On random stable partitions

Abstract

It is well known that the one-sided stable matching problem (“stable roommates problem”) does not necessarily have a solution. We had found that, for the independent, uniformly random preference lists, the expected number of solutions converges to $e^{1/2}$ as n, the number of members, grows, and with Rob Irving we proved that the limiting probability of solvability is below $e^{1/2}/2$, at most. Stephan Mertens’s extensive numerics compelled him to conjecture that this probability is of order $n^{-1/4}$. Jimmy Tan introduced a notion of a stable cyclic partition, and proved existence of such a partition for every system of members’ preferences, discovering that presence of odd cycles in a stable partition is equivalent to absence of a stable matching. In this paper we show that the expected number of stable partitions with odd cycles grows as $n^{1/4}$. However the standard deviation of that number is of order $n^{3/8}\gg n^{1/4}$, i.e. too large to conclude that the odd cycles exist with probability $1-o(1)$. Still, as a byproduct, we show that with probability $1-o(1)$ the fraction of members with more than one stable “predecessor” is of order $n^{-1/2+o(1)}$. Furthermore, with probability $1-o(1)$ the average rank of a predecessor in every stable partition is of order $n^{1/2}$. The likely size of the largest stable matching is $n/2-O(n^{1/4+o(1)})$, and the likely number of pairs of unmatched members blocking the optimal complete matching is $O(n^{3/4+o(1)})$.

Appendix

Proof of Lemma 5.3

We already observed, (5.8), that $\sum _i (x_i\vee y_i)=\sum _j\xi _j$. So, similarly to (5.9)–(5.10) we have:

$$\begin{aligned}&\text{ P }(\varvec{\Pi }_1,\varvec{\Pi }_2)- \text{ P }_{\mathcal {C}_1}(\varvec{\Pi }_1,\varvec{\Pi }_2)\nonumber \\&\quad \le _b 2^{\mu }\iiint \limits _{\xi _1+\xi _2+\xi _3\ge s_n}\exp \Biggl (-\frac{1}{2}\Bigl (\sum _j\xi _j\Bigr )^2+ \xi _2\xi _3\Biggr ) \frac{\xi _1^{n+m-1}}{(n+m-1)!}\cdot \frac{(\xi _2\xi _3)^{\nu -1}}{[(\nu -1)!]^2}\,d\varvec{\xi }\nonumber \\&\quad = 2^{\mu }\sum _{k\ge 0}\frac{[(\nu -1+k)!]^2}{[(\nu -1)!]^2k!(n+m+2(\nu +k)-1)!}\nonumber \\&\qquad \quad \times \int _{s\ge s_n}\exp \Bigl (-\frac{s^2}{2}\Bigr ) s^{n+m+2(\nu +k)-1}\,ds. \end{aligned}$$

(6.1)

(Relaxing the constraint on s to $s\ge 0$ we get back to (5.10)). The last integrand attains its maximum at

$$\begin{aligned} s_{\text {max}}=\bigl (n+m+2(\nu +k)-1\bigr )^{1/2}, \end{aligned}$$

which is below $s_n-3\log n$ if $k\le m_n$. Let $S_{\le m_n}$ and $S_{>m_n}$ denote the sub-sums of the sum above, for $k\le m_n$ and $k>m_n$ respectively. Then, expanding integration to $[0,\infty )$, we obtain

$$\begin{aligned} S_{>m_n}&\le \sum _{k>m_n}\frac{[(\nu -1+k)!]^2}{[(\nu -1)!]^2\,k!\,(n+m+2(\nu +k)-1)!!}\\&\le _b \frac{[(\nu +m_n)!]^2}{[(\nu -1)!]^2\,m_n!\,(n+m+2(\nu +m_n)+1)!!}; \end{aligned}$$

since $\nu \le m_n$, the ratio of the consecutive terms in the sum is below 2 / 3. Droping $[(\nu -1)!]^2$ in the denominator and using the Stirling formula for the other factorials, we simplify the bound to

$$\begin{aligned} S_{>m_n}\le _b \left( \frac{e}{n+m}\right) ^{\frac{n+m}{2}} (n+m)^{-(\nu +m_n)}. \end{aligned}$$

(6.2)

The bound is smaller than the bound (5.11) for the full sum of $s(n+m,\nu ,k)$ by the factor $(n+m)^{m_n}$. Turn to $S_{\le m_n}$. This time the bottom integral over $s\ge s_n$ in (6.1) is small, compared to the integral over all $s\ge 0$, because for $k\le m_n$ the maximum point of the integrand is at distance $3\log n$, at least, from the interval $[s_n,\infty )$. More precisely, using the argument in the proof of Lemma 4.5, we have

$$\begin{aligned} \int _{s\ge s_n}\exp \left( -\frac{s^2}{2}\right) s^{n+m+2(\nu +k)-1}\,ds\le _b e^{-8\log ^2 n}\bigl (n+m +2(\nu +k)-2\bigr )!!. \end{aligned}$$

Therefore

$$\begin{aligned} \begin{aligned} S_{\le m_n}&\le _b e^{-8\log ^2 n}\sum _{k\le m_n}\frac{[(\nu -1+k)!]^2}{[(\nu -1)!]^2k!(n+m+2(\nu +k)-1)!!}\\&\le e^{-8\log ^2 n}\sum _{k\ge 0}\frac{[(\nu -1+k)!]^2}{[(\nu -1)!]^2k!(n+m+2(\nu +k)-1)!!}\\&=e^{-8\log ^2 n}\sum _{k\ge 0}s(n+m,\nu ,k). \end{aligned} \end{aligned}$$

(6.3)

Combining (6.2), (6.3) and (5.11) we transform the inequality (6.1) into

$$\begin{aligned} P(\varvec{\Pi }_1,\varvec{\Pi }_2)-P_{\mathcal {C}_1}(\varvec{\Pi }_1,\varvec{\Pi }_2)\le e^{-\Theta (\log ^2 n)}\, 2^\mu \left( \frac{e}{n+m}\right) ^{\frac{n+m}{2}} (n+m)^{-\nu }. \end{aligned}$$

(6.4)

So, like the part (1) in the proof of Lemma 5.1, we obtain

$$\begin{aligned} \begin{aligned}&\sum _{\varvec{\Pi }_1,\varvec{\Pi }_2}\bigl [P(\varvec{\Pi }_1,\varvec{\Pi }_2)-P_{\mathcal {C}_1}(\varvec{\Pi }_1,\varvec{\Pi }_2)] \le e^{-\Theta (\log ^2n)}n!\,m_n\\&\quad \times \sum _{m\le m_n}\left( \frac{e}{n+m}\right) ^{\frac{n+m}{2}} \left[ (n+m)^{0}\, 2^{\frac{n-m-2\cdot 0}{2}}\left( \frac{n-m-2\cdot 0}{2} \right) ! \right] ^{-1}\\&=e^{-\Theta (\log ^2n)}. \end{aligned} \end{aligned}$$

(6.5)

$\square$

Proof of Lemma 5.4

Introduce $L_1',\dots ,L_{n+2\nu }'$, the intervals lengths in the random partition of [0, 1] by the $n+2\nu -1$ random points. Analogously to (5.9), but using the sharper inequality in Lemma 3.1, (3.1), we have: with $s:=\xi _1+\xi _2+\xi _3=\sum _{i\in [n]}x_i'+\sum _{i\in A\cup B}y_i'$ (see (5.5) and (5.6)),

$$\begin{aligned}&\text{ P }_{\mathcal {C}_1}(\varvec{\Pi }_1,\varvec{\Pi }_2)- \text{ P }_{\mathcal {C}_2}(\varvec{\Pi }_1,\varvec{\Pi }_2)\\&\le _b\,2^{\mu } \idotsint \limits _{\begin{array}{c} {\mathbf {x}}', {\mathbf {y}}'\ge {\mathbf {0}}\\ T_1(\mathbf {u}')>1.01\frac{\log ^2 n}{n} \end{array}} e^{-\frac{s^2}{2}+\xi _2\xi _3} \prod _{h\in \text {Odd}_{1,2}}x_h\,\prod _{i\in [n]} dx'_i\prod _{j\in A\cup B} dy'_j\\&\le \frac{2^{\mu }}{(n+2\nu -1)!}\,\text{ E }\Biggl [\chi \Bigl (T_1({\mathbf {L}}')\ge \frac{\log ^2n}{n}\Bigr )\prod _{h\in \text {Odd}_{1,2}}L_h'\Biggr ]\\&\quad \times \idotsint \limits _{{\mathbf {x}}', {\mathbf {y}}'\ge {\mathbf {0}}} e^{-\frac{s^2}{2}+\xi _2\xi _3}\,s^m\prod _{i\in [n]} dx'_i\prod _{j\in A\cup B} dy'_j. \end{aligned}$$

Arguing as in (4.12), the expectation factor is less than

$$\begin{aligned} e^{-\Theta (\log ^2 n)}\frac{(n+2\nu )!}{(n+m+2\nu -2)!}. \end{aligned}$$

The integral is less than

$$\begin{aligned} I_n(m,\nu )&:=\iiint \limits _{\xi _j\ge 0} e^{-\frac{s^2}{2}+\xi _2\xi _3} s^m\frac{\xi _1^{n-1}}{(n-1)!} \cdot \frac{(\xi _2\xi _3)^{\nu -1}}{\bigl [(\nu -1)!\bigr ]^2}\,d\varvec{\xi }\\&=\sum _{k\ge 0}\frac{\bigl [(\nu -1+k)!\bigr ]^2}{(n-1)!\,\bigl [(\nu -1)!\bigr ]^2\,k!\,\bigl (2(\nu +k)-1\bigr )!}\\&\qquad \times \iint \limits _{\xi _1,\xi _4\ge 0}e^{-\frac{(\xi _1+\xi _4)^2}{2}} (\xi _1+\xi _4)^m\, \xi _1^{n-1}\xi _4^{2(\nu +k)-1}\,d\xi _1 d\xi _4. \end{aligned}$$

Here the double integral equals

$$\begin{aligned} \frac{(n-1)!\,\bigl (2(\nu +k)-1\bigr )!\,\bigl (n+m+2(\nu +k)-2\bigr )!!}{\bigl (n+2(\nu +k)-1\bigr )!}. \end{aligned}$$

So

$$\begin{aligned} I_n(m,\nu )=\sum _{k\ge 0}\frac{\bigl [(\nu -1+k)!\bigr ]^2\,\bigl (n+m+2(\nu +k)-2\bigr )!!}{\bigl [(\nu -1)!\bigr ]^2\,k!\,\bigl (n+2(\nu +k)-1\bigr )!} \end{aligned}$$

Therefore

$$\begin{aligned}&\quad \text{ P }_{\mathcal {C}_1}(\varvec{\Pi }_1,\varvec{\Pi }_2)- \text{ P }_{\mathcal {C}_2}(\varvec{\Pi }_1,\varvec{\Pi }_2) \le _b 2^{\mu } e^{-\Theta (\log ^2 n)}\sum _{k\ge 0} s'(n,m,\nu ,k),\\&s'(n,m,\nu ,k):=\frac{\bigl [(\nu -1+k)!\bigr ]^2\bigl (n+m+2(\nu +k)-2\bigr )!!\,(n+2\nu )!}{(n+m+2\nu -2)!\,\bigl [(\nu -1)!\bigr ]^2\,k!\,\bigl (n+2(\nu +k)-1\bigr )!}. \end{aligned}$$

The summand $s'(n,m,\nu ,k)$ is similar to the summand $s(n+m,\nu ,k)$ defined in (5.10). Closely following the derivation of the bound for $\sum _{k\ge 0}s(n,\nu ,k)$ in Pittel (1993b), we obtain

$$\begin{aligned} \sum _{k\ge 0} s'(n,m,\nu ,k)\le _b n^2\left( \frac{e}{n+m}\right) ^{\frac{n+m}{2}}n^{-\nu }, \end{aligned}$$

compare to (5.11). The last two bounds complete the proof. $\square$