−z[Q0]
ifz0>n/2then return‘yes’
else return‘no’
We remark that a straightforward implementation of tablezwould use O(bmax·mbmax) registers since a crucial ballotQ0⊆P can have up tobmax
proposals. Furthermore, in the nondeterministic phase we assume thatP can decide whether a ballot is crucial inf(bmax) steps; we address both these issues at the end of the membership proof.
As for the nondeterministic phase ofP (seeAlgorithm 1for the pseu-docode), note that any ballot making more than half of the voters happy contains at most 2bmaxproposals. Hence, the programPguesses a ballotQ with|Q+∩Q| ≥q+and|Q| ≤2bmax. Checking whetherQis a solution works as follows.
(i) Take acrucialballotQ0⊆Qwith|Q0| > |Q|/2 (note that|Q0| ≤bmax), (ii) increase the counterz0byz[Q0], and
(iii) decrease z[Q00] for eachcrucial ballotQ00(Q0 with|Q00| > |Q|/2 by z[Q0].
The program repeats Steps (i)–(iii), considering all crucial ballotsQ0⊆Q satisfying the condition in Step (i) ordered by decreasing size and starting with aQ0of size min(bmax,|Q|). The number of steps needed in the nonde-terministic part is inO(4bmax) since the number of setsQ0fulfilling (i) and the number of subsetsQ00⊆Q0 with|Q00| > |Q|/2 are both upper-bounded by 2bmax.
3.2 Theoretical Results
Finally, the program decides whether I is a yes-instance by checking whetherz0≥ bn/2c +1. To show the correctness ofP, it remains to show that z0indeed equals the number of voters happy withQ. Let the happy voters bev1, . . . ,vr. By accounting for every possible crucial subsetQ0⊆Qordered by decreasing size and reducing the entry z[Q00] of any subsetQ00(Q0 byz[Q0], we count every happy voter exactly once. That is, we partition the happy votersv1, . . . ,vrinto subsetsV1, . . . ,Vssuch that the favorite ballots of any two voters from the same subset have the same intersection withQ and the subsets with lower indices have larger intersections withQ. Then, in theith iteration, programPadds the size of setVito counterz0. In this way, counterz0sums up to the number of voters happy withQ.
Let us now explain how to implement tablezusing only the firstf(bmax)· poly(|I|) registers and how to decide whether a ballot is crucial. For detecting crucial ballots we show that it is sufficient to be able to have certificates for crucial ballots and for “pseudo-crucial” ballots. To this end, we additionally storez[Q0]=0 for every ballotQ0that ispseudo-crucial, that is, for every ballotQ0 that is not crucial itself but contains a proposal whose (single) removal would yield a crucial ballot. This enlarges the tablezby a factor of at mostm(as all candidates for being pseudo-crucial can be obtained by adding one of themproposals to some crucial ballot). Note that a ballot is non-crucial if and only if it has a subset that is pseudo-crucial, because we can assume without loss of generality that every singleton fromP is crucial. Thus, a pseudo-crucial subsetQ00⊆Q0is a certificate forQ0being non-crucial.
The table implementation is based on the fact thatPactually uses only f(bmax)·poly(|I|) entries. To ensure that we only use the firstf(bmax)·poly(|I|) registers we store the entries of table z in an unordered fashion in the first registers, but we augment the entry for each of such ballotsQ0 by additionally storing Q0 itself; each of these augmentations has size at most O(bmax·logm) for encoding up to bmax proposals in each ballot Q0. Note that, for simplicity, we could just use a pair of registers for each entry.
Throughout the filling of the table, this makes no difference as we have f(bmax)·poly(|I|) time available and can afford sequential search for entries, if needed.
The essential idea for queryingz[Q0] is to use nondeterminism to guess the position of a required table entry in the unordered sequence of entries and to check whether the augmented entry stored in this entry corresponds
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’.