Brain Teasers
Water Juggling
Here's what you have:
-Two 8-liter jugs, filled with water
-One 3-liter 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 8-liter 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.
-Two 8-liter jugs, filled with water
-One 3-liter 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 8-liter 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.
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Comments
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 8-liter jug to
help achieve a particular measure). Looking at what
distributions of 10 and 11 liters can be achieved
can lead to the solution.
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 8-liter jug to
help achieve a particular measure). Looking at what
distributions of 10 and 11 liters can be achieved
can lead to the solution.
In every way this Logic puzzle was confusing but I still didn't get the right answer no mater what I did
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.
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 5-3 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? 
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.
1. Fill 3 litre jug from 8 litre jugleaving 5l in jug
2. Empty 3l
3. repeat so 5l-3l =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.
2. Empty 3l
3. repeat so 5l-3l =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.
exactly! i wuz just about to say what big_papa_hfx said...
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.
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.
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?
This is in response to your last comment "just_moi".
As it states in the given, there is ONLY ONE 3-liter 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 8-liter jugs filled with "special cleaning acid". That can only be contained in either one of the two jugs or the third (originally) empty 3-liter 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?
As it states in the given, there is ONLY ONE 3-liter 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 8-liter jugs filled with "special cleaning acid". That can only be contained in either one of the two jugs or the third (originally) empty 3-liter 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?
This is similar to the situation given in the movie Die Hard 3
lol... oops...i get it now...
nice teaser!
nice teaser!
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.
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.
This is truly a legendary teaser.
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.
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
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 e-mail it to me
accidentalentry@yahoo.com
Thanks
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 e-mail it to me
accidentalentry@yahoo.com
Thanks
Good one! Kept me entertained for a while (did it in 26 steps)
Excellent teaser. I was getting very frustrated and starting to manually recede my hairline, but I finally solved it.
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
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
Great.
One day I gonna write a program to solve these and find the most difficult water-juggling 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 3-Jug 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
One day I gonna write a program to solve these and find the most difficult water-juggling 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 3-Jug 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
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
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 -- 3 309) 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
NOPE, JUST KIDDING. I thought it said two pools. It says four pools. Ignore me
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
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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