On Finding Constrained Independent Sets in Cycles

07/01/2023
by   Ishay Haviv, et al.
0

A subset of [n] = {1,2,…,n} is called stable if it forms an independent set in the cycle on the vertex set [n]. In 1978, Schrijver proved via a topological argument that for all integers n and k with n ≥ 2k, the family of stable k-subsets of [n] cannot be covered by n-2k+1 intersecting families. We study two total search problems whose totality relies on this result. In the first problem, denoted by 𝖲𝖼𝗁𝗋𝗂𝗃𝗏𝖾r(n,k,m), we are given an access to a coloring of the stable k-subsets of [n] with m = m(n,k) colors, where m ≤ n-2k+1, and the goal is to find a pair of disjoint subsets that are assigned the same color. While for m = n-2k+1 the problem is known to be 𝖯𝖯𝖠-complete, we prove that for m < d ·⌊n/2k+d-2⌋, with d being any fixed constant, the problem admits an efficient algorithm. For m = ⌊ n/2 ⌋-2k+1, we prove that the problem is efficiently reducible to the 𝖪𝗇𝖾𝗌𝖾𝗋 problem. Motivated by the relation between the problems, we investigate the family of unstable k-subsets of [n], which might be of independent interest. In the second problem, called Unfair Independent Set in Cycle, we are given ℓ subsets V_1, …, V_ℓ of [n], where ℓ≤ n-2k+1 and |V_i| ≥ 2 for all i ∈ [ℓ], and the goal is to find a stable k-subset S of [n] satisfying the constraints |S ∩ V_i| ≤ |V_i|/2 for i ∈ [ℓ]. We prove that the problem is 𝖯𝖯𝖠-complete and that its restriction to instances with n=3k is at least as hard as the Cycle plus Triangles problem, for which no efficient algorithm is known. On the contrary, we prove that there exists a constant c for which the restriction of the problem to instances with n ≥ c · k can be solved in polynomial time.

READ FULL TEXT

Please sign up or login with your details

Forgot password? Click here to reset
Success!
Error Icon An error occurred

Sign in with Google

×

Use your Google Account to sign in to DeepAI

×

Consider DeepAI Pro