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

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.


Introduction
The length of the longest increasing subsequences (LISs) of a uniform random permutation σ ∈ S n (where 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 σ 1 , σ 2 ∈ S n , and as long as σ 1 is uniformly distributed and regardless of the distribution of σ 2 , the length of the longest common subsequences (LCSs) of the two permutations is identical in law to the length of the LISs of σ 1 , i.e. LCS(σ 1 , σ 2 ) = L LIS(σ 1 ). This equality ensures, in particular, that when σ 1 and σ 2 are uniformly distributed, ELCS(σ 1 , σ 2 ) is upper bounded by 2 √ n, for any n, (see [4]) and asymptotically of order 2 √ n ( [5]). It is then rather natural to study the behavior of LCS(σ 1 , σ 2 ), when σ 1 and σ 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 S n . Let σ 1 and σ 2 be two i.i.d. permutations sampled from P . Then E P [LCS(σ 1 , σ 2 )] √ n. It might even be true that the uniform distribution U on 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.

Main Results
We begin with a few notations. Throughout, σ and π are, respectively, used for random and deterministic permutations. By convention, [n] := {1, 2, 3, ..., n} and so {π i } i∈[n!] = S n is a particular ordered enumeration of S n . (Some other orderings of S n will be given when necessary.) Next, a random permutation σ is said to be sampled from P = (p i ) i∈[n!] , if P P (σ = π i ) = p i . The uniform distribution is therefore U = (1/n!) i∈[n!] and, for simplification, it is denoted by E/n!, where E = (1) i∈[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., σ (n) and P (n) are respectively a random permutation and distribution from S n .
Let us now formalize the expectation as a quadratic form: where ℓ ij := LCS(π i , π j ) and L (n) := {ℓ ij } (i,j)∈[n!]×[n!] . It is clear that ℓ ij = ℓ ji and that ℓ 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: Proof. Linear algebra gives λ 1 . To reveal the connection between L (k+1) and L (k) , the enumeration of S k+1 is iteratively built on that of S k by inserting the new element (k + 1) into the permutations from 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 S k . ( Via this enumeration, the principal minor of size k! × k! is row and column indexed by the enumeration of the permutations {π (k) i } i∈[k!] from S k with (k + 1) as the last digit, i.e., {[π (k) i (k + 1)]} i∈[k!] ⊆ S k+1 . Then the (i, j) entry of the submatrix is LCS([π i (k + 1)], [π j (k + 1)]) = LCS(π i , π j ) + 1, since the last digit (k + 1) adds an extra element into the longest common subsequences.
since simultaneously relabeling π i and π j does not change the length of the LCSs and also since a particular relabeling to make π i to be the identity permutation, which is equivalent to left composition by π −1 i , is applied here. Further, any LCS of the identity permutation and of π −1 i π j is a LIS of π −1 i π j and vice versa. So the row sum is equal to since left composition by π −1 i is a bijection from S k to 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 π∈S k LIS(π) > 0 as its eigenvalue, which is distinct from the smallest eigenvalue λ (k) 1

0.
On the other hand, since L (k) is symmetric, the eigenvectors R (k) 1 and E (k) associated with the eigenvalues λ (k) 1 and π∈S k LIS(π) are orthogonal, i.e., Without loss of generality, let R (k) 1 be a unit vector, then from (2), As the electronic journal of combinatorics 25 (2018), #P00 where R denotes the corresponding Euclidean norm. Moreover, equality in (4) holds if and only if R . We show next, by contradiction, that this cannot be the case. Indeed, assume that 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 1 , which is a non-zero vector. However, on the right-hand-side, the corresponding bottom k! elements of the vector R 1 > λ 1 > λ 1 ...
The above result on the smallest negative eigenvalue, and its associated eigenvector, will help build a distribution on S n , for which the LCSs have a smaller expectation than for the uniform one.
Theorem 3. Let σ 1 and σ 2 be two i.i.d. random permutations sampled from a distribution P on the symmetric group S n . Then, for n 3, the uniform distribution U minimizes E p [LCS(σ 1 , σ 2 )], while, for n 4, U is sub-optimal.
Proof. As we have seen in (1), where P T LU = U T LU, since U is an eigenvector of L and P T U = 1.
However, when n 4, by Lemma 2, the smallest eigenvalue λ (n) 1 is strictly negative and the associated eigenvector R 1 . Hence, there exists a positive constant c such that cR (n) 1 −1/n!, where stands for componentwise inequality. Let P 0 be such that P 0 − U = cR (n) 1 , then it is immediate that Therefore, P 0 is a well-defined distribution on S n . On the other hand, by (6), the expectation under P 0 is such that However, the right-hand side of (7) is nothing but the expectation under the uniform distribution, namely, 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.

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 π∈Sn LIS(π). It is not hard to show that this eigenvalue is, in fact, the spectral radius of L (n) . Moreover, since the expectation of the longest increasing subsequence of a uniform random permutation is asymptotically 2 √ n, this gives an asymptotic order of −2n! √ n for the lower bound. On the other hand, we are interested in an upper bound for λ (n) 1 . The next result shows that λ (n) 1 decreases at least exponentially fast, in n.
Proof. This is proved by showing that λ (n+1) 1 2λ (n) 1 . As well known, Let λ (n) 1 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! × n! principal minor of L (n+1) is L (n) + EE T , while its bottom-left n! × n! submatrix is L (n) . Symmetrically, it can be proved that the top-right n! × n! submatrix is also L (n) , while the bottom-right n! × n! submatrix is L (n) + EE T , i.e., L (n+1) is Further, let the electronic journal of combinatorics 25 (2018), #P00 where, by an abuse of notation, E denotes the vector only made up of ones and of the appropriate dimension. Also, In (8), the corresponding numerator By a very similar method, it can also be proved, as shown next, that the second largest eigenvalue λ (n) n!−1 , which is positive, grows at least exponentially fast. Proof. Using the identity with a particular choice of n!−1 is the eigenvector associated with the second largest eigenvalue λ  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

Concluding Remarks
The √ n lower-bound conjecture of Bukh and Zhou is still open and seems quite reasonable in view of the fact that ELCS(σ 1 , σ 2 ) ∼ 2 √ n, in case σ 1 is uniform and σ 2 arbitrary (again, see [3]). We do not have a proof of this conjecture, but let us nevertheless present, next, a quick 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]).
We are now ready for the cubic root lower bound.
Combining this last identity with (10) proves the result.
The above proof is simple; it basically averages out each LCS(·, ·) as 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 √ n conjecture, it would be interesting to know the exact asymptotic order of the smallest eigenvalue λ (n) 1 of L (n) . In contrast, the largest eigenvalue λ (n) n! corresponding to the uniform distribution is known to be asymptotically of order 2n! √ 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.