Water Juggling
Logic puzzles require you to think. You will have to be logical in your reasoning.
Here's what you have:
Two 8liter jugs, filled with water
One 3liter jug, empty
Four infinite size, empty pools
Here's what your objective is:
Fill each of the four pools with exactly 4 liters of water.
Here are your constraints:
You have nothing else at your disposal.
There is no other water aside from the two 8liter water filled jugs.
Once water is poured into any of the 4 pools it cannot be removed from there.
The jugs are not symmetric so you cannot measure amount by eye or judge based on shape.
HintI am not sure how to help you :). I solved this in 24 steps (you may have more or less). It helps to label the jugs and pools, and then draw. Oh, main key, try to work backwards, from filled pools, and see what final steps are even possible.
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Answer
It's not short but it's right  use a piece of paper and follow the steps using A, B, C and pools. Change the quantities in each as you complete each step  it is worth it to see it come out right.
Lets label the jugs.
Jug A  first 8liter
Jug B  second 8liter
Jug C  3liter.
The four infinites are pool 1, pool 2, pool 3, and pool 4.
1. Jug A to Jug C
2. Jug C to pool 1
3. Jug A to Jug C
4. Jug A to pool 2
5. Jug C to Jug A
6. Jug B to Jug C
7. Jug C to Jug A
8. Jug B to Jug C
9. Jug C to Jug A
At this point, we have:
Jug A  8
Jug B  2
Jug C  1
Pool 1  3
Pool 2  2
Pools 3&4  empty
10. Jug C to pool 3
11. Jug B to Jug C
12. Jug A to Jug C
13. Jug C to Jug B
14. Jug A to Jug C
15. Jug C to Jug B
16. Jug A to Jug C
17. Jug A to pool 4
At this point, we have:
Jug A  0
Jug B  6
Jug C  3
Pool 1  3
Pool 2  2
Pool 3  1
Pool 4  1.
18. Jug C to Jug B
19. Jug C to pool 1
20. Jug B to Jug C
21. Jug C to pool 3
22. Jug B to Jug C
23. Jug C to pool 4
24. Jug B to pool 2
... And we end up with the desired result:
Jug A  0
Jug B  0
Jug C  0
Pool 1  4
Pool 2  4
Pool 3  4
Pool 4  4
Tough, but workable.
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Comments
dewtell
Nov 18, 2002
 Nice one. The key is realizing that you can't
do very much when the jugs are mostly full or less
than half full. Most of the manipulations needed occur
when you have 10 or 11 liters total between the three
jugs (so you can use the topping off of an 8liter jug to
help achieve a particular measure). Looking at what
distributions of 10 and 11 liters can be achieved
can lead to the solution. 
LimnShicks2
Nov 18, 2002
 In every way this Logic puzzle was confusing but I still didn't get the right answer no mater what I did 
sakirski
Nov 19, 2002
 Thank you both for the comments. And yes, I realize that it is a bit confusing, but if you care to work through the steps in the solution you can see what you missed and why it worked out. 
just_moi
Nov 20, 2002
 hey there is an easier way to do this... u can fill the 8 l jug with water and then empty it out into the 3 l jug until its full... in htis way u end up with 5 l in the 8 l jug. If u throw away the water in the 3 l jug and fill it up again as before, you will end up with 53 l ie 2 litres in the 8 l jug. You can use this procedure ot obtain 2 l again and again and fill up the pools using it ( 2 X the water left in the 8l jug whihc is 2 l) dumped into each pool will give the required amt in each pool. sorry did i make sense?

sakirski
Nov 20, 2002
 In response to "just_moi": Your solution would be perfect if I had an unlimited amount of water. BUT, all I have is the two 8 liter jugs filled with water (as it says in the constraints). So all I have is exactly what I need for the pools 4x4 liters. If I throw any of it away, I can't fill the pools. Let me know if you have questions. 
big_papa_hfx
Nov 20, 2002
 1. Fill 3 litre jug from 8 litre jugleaving 5l in jug
2. Empty 3l
3. repeat so 5l3l =2l in the 8l jug
4. Do the same with the other jug leaving 2l in each jug.
2l + 2l = 4l
No infinite water supply required.

just_moi
Nov 21, 2002
 exactly! i wuz just about to say what big_papa_hfx said... 
sakirski
Nov 21, 2002
 Guys, "big_papa_hfx" and "just_moi" , you are not hearing me or the problem. You don't need to make 4 liters once, you need to make 4 liters 4 times. There are 4 pools to fill, each with 4 liters. If you loose any of the water from the 16 liters that you have, you cannot do it.
In your solution you can only fill one pool since you are getting rid of 3+3+3+3 which is 12 liters. Sure if you pour 12 liters out into nowhere you will fill one pool with the remaining 4 liters, but that will only solve 1/4 of the problem. And you do not have an extra container to put those 12 liters of water in to repeat the process. 
just_moi
Nov 21, 2002
 it does work out! once you make 2 l, you end up with both 3l jugs full... u can transfer the water in both of these to the other 8l jug. Then instead of filling the 1st 8l jug with fresh water form the source, you fill it with the 6 l in the other 8l jug + 2l of water from the source... and so forth... it would work then rite?

sakirski
Nov 21, 2002
 This is in response to your last comment "just_moi".
As it states in the given, there is ONLY ONE 3liter jug. There is also no avaliable water source (as it also states in the given).
I think the problem would be clearer to you if you were not dealing with water but rather with two 8liter jugs filled with "special cleaning acid". That can only be contained in either one of the two jugs or the third (originally) empty 3liter jug (just one jug). You also have 4 huge swimming pools with water. You need to clean them using exactly 4 liters of this special acid in each pool (so 4 liters for one, 4 for the other, and so on for a total of 16 liters, which is exactly how much you have). Obviously you cannot remove the acid from the pool once you place it there. I just thought that telling it like this would be more complicated, but maybe it helps assess the situation. Do you understand it now? 
stonemug
Nov 21, 2002
 This is similar to the situation given in the movie Die Hard 3 
just_moi
Nov 22, 2002
 lol... oops...i get it now...
nice teaser! 
sakirski
Nov 22, 2002
 Ok, good, I am glad I could clear it up (cause some times I confuse even myself
Oh and it is exactly like the die hard 3 problem, except that one was probably the easiest of it's type, and this is the hardest that I have seen. 
dsquared
Feb 13, 2003
 This is truly a legendary teaser. 
blue82785
Feb 17, 2003
 Hey, I can do this in half the steps, if u feel like having your iqs dwarfed feel free to post a comment and ask. I'll check for it the first day of March. Don't email me. If it makes u feel better, I got it in five minutes. 
gooblah
Feb 26, 2003
 great teaser I got the idea of it but didn't bother solving it all out soo tired, and sure send me an email, gooblah2@hotmail.com 
sakirski
Feb 26, 2003
 To 'blue82785'
I know for a fact that this problem can be solved in less steps than what I gave. But just to verify that your way is not wrong I'll take a look at it if you care to email it to me
accidentalentry@yahoo.com
Thanks 
naiem
Mar 12, 2003
 Good one! Kept me entertained for a while (did it in 26 steps) 
CNMNE
Jun 24, 2004
 Excellent teaser. I was getting very frustrated and starting to manually recede my hairline, but I finally solved it. 
(user deleted)
Dec 21, 2005
 Good One!!!
Here is my solution
5.8.3  0.0.0.0
5.8.0 > 3.0.0.0
2.8.3  3.0.0.0
0.8.3 > 3.2.0.0
3.8.0  3.2.0.0
3.5.3  3.2.0.0
6.5.0  3.2.0.0
6.2.3  3.2.0.0
8.2.1  3.2.0.0
8.2.0 > 3.2.1.0
5.2.3  3.2.1.0
7.0.3  3.2.1.0
7.3.0  3.2.1.0
4.3.3  3.2.1.0
4.6.0  3.2.1.0
1.6.3  3.2.1.0
1.8.1  3.2.1.0
1.8.0 > 4.2.1.0
1.5.3  4.2.1.0
1.5.0 > 4.2.4.0
1.2.3  4.2.4.0
1.0.3 > 4.4.4.0
4.0.0  4.4.4.0
0.0.0 > 4.4.4.4 
bfriedfischtom
Apr 16, 2008
 Great.
One day I gonna write a program to solve these and find the most difficult waterjuggling problem ever ;)
Used some EXCEL to speed up the solution process.
Here is another solution, different to the one above after the first 4200.
Also 24 steps.
The 3Jug is left here.
aBC**1234
088 0000
358>0000
058 3000
328>3000
308 3200
038 3200
335 3200
065 3200
362 3200
182>3200
082 4200
352 4200
370 4200
073 4200
343 4200
046 4200
316 4200
118>4200
018>4210
008 4211
305>4211
005 4241
302>4241
002>4244
000>4444 
Punk_Rocker
Nov 20, 2011
 I got it in 14 steps. Here's what I did:
8L 8L 3L  P1 P2
01) 8 8 0  0 0
02) 8 5 3  0 0
03) 8 5 0  3 0
04) 5 5 3  3 0
05) 5 5 0  3 3
06) 5 2 3  3 3
07) 7 0 3  3 3
0 7 3 0  3 3
09) 4 3 3  3 3
10) 4 6 0  3 3
11) 1 6 3  3 3
12) 0 6 3  4 3
13) 0 8 1  4 3
14) 0 8 0  4 4 
Punk_Rocker
Nov 20, 2011
 NOPE, JUST KIDDING. I thought it said two pools. It says four pools. Ignore me 
lvla
Apr 03, 2012
 Here is mine, I think its 23 steps. The hint I used is to try and make as much as possible 1liter transfers as you only have an easy 3 liter available to sum up to the 4 required.
8,8,0
8,5,3
8,5,0 3
8,2,3
8,0,3 3,2
8,3,0
5,3,3
5,6,0
2,6,3
2,8,1
2,8,0 3,2,1
0,8,2
0,7,3
3,7,0
3,4,3
6,4,0
6,1,3
6,0,3 3,2,1,1
8,0,1 3,2,1,1
8,0,0 4,2,1,1
5,0,3
5,0,0 4,2,4,1
2,0,3 > 4,4,4,4 
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