On Finding Constrained Independent Sets in Cycles

Abstract

A subset of $[n] = \{1,2,\ldots ,n\}$ 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 $n \ge 2k$, 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 $\textsc {Schrijver}(n,k,m)$, we are given an access to a coloring of the stable k-subsets of [n] with $m = m(n,k)$ colors, where $m \le n-2k+1$, 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 $\textsf{PPA}$-complete, we prove that for $m < d \cdot \lfloor \frac{n}{2k+d-2} \rfloor$, with d being any fixed constant, the problem admits an efficient algorithm. For $m = \lfloor n/2 \rfloor -2k+1$, we prove that the problem is efficiently reducible to the $\textsc {Kneser}$ 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 $\ell$ subsets $V_1, \ldots , V_\ell$ of [n], where $\ell \le n-2k+1$ and $|V_i| \ge 2$ for all $i \in [\ell ]$, and the goal is to find a stable k-subset S of [n] satisfying the constraints $|S \cap V_i| \le |V_i|/2$ for $i \in [\ell ]$. We prove that the problem is $\textsf{PPA}$-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 $n \ge c \cdot k$ can be solved in polynomial time.

On the Tightness of Corollary 5.5

We prove the following result.

Theorem A.1

For all integers n and k such that n is congruent to 1 modulo 4 and $n \ge 2k$,

$$\begin{aligned} \chi (U(n,k)) = \min (n-2k+2, \lceil n/2 \rceil ). \end{aligned}$$

Theorem A.1 shows that the upper bound in Corollary 5.5 is tight whenever n is congruent to 1 modulo 4. Its proof relies on the following lemma whose proof resembles that of Lemma 5.4.

Lemma A.2

For integers n and k for which it holds that n is congruent to 1 modulo 4 and $n \ge 2k$, let $t = \min (n-2k+2,\lceil n/2 \rceil )-1$. Then, there exist n points $y_1,\ldots ,y_n \in {{{\mathbb {S}}}}^t$ such that for every hyperplane h in ${{\mathbb {R}}}^{t+1}$ that passes through the origin, at least one of the two open hemispheres that h determines contains the points of $\{y_i \mid i \in A\}$ for some vertex A of U(n, k).

Proof

Let n, k, and t be integers as in the statement of the lemma. Observe that the assumption that n is congruent to 1 modulo 4 implies that t is even. Let $\gamma : {{\mathbb {R}}}\rightarrow {{\mathbb {R}}}^{t+1}$ denote the function defined by $\gamma (x) = (1,x,x^2,\ldots ,x^t)$. For every $i \in [n]$, consider the point $w_i = \gamma (i) \in {{\mathbb {R}}}^{t+1}$. It suffices to show that for every hyperplane h in ${{\mathbb {R}}}^{t+1}$ that passes through the origin, at least one of the two open half-spaces that h determines contains the points of $\{w_i \mid i \in A\}$ for some vertex A of U(n, k).

Consider an arbitrary hyperplane h in ${{\mathbb {R}}}^{t+1}$ that passes through the origin. Every point $w_i$ either lies on h or belongs to one of the two open half-spaces determined by h. Let $W_{on}$ denote the set of indices $i \in [n]$ for which the point $w_i$ lies on h, and let $W_1$ and $W_2$ denote the sets of indices $i \in [n]$ of the points $w_i$ that belong to the two open half-spaces determined by h. Our goal is to show that at least one of the sets $W_1$ and $W_2$ contains a vertex of U(n, k).

The definition of the points $w_1,\ldots ,w_n$ implies that the size of $W_{on}$ does not exceed the number of roots of some nonzero polynomial of degree at most t, hence $|W_{on}| \le t$, and it can be assumed that $|W_{on}|=t$. The points $w_i$ with $i \in W_{on}$ divide the image $\mathop {\textrm{Im}}(\gamma )$ of the function $\gamma$ into $t+1$ open continuous parts that alternate between the two open half-spaces determined by h. Since t is even, the first and last parts lie on the same open half-space determined by h. We merge these two parts into a single part. Observe that all the indices i of the points $w_i$ that belong to each of the obtained t parts of $\mathop {\textrm{Im}}(\gamma )$ are either in $W_1$ or in $W_2$. It follows that each of the sets $W_1$ and $W_2$ is associated with t/2 of these parts.

Suppose without loss of generality that $|W_1| \ge |W_2|$. We claim that $W_1$ contains a vertex of U(n, k). To this end, we show that it includes two consecutive elements modulo n and has size at least k. Indeed, it holds that $|W_1| > t/2$, as otherwise $n-t = |W_1|+|W_2| \le t$, which implies that $n \le 2 t$. Since n is odd, this contradicts the definition of t which guarantees that $t \le \lceil n/2 \rceil -1$. It thus follows that at least one of the parts of $\mathop {\textrm{Im}}(\gamma )$ associated with $W_1$ contains at least two of the points $w_1,\ldots , w_n$, hence $W_1$ includes two consecutive elements modulo n. Further, by the definition of t, we have $t \le n-2k+1$, which implies that $|W_1| \ge \lceil \frac{n-t}{2} \rceil \ge \lceil \frac{2k-1}{2} \rceil = k$, completing the proof. $\square$

Proof of Theorem A.1

Let n and k be integers such that n is congruent to 1 modulo 4 and $n \ge 2k$. The upper bound on $\chi (U(n,k))$ follows from Corollary 5.5. For the lower bound, let

$$\begin{aligned} t = \min (n-2k+2, \lceil n/2 \rceil ) - 1, \end{aligned}$$

and suppose for the sake of contradiction that there exists a proper coloring of U(n, k) with t colors. Let $y_1,\ldots ,y_n \in {{{\mathbb {S}}}}^t$ denote the points given by Lemma A.2. We define t sets $F_1,\ldots ,F_t \subseteq {{{\mathbb {S}}}}^t$ as follows. A point $x \in {{{\mathbb {S}}}}^t$ is included in $F_j$ with $j \in [t]$ if there exists a vertex A of U(n, k) colored j such that $\{ y_i \mid i \in A\} \subseteq H(x)$, where $H(x) = \{ z \in {{{\mathbb {S}}}}^t \mid \langle x,z\rangle >0 \}$. We further define $F_{t+1} = {{{\mathbb {S}}}}^t \setminus (F_1 \cup \cdots \cup F_t)$. Note that the sets $F_1,\ldots ,F_{t+1}$ cover ${{{\mathbb {S}}}}^t$. By Theorem 5.3, there exist an index $j \in [t+1]$ and a point $x \in {{{\mathbb {S}}}}^t$ such that both x and $-x$ belong to $F_j$. If $j \in [t]$, then it follows from the definition of $F_j$ that there exist two vertices of U(n, k) with color j that correspond to disjoint sets, contradicting the assumption that the given coloring is proper. If $j = t+1$ then neither H(x) nor $H(-x)$ contains $\{ y_i \mid i \in A\}$ for a vertex A of U(n, k), contradicting Lemma A.2. This completes the proof. $\square$