The problem D in the latest Open Cup involves function f(n) which is defined as the minimum sum of sequence b₁,...,b_k such that any sequence a₁,...,a_l with sum less than or equal to n can be dominated by some subsequence of b. It turns out that f(n) = n + f(k) + f(n − 1 − k) where k = [(n − 1) / 2]. If this equation still keeps you up at night, you can finally sleep well now. I have found a wonderful proof of this statement which fits the bounds of this site.

Two sides of the coin

Let us construct the b sequence first.

Theorem 1. Let B₁ be a sequence that covers all sequences with sum less than or equal to k. Let B₂ be a sequence that covers all sequences with sum less than or equal to l. Let n = k + l + 1. Then sequence B₁, n, B₂ covers all sequences with sum less than or equal to n.

Proof. Let a₁, ..., a_s be a sequence with sum less than or equal to n. Let t be the largest index such that a₁ + ... + a_t ≤ k. Then the first t elements of sequence a can be dominated by some subsequence of B₁. If t = s we don't need to go any further. Otherwise, the element a_t + 1 can be dominated by n and the rest of the sequence can be dominated by B₂ since its sum is less than of equal to n − (a₁ + ... + a_t + 1) ≤ n − k − 1 = l since the sum of first t + 1 elements of a is strictly greater than k by definition of t.

From Theorem 1 follows that f(n) ≤ min_k(n + f(k) + f(n − 1 − k)).

Simple enough, but how to prove the lower bound?

First of all, any covering sequence must cover sequence with single element a₁ = n and so must have element greater than or equal to n.

Theorem 2. Let B = B₁, m, B₂ be an arbitrary split of a sequence that covers all sequences with sum less than or equal to n. Let k be the largest number such that all sequences with sum less than or equal to k are covered by B₁. Let l be the largest number such that all sequences with sum less than of equal to l are covered by B₂. Then k + l ≥ n − 1.

Proof. Assume k + l ≤ n − 2. By definition of k there is sequence A₁ with sum k + 1 which is not covered by B₁. By definition of l there is sequence A₂ with sum l + 1 which is not covered by B₂. Let us consider sequence A₁, A₂. It has sum k + 1 + l + 1 ≤ n. Since A₁ is not covered by B₁, the latest element of A₁ must be covered by m or element to the right of it. By the same logic, the first element of A₂ must be covered by m or element to the left of it. But the latest element of A₁ must be covered by element of B to the left of element covering the first element of A₂. We arrive at the contradiction and so k + l ≥ n − 1.

If we split the sequence B at element greater than or equal to n, from Theorem 2 follows f(n) ≥ n + f(k) + f(l) where k + l ≥ n − 1. Since f(n) is non-decreasing we can set l = n − 1 − k and get f(n) ≥ min_k(n + f(k) + f(n − 1 − k)).

Combining results from both theorems, we have f(n) = min_k(n + f(k) + f(n − 1 − k)). QED.

Piece of convex cake