A Levelset Algorithm for 3D-Tarski

TL;DR

We study the Tarski fixed-point problem for monotone $F: [N]^3 \rightarrow [N]^3$ and introduce a levelset-based algorithm that operates on level sets $L_k = \{ x \in G : |x| = k \}$ to locate points with $F(x) \le x$ or $F(x) \ge x$. The method maintains six bounding points on $L_k$ and classifies candidates into three configurations, ensuring progress in $O(\log N)$ per level, which yields a total $O(\log^2 N)$ time and query bound. This matches the best known 3D bound and, via a decomposition lemma, extends to general $d$, with potential applicability to restricted variants of Tarski. The approach offers a conceptually symmetric and potentially extensible framework for higher dimensions and related fixed-point problems.

Abstract

We present a simple new algorithm for finding a Tarski fixed point of a monotone function $F : [N]^3 \rightarrow [N]^3$. Our algorithm runs in $O(\log^2 N)$ time and makes $O(\log^2 N)$ queries to $F$, matching the $Ω(\log^2 N)$ query lower bound due to Etessami et al. as well as the existing state-of-the-art algorithm due to Fearnley et al.

Figures (6)

• Figure 1: Levelset $L_k$ of a 3D-instance (on the left) projected to 2D (on the right). On the right, the three dimensions are indicated in red, orange, and blue, respectively. For both points $x$ and $y$, we indicate the function $F$ in each dimension separately. For example, $x$ is an upward point with $F(x)_i > x_i$ for all $i \in [3]$ while $y$ satisfies $F(y)_1 < y_1, F(y)_3 > y_3$, and $F(y)_2 = y_2$ (indicated by the absence of an orange arrow).
• Figure 2: The three points $x, y, z$ are examples of $1$-upward points. Note that $x$ is in fact an overall upward point since it satisfies $F(x)_i \geq x_i$ for all $i \in [3]$.
• Figure 3: Example of the remaining search space $S$ (the region bounded by the black pentagon), induced by the six drawn bounding points. For simplicity, we only drew the arrow in dimension $i$ for each point $u^{(i)}$ or $d^{(i)}$. Observe in particular how $d^{(1)}$ does not currently contribute to the boundary of $S$, and in particular we have $\mathop{\mathrm{diam}}\nolimits(S)_1 < d^{(1)}_1 - u^{(1)}_1$. Still, there is a point $q$ such that querying $q$ guarantees significant progress: Either $q$ is upward or downward which allows us to terminate, or it is $i$-upward or $i$-downward for some $i \in [3]$ which allows us to considerably shrink the diameter in one of the three dimensions.
• Figure 4: Illustrations for \ref{['lemma:implied_configuration']}. On the left, the two points $u^{(1)}$ and $u^{(2)}$ are in first configuration. In the middle, the three points $u^{(1)}, u^{(2)}, u^{(3)}$ are in second configuration. On the right, the two points $u^{(1)}, d^{(1)}$ are in third configuration. For simpilicity, we only show the arrow in dimension $i$ for points $u^{(i)}$ and $d^{(i)}$.
• Figure 5: An example of the procedure that we use to deal with two points $x, y$ in third configuration. The point $r$ was initialized as the unique point with $r_1 = y_1$ and $r_2 = x_2 - 1$ while $\ell$ was initialized to be $y$ but then updated after one binary search step. Within $\mathcal{O}(\log N)$ steps, we will either find an upward or downward point, or shrink the interval between $\ell$ and $r$ to only contain two points, which again implies an upward or downward point, as desired.

Theorems & Definitions (17)

• Definition 3.1: Levelset
• proof
• Definition 3.3: $i$-Upward and $i$-Downward Points
• Definition 3.4: Remaining Search Space
• Lemma 3.5: Shrinking Remaining Search Space I
• proof
• Lemma 3.6: Shrinking Remaining Search Space II
• proof
• Definition 3.7: First Configuration
• Definition 3.8: Second Configuration