We are approaching the ever-present December dilemma: should you study for exams or celebrate Christmas? Let us contribute as well, and turn this dilemma into a trilemma: why not try to solve the puzzle below?
On a cold afternoon, six plus operators and six minus operators want to get to their beloved equation, which is on the north side of a river called the pi. The operators are on the south side and none of them has the ability to swim. Luckily, there is a ferry with a friendly skipper, who is willing to transfer them to the north side.
The ferry can only carry three operators at a time, and it turns out the operators are not the easiest creatures to deal with either. To keep the plus operators in a positive mood, neither one of the three locations – the north side, the south side and the ferry itself – should be occupied by more minus operators than plus operators, unless there are no plus operators at that location at all. This include change moments at the shores, as all operators have to leave the ferry before others are allowed to board. Lastly, on every ride at least one operator should be on board. Because these operators are very elementary, they require you to devise a plan so that they can make it to their equation.
Is there a solution such that the friendly skipper can transfer all twelve operators to the north shore, whilst keeping the plus operators in a positive mood?
Please send your solution to Nekst@Asset-Econometrics.nl before March 18. 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!
Robin Buijs 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 previous puzzle had several solutions. If you want to know more, please read the explanation below.
Explanation Puzzle 1 2015-2016
The solution is as follows: The professor should start on floor 2, and move up one floor every minute, until he is at floor 6. He should stay at floor 6 for one minute. After this, he should move down one floor every minute until he is at floor 2. An alternative approach is to start over from floor 2 when he has waited at floor 6 for one minute. These strategies result in the sequences 2-3-4-5-6-6-5-4-3-2 and 2-3-4-5-6-2-3-4-5-6, respectively. After these 10 steps, he is sure to have found the student.
Suppose the professors adopts the strategy 1-2-3-4-5-6-6-5-4-3-2. If he does not find the student on the first floor, he is sure that the student is on a higher floor than he is. Let us distinguish two cases: The student is an even or an odd number of floors away from the professor.
First, assume the student is an even number of floors away from the professor at the start. Now the professor can simply move up one floor every minute. There is no way in which the student can go to a lower floor than the professor without being on the same floor as the professor at a certain time. Thus, once the professor arrives at floor 6 the student is either found or at floor 7. Now, by waiting 1 minute on floor 6, the student will come down from floor 7, in case he was not found yet. This is because the student cannot wait at a certain floor, unlike the professor. So, under this assumption the strategy is sure to find the student.
For the second case, assume the student is an odd number of floors away from the professor at the start. We adopt the same strategy: starting from floor 1 the professor still moves up one floor every minute. In this case however, they might switch places at a certain moment in time. E.g. the professor was at floor 4 and moves to floor 5, while the student was at floor 5 an moves to floor 4. Thus, once the professor arrives at floor 6, the student might actually be on a floor below floor 6. By waiting on floor 6 for one minute, the professor makes sure the distance between the professor and the student changes to an even number, and thus he eliminates the possibility for the student to evade him once again by swapping floors. Thus, once the professor starts moving down, he will for sure close in on the student, and the student cannot be at floor 1 once the professor is at floor 2, as the difference is even for sure. In a similar way, instead of waiting on floor 6 for one minute the professor can also continue back to floor 2. He now knows that the difference is again even, and he can continue his way up again and find the student, just like explained in case 1.
Instead of starting on floor 1, the professor can also start at floor 2. In case the student is at floor 1 at the start, the difference is odd, so once the professor starts moving down it will be even and he will be found by the time the professor is back at floor 2, as described above. Thus, strategies 2-3-4-5-6-6-5-4-3-2 and 2-3-4-5-6-2-3-4-5-6 guarantee finding the student in 10 moves.