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No matter...whether Ascending or Descending order



Hi dear friends...I missed you so much....I love you all... :wub:

I have today a nice puzzle for you...hope to enjoy solving it ^_^ ..

We have 4 beakers( A,B,C, and D)

(A) can take only 3 gallons of water,(B) can take 4 gallons...they are both empty at start position.

© and (D) have the same size , each can take exactly 5 gallons, both are full with water to start with.

Use these beakers only to make an ascending or descending order of water amount in each beaker...

i.e. , either...1 gallon, 2,3,and 4 gallons,

or 4 gallons, 3,2 and one gallon...

Have a nice time.. :thumbsup:

Edited by wolfgang
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Can we make the assumption that the beakers are either perfect cylinders or perfect rectangular prisms??

If we can assume that, then this can be done in 5 steps (might be better, I just haven't spent time looking) because of the fact that we can easily pour out enough liquid in any beaker to leave it EXACTLY half full (just start pouring, carefully, until the bottom corner of the beaker is level with the water):

A0, B0, C5, D5 -- Initial configuration

  1. A0, B4, C1, D5 -- Pour C into B, filling B
  2. A0, B2, C3, D5 -- Pour EXACTLY half of B back into C using the method above
  3. A3, B2, C3, D2 -- Pour D into A, filling A
  4. A1, B4, C3, D2 -- Pour A into B, filling B
  5. A1, B2, C3, D4 -- Pour EXACTLY half of B into D using the method above.

** Note the wording that I used above about leaving half of the liquid in the beaker...not pouring half of the liquid out. This is because if we had 3 gallons in B, and used the method I described above, we would still be left with 2 gallons in B, and only poured 1 out.

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I don't think it can be done.

assuming we can only pour the total amount of liquid a container can take e.g. if A can take 3 and has 1 g in it, then i must pour 2 gal into the container (not one, 1/2, etc.).

To get to the ending state of A1 ,B2 , C3 , D4, then A must have poured into another container since it is the smallest container and this would not have the issue of overfilling the other containers. But to have accomplished this and still keep 1gallon in the container, the other containers must have been full. As we can see, neither B2, C3, and D4 are full. Meaning that A would have been completely empty and not ended with 1 gallon. Therefore A could not have been the last pouring container. But the last idea can be extended to any of the other containers showing that if any of them poured into another container then that container must have filled up the others.

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hmmm @Pickett i guess i assumed that the containers were unmarked and only the total amount was known

Even unmarked containers can be poured out exactly half way if they are cylinders or rectangular prisms...so as long as you know the total volume (which we do), my steps still work.

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But how can you be sure that it is exactly half? I mean there will some lost of precision due to estimation.

I'd say there's as much loss of precision as there is in pouring from one container to the other (there's always left over water in the original container...even if it's just drops)...

I fully admit that my method has a number of built in assumptions:

  1. My original assumption that the beakers are perfect cylinders/rectangular prisms
  2. The containers hold EXACTLY the number of gallons represented...not a drop more, nor less...(of course, this causes problems in that the second you MOVE a container to pour into another, you would lose some...unless you had an infinitely steady hand...)
  3. You have a perfect eye and can tell the exact second the bottom corner is visible and stop pouring at exactly that time
  4. I'm treating this as a purely mathematical problem...meaning I'm ignoring the surface tension of water, any distortion caused by the beakers in your reading, etc..

Definitely not a perfect answer, but I'd say you could get "close enough" with it...much closer than just guessing the amounts :thumbsup:

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