Chvátal–Sankoff constant for a binary alphabet

Description of constant

Let $\lambda_{n,2}$ be the random variable assigning two uniformly random binary strings of length $n$ the length of their longest common subsequence. Then $C_{31a}$ is the (well-defined) limit $C_{31a} := \lim_{n \to \infty}\frac{\mathbb{E}[\lambda_{n,2}]}{n}$.

Known upper bounds

Bound Reference Comments
$1$ Trivial  
$0.837623$ [DP1995]  
$0.826280$ [L2009] Computer assisted
$0.825101$* [J2026] Computer assisted. Lueker’s automaton $M_h$ of [L2009] (Section 3.2) run with history length $h=9$: 51,425,006 accepting states, $z=0.3298$, certified $\lambda(z)\le 0.364708361184292$ (upward rounding), bound $2\log(2z)/\log\lambda(z)$. The same code reproduces Table II of [L2009] for $h\le 8$.
$0.824120$* [J2026] Same method with $h=10$: 400,253,317 accepting states, $z=0.3376$, certified $\lambda(z)\le 0.385532705434511$.

Known lower bounds

Bound Reference Comments
$0$ Trivial  
$> 0$ [CS1975] Showed existence of limit
$0.773911$ [D1994] Computer assisted
$0.788071$ [L2009] Computer assisted
$0.792665992$ [H2024] Computer assisted
$0.79970$ [AA2026] AI and computer assisted (unverified). Monte Carlo with a Hoeffding margin: the mean length of a width-100 beam-search common subsequence over $10^6$ random pairs of binary strings of length $1000$ is $0.803417$, minus $0.003717$ from Hoeffding’s inequality with $\delta = 10^{-12}$. So the statement holds with probability $1 - 10^{-12}$ over the sampled strings. It is not a deterministic lower bound.

Additional comments

References

Contribution notes

The [J2026] rows were prepared with the help of a large language model; the computation was checked as described in the linked repository.