Hi muffin choppers, I've been thinking about the muffin problem, too, though I've only just read most of the posts, since I wanted to work things out for myself. Before I talk specifics, I'll say say that I'm using Veit's U(m, n) notation (with each muffin having size n, which is the same as looking at a partition of mn that refines both mn = m + m + ... + m and mn = n + n + ... + n). I agree that, from a mathematical point of view, Mike Reid's T(m, n) is more natural, but it makes the denominators much bigger and seems to needlessly complicate the arguments. And I don't like the asymmetry of S(m, p). Also, I'll point out that U(m, n, p), for instance, can be viewed as a partition of mnp. Another thing, I don't like Scott's "piece optimal" or "ratio optimal" as secondary objectives, since they're sort of ad hoc and still don't give a unique optimum. I figure that, since the primary objective is to maximize the size of the smallest piece, the next objective should be to maximize the size of the second smallest piece, and so on. That is, I want the lexically largest multiset of pieces, when arranged in increasing order. Then there is always a unique optimal set of pieces. Finally, I prefer to write multiplicities _after_ parts, so I write "(1/2) * 2" to mean two parts of size 1/2. Now I came up with essentially the same arguments for computing U(3, 3k+-1) that Mike Reid did (for T), but I have a much simpler proof for T(4, 2k+1) = (2k-1)/k (with k>=1) than his: Recall that we want a corefinement of 4 * (2k+1) and (2k+1) * 4. We have the partition 4(2k+1) = (2k-1)/k * 2k + 2 * 2 + (2k+1)/k * 2k, which we want to prove is optimal (in my strong sense above). We may assume k>=2, since the case k=1 is easy. Then U(4, 2k+1) >= (2k-1)/k > 4/3, so no 4 can be divided into more than two pieces, which means that there are at most 2(2k+1) pieces. Then one of the four 2k+1's must be divided into at most (2k+1)/2 pieces, meaning at most k, one of which has size at least (2k+1)/k. This piece is part of a 4, so the other part of that 4 has size at most 4 - (2k+1)/k = (2k-1)/k, which proves U(4, 2k+1) = (2k-1)/k. Now for the optimal solution, we must have equality, which means that 2k+1 is divided into exactly k pieces, which means a second 2k+1 is divided into k equal pieces. So now we are forced to have (2k-1)/k * 2k and (2k+1)/k * 2k, with the latter bunch the constituents of two of the 2k+1's. Now these account for 8k of the total 8k+4, so the best possibilies for the rest would be 4 or 2 * 2. In either case, these pieces have integer sizes, so when making the other two 2k+1's, we have to use k of the (2k-1)/k's at a time, giving 2k-1. Then we can't use the 4, so 2 * 2 is optimal, proving the result. I think I have the optimal partition for all cases of (m, m+1), and I have calculated optimal partitions for many of the values of U(m, n) given earlier, but I'll give those later. David P. Moulton