In this paper we study the relation of two fundamental problems in scheduling and fair allocation: makespan minimization on unrelated parallel machines and max-min fair allocation, also known as the Santa Claus problem. For both of these problems the best approximation factor is a notorious open question; more precisely, whether there is a better-than-2 approximation for the former problem and whether there is a constant approximation for the latter. While the two problems are intuitively related and history has shown that techniques can often be transferred between them, no formal reductions are known. We first show that an affirmative answer to the open question for makespan minimization implies the same for the Santa Claus problem by reducing the latter problem to the former. We also prove that for problem instances with only two input values both questions are equivalent. We then move to a special case called ``restricted assignment'', which is well studied in both problems. Although our reductions do not maintain the characteristics of this special case, we give a reduction in a slight generalization, where the jobs or resources are assigned to multiple machines or players subject to a matroid constraint and in addition we have only two values. This draws a similar picture as before: equivalence for two values and the general case of Santa Claus can only be easier than makespan minimization. To complete the picture, we give an algorithm for our new matroid variant of the Santa Claus problem using a non-trivial extension of the local search method from restricted assignment. Thereby we unify, generalize, and improve several previous results. We believe that this matroid generalization may be of independent interest and provide several sample applications. As corollaries, we obtain a polynomial-time (2−1/nǫ)-approximation for two-value makespanminimization for every ǫ > 0, improving on the previous (2 − 1/m) approximation, and a polynomial-time (1.75 + ǫ)-approximation for makespan minimization in the restricted assignment case with two values, improving the previous best rate of 1 + 2/√ 5 + ǫ ≈ 1.8945
