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The Gale-Shapley algorithm finds a stable matching in the complete bipartite graph, for any preference matrix. It's also well-known that stable matchings don't always exist in the complete graph (roommates problem), and it seems to be still open whether they almost surely don't exist in a uniformly random instance of the latter. See

http://arxiv.org/pdf/cond-mat/0509221.pdf

Given an instance of the marriage or roommates problem, call a matching \emph{$t$-stable} if there is no unmatched pair $\{x,y\}$ such that $x$ ranks $y$ at least $t$ places higher than his match, and similarly for $y$. There are a couple of obvious motivations for considering this extension: (i) imagine that there is a cost associated with switching partners (e.g.: the time and money it takes to physically move, or the "social cost" incurred to one's reputation as an unreliable roommate, or the cost imposed by the laws of the society - in Ireland, where I'm from, divorce was illegal up to the mid-1990s), (ii) maybe each person has a bounded ability to distinguish beteen alternative partners. Hence, someone won't switch unless the alternative is significantly better.

Now there are some obvious questions one can ask, in particular in the roommates problem, where $1$-stable matchings don't always exist. For example:

Question 1: What is the smallest function $t=t(n)$ such that every instance of the roommates problem on $n$ nodes has a $t$-stable matching ?

Question 2: Determine a threshold function $t=t(n)$ such that a random instance of the roommates problem on $n$ nodes almost surely has a $t$-stable matching when $t \gg t(n)$, and almost surely doesn't when $t \ll t(n)$.

Question 3: Given $t,n$, say $n$ even, describe an algorithm which decides whether an instance of the roommates problem on $K_n$ has a $t$-stable matching or not, and finds one if it has.

I have not been able to locate anything at all on this notion in the literature. Remarks ? By the way, regarding Question 1, I think I can construct an instance on $K_{n^2}$ where there is no $n$-stable matching. Is this best-possible ? Regarding Question 2, since it's not even known if $1$-stable matchings a.s. don't exist, finding the right threshold is probably hard. But can we bound it from above ? Regarding Question 3, Irving's algorithm for $1$-stable matchings doesn't seem to extend trivially to $t$-stability.

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    $\begingroup$ The similar problem was mentioned in cs.helsinki.fi/u/polishch/pages/geomSR.pdf as $\alpha$-stable matching. The basic setting is based on multiplicative improvements. It is easy to apply this setting to an additive version where “a participant prefers to break partnership only when the new partner is at least $\alpha$ positions higher, in the participant's preference list, than the current partner.” Thus, it is equivalent to a $t$-stable matching. They prove “that, in general, finding $\alpha$-stable matchings is not easier than finding matchings that are stable in the usual sense” $\endgroup$
    – Waldemar
    Nov 26, 2013 at 12:52
  • $\begingroup$ Contrary to point (i), point (ii) does not seem very convincing. Such situations would probably be better modelled with “The stable roommates problem with ties” than with $t$-stable matchings. $\endgroup$
    – Waldemar
    Nov 27, 2013 at 13:52
  • $\begingroup$ @Waldemar: Thanks for that link! I looked through the paper and I don't think they prove anything about $\alpha$-stability in the sense I define it, but they refer to this definition in Remarks at the end and pose essentially the same questions as I do. Their result that finding an $\alpha$-stable matching is at least as hard as usual s.m. is for the case where each match comes with a "utility" and $\alpha$ is to be a multiplicative improvement of utility. It seems quite a formal fact - see proof of Thm 4.2. I don't think it translates to anything directly in the "additive" case. $\endgroup$ Nov 27, 2013 at 17:42
  • $\begingroup$ The same “formal fact” holds for a $t$-stable matching as with $t=1$ it’s just a usual s.m. The whole proof of thm 4.2. is simply saying that a usual stable matching is a special case of $\alpha$-stable matching. Nevertheless, it is interesting to know that the questions similar to yours were posed before. An additive version is a special case of the multiplicative one. Let $p(i)$ be the $i$th entry in $p$'s preference list. Assign regret $\alpha ^{\frac{i}{t}}$ to the match of $p$ with $p(i)$. This $\alpha$-stable matching is equivalent to a $t$-stable matching. $\endgroup$
    – Waldemar
    Nov 27, 2013 at 23:13
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    $\begingroup$ The word from two of the authors of that paper is that there has been no progress on these questions. Btw, regarding (ii) as a motivation: I am thinking of "indistinguishability" as something that need not be transitive, whereas "being tied" is transitive. You can imagine, for example, that someone might be more likely to accept a substantially worse situation if they can make the transition gradually, rather than all in one go. $\endgroup$ Dec 3, 2013 at 13:52

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As a first approach to answering Question 3 we can consider the following integer programming formulation:

$\sum _{j}x_{i,j}=1\; \; \; \forall i,$

$x_{i,j}+\sum _{l:l<_{j}i}x_{l,j}+\sum _{l:l<_{i}j}x_{i,l}\leq 1\; \; \; \forall i,j,$

$x_{i,j}\in \left \{ 0,1 \right \}\; \; \; \forall i,j.$

See: http://www.columbia.edu/~js1353/marriage.ps p.18

The only difference between the IP formulation of the stable roommates problem (SRP) and its $t$-stable variant is in an interpretation. The expression "$l<_{j}i$" is interpreted as “$j$ prefers $i$ to $l$” or equivalently: “$i$ is higher, in the $j$'s preference list, than $l$” in the SRP. In the $t$-stable variant of the SRP it should be interpreted as “$i$ is at least $t$ positions higher, in the $j$'s preference list, than $l$".

There are some similarities between $t$-stable matchings and weakly stable matchings (as defined in Irving, R. W. and Manlove, D. F. (2002) The stable roommates problem with ties). It is known that deciding whether an instance of stable roommates problem admits a weakly stable matching is NP-complete. Thus, the above IP formulation may not be as bad as it seems at first.

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