Hi! Getting Closer

So how close, and how quickly, can we get back to exactly one glass of water, adding and subtracting 50% of the total at each step. And what is happening with the “reverse of the square is the square of the reverse” property of 2012, 2011 and 2010?


  1. strauss said,

    January 16, 2012 at 3:32 pm

    … and of course 33^2 reversed is not, in fact, (the reverse of 33)^2….

  2. Byon said,

    January 16, 2012 at 5:23 pm

    Cool puzzle.  According to a quick Excel spreadsheet, you get within 0.21% after 53 adds and 31 subtracts.  That doesn’t get beat until it reaches 0.1% at 306 & 179.  Not sure how to extend that to molecules, though.

    Also, 2^49 is not very close to 3^31.   Are those the numbers you meant?  Thanks!

  3. strauss said,

    January 16, 2012 at 6:33 pm

    It’s not, is it! Well, what the heck did I mean? Thanks for checking!

    [Edit: Now, wait a sec. These were what I meant. It’s not “close” I guess, but it’s “closer”: 3^31/2^49 is about 1.097, which is about as close as you can get up to that size. Byon’s right of course, that you can get a lot closer going about twice as long…]

  4. strauss said,

    January 16, 2012 at 6:37 pm

    But you also give an important clue about the feasibility of actually accomplishing this… interesting how slowly the steps seem to get there!

  5. strauss said,

    January 18, 2012 at 9:24 am

    The first clue here is to factor the number of molecules involved at each step. In particular, how many 2’s there are in this number

    [spoiler] If we have, say, 1,000,000 molecules at one step, well, 1,000,000 = 2^6 x some junk; at the next step, whether or not we multiply by 1/2 or 3/2, we will have one less factor of 2, namely 2^5 * some other junk. So when we’re thinking about the physical problem, we’re limited by how many “factors of two we can get into a glass of water”

  6. Harry Kaplan said,

    January 18, 2012 at 9:14 pm

    On another note:  Chaim didn’t give the proof for last week’s wine-water mixing problem, because it was too much for a podcast.  After going down several blind alleys, I came up with the following proof of the impossibility of ever reaching 5 oz. wine and 5 oz. water in each bottle.  I’d appreciate knowing if I’m right or if I have erred.
    Let W(0) be the amount of wine in the wine bottle before the dual 3 oz. transfers, and let w(0) be the amount of wine in the water bottle at the same time.  Similarly, let W(1) and w(1) be the amount of wine in those two bottles after the two transfers are complete.  If the first transfer is from wine bottle to water bottle, it’s easy to derive:
          10/13 W(0) + 3/13 w(0) = W(1)
          10/13 w(0) + 3/13 W(0) = w(1)
    The problem is to determine whether we can ever end up with W(1) = w(1) = 5 after any series of dual transfers.  Assuming we can, let’s consider the final pair of transfers that created the desired outcome.  We can arbitrarily claim that the first of the transfers is out of the wine bottle.  The difficulty is that if we set W(1) = w(1) = 5, our equations tell us that W(0) = w(0) = 5 as well.  Therefore our assumption is wrong, and we can only end up with two identical bottles post-transfers if we started with two identical bottles.  Yes or no?

  7. Evan Romer said,

    January 24, 2012 at 1:22 am

    A generalization (?) of the mixing problem:

    I was a little confused by the mixing problem because of the phrase, “any sequence of transfers ….” Are the transfers always A-to-B followed by B-to-A, or can we use ANY sequence of transfers, such as A-to-B, then A-to-B again, then B-to-A, …? I think the former is what was intended (?); the latter is a more general question.

    I think I have a proof that even in this more general case, it impossible to ever get a 50-50 mixture. First note that if at any point one bottle has a 50-50 mixture, then the other bottle does also. (If A has 2 oz wine and 2 oz water, then B has 8 oz wine and 8 oz water.) So if one bottle is not 50-50, neither is the other. 

    Second, note that if, say, we’re transferring from A to B, the ratio of water to wine in A after the transfer will be the same as the ratio before the transfer. (For example, if A is 63% wine, and I scoop out 3 ounces, 63% of what’s left will be wine.) The ratio in the bottle I’m scooping from can’t be changed by the transfer. (The ratio in to bottle I’m transferring to can be changed.) So if the “from” bottle was not 50-50 before the transfer, it can’t be afterward. And if the “from” bottle is not 50-50 after the transfer, then neither is the “to” bottle.

    The only exception to the above argument is if I scoop out ALL of the liquid from one bottle: then I can’t say that if A was 63% wine before the transfer, it is still 63% wine afterward. If I manage to scoop out all of the liquid from A, then B will be a 50-50 mixture. But in this problem we have 10-ounce bottles and a 3-ounce scoop, so I can never empty a bottle. (If it were 12-ounce bottles and a 3-ounce scoop, then I could.)

    So no sequnece of scoops can result in a 50-50 mixture.

    (I think Harry Kaplan’s proof could be modified to cover this more general problem as well.)

    Evan Romer
    Windsor NY


  8. strauss said,

    January 24, 2012 at 5:55 pm

    ….yep, any desired sequence of transfers…

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