On the set of elasticities in numerical monoids

Abstract

In an atomic, cancellative, commutative monoid $S$, the elasticity of an element provides a coarse measure of its non-unique factorizations by comparing the largest and smallest values in its set of factorization lengths (called its length set). In this paper, we show that the set of length sets $\mathcal L(S)$ for any arithmetical numerical monoid $S$ can be completely recovered from its set of elasticities $R(S)$; therefore, $R(S)$ is as strong a factorization invariant as $\mathcal L(S)$ in this setting. For general numerical monoids, we describe the set of elasticities as a specific collection of monotone increasing sequences with a common limit point of $\max R(S)$

Factoring in the Chicken McNugget Monoid

“…What other ways are there to order that same number of Chicken McNuggets? Well, using the relation (3, 0, 0) ∼ (0, 2, 0), we obtain at least 3 more ways, yielding (10, 0, 3), (7, 2, 3), (4,4,3), and (1,6,3).…”

“…Intuitively, elasticity measures how "spread out" a monoid's factorization lengths are. The interested reader can find numerous papers in the recent literature that study problems related to elasticity, both in numerical monoids [4,5,14] and more broadly [2]. The delta set of a McNugget number n is defined by…”

“…Starting with the initial factorization (10, 0, 3), we can trade 6's for 20's to obtain (0, 0, 6). Moreover, we can instead trade 20's for 6's in (10, 0, 3) and obtain (20, 0, 0), (17, 2, 0), (14,4,0), (11,6,0), (8,8,0), (5, 10, 0), and (2,12,0) by subsequently trading 6's for 9's using (3, 0, 0) ∼ (0, 2, 0). These turn out to be the final factorizations of 120 ∈ .…”

“…These turn out to be the final factorizations of 120 ∈ . Had we instead chosen to use the relation (4, 4, 0) ∼ (0, 0, 3), we can still obtain the factorization (0, 0, 6), this time starting with (4,4,3) and trading all of the 6's and 9's for 20's, and the remaining factorizations in the centered expression above can be obtained by swapping out the 20's in (10, 0, 3), and then once again repeatedly applying (3, 0, 0) ∼ (0, 2, 0).…”

“…Notice that if we want to go from (10, 0, 0) to (4,4,0), the common part is (4, 0, 0) and the new "bridge" is 6 · 6 = 4 · 9 = 36, with factorizations Z(36) = {(6, 0, 0), (3, 2, 0), (0, 4, 0)}.…”