Splitting authentication codes with perfect secrecy: new results, constructions and connections with algebraic manipulation detection codes

TL;DR

This paper explores the use of combinatorial design theory to construct splitting authentication codes with perfect secrecy, establishing connections with algebraic manipulation detection codes and providing existence conditions for certain splitting BIBDs.

Contribution

It introduces group-generated splitting authentication codes with perfect secrecy and analyzes conditions for the existence of equitably ordered splitting BIBDs.

Findings

All group-generated authentication codes have perfect secrecy.

Algebraic manipulation detection codes are a special case of authentication codes with perfect secrecy.

Necessary and sufficient conditions are provided for the existence of certain splitting BIBDs.

Abstract

A splitting BIBD is a type of combinatorial design that can be used to construct splitting authentication codes with good properties. In this paper we show that a design-theoretic approach is useful in the analysis of more general splitting authentication codes. Motivated by the study of algebraic manipulation detection (AMD) codes, we define the concept of a group generated splitting authentication code. We show that all group-generated authentication codes have perfect secrecy, which allows us to demonstrate that algebraic manipulation detection codes can be considered to be a special case of an authentication code with perfect secrecy. We also investigate splitting BIBDs that can be "equitably ordered". These splitting BIBDs yield authentication codes with splitting that also have perfect secrecy. We show that, while group generated BIBDs are inherently equitably ordered, the concept is applicable to more general splitting BIBDs. For various pairs $(k,c)$, we determine necessary and sufficient (or almost sufficient) conditions for the existence of $(v, k \times c,1)$-splitting BIBDs that can be equitably ordered. The pairs for which we can solve this problem are $(k,c) = (3,2), (4,2), (3,3)$ and $(3,4)$, as well as all cases with $k = 2$.