It was a silly joke. A “regularizer” is a penalty term added to optimization problems to avoid overly elaborate solutions. A common regularizer in machine learning is the Lasso (not Lassi) which is the L_1 norm of the variables. C
On Apr 10, 2018, at 8:26 PM, James Propp <jamespropp@gmail.com> wrote:
I don’t understand what you mean. Can you please say more?
On Tuesday, April 10, 2018, Cris Moore <moore@santafe.edu> wrote:
People in machine learning often use the lassi as a regularizer. (I assume that has something to do with digestion.)
- Cris
On Mar 20, 2018, at 11:40 AM, James Propp <jamespropp@gmail.com> wrote:
I had dinner last night with Alan Frank (the inventor of the muffin problem that was discussed on math-fun a few years back and which Bill Gasarch will be talking about at the MIT Combinatorics Seminar tomorrow), and we encountered a problem about sharing that might turn out to be a sequel to the muffin problem, except that I don't quite know what the right problem is. (I solved the special case I needed to solve, but I'm not sure what the right general setting is.)
Dining with me and Alan were three kids. I ordered two mango lassis to split between myself and the three kids, along with two extra glasses, so that each of the four lassi-drinkers would get half a serving.
The two glasses containing (identical amounts of) mango lassi were the same shape as one another, and the two empty glasses were the same shape as one another, but the empty glasses were NOT the same shape as the other two glasses.
Not a problem! I took a full glass of lassi and divided it evenly between the two initially empty glasses, and then I took the remaining full glass and divided the lassi it contained evenly between it and the glass I'd drained in the first step.
But what if the two empty glasses had NOT been the same shape as each other? (We may assume that each is big enough to hold a full serving of mango lassi.)
PROBLEM: Show that even if the two empty glasses were different shapes from one another (and from the identical full glasses), one can still divide the lassi into four equal portions. The only allowed operation is to equalize the amount of lassi in two glasses of the same shape by equalizing the height of the lassi. (One might quibble that in practice you can only do this approximately, but for purposes of this puzzle, ignore that nicety.)
META-PROBLEM: What's the best way to alter and/or extend this problem, where "best" means "most fun"?
Jim Propp
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