[math-fun] The mango lassi problem (and meta-problem)
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 Jim Propp
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.)
I presume there is a second allowed operation, which is to transfer all of the fluid from one glass into another (irrespective of the shape of the source and target glasses), provided doing so does not cause an overflow error in the target glass...? -- APG.
Oh yeah, I should've included that operation in the repertoire of permitted moves. One might also wish to allow a version of this move in which the second glass is smaller than the first, and one is allowed to fill the second glass. (As in those well-known problems in which one implements a constrained version of the Euclidean algorithm by passing fluid back and forth between containers.) Jim Propp On Tue, Mar 20, 2018 at 2:59 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
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.)
I presume there is a second allowed operation, which is to transfer all of the fluid from one glass into another (irrespective of the shape of the source and target glasses), provided doing so does not cause an overflow error in the target glass...?
-- APG.
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Andy Latto asks whether the operation "pour from glass B into glass A until A is at the same level as A' (where A and A' are identical)" is permitted; yes, I should have included it. Jim Propp On Tue, Mar 20, 2018 at 3:11 PM, James Propp <jamespropp@gmail.com> wrote:
Oh yeah, I should've included that operation in the repertoire of permitted moves.
One might also wish to allow a version of this move in which the second glass is smaller than the first, and one is allowed to fill the second glass. (As in those well-known problems in which one implements a constrained version of the Euclidean algorithm by passing fluid back and forth between containers.)
Jim Propp
On Tue, Mar 20, 2018 at 2:59 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
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.)
I presume there is a second allowed operation, which is to transfer all of the fluid from one glass into another (irrespective of the shape of the source and target glasses), provided doing so does not cause an overflow error in the target glass...?
-- APG.
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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
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Thanks, Cris. Inasmuch as it contains dairy, I don’t think lassi would make a very good regularizer. :-) Jim On Wednesday, April 11, 2018, Cris Moore <moore@santafe.edu> wrote:
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
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (3)
-
Adam P. Goucher -
Cris Moore -
James Propp