# Water jug for gaining 8 gallons of water? [duplicate]

I have gone through simple water jug problems like this link but when there are 3 jugs:

• 16 Galoon jug with water filled.
• 11 Galoon empty jug.
• 6 Galoon empty jug.

Find a solution to gain 8 Galoon, using these 3 jugs.

Assuming you may dispose of water...

Call the jugs A, B, C respectively.

• fill C from A. (10,0,6)
• empty C into B. (10,6,0)
• fill C from A. (4,6,6)
• fill B from C. (4,11,1)
• pour out the litre from C (4,11,0)
• empty B into A. (15,0,0)
• fill C from A (9,0,6)
• empty C into B (9,6,0)
• fill C from A. (3,6,6)
• fill B from C. (3,11,1)
• pour out the litre from C (3,11,0)
• empty B into A (14,0,0)
• fill C from A (8,0,6)

Jug A now has 8 litres.

If we may not discard water:

• A to C (10,0,6)
• A to B (0,10,6)
• C to A (6,10,0)
• B to C (6,4,6)
• C to A (12,4,0)
• B to C (12,0,4)
• A to B (1,11,4)
• B to C (1,9,6)
• C to A (7,9,0)
• B to C (7,3,6)
• C to A (13,3,0)
• B to C (13,0,3)
• A to B (2,11,3)
• B to C (2,8,6)

Jug B now has 8 litres.

• Darn it. Late by a minute. Well done @frodoskywalker. – CodeNewbie Jun 19 '15 at 9:52
• also late by a min – Wouter Jun 19 '15 at 9:53
• can not dispose the water :( – KRU Jun 19 '15 at 10:03
• @KRU that's fine, I've added a method which doesn't need you to dispose of water. – frodoskywalker Jun 19 '15 at 10:19

Another way of producing same result with minimum steps would be as below:

Lets assume 16 Litre jug as A, 11 Litre jug as B and 6 Litre jug as C.

Initial Configuration : (16 0 0)

1. A to C (10 0 6).
2. C to B (10 6 0).
3. A to C ( 4 6 6).
4. C to B (4 11 1).
5. B to A (15 0 1).
6. C to B (15 1 0).
7. A to C (9 1 6).
8. C to B (9 7 0).
9. A to C (3 7 6).
10. C to B (3 11 2).
11. B to A (14 0 2).
12. C to B (14 2 0).
13. A to C (8 2 6).