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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.

boboquack
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KRU
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2 Answers2

7

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.

frodoskywalker
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1

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).
Curious
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KRU
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