Even though Easter has already passed, most people will agree that chocolate Easter eggs are among the most pleasant parts of this tradition. Therefore, this puzzle does not consider just a single chocolate Easter egg, but ten bowls full of the tasty chocolate treats!
You have been given ten bowls of chocolate eggs, each containing twenty eggs. Also, you have been informed that nine of the bowls contain eggs with a weight of ten grams. The remaining bowl contains eggs with a weight of just nine grams. However, which bowl is which is left in the dark; you cannot tell with the naked eye which bowl contains the lighter eggs. You have been given an electronic scale to help you determine which of the ten bowls contains the eggs of nine grams. Unfortunately, the scale is almost out of battery and can only be used exactly once before you run out of battery completely. How can you determine which bowl is the one with the lighter chocolate eggs, using only one measurement on the scale?
Please send your solution to Nekst@Asset-Econometrics.nl before June 3. A crate of beer or a delicious pie, whichever the winner prefers, will be waiting for whoever has the best (partial) solution. Please note that, as before, every recipient of this magazine is eligible to send in their solution, so members of the department are invited to participate as well. Good luck!
Martijn Tervelde is the winner of the previous puzzle. As a reward, he can come and pick up a crate of beer or a pie at room E1.10. The answer to the previous puzzle was no. If you want to know more, please read the explanation below.
Explanation Puzzle 2 2015-2016
It is not possible for the skipper to transfer all twelve operators to the north shore, whilst keeping the plus operators in a positive mood. Below you can find a schematic representation of the problem. On the horizontal axis you can find the number of plus operators on the south shore, and on the vertical axis you can find the number of minus operators on the south shore. Hence, the diagram displays all possible states of the problem. The starting state is the top right corner, denote this with (6,6). Our goal is to transfer all twelve operators to the north shore, so we want to end up in state (0,0). However, not all intermediate states are allowed. The ineligible states are colored black. Note that these are not only the states in which we have strictly more minus operators than plus operators, but also those where we have strictly more plus operators than minus operators. This is because such states would lead to situations with more minus operators than plus operators on the north shore.
We must now try to find a path from Start (6,6) to Finish (0,0) whilst avoiding the black cells. As we move between south shore and north shore, we alternate between two sets of eligible moves. If we are on the south shore, we can perform the following moves:
(-3,0), (-2,0), (-1,0), (-2,-1), (-1,-1), (0,-1), (0,-2), (0,-3)
where (-a,-b) means: move a plus operators and b minus operators from the south shore to the north shore. Clearly, a has to be larger than b. If we are on the north shore, we can perform the following moves:
(3,0), (2,0), (1,0), (2,1), (1,1), (0,1), (0,2), (0,3).
Now, we have to alternate between selecting a move from the first list and a move from the second list. One can use trial and error to figure out that it is indeed not possible to go from (6,6) to (0,0). The following moves can be performed, see also Figure 2. The first move we perform is (0,-3) and end up in (6,3). We then perform (0,1) and move to (6,4). Subsequently, we perform (0,-2) and move to (6,2). We now perform (0,1) and end up in (6,3). We can now start moving plus operators, and perform (-3,0) and end up in (3,3). We now have run into a dead-end, as it turns out. The maximum horizontal distance that we can move, is 3 cells. Hence, (3,3) is the most ‘left’ state we can reach. From this state, we would like to move to (2,2) or (3,0), but this would require a move from list 1, and we have just performed (-3,0) so we have to make a move from list 2 first. The only option is (1,1) (or return back to (6,3)), and this would only lead us back to (4,4), from where we cannot move closer towards (0,0) either.
Therefore, we cannot continue past the state (3,3), and can never reach either (1,0), (2,0), (3,0) or (1,1), the states from which we can reach (0,0). Hence, this puzzle has no solution in which all plus operators can be kept happy.