You have a container of a capacity of 12L filled with water. Your task is to split that quantity (12L) of water into two equal quantities (6L) using only two additional containers: one of a capacity of 7L and another of a capacity of 4L.

How would you acheive that?

  • $\begingroup$ are the containers perfect cylinders? $\endgroup$ – Kevin Nov 24 '17 at 16:53
  • $\begingroup$ @Kevin not necessarly. They could be any shape. $\endgroup$ – ibrahim mahrir Nov 24 '17 at 16:55
  • 1
    $\begingroup$ oh well, I could do it in 1 step if they were. As is, I can't beat Apep's answer. $\endgroup$ – Kevin Nov 24 '17 at 16:59
  • $\begingroup$ @Kevin Feel free to post your answer. Just state the condition at the begining of the answer. (I won't promise to accept it as the correct one though haha). $\endgroup$ – ibrahim mahrir Nov 24 '17 at 17:01
  • $\begingroup$ Are we able to fill half of the 4L container? $\endgroup$ – Sigur Nov 24 '17 at 20:20

A 7 move solution:

Fill 7 from 12. (5, 7, 0)
Fill 4 from 7. (5, 3, 4)
Empty 4 into 12. (9, 3, 0)
Empty 7 into 4. (9, 0, 3)
Fill 7 from 12. (2, 7, 3)
Fill 4 from 7. (2, 6, 4)
Empty 4 to 12. (6, 6, 0)

  • $\begingroup$ Argh, bested by 30 seconds. You have won this round, Apep. $\endgroup$ – mr23ceec Nov 24 '17 at 15:05

IF the containers are perfect cylinders, you can do it in one step.

Pour slowly from the 12 to the 7. Stop when the waterline in the 12 goes from the top of one side to the exact bottom on the other, like in the picture below. This is a trick I use when cooking if I don't have the right size measuring cup available, but have one twice as big. E.g all my half-cup measuring cups are dirty but I have a one cup.

Here is a (badly made) image.

Badly made image

  • 1
    $\begingroup$ Poor useless 4L container. $\endgroup$ – ibrahim mahrir Nov 24 '17 at 17:21
  • $\begingroup$ Thinking about it not only perfect cylinders work for this. ِCubes, cuboids can acheive this too. $\endgroup$ – ibrahim mahrir Nov 24 '17 at 23:53

R(A) notation, where:

R= IV| VII| XII capacity

A= 0..12 filling (informational only)

-> direction of flow

XII(12)->VII(0), VII(7)->IV(0), IV(4)->XII(5), VII(3)->IV(0), XII(8)->VII(0), VII(7)->IV(3) and finally IV(4)->XII(2). This leaves us with XII(6) and VII(6).

  • $\begingroup$ This move XII(8)->VII(0) should be XII(9)->VII(0). It's hard to track those numbers representing them like that. $\endgroup$ – ibrahim mahrir Nov 24 '17 at 15:12

It's quite possibly not the fastest way, but here's my solution:

12    0   0    Base situation
 8    0   4  Fill the 4 liter container (C4) from the 12 liter container (C12)
 8    4   0  Empty C4 into C7
 4    4   4  Fill C4 from C12
 4    7   1  Fill C7 from C4 - 1L left
 11   0   1  Empty C7 into C12
 11   1   0  Empty C4 into C7
 EDIT: Everything above here is useless, obviously... I was just experimenting a bit and didn't pay enough attention.
 5    7   0  Fill up C7 from C12 
 5    3   4  Fill C4 from C7 - 3L left
 9    3   0  Empty C4 into C12
 9    0   3  Empty C7 into C4
 2    7   3  Fill C7 from C12 - 2L left
 2    6   4  Fill up C4 from C7
 6    6   0  Empty C4 into C12 - Done!

Where the three columns represent the amount of water in the 12, 7 and 4 liter container respectively.


Another non-standard answer (inspired by the perfect cylinders.)

  • 4L fits entirely within 7L
  • The top of 4L will create a water-tight seal with the interior of 7L's bottom when placed upside-down inside 7L
  • The thickness of the walls of 4L is zero
  • You have a zero volume tool that allows you to reach down to the bottom of 7L when it is full of water and break the aforementioned seal that 4L makes with the interior of 7L's bottom without spilling any water over the edge.

The solution:

  1. Place 4L upside-down inside 7L
  2. Pour 12L into 7L (only 3L will be poured because 4L is inside taking up the remaining space)
  3. Using the zero volume tool, reach into 7L and break the seal between 4L and the bottom -- then remove 4L from 7L and place it upright.
  4. Pour the 3L from 7L into 4L
  5. Take 7L and place it upside-down on top of 4L such that interior of 7L's bottom seals against the top of 4L -- then invert the pair so that 7L is right-side-up.
  6. Pour 12L into 7L (only 3L will be poured because 4L is inside taking up the remaining space)
  7. Using the zero volume tool, reach into 7L and break the seal between 4L and the bottom -- then remove the empty 4L from 7L and discard.

Final state will be 6L in 12L and 6L in 7L.

Before I wrote this out I thought it might require fewer steps than the official answer, but, alas, it too requires 7 steps (plus some requirements that are a little far-fetched.)


Just for fun, here is a different solution to the one given by Apep:

The jars are as follows: (12L, 4L, 7L)



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