Mutually exclusive edge pairs

Let us define any matching that satisfies Conditions 1 and 2 of Theorem 6.3.1 to be well- formed. Clearly, all well-formed matchings must be contained in G0. Let a be any even agent in I, and let b be any even agent that precedes the members of s(a) in a’s preference list. Clearly, b /∈ f (a) for otherwise we have an EE edge in G1, a contradiction by Lemma

1.2.1(c). Then, we define the edge pair {(b, b0), (a0, a)} to be a mutually exclusive edge pair, or mutex edge pair for short, if b0 ∈ f (b) and a0∈ s(a). The next theorem gives us an important characterisation of popular matchings in SMTI-SYM with respect to mutex edge pairs.

Theorem 6.4.1. Let M be a well-formed matching in any given SMTI-SYM instance I. Then, M is popular if and only if M does not contain any mutex edge pairs.

Proof. We first show that if M is popular, then M contains no mutex edge pairs. Suppose for a contradiction that M contains the mutex edge pair {(ul, wj), (ui, wp)} such that ui is

an even agent, wp∈ s(ui) and wj is an even agent preceding wp in ui’s preference list and

ul ∈ f (wj). We have that ui prefers wj to wp but wj prefers ul to ui. Since wj is an even

agent and wj is assigned in M , it follows by the definitions of a well-formed matching and

an even vertex that there exists an even length alternating path P in G1 to wj from an

it follows that the sequence of agents hui, wj, ..., wki then forms a cycle C such that M ⊕ C

gives us a more popular matching than M , a contradiction.

Hence, suppose that wk 6= wp. Let M (wk) = ua. Now, if wp = l(ui) and ua = l(wk),

then we can unassign uiand wkfrom their last resort partners, and use (M ⊕P )∪{(ui, wj)}

to give us a more popular matching than M . Let us thus suppose that wp = l(ui) and

ua 6= l(wk). Let wb ∈ f (ua) and let M (wb) = uc. Now, if wb ∈ P , then the sequence of

agents hwb, M (wb), ..., wk, uai form a cycle C such that M ⊕ C gives us a more popular

matching than M . Hence, wb ∈ P . Since w/ k ∈ s(ua), it follows that ua is an even agent

so that wb is odd. By Corollary 6.3.1, it must then be the case that uc 6= l(wb). Let M0

be the matching obtained by

((M \ {(ui, wp), (ua, wk), (uc, wb)}) ⊕ P ) ∪ {(ui, wj), (ua, wb), (uc, l(uc))}

It follows that M0 is more popular than M . On the other hand, if wp 6= l(ui) and

ua= l(wk), then let uq∈ f (wp) and let M (uq) = wr. Now, if uq ∈ P , then the sequence of

agents hwp, ui, wj, ..., uqi form a cycle C such that M ⊕C gives us a more popular matching

than M , a contradiction. Hence, uq ∈ P . Reusing a similar argument to the above, we/

have that wr6= l(uq). However, let M0 be the matching obtained by

((M \ {(uq, wr), (ui, wp), (ua, wk)}) ⊕ P ) ∪ {(uq, wp), (ui, wj), (l(wr), wr)}

It follows that M0 is more popular than M . Hence, suppose that neither wp nor ua

is a last resort agent. Then, let uq, wr, wb and uc be defined as before. Let P0 =

hl(wr), wr, uq, wp, ui, wj, ..., wk, ua, wb, uc, l(uc)i. Then, M ⊕ P0 gives us a more popular

matching than M . It follows that we obtain a contradiction in all cases so that if M contains a mutex edge pair, then M cannot be popular.

Conversely, let M be a well-formed matching that contains no mutex edge pairs. Sup- pose for a contradiction that there exists another matching M0 = {(u1, w1), ..., (ur, wr)}

such that M0 is more popular than M . We firstly observe that if, for every agent ai who

prefers M0 to M (1 ≤ i ≤ r), his partner in M0 prefers M to M0, then M0 cannot be more popular than M . Hence, there exists at least one ai who prefers M0 to M and his partner

in M0 either (i) also prefers M0 to M or (ii) is indifferent between the two matchings. Without loss of generality, let ai be a man whom we denote by ui and hence, M0(ui) = wi

by definition of M0. By Theorem 6.3.1 and the definition of a well-formed matching, we can partition the set of agents who are assigned in M into the disjoint sets F and S, where agents in F are assigned to their f -partners in M , and agents in S are assigned to their

s-partners in M respectively. It is clear to see that agents in F cannot improve in M0 relative to M but can either become worse off or remain indifferent. On the other hand, agents in S can either improve, become worse off in M0 relative to M or remain indifferent. In case (i), it must be the case that each of ui and wi can only belong to S. It follows

that ui and wi are both even agents because only even agents have s-partners defined.

However, this gives a contradiction since Algorithm Label-s would have defined ui and wi

to be one of each other’s s-partners because (ui, wi) is then an E E edge such that ui prefers

wi to any member of s(ui) and vice versa.

Hence, it remains to consider case (ii). It is clear that ui ∈ S. Now, if wi ∈ S and

wi is indifferent between ui and M (wi), we obtain a contradiction as in case (i). Hence,

wi ∈ F . Consider H0 = (M0⊕ M ) ∩ E1. It follows that the only connected components

of H0 where an agent in S can become assigned to an agent in F who remains indifferent between M and M0 are even length alternating paths. Let ui and wi belong to such a

component P . Since ui improves to wi ∈ F , and wi is indifferent between ui and M (wi),

it follows that ui is the end vertex of the end edge of P that is in M0. It also follows that

we have a uj who is the end vertex of the end edge of P that is in M . Clearly, uj becomes

worse off in M0 relative to M . Now, suppose that wj, who is uj’s partner in M0, prefers

M0 to M . By the structure of P , wj ∈ f (u/ j). Now, if wj also improves in M0 by becoming

assigned to uj, it follows that wj ∈ S and uj is an even agent who lies between f (wj)

and s(wj) in wj’s preference list. However then, it follows that {(uj, M (uj)), (M (wj), wj)}

constitutes a mutex edge pair in M , a contradiction. Hence, wj either becomes worse off

in M0 relative to M or is indifferent between the two matchings. However, it then follows that for every edge (ui, wi) where one of the agents improves in M0 relative to M , exactly

one of these agents prefer M0 relative to M and the other remains indifferent. Moreover, we have a unique corresponding edge (uj, wj) in which at least one of the agents prefers

M relative to M0 and neither agents prefers M0 to M . It cannot then be the case that M0 is more popular than M .

What Theorem 6.4.1 thus implies is that a well-formed matching M in G0 is popular if and only if M contains only one or none of the edges in any mutex edge pair. To illustrate this concept, let us return to instance I5 in Figure 6.5. Let G0_I5 be the underlying graph

of I5 which contain edges incident to only f - and s-partners. Then, it may be verified

that G0_I5 contains one mutex edge pair, namely {(u2, w1), (u3, w2)}. Here, w2 and u2 are

straightforward to see that the matching M is popular in I5 because it contains no edges

of this mutex edge pair, while M0 is not popular because it contains both edges of the mutex edge pair.