The formulation of the muiffin problem was: you have m muffins (of equal size) and n people. You want to divide the muffins so that each person gets an equal share. In the original problem one sought such a division so that the smallest piece was as large as possible. In modifications of it, one looked at the sequence of piece sizes ordered from smallest to largest, and then one wanted the lexicographically largest such sequence (i.e. among all the sequences with largest smallest piece, one wanted to maximize the size of the next largest piece, etc.) On Fri, Dec 15, 2017 at 1:00 PM, Allan Wechsler <acwacw@gmail.com> wrote:
This is so similar in feel to the "muffin problem" that we talked about a couple of years ago, that I am convinced there must be a relationship between the problems. But I can't recall the original formulation of the muffin problem.
I think we can use reasoning like Tom Karzes's to show that f(5) > 7. Five fifths must be possible, and if we have only 7 pieces, then when the five fifths are arranged, there can be only two internal divisions. If those are in the same fifth, then we can't achieve four quarters; if they are in different fifths, then the three atomic fifths must be combined with 1/20 to get to 1/4, and there's no way to generate the required three 1/20 pieces.
On Fri, Dec 15, 2017 at 12:50 PM, Tom Karzes <karzes@sonic.net> wrote:
I think f(4) has to be 6. Consider that 1/4 + 1/4 + 1/4 + 1/4 must be possible. If f(4) is 5, then that means three of those 1/4 are indivisible, and the fourth is divisible once, into two fractions that sum to 1/4. So even if those two fractions are added to two of the other 1/4 values, it still leaves one bare 1/4 (and the other two couldn't both be increased to 1/3 anyway).
With 6 values, it's easy: 1/12, 1/12, 1/12, 1/4, 1/4, 1/4 works.
Tom
James Propp writes:
Perhaps my question has been considered in the past as a question about cutting an interval into pieces, since the circularity of the pie/pizza/whatever is irrelevant. (The very first radial cut effectively turns a problem about cutting a disk into wedges into a problem about cutting an interval into subintervals.)
Or maybe we should get rid of geometry entirely, and just ask: What is the smallest collection of fractions (with repetitions allowed), summing to 1, such that by combining fractions in the collection we can write 1 as 1/2 + 1/2, or as 1/3 + 1/3 + 1/3, or ..., or as 1/n + 1/n + ... + 1/n?
Let f(n) be the smallest possible number of such fractions. Clearly f(1) = 1 and f(2) = 2, and it's not hard to show that f(3) = 4. I haven't figured out f(4) (it's either 5 or 6). Has anyone seen this sequence before?
Jim
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