Rumors swirl about a clandestine group within the festival, known only as The Curators. Allegedly responsible for Missax’s enigmatic design, they operate from a hidden chamber called the “Vault of 372,” referencing the festival’s ancient origin date. While organizers deny its existence, many point to recurring symbols in the festival—spirals, three-star patterns, and riddles etched into lanterns—as clues left by this elusive collective.
Missax occurs biennially and is invitation-only, with applications accepted via a cryptic online riddle. Prospective attendees are advised to:
The algorithm works for both increasing and decreasing variants; the final answer is the maximum of the two.
In the Missax community, episode 372 is known as the "Tattoo Tour." The first five minutes of the video are dedicated to a slow pan of the model’s ink. For fans of alternative body art, this specific episode is considered a gold standard. 372. Missax
Title (Unofficial): The Rainy Day Tattoo Check Runtime: Approximately 25 minutes. Setting: A dimly lit, messy bedroom with neon LED strip lighting (green and pink hues). Plot Context: The scene opens with the model looking out a window at a storm. She is wearing oversized band merch and thigh-high socks. The "plot" is minimal—she is bored, checks her phone, and then begins to explore her own body while showing off her extensive chest and arm tattoos.
We prove by induction on i that the tree always stores, for every possible value v, the length of the longest Missax‑feasible subsequence that ends with v and uses only elements among the first i positions.
Base: For i = 1, the tree contains the single entry (1, a₁), which is trivially optimal. Rumors swirl about a clandestine group within the
Inductive step: Assume the invariant holds after processing a₁,…,a_i‑1.
When processing a_i, any feasible extension must end with a value v that satisfies the monotonicity and Δ‑gap constraints. By the inductive hypothesis, the tree already stores the optimal length ℓ(v) for each such v. The RMQ returns
[ ℓ_\max= \max_v\text feasible ℓ(v) . ]
Appending a_i yields a subsequence of length ℓ_\max+1. No longer subsequence ending at a_i can exist, because any such subsequence must have a predecessor that is feasible and therefore its length is bounded by ℓ_\max. Updating the tree with (ℓ_\max+1, a_i) preserves the invariant. ∎ The algorithm works for both increasing and decreasing
Find a subsequence B* that satisfies the Missax constraint and maximises its length k. Equivalently, minimise the number of deletions
[ \operatornamedel(A) = n - |B*| . ]
The decision version asks: Given an integer K, does there exist a Missax‑feasible subsequence of length at least K?