A Note on the Expected Length of the Longest Common Subsequences of two i.i.d. Random Permutations

TL;DR

This paper investigates the expected length of the longest common subsequences of two independent random permutations, disproving a conjecture that it is at least proportional to the square root of n, and establishing a lower bound proportional to the cube root of n.

Contribution

It resolves a question about the minimal expectation not occurring in the uniform case and provides a new lower bound of order n^{1/3} for the expected LCS length.

Findings

Minimal expectation not attained in the uniform case

Established a lower bound of order n^{1/3} for the expected LCS length

Disproved the conjecture that the expectation is at least proportional to √n

Abstract

We address a question and a conjecture on the expected length of the longest common subsequences of two i.i.d.$\ $random permutations of $[n]:=\{1,2,...,n\}$. The question is resolved by showing that the minimal expectation is not attained in the uniform case. The conjecture asserts that $\sqrt{n}$ is a lower bound on this expectation, but we only obtain $\sqrt[3]{n}$ for it.

Equations

$${\mathbb{E}}_P[LCS({\sigma }_1,{\sigma }_2)]$$

$$=\sum _{i,j\in [n!]}{p}_i{\ell }_{ij}{p}_j=P^T{L}^{(n)}P,$$

$$LCS([{\pi }_i(k+1)],[{\pi }_j(k+1)])=LCS({\pi }_i,{\pi }_j)+1,$$

$$\sum _{j\in [k!]}LCS({\pi }_i,{\pi }_j)$$

$$=\sum _{j\in [k!]}LIS({\pi }_i^{-1}{\pi }_j),$$

$$\sum _{j\in [k!]}LIS({\pi }_i^{-1}{\pi }_j)=\sum _{\pi \in {\mathcal{S}}_k}LIS(\pi ),$$

$$(E^{(k)}{)}^T{R}_1^{(k)}=0.$$

$${\lambda }_1^{(k)}$$

$$=(R_1^{(k)}{)}^T(L^{(k)}+E^{(k)}(E^{(k)}{)}^T)R_1^{(k)}.$$

$$\left[\begin{array}[]{c}R_{1}^{(k)}\\ 0\end{array}\right]^{T}L^{(k+1)}\left[\begin{array}[]{c}R_{1}^{(k)}\\ 0\end{array}\right]\geqslant\min_{R^{T}E=0,||R||=1}R^{T}L^{(k+1)}R=\lambda_{1}^{(k+1)},$$

$$L^{(k+1)}\left[\begin{array}[]{c}R_{1}^{(k)}\\ 0\end{array}\right]=\lambda_{1}^{(k+1)}\left[\begin{array}[]{c}R_{1}^{(k)}\\ 0\end{array}\right].$$

$$LCS([(k+1){\pi }_i],[{\pi }_j(k+1)])=LCS({\pi }_i,{\pi }_j),$$

$${\lambda }_1^{(2)}=1>0={\lambda }_1^{(3)}>{\lambda }_1^{(4)}>{\lambda }_1^{(5)}...$$

$${\mathbb{E}}_P[LCS({\sigma }_1,{\sigma }_2)]$$

$$=(P-U{)}^{T}L(P-U)+2P^{T}LU-U^{T}LU$$

$$=(P-U{)}^{T}L(P-U)+2U^{T}LU-U^{T}LU$$

$$=(P-U{)}^{T}L(P-U)+U^{T}LU,$$

$$E^T{P}_0=E^T(U+c{R}_1^{(n)})=1+0=1,$$

$$P_0=U+c{R}_1^{(n)}⪰0.$$

$${\mathbb{E}}_{P_0}[LCS({\sigma }_1,{\sigma }_2)]$$

$$=c^2(R_1^{(n)}{)}^{T}L(R_1^{(n)})+U^{T}LU$$

$$=c^2{\lambda }_1^{(n)}+U^{T}LU$$

$$<U^{T}LU.$$

$$|\lambda |$$

$${\lambda }_1^{(n)}⩾-\sum _{\pi \in S_n}LIS(\pi ).$$

$${\lambda }_1^{(n+1)}=\underset{E^{T}R=0}{\min }\frac{R^T{L}^{(n+1)}R}{R^{T}R}.$$

$$\left[\begin{array}[]{ccc}L^{(n)}+EE^{T}&\cdots&L^{(n)}\\ \vdots&\ddots&\vdots\\ L^{(n)}&\cdots&L^{(n)}+EE^{T}\end{array}\right].$$

$$R=\left[\begin{array}[]{c}R_{1}^{(n)}\\ 0\\ \vdots\\ 0\\ R_{1}^{(n)}\end{array}\right].$$

$${\Vert R\Vert }^2$$

$$\displaystyle\left[\begin{array}[]{c}R_{1}^{(n)}\\ 0\\ \vdots\\ 0\\ R_{1}^{(n)}\end{array}\right]^{T}$$

$$=2{\left(R_1^{(n)}\right)}^T\left(L^{(n)}+E{E}^T\right)\left(R_1^{(n)}\right)+2{\left(R_1^{(n)}\right)}^T{L}^{(n)}\left(R_1^{(n)}\right)$$

$$=4{\left(R_1^{(n)}\right)}^T{L}^{(n)}\left(R_1^{(n)}\right)=4{\lambda }_1^{(n)}.$$

$${\lambda }_1^{(n+1)}⩽2{\lambda }_1^{(n)}.$$

$${\lambda }_{(n+1)!-1}^{(n+1)}=\underset{E^{T}R=0}{\max }\frac{R^T{L}^{(n+1)}R}{R^{T}R},$$

$$R=\left[\begin{array}[]{c}R_{n!-1}^{(n)}\\ 0\\ \vdots\\ 0\\ R_{n!-1}^{(n)}\end{array}\right],$$

$$\underset{n\to +\infty }{\lim }\frac{{\lambda }_1^{(n+1)}}{{\lambda }_1^{(n)}(n-1)}=c_1⩾1,$$

Full text

1 Introduction

The length of the longest increasing subsequences ($\mathit{LISs}$) of a uniform random permutation $\sigma\in{\cal S}_{n}$ (where ${\cal S}_{n}$ is the symmetric group) is well studied and we refer to the monograph [5] for precise results and a comprehensive bibliography on this subject. Recently, [3] showed that for two independent random permutations $\sigma_{1},\ \sigma_{2}\in{\cal S}_{n}$, and as long as $\sigma_{1}$ is uniformly distributed and regardless of the distribution of $\sigma_{2}$, the length of the longest common subsequences ($LCSs$) of the two permutations is identical in law to the length of the $LISs$ of $\sigma_{1}$, i.e. $LCS(\sigma_{1},\sigma_{2})=^{\mathcal{L}}LIS(\sigma_{1})$. This equality ensures, in particular, that when $\sigma_{1}$ and $\sigma_{2}$ are uniformly distributed, $\mathbb{E}LCS(\sigma_{1},\sigma_{2})$ is upper bounded by $2\sqrt{n}$, for any $n$, (see [4]) and asymptotically of order $2\sqrt{n}$ ([5]). It is then rather natural to study the behavior of $LCS(\sigma_{1},\sigma_{2})$, when $\sigma_{1}$ and $\sigma_{2}$ are i.i.d.  but not necessarily uniform. In this respect, Bukh and Zhou raised, in [2], two issues which can be rephrased as follows:

Conjecture/Question 1*.*

Let $P$ be an arbitrary probability distribution on ${\cal S}_{n}$. Let $\sigma_{1}$ and $\sigma_{2}$ be two i.i.d. permutations sampled from $P$. Then $\mathbb{E}_{P}[LCS(\sigma_{1},\sigma_{2})]\geqslant\sqrt{n}$. It might even be true that the uniform distribution $U$ on ${\cal S}_{n}$ gives a minimizer.

Below we prove the suboptimality of the uniform distribution by explicitly building a distribution having a smaller expectation. In the next section, before presenting and proving our main result, we give a few definitions and formalize this minimizing problem as a quadratic programming one. Section 3 further explore some properties of the spectrum of the coefficient matrix of our quadratic program. In the concluding section, a quick cubic root lower bound is given along with a few pointers for future research.

2 Main Results

We begin with a few notations. Throughout, $\sigma$ and $\pi$ are, respectively, used for random and deterministic permutations. By convention, $[n]:=\{1,2,3,...,n\}$ and so $\{\pi_{i}\}_{i\in[n!]}={\cal S}_{n}$ is a particular ordered enumeration of ${\cal S}_{n}$. (Some other orderings of ${\cal S}_{n}$ will be given when necessary.) Next, a random permutation $\sigma$ is said to be sampled from $P=(p_{i})_{i\in[n!]}$, if $\mathbb{P}_{P}(\sigma=\pi_{i})=p_{i}$. The uniform distribution is therefore $U=(1/n!)_{i\in[n!]}$ and, for simplification, it is denoted by $E/n!$, where $E=(1)_{i\in[n!]}$ is the $n$-tuple only made up of ones. When needed, a superscript will indicate the degree of the symmetric group we are studying, e.g., $\sigma^{(n)}$ and $P^{(n)}$ are respectively a random permutation and distribution from ${\cal S}_{n}$.

Let us now formalize the expectation as a quadratic form:

[TABLE]

where $\ell_{ij}:=LCS(\pi_{i},\pi_{j})$ and $L^{(n)}:=\{\ell_{ij}\}_{(i,j)\in[n!]\times[n!]}$. It is clear that $\ell_{ij}=\ell_{ji}$ and that $\ell_{ii}=n$. A quick analysis of the cases $n=2$ or $3$ shows that both $L^{(2)}$ and $L^{(3)}$ are positive semi-definite. However, this property does not hold further:

Lemma 2**.**

For $n\geqslant 4,$ the smallest eigenvalue $\lambda_{1}^{(n)}$ of $L^{(n)}$ is negative.

Proof.

Linear algebra gives $\lambda_{1}^{(2)}=1$ and $\lambda_{1}^{(3)}=0$. So to prove the result, it suffices to show that $\lambda_{1}^{(k+1)}<\lambda_{1}^{(k)}$, $k\geqslant 1$ and this is done by induction. The base case is true, since $\lambda_{1}^{(2)}=1>0=\lambda_{1}^{(3)}$. To reveal the connection between $L^{(k+1)}$ and $L^{(k)}$, the enumeration of ${\cal S}_{k+1}$ is iteratively built on that of ${\cal S}_{k}$ by inserting the new element $(k+1)$ into the permutations from ${\cal S}_{k}$ in the following way: the enumeration of the $(k+1)!$ permutations is split into $(k+1)$ trunks of equal size $k!$. In the $ith$ trunk, the new element $(k+1)$ is inserted behind the $(k+1-i)th$ digit in the permutation from ${\cal S}_{k}$. (For example, if ${\cal S}_{2}$ is enumerated as $\{[12],[21]\}$, then the enumeration of the first trunk in ${\cal S}_{3}$ is $\{[123],[213]\}$, the second is $\{[132],[231]\}$ and the third is $\{[312],[321]\}$. Then the overall enumeration for ${\cal S}_{3}$ is $\{[123],[213],[132],[231],[312],[321]\}$.)

Via this enumeration, the principal minor of size $k!\times k!$ is row and column indexed by the enumeration of the permutations $\{\pi_{i}^{(k)}\}_{i\in[k!]}$ from ${\cal S}_{k}$ with $(k+1)$ as the last digit, i.e., $\{[\pi_{i}^{(k)}(k+1)]\}_{i\in[k!]}\subseteq{\cal S}_{k+1}$. Then the $(i,j)$ entry of the submatrix is

[TABLE]

since the last digit $(k+1)$ adds an extra element into the longest common subsequences. Hence, the $k!\times k!$ principal minor of $L^{(k+1)}$ is $L^{(k)}+E^{(k)}(E^{(k)})^{T}$, where $E^{(k)}$ is the vector of $\mathbb{R}^{k!}$ only made up of ones. Moreover, notice that the sum of the $\pi_{i}$-indexed row of $L^{(k)}$ is

[TABLE]

since simultaneously relabeling $\pi_{i}$ and $\pi_{j}$ does not change the length of the $LCSs$ and also since a particular relabeling to make $\pi_{i}$ to be the identity permutation, which is equivalent to left composition by $\pi_{i}^{-1}$, is applied here. Further, any $LCS$ of the identity permutation and of $\pi_{i}^{-1}\pi_{j}$ is a $LIS$ of $\pi_{i}^{-1}\pi_{j}$ and vice versa. So the row sum is equal to

[TABLE]

since left composition by $\pi_{i}^{-1}$ is a bijection from ${\cal S}_{k}$ to ${\cal S}_{k}$. This indicates that all the row sums of $L^{(k)}$ are equal. Hence, $E^{(k)}$ is actually a right eigenvector of $L^{(k)}$ and is associated with the row sum $\sum_{\pi\in{\cal S}_{k}}LIS(\pi)>0$ as its eigenvalue, which is distinct from the smallest eigenvalue $\lambda_{1}^{(k)}\leqslant 0$.

On the other hand, since $L^{(k)}$ is symmetric, the eigenvectors $R_{1}^{(k)}$ and $E^{(k)}$ associated with the eigenvalues $\lambda_{1}^{(k)}$ and $\sum_{\pi\in{\cal S}_{k}}LIS(\pi)$ are orthogonal, i.e.,

[TABLE]

Without loss of generality, let $R_{1}^{(k)}$ be a unit vector, then from (2),

[TABLE]

As $L^{(k)}+E^{(k)}(E^{(k)})^{T}$ is the $k!\times k!$ principal minor of $L^{(k+1)}$, (3) becomes

[TABLE]

where $R_{1}^{(k)}$ is properly extended to \left[\begin{array}[]{c}R_{1}^{(k)}\\ 0\end{array}\right]\in\mathbb{R}^{(k+1)!} and where the above inequality holds true since \left[\begin{array}[]{c}R_{1}^{(k)}\\ 0\end{array}\right]^{T}E^{(k+1)}=(R_{1}^{(k)})^{T}E^{(k)}=0 and \left\|\left[\begin{array}[]{c}R_{1}^{(k)}\\ 0\end{array}\right]\right\|=\left\|R_{1}^{(k)}\right\|=1, where $\left\|\cdot\right\|$ denotes the corresponding Euclidean norm. Moreover, equality in (4) holds if and only if \left[\begin{array}[]{c}R_{1}^{(k)}\\ 0\end{array}\right] is a eigenvector of $L^{(k+1)}$ associated with $\lambda_{1}^{(k+1)}$. We show next, by contradiction, that this cannot be the case. Indeed, assume that

[TABLE]

Now, consider the $k!\times k!$ submatrix at the bottom-left corner of $L^{(k+1)}$, which is row-indexed by $\{[(k+1)\pi_{i}]\}_{i\in[k!]}$ and column-indexed by $\{[\pi_{i}(k+1)]\}_{i\in[k!]}$. Notice that the $(i,j)$-entry of this submatrix is

[TABLE]

since $(k+1)$ can be in some $LCS$ only if the length of this $LCS$ is $1$. So this submatrix is in fact equal to $L^{(k)}$. Further, the vector consisting of the bottom $k!$ elements on the left-hand-side of (5) is $L^{(k)}R_{1}^{(k)}=\lambda_{1}^{(k)}R_{1}^{(k)}$, which is a non-zero vector. However, on the right-hand-side, the corresponding bottom $k!$ elements of the vector \left[\begin{array}[]{c}R_{1}^{(k)}\\ 0\end{array}\right] form the zero vector. This leads to a contradiction. So,

[TABLE]

∎

The above result on the smallest negative eigenvalue, and its associated eigenvector, will help build a distribution on ${\cal S}_{n}$, for which the $LCSs$ have a smaller expectation than for the uniform one.

Theorem 3**.**

Let $\sigma_{1}$ and $\sigma_{2}$ be two i.i.d. random permutations sampled from a distribution $P$ on the symmetric group ${\cal S}_{n}$. Then, for $n\leqslant 3$, the uniform distribution $U$ minimizes $\mathbb{E}_{p}[LCS(\sigma_{1},\sigma_{2})]$, while, for $n\geqslant 4$, $U$ is sub-optimal.

Proof.

As we have seen in (1),

[TABLE]

where $P^{T}LU=U^{T}LU$, since $U$ is an eigenvector of $L$ and $P^{T}U=1$.

When $n=2,3$, $L^{(n)}$ is positive semi-definite and therefore $(P-U)^{T}L(P-U)\geqslant 0$. So, $P^{T}LP\geqslant U^{T}LU$.

However, when $n\geqslant 4$, by Lemma 2, the smallest eigenvalue $\lambda_{1}^{(n)}$ is strictly negative and the associated eigenvector $R_{1}^{(n)}$ is such that $U^{T}R_{1}^{(n)}=0=E^{T}R_{1}^{(n)}$. Hence, there exists a positive constant $c$ such that $cR_{1}^{(n)}\succeq-1/n!$, where $\succeq$ stands for componentwise inequality. Let $P_{0}$ be such that $P_{0}-U=cR_{1}^{(n)}$, then it is immediate that

[TABLE]

and that

[TABLE]

Therefore, $P_{0}$ is a well-defined distribution on ${\cal S}_{n}$. On the other hand, by (6), the expectation under $P_{0}$ is such that

[TABLE]

However, the right-hand side of (7) is nothing but the expectation under the uniform distribution, namely, $\mathbb{E}_{U}[LCS(\sigma_{1},\sigma_{2})]$. ∎

The existence of negative eigenvalues contributes to the above construction and to the corresponding counterexample. So, as a next step, properties of this smallest negative eigenvalue and of the spectrum of the coefficient matrix $L^{(n)}$ are explored.

3 Further Properties of $L^{(n)}$

As we have seen, the vector $E^{(n)}$ which is made up of only ones is an eigenvector associated with the eigenvalue $\sum_{\pi\in S_{n}}LIS(\pi)$. It is not hard to show that this eigenvalue is, in fact, the spectral radius of $L^{(n)}$.

Proposition 4**.**

$\sum_{\pi\in S_{k}}LIS(\pi)$* is the spectral radius of $L^{(n)}$.*

Proof.

Without loss of generality, let $(\lambda,R)$ be a pair of eigenvalue and corresponding eigenvector of $L^{(n)}$ such that $\max_{i\in[n!]}|r_{i}|=1$, where $R=(r_{1},...,r_{n!})^{T}$, and let $i_{0}$ be the index such that $|r_{i_{0}}|=1$. Let us focus now on the $i_{0}th$ element of $\lambda R$. Then, since $L^{(n)}R=\lambda R$,

[TABLE]

with equality if and only if all the $r_{j}$’s have the same sign and have absolute value equal to $1$. ∎

This gives a trivial bound on the smallest negative value $\lambda_{1}^{(n)}$: namely,

[TABLE]

Moreover, since the expectation of the longest increasing subsequence of a uniform random permutation is asymptotically $2\sqrt{n}$, this gives an asymptotic order of $-2n!\sqrt{n}$ for the lower bound. On the other hand, we are interested in an upper bound for $\lambda_{1}^{(n)}$. The next result shows that $\lambda_{1}^{(n)}$ decreases at least exponentially fast, in $n$.

Proposition 5**.**

$\lambda_{1}^{(n)}\leqslant 2^{n-4}\lambda_{1}^{(4)}=-2^{n-3}<0$.

Proof.

This is proved by showing that $\lambda_{1}^{(n+1)}\leqslant 2\lambda_{1}^{(n)}$. As well known,

[TABLE]

Let $\lambda_{1}^{(n)}$ be the smallest eigenvalues of $L^{(n)}$ and let $R^{(n)}$ be the corresponding eigenvector. Then, in generating $L^{(n+1)}$ from $L^{(n)}$ as done in the proof of Lemma 2, the $n!\times n!$ principal minor of $L^{(n+1)}$ is $L^{(n)}+EE^{T}$, while its bottom-left $n!\times n!$ submatrix is $L^{(n)}$. Symmetrically, it can be proved that the top-right $n!\times n!$ submatrix is also $L^{(n)}$, while the bottom-right $n!\times n!$ submatrix is $L^{(n)}+EE^{T}$, i.e., $L^{(n+1)}$ is

[TABLE]

Further, let

[TABLE]

Then $E^{T}R=E^{T}R_{1}^{(n)}+E^{T}R_{1}^{(n)}=0$, where, by an abuse of notation, $E$ denotes the vector only made up of ones and of the appropriate dimension. Also,

[TABLE]

In (8), the corresponding numerator $R^{T}L^{(n+1)}R$ is

[TABLE]

Thus,

[TABLE]

∎

By a very similar method, it can also be proved, as shown next, that the second largest eigenvalue $\lambda_{n!-1}^{(n)}$, which is positive, grows at least exponentially fast.

Proposition 6**.**

$\lambda_{n!-1}^{(n)}\geqslant 2^{n-2}\lambda_{1}^{(2)}=2^{n-2}>0$.

Proof.

Using the identity

[TABLE]

with a particular choice of

[TABLE]

where $R_{n!-1}^{(n)}$ is the eigenvector associated with the second largest eigenvalue $\lambda_{n!-1}^{(n)}$ of $L^{(n)}$, leads to $\lambda_{(n+1)!-1}^{(n+1)}\geqslant 2\lambda_{n!-1}^{(n)}$ and thus proves the result. ∎

The above bounds for $\lambda_{1}^{(n)}$ and $\lambda_{n!-1}^{(n)}$ are far from tight even as far as their asymptotic orders are concerned. Numerical evidence is collected in the following table:

[TABLE]

.

A reasonable conjecture will be that both the smallest and the second largest eigenvalues grow at a factorial-like speed. More precisely, we believe that

[TABLE]

and that

[TABLE]

4 Concluding Remarks

The $\sqrt{n}$ lower-bound conjecture of Bukh and Zhou is still open and seems quite reasonable in view of the fact that $\mathbb{E}LCS(\sigma_{1},\sigma_{2})\sim 2\sqrt{n}$, in case $\sigma_{1}$ is uniform and $\sigma_{2}$ arbitrary (again, see [3]). We do not have a proof of this conjecture, but let us nevertheless present, next, a quick $\sqrt[3]{n}$ lower bound result.

We start with a lemma describing a balanced property among the lengths of the $LCSs$ of pairs of any three arbitrary deterministic permutations. This result is essentially due to Beame and Huynh-Ngoc ([1]).

Lemma 7**.**

For any $\pi_{i}\in{\cal S}_{n}$ $(i=1,2,3$),

[TABLE]

Proof.

The proof of Lemma $5.9$ in [1] applies here with slight modification. We further note that this inequality is tight, since letting $\pi_{1}=\pi_{2}=id$ and $\pi_{3}=rev(id)$, which is the reversal of the identity permutation gives, $LCS(\pi_{1},\pi_{2})LCS(\pi_{2},\pi_{3})LCS(\pi_{3},\pi_{1})=n$. ∎

In Lemma 7, taking $(\pi_{1},\pi_{2})=(id,rev(id))$ gives, for any third permutation $\pi_{3}$, $LCS(id,\pi_{3})LCS(rev(id),\pi_{3})\geqslant n/LCS(id,rev(id))=n$. But, since $LCS(id,\pi_{3})$ and $LCS(rev(id),\pi_{3})$ are respectively the lengths of the longest increasing/decreasing subsequences of $\pi_{3}$, this lemma can be considered to be a generalization of a well-known classical result of Erdös and Szekeres (see [5]).

We are now ready for the cubic root lower bound.

Proposition 8**.**

Let $P$ be an arbitrary probability distribution on ${\cal S}_{n}$ and let $\sigma_{1}$ and $\sigma_{2}$ be two i.i.d.  random permutations sampled from $P$. Then, for any $n\geqslant 1$, $\mathbb{E}_{P}[LCS(\sigma_{1},\sigma_{2})]\geqslant\sqrt[3]{n}$.

Proof.

Let $\pi_{1},\ \pi_{2}$ and $\pi_{3}\in S_{n}$ and set

[TABLE]

for $i=2,3$. Then,

[TABLE]

by the arithmetic mean-geometric mean inequality and the previous lemma. Further, summing over $p(\pi_{2})$ in (9) gives:

[TABLE]

Repeating this last procedure but with weights over $p(\pi_{3})$ leads to

[TABLE]

However,

[TABLE]

Combining this last identity with (10) proves the result. ∎

The above proof is simple; it basically averages out each $LCS(\cdot,\cdot)$ as $\sqrt[3]{n}$ on the summation weighted by $P$. However, in view of the original conjecture, our partial results, as well as those mentioned in the introductory section, the cubic root lower-bound is not tight. Apart from our curiosity concerning this $\sqrt{n}$ conjecture, it would be interesting to know the exact asymptotic order of the smallest eigenvalue $\lambda_{1}^{(n)}$ of $L^{(n)}$. In contrast, the largest eigenvalue $\lambda_{n!}^{(n)}$ corresponding to the uniform distribution is known to be asymptotically of order $2n!\sqrt{n}$, since it is equal to the length of the $LISs$ of a uniform random permutation of $[n]$ scaled by $n!$. In this sense, the study of the length of the $LCSs$ between a pair of i.i.d.  random permutations having an arbitrary distribution, or equivalently, the study of $L^{(n)}$, can be viewed as an extension of the study of the length of the $LISs$ of a uniform random permutation of $[n]$. Having a complete knowledge of the distribution of all the eigenvalues of $L^{(n)}$ would be a nice achievement.