Jim Propp writes:

Scott Huddleston writes:

...

The muffin problem should really be considered in projective space r = m/p, and I'll henceforth usually write S(r) instead of S(m,p).

I don't understand; if we haven't proved that S(km,kp) = S(m,p) for all k, then how can this be right?

Even if there were cases with S(km,kp) < S(m,p), the non-projective view doesn't encourage and might even preclude asking about and exploring the (many) cases where the projective view S(r) has continuous intervals. For the original 2 partition problem, no one has seen a case where S(km,kp) < S(m,p), or suggested a plausible way one could arise. My ongoing 2 partition structure explorations might actually be approaching a complete solution. At this point I consider it extremely remote that S(km,kp) ever improves S(m,p). On the other hand, we've already seen examples S(ka,kb,kc) < S(a,b,c) in the 3 partition case. I expect the projective view will be the more insightful view even here. It's useful and helpful to normalize a < b < c in S(a,b,c), and we could consider projective S_0(r_1,r_2) = non-projective S(m, m*r_1, m*r_2), where the non-projective arguments are integers in lowest terms. But I expect the more interesting and "universal" projective quantity is S^*(r_1,r_2) = infimum_{k>0} S(k*m, k*m*r_1, k*m*r_2). It's surely the case, for given a<b<c, that S^*(b/a,c/a) = S(ka,kb,kc) for some finite k. I conjecture there's a single finite K_3 which achieves the S^* bound for all 3 partition instances, S*(b/a,c/a) = S(K_3*a, K_3*b, K_3*c), and similarly, for all j partition instances, a single finite K_j that achieves S^*(a_2/a_1, a_3/a_1, ..., a_j/a_1) = S(K_j*a_1, K_j*a_2, ..., K_j*a_j). I predict for the 3 partition case that we'll eventually find interesting regions and/or linear submanifolds where S^*(r_1,r_2) is continuous. Even stating this in a non-projective notation is clumsy at best. - Scott