# Tag Info

1

Let's assume for the moment that the judge chooses the starting location at random uniformly, and similarly the direction of travel. In that case, the location at either end of the row is the best for the Janets. Clearly every one of them is equally likely to be the first to be examined. The ones at the ends have only one neighbour, while all the rest have ...

3

This is only solvable in the sense that no solution is better than another. Since the judge has no prior knowledge, starting at any position is just as good, so the judge might as well toss a (38-sided) coin to decide, (treating the line as if its ends were connected to each other) and this is still optimal for the judge. If the judge does that, it becomes ...

1

Here a possible a possible solution to the 4 explorer case, inspired by hdsdv's answer:

0

I will answer this puzzle based on the assumption that when the travelers need one day of water to go from point A to point B they also need the same amount of water to go from point B to point A. Let's have the travelers a,b,c,d . The travelers go 2 days into the desert then traveler a gives 2 days of water to travelers b,c,d and goes back to the ...

1

Assuming that water cannot be dropped, and return trips cannot be made, which leaves the puzzle as the simplest form described in the question: Reasoning:

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Improvement on the 2-explorer case: First attempt at 3-explorer case, using the same strategy and formatting:

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Question 3 (2 Explorers): Question 2 (3 Explorers): Question 1 (4 Explorers):

4

I asked a question over on Math.SE, pointing to this puzzle, and @Elaqqad gave a very long answer with lots of links to math that might help. @Elaqqad also produced a concrete solution with only 59 tests — beating @noedne's long-standing solution of 63 tests! Background math The "wolves and sheep" puzzle is a specific case of non-adaptive group testing. ...

2

The answer is Unfortunately, this answer is neither elegant nor easy to explain since I found it via brute force. It's pretty disappointing to solve a puzzle this way, but I don't think anyone explained a correct answer yet (at least before I was sniped by Charles Gleason!). The General Approach Consider the case of 9 coins with one heavier than the rest....

8

This can be done in Steps

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With the assumption that you can distinguish the coins by either marking them or retrieving them in the reverse order by stacking them on the scale, then worst case is seven weighings: Arrange the coins into a grid of 4 rows and 5 columns Weigh Row1 against Row2 Weigh Row3 against Row4 At this point, you know either: Which row has the heavy coin AND ...

3

Reasoning:

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I think it can be done in: Method:

4

This can be done in Steps

0

You make four stacks of 5 coins each. You put one stack on one side and one on the other side. You have two possibilities. a) They balance the scale. b) They do not balance the scale. If the scale is balanced then you have the smallest number of steps. If the scale is not balanced, we have the following combinations 9[10-11-10]. Remove the stack with the 9 ...

2

Answer Steps If there is exactly one imbalance $a<b$, then coin $a$ weighs 9 g and coin $b$ weighs 11 g. If there are two imbalances $a<b$ and $c<d$, then weigh only coins $a$ and >!$c$ (one extra weighing) Case 1: $a<c \implies$ coin $a$ weighs 9 g and coin $d$ weighs 11 g Case 2: $c<a \implies$ coin $c$ weighs 9 g and coin $b$ weighs 11 g ...

-1

Weigh two random coins. Luckily, one coin is heavier than the other. Could be: 11 g and 10 g 10 g and 9 g 11 g and 9 g Take both coins (just weighed) and put them on same side (keeping track of the heavier coin) and weigh them against another two coins. You get lucky and both sides weigh the same. If if was the 11 g and 10 g on one side this is ...

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