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.
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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.
answered Jun 19 '15 at 9:51
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$\begingroup$ Darn it. Late by a minute. Well done @frodoskywalker. $\endgroup$ – CodeNewbie Jun 19 '15 at 9:52
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$\begingroup$ @KRU that's fine, I've added a method which doesn't need you to dispose of water. $\endgroup$ – frodoskywalker Jun 19 '15 at 10:19
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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)
- A to C (10 0 6).
- C to B (10 6 0).
- A to C ( 4 6 6).
- C to B (4 11 1).
- B to A (15 0 1).
- C to B (15 1 0).
- A to C (9 1 6).
- C to B (9 7 0).
- A to C (3 7 6).
- C to B (3 11 2).
- B to A (14 0 2).
- C to B (14 2 0).
- A to C (8 2 6).
