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A person has 3 cups with him , he has a 8L cup filled with coke , and an empty 5L and 3L cups, using these three can anyone make 2 cups each with 4L to be served to two people. This person has no measuring instruments and assume that things like coke doesn't spill or anything, and you should make a cup with 4L and another cup with 4L. I believe I did get the answer to this back then when I spent one whole day with my friends in school trying to solve this. But now I can't remember it , and it's for the others too I'm sharing this question.

M.S.E
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  • Are you sure you remember this right? Put 2 in the 2L put these in the 5L. Repeat once more. – quid May 31 '14 at 13:02
  • If you have 3L cup instead of a 2L cup like your comment below suggests, there are 2 solutions, {008, 305, 035, 332, 152, 107, 017, 314, 044} or {008, 053, 323, 026, 206, 251, 341, 044} – DanielV May 31 '14 at 13:32

2 Answers2

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This is a classic problem. I think Lloyd and Dudeney had similar (and much more difficult) problems in their puzzle books.

Note $L$ - cup with 8L of coke, $M$ - cup of 5L, $S$ - cup of 3L.

Pour $L\to S$, then $S\to M$. Pour $L\to S$, then $S\to M$ until $M$ fills up. Now you have 5L in $M$ and 1L in $S$. Pour $M\to L$, then $L\to S$, $S\to M$.

Now you have 4L of coke in $L$ and 4L of coke in $M$.

TZakrevskiy
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  • Omg I'm so sorry it's a 3L cup , made a mistake S is 3L cup – M.S.E May 31 '14 at 13:04
  • @Tharindu edited the answer. – TZakrevskiy May 31 '14 at 13:06
  • There is a mistake, @TZakrrvskiy when u said there is now 5L in M and 1L in S , u forgot about the 2L left in L. From that point your answer has mistakes. – M.S.E May 31 '14 at 13:10
  • @Tharindu how's that a mistake? I simply didn't mention the state of $L$. Follow the reasoning after that point. – TZakrevskiy May 31 '14 at 13:12
  • Oh wow ya ur right :o that was fast how did you do that so fast ? – M.S.E May 31 '14 at 13:14
  • @Tharindu Martin Gardner described a general algorithm to solve similar problems in one of his books. And, back in prehistoric times when I was a kid, I solved the exactly same problem (up to a choice of liquid; as far as I remember, it was water). – TZakrevskiy May 31 '14 at 13:17
  • Although I greatly favor intelligent solutions, I think I should point out that in this problem there are only 24 possible "states" of the system, namely any integer number of liters from 0 to 3 in the first cup, any integer number from 0 to 5 in the second, and whatever's left in the third. You can draw a directed graph by joining each state to the ones reachable from it in one step, and then the problem is to find a path from (0,0,8) to (0,4,4). That looks as if it can be done reasonably quickly by hand using standard digraph connectivity algorithms (in effect exploring all possibilities). – Andreas Blass May 31 '14 at 14:18
  • @AndreasBlass that's essentially a method published by Martin Gardner I mentioned earlier, though it was presented in more accesible language. Furthermore, his method didn't require to explore all possibilities, it was sufficient to chose on of two so-called "directions". Unfortunately, I have neither the illustration from that book, nor the reference to that book. – TZakrevskiy May 31 '14 at 19:25
  • @TZakrevskiy In which book can I find the general algorithm by Gardner? – naslundx Jul 15 '14 at 11:42
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we can pour coke to 2L.then we can transfer that into 4L.do this process twice a time.then we have 8L,4L cups each with 4L of coke.

revathy
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