| Step no | F5 | F3 | Description |
|---|---|---|---|
| 1 | 5 | 0 | Fill F5 |
| 2 | 2 | 3 | Fill F3 from F5 |
| 3 | 2 | 0 | Empty F3 |
| 4 | 0 | 2 | Pour F5 into F3 |
| 5 | 5 | 2 | Fill F5 |
| 6 | 4 | 3 | Fill F3 from F5. F5 now contains the desired 4 liters! |
For the more mathematically minded reader the problem can be formulated as follows: Find Sn = 4 given that
Sn = Sn-1 + tn
where S0 = 0 and tn represents the change and thus only can take the values ±5 or ±3 . A condition is that for each n, Sn must be non-negative not greater than 8 (max. volume in the two bottles).
The solution is: Sn = 4 = 3+3-5+3, i.e.
S1 = 0 + 3
(F3 is filled)
S2 = 3 + 3 = 6
(F3 is emptied in F5 and filled again; F5=3, F3=3)
S3 = 6 - 5 = 1
(F3 is filled in F5 that is poured out; F5=0, F3=1)
S4 = 1 + 3 = 4
(F3 is emptied in F5 and filled again; F5=1, F3=3)
Remark: The two solutions aren't equal; in the first case only 3 liters of water is wasted, while 5 liters is wasted in the second solution !
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Last updated: 2003-04-13 by Ingvar
Jönsson