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An unsolvable instance

We firstly show that, by completing the preference lists ofR6 in an arbitrary way (i.e. by appending agents not on the lists in an arbitrary order to the tail of the original lists), the resulting instance of cyclic 3DSM, denoted byR6, does not admit any strongly stable matching. The subinstanceR6 ofR6 is called the suitable part of R6, the original entries of an agentxinR6 are thesuitable partners ofxand the families ofR6 are calledsuitable families.

Lemma 5. The instanceR6of cyclic 3DSM admits no strongly stable matching.

Proof. Suppose for contradiction thatFis a strongly stable matching. As the 9 inner agents form a 9-cycle in the underlying directed graph, the 9 suitable families have a natural cyclic order. We show that if a suitable family, say (m1, w1, d′1) is not in F, then the successor suitable family (m′

2, w1, d1) must be inF, which would imply a contradiction given that the number of these suitable families is odd. If (m1, w1, d′1) ∈ F/ then F(w1) = d1, since otherwise (m1, w1, d′1) would be weakly blocking. Similarly, (m′2, w1, d1)∈ F/ implies

F(d1) =m2. But this means that (m2, w1, d1)∈ F, so (m2, w′2, d1) is weakly blocking. Recall that FR6\at is the unique stable matching forR6\at. Let R6\at denote the

instance created by removing an inner agent atfrom R6. We denote byCR6\at the subset

of agents ofR6\atthat are covered byFR6\at, and byUR6\at those who are uncovered by

FR6\at, respectively.

Lemma 6. Letatbe an inner agent ofR6. For every matchingF∗⊇ FR6\at ofR6\at, no suitable family can be weakly blocking, and therefore no agent from CR6\at can be involved in a weakly blocking family. For any other matching, at least one suitable family is weakly blocking.

Proof. It is straightforward to verify thatFR6\at is a strongly stable matching forR6\at,

so no suitable family in R6\at can weakly block F∗ ⊇ FR6\at. Moreover, no agentxof

CR6\at can be involved in a non-suitable weakly blocking family either, sincexhas a suitable

partner inF∗.

Suppose that F′ is a matching ofR6\a

t which is not a superset of FR6\at. As in the

proof of Lemma 5, we use the fact that if a suitable family is not in F′, then the successor

suitable family is either in F′ or weakly blocking. Therefore, if we do not include four

from the seven suitable families ofR6\atin a matching then one of them would be weakly

blocking.

The NP-completeness proof

The reduction we describe in this section again begins with an instance of Restricted SMTI, only we assume without loss of generality the role of the men and women of the instance to be “reversed”. To be precise, we assume a given instance of Restricted SMTI I that its vertex set ((A1∪A2)∪B) consists of a setA1 ={a1, a2, . . . , an1} of men with strictly

ordered preference lists, andA2={aT1, aT2, . . . , aTn2}of men with preference lists consisting

of a single tie of length 2, and n1+n2=n. The setB ={b1, b2, . . . , bn} consists entirely of

women with strictly ordered preference lists.

Given an instance I of Restricted SMTI as defined above, we create an instance I′ of

cyclic 3DSM. First we create a proper instance I′

p of cyclic 3DSMI as a subinstance of I′

with agents Mp∪Wp∪Dpin the following way.

First we create a setWp of n women {w1, w2, . . . , wn} such that the preference list of

womanwjis a single entry, dogdj ∈Dp. The preference list ofdjis such that ifP(bj)[l] =ai,

then P(dj)[l] =mi, otherwise ifP(bj)[l] =aTi , thenP(dj)[l] =m′i,j for 1≤l ≤r, wherer

is the length ofbj’s list. So the preference list of dogdj is essentially the “same” as that of

womanbj, only with men inMp rather thanA.

For each man ai ∈ A1, create a man mi ∈ Mp, such that if P(ai)[l] = bj, then let

P(mi)[l] = wj for 1 ≤ l ≤ r, where r is the length of ai’s list. So the preference list

of man mi is essentially the “same” as that of man ai. For each man aTi ∈ A2, with

a preference list consisting of a single tie of length two, say (br, bs), we create five men

mT

i , m′i,r, m′′i,r, mi,s′ , m′′i,s, four women w′i,r, wi,r′′ , wi,s′ , w′′i,s and four dogs di,r′ , d′′i,r, d′i,s, d′′i,s

where the preference list ofmT

i containswi,r′ andw′i,s in an arbitrary order, and the other

preference lists are as shown below.

m′ i,r : w′i,r wr m′′ i,r : w′′i,r m′ i,s : w′i,s ws m′′ i,s : w′′i,s w′

i,r : d′i,r d′′i,r

w′′

i,r : d′′i,r d′i,r

w′

i,s : d′i,s d′′i,s

w′′

i,s : d′′i,s d′i,s

d′

i,r : m′′i,r m T i

d′′

i,r : m′i,r m′′i,r

d′

i,s : m′′i,s m T i

d′′

i,s : m′i,s m′′i,s

We also add these agents toMp, Wp andDp, respectively. Note that inIp′ every set of

agents has the same cardinality: np=|Mp|=|Wp|=|Dp|=n+ 4n2. The notions ofproper agent, proper partner andproper family are defined in the obvious way.

The additional part of instance I′ contains three subinstances. The suitable part of I′ is the disjoint union of 3n

p copies of R6, such that the ith copy of R6, denoted R6i,

incorporates the ith agent of I′

p, as described in the previous reduction in the proof of

Theorem 1 (we omit the full description of this process again). The new agents are referred to asadditional agents.

LetFs=∪i∈{1,...3np}FR6i\ai be the so-called suitable matching of the additional part,

where ai is the proper agent of R6i. We call the set C = ∪i∈{1,...3np}CR6i\ai covered additional agents, as these additional agents are covered by Fs, and we call the set U =

∪i∈{1,...3np}UR6i\ai uncovered additional agents, as these additional agents are not covered

byFs.

Thefitting part of I′ is constructed onU as follows. Note that U has equal numbers

of men, women and dogs. The fitting part consists of disjoint families that covers U, so that every agent has exactly one agent in his/her/its list, i.e. the fitting part is a complete matching of U, denoted by Ff.

Finally, thedummy part is obtained by an arbitrary extension of the preference lists, so that by putting together the four subinstances, the proper and the three additional parts, we get the complete instanceI′. The preferences of the agents over the partners in different

parts respect the order in which we defined these parts: the list of a proper agent contains the proper partners first, then the suitable partners, and finally the dummy partners; the list of a covered additional agent contains the suitable partners first, then the dummy partners; the list of an uncovered additional agent contains the suitable partners first, then the fitting partner, and finally the dummy partners.

First we show that there is a one-to-one correspondence between the complete stable matchings of I and the complete strongly stable matchings of I′

p. The stability is pre-

served via the following one-to-one correspondence between the complete matchings of I

and complete matchings ofI′:

(ai, bj)∈ M ⇐⇒ (mi, wj, dj)∈ Fp

(aT

i, bs)∈ M ⇐⇒ (mTi, wi,s′ , d′i,s),(mi,s′′ , w′′i,s, d′′i,s),(m′i,s, ws, ds)∈ Fp

(aTi , bs)∈ M ⇐⇒/ (m′i,s, w′i,s, d′′i,s),(m′′i,s, wi,s′′ , d′i,s)∈ Fp

Lemma 7. A complete matchingMofI is stable if and only if the corresponding complete

matching Fp ofIp′ is strongly stable. Proof. As a man aT

i cannot belong to a blocking pair in I, it may be verified that his

corresponding copymT

i cannot belong to a weakly blocking family inIp either. Therefore,

it is enough to show that a pair (ai, bj) is blocking forMif and only if the corresponding

family (mi, wj, dj) is blocking forFp. But this is obvious, because the preference lists ofai

andmiare essentially the same, and the preference lists ofbj anddj are also essentially the

same.

Now, given a matching M of I let us create the corresponding matching F of I′ by addingFsandFf toFp, soF =Fp∪ Fs∪ Ff.

Lemma 8. The instance I admits a complete stable matchingMif and only if the reduced

instance I′ admits a strongly stable matching F, whereF is the corresponding matching of

M.

Proof. Suppose that we have a complete stable matchingMofI, andFis the corresponding matching in I′. Lemma 7 implies that every proper agent has a proper partner in F and

no proper family is weakly blocking. Therefore, no proper agent can be involved in any weakly blocking family either. By construction of Fs, every covered additional agent has a

suitable partner inFand by Lemma 6, no suitable family is weakly blocking. Therefore, no such agent can be part of any weakly blocking family. Finally, every uncovered additional agent has a fitting partner inF, so these agent cannot form a weakly blocking family either, since an uncovered additional agent prefers only suitable partners to fitting partners, which cannot be involved in a weakly blocking family. HenceF is strongly stable.

In the other direction, suppose thatF is a strongly stable matching ofI′. Every proper

agent must have a proper partner, since otherwise if athad no proper partner in F, then

R6t would contain a suitable weakly blocking family by Lemma 5. So the corresponding

matching M in I is complete. The stability of M is a consequence of Lemma 7. Finally, we note that the additional agents must be matched in the unique strongly stable way in

F, namely, the covered additional agents must be covered by matching Fs by Lemma 6,

and the uncovered additional agents must be covered byFf, since otherwise a fitting family

would weakly blockF. Therefore, we have a one-to-one correspondence as was claimed.

Theorem 2. Determining the existence of a strongly stable matching in a given instance

of cyclic 3DSMis NP-complete.