Implicit representation conjecture for semi-algebraic graphs

TL;DR

This paper investigates the implicit representation conjecture for semi-algebraic graphs, demonstrating limitations of naive coordinate storage and proposing a more efficient encoding method that improves upon trivial bounds.

Contribution

It provides a new encoding approach for semi-algebraic graphs that reduces bits per vertex below trivial bounds, advancing understanding of the conjecture.

Findings

Naive coordinate approximation storage is ineffective.

Proposed encoding requires O(n^{1-ε}) bits per vertex.

Results offer a slight improvement over trivial bounds.

Abstract

The implicit representation conjecture concerns hereditary families of graphs. Given a graph in such a family, we want to assign some string of bits to each vertex in such a way that we can recover the information about whether 2 vertices are connected or not using only the 2 strings of bits associated with those two vertices. We then want to minimise the length of this string. The conjecture states that if the family is hereditary and small enough (it only has $2^{O(n\ln(n))}$ graphs of size $n$), then $O(\ln(n))$ bits per vertex should be sufficient. The trivial bounds on this problem are that: (1) some families require at least $\ln_2(n)$ bits per vertex ; (2) $(n-1)/2+\ln_2(n)$ bits per vertex are sufficient for all families. In this paper, we will be talking about a special case of the implicit representation conjecture, where the family is semi-algebraic (which roughly means that the vertices are points in some euclidean space, and the edges are defined geometrically, or according to some polynomials). We will first prove that the `obvious' way of storing the information, where we store an approximation of the coordinates of each vertex, doesn't work. Then we will come up with a way of storing the information that requires $O(n^{1-\epsilon})$ bits per vertex, where $\epsilon$ is some small constant depending only on the family. This is a slight improvement over the trivial bound, but is still a long way from proving the conjecture.