On the impossibility of a perfect counting method to allocate the credits of multi-authored publications

Abstract

The problem on how to distribute the publication credits among ordered coauthors has been extensively discussed in the literature. However, there is no consensus about what is the most adequate procedure. This paper studies the properties of the existing counting methods and shows an impossibility result regarding the existence of a general counting method able to satisfy no advantageous merging and no advantageous splitting simultaneously—two properties that we consider fundamental. Our results suggest that the generalized variations of the geometric and the harmonic counting methods are the most flexible and robust in theoretical terms.

Appendix

Proof of Proposition 1

Following the discussion preceding Proposition 1, start by considering NAS (i.e., the lower bound). In this case, for $n=1$ we would have $c_{1}^{2}+c_{2}^{2}\leqslant c_{1}^{1},$ which is feasible because $c_{1}^{n}+\cdots +c_{n}^{n}=1,$ and we have $c_{1}^{2}\geqslant c_{2}^{2}\geqslant 0$. For $n=2,$ we would have $c_{2}^{3}+c_{3}^{3}\leqslant c_{2}^{2}$ and $c_{1}^{3}+c_{3}^{3}\leqslant c_{1}^{2},$ that would imply $c_{1}^{3}+c_{3}^{3}+c_{2}^{3}+c_{3}^{3}\leqslant c_{1}^{1}=1$ (by $c_{1}^{n}+\cdots +c_{n}^{n}=1$). This inequality is satisfied if $c_{3}^{3}=0.$ In this case, in order to satisfy OrM, we can have $c_{1}^{3}\geqslant c_{2}^{3}\geqslant c_{3}^{3}=0,$ but to satisfy CoM, we must also have $c_{2}^{3}=c_{2}^{2}$ and $c_{1}^{3}=c_{1}^{2}.$ For $n=3,$ we would have $c_{3}^{4}+c_{4}^{4}\leqslant c_{3}^{3}, c_{2}^{4}+c_{4}^{4}\leqslant c_{2}^{3}$ and $c_{1}^{4}+c_{4}^{4}\leqslant c_{1}^{3},$ that would imply $c_{1}^{4}+c_{4}^{4}+c_{2}^{4}+c_{4}^{4}+c_{3}^{4}+c_{4}^{4}\leqslant c_{1}^{1}=1$ (by $c_{1}^{n}+\cdots +c_{n}^{n}=1$). This inequality is satisfied if $c_{4}^{4}=0.$ In this case, in order to satisfy OrM, we can have $c_{1}^{4}\geqslant c_{2}^{4}\geqslant c_{3}^{4}\geqslant c_{4}^{4}=0,$ but to satisfy CoM, we must also have $c_{3}^{4}=c_{3}^{3}=0, c_{2}^{4}=c_{2}^{3}$ and $c_{1}^{4}=c_{1}^{3}.$ This construction can be generalized for every n, to obtain that any counting method satisfying NAS (i.e., the lower bound) is required to satisfy $c_{1}^{n}\geqslant c_{2}^{n}\geqslant 0$ and $c_{i}^{n}=0$ for $i>2$ with $c_{1}^{n}=c_{1}$ and $c_{2}^{n}=c_{2}$ constant for all $n\geqslant 2.$

Now, consider NAM (i.e., the upper bound). In this case, $n=1$ is not defined. For $n=2,$ we would have $c_{2}^{2}\leqslant c_{2}^{3}+c_{j}^{3}$ and $c_{1}^{2}\leqslant c_{1}^{3}+c_{j}^{3},$ where $j=2,3$ denote the possible values that $j>i$ can take. However, by OrM, the case that is most difficult to satisfy NAM occurs when $c_{j}^{3}$ takes the smallest value, i.e., at $j=3.$ In this case, the sum of the previous inequalities would imply $c_{1}^{1}=1\leqslant c_{1}^{3}+c_{3}^{3}+c_{2}^{3}+c_{3}^{3}$ (by $c_{1}^{n}+\cdots+c_{n}^{n}=1$). This inequality is always satisfied. The consideration of each of these two inequalities under the previously found NAS condition $c_{3}^{3}=0$ implies that $c_{2}^{2}\leqslant c_{2}^{3}$ and $c_{1}^{2}\leqslant c_{1}^{3},$ which by CoM require that $c_{2}^{2}=c_{2}^{3}$ and $c_{1}^{2}=c_{1}^{3}.$ Similarly, for $n=3,$ the restriction is imposed by each inequality $c_{3}^{3}\leqslant c_{3}^{4}+c_{4}^{4}, c_{2}^{3}\leqslant c_{2}^{4}+c_{4}^{4}$ and $c_{1}^{3}\leqslant c_{1}^{4}+c_{4}^{4},$ which must be satisfied with equality if the previously found NAS condition $c_{4}^{4}=0$ and CoM are satisfied. This argument can be generalized for every n, to obtain that any counting method satisfying NAM does not add further restrictions other than NAS. $\square$