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88

No real-world car can accelerate faster than $5g$, so the maximum acceleration of the car is under $50m/s^2$. Even if the car instantly decelerates (against a wall or something), the fact that $P$ can do the trip in $1$ second means that $A$ and $B$ are at most $25m$ apart (more realistically $1m$), so running this distance takes less than five seconds at a ...

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The quickest way may be by some form of intelligent trial and error, but there is one systematic method which will always work, and that is to translate the problem into the language of graph theory. Start by making a complete list of all possible configurations of people and items. In this case, the boat will always be together with the farmer, so there ...

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I think this fits the criteria:

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Total time: 29 seconds

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A possible way

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Solution:

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You'll want the two slowest people to cross together, but you'll also want one fast person on either side to bring the boat back quickly. So I believe this will do it (I labeled the people from A (fastest) to D (slowest)): A and B cross (7 mins) A returns (3 mins) C and D cross (15 mins) B returns (7 mins) A and B cross (7 mins) This takes a total of 39 ...

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Yikes.   Labeling us 10, 5, 2 and 1 by how long we take to cross the bridge — always with lantern L — each forward crossing will have two of us, taking as long as the slower one. Each return crossing will be made by someone alone. Whew!   Now safe, the story of temptation and resistance may be told. ...

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Note: This answers the question before the question was changed from "left alone ... with" to just "left ... with". Definitions: trip across means going from the original side to the destination side woman left alone with a man means those individuals are the only two people in that place. Send all the women across first. This requires $n-1$ trips, ...

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Here’s my quadruple-checked, optimality-guaranteed solution with (Glorfindel’s initial answer had the same crossing count earlier, but UselessInfoMine discovered an error in that method. The updated version of that method works, but uses 4 crossings more than the optimal method.) You can get any 2 consecutive thieves over like this Repeat four times, ...

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It can be done with a maximum of $n=5$ couples in $m=11$ trips, for a product of $55$. First, let's show that more than $5$ couples cannot cross. If both banks have a man, then every woman must be with her husband so that he won't be jealous. So, if a crossing step starts and ends with men on both banks, it must keep couples together, so it must move ...

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Here is a solution with Explanation:

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The minimum is This is achieved by:

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You can do it like this: On each line other than the last, parentheses indicate what is to be moved to the other side.

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Amazing puzzle! Found a solution after a couple of hours of hunting (I hope that all the words I used are allowed in this puzzle)

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It is possible. Start: BD MD SD BC MC SC X on left, none on right (X is the boat) MD MC go to the right. BD SD BC SC on left, MD MC X on right MC returns to the left. BD SD BC MC SC X on left, MD on right BD SD go to the right. BC MC SC on left, BD MD SD X on right MD returns to the left. MD BC MC SC X on left, BD SD on right BC SC go to the right. ...

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If you have just items to transport, infinity. But it's no riddle, then. But in 'classical' problem, every item except 1 is endangered by another item, every item except 1 is threatened by another item. Under that circumstances you can't even solve it with 4, because no matter what item you take, there will be at least one 'colission' pair: A -> B -> ...

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You can get one grown person across as follows To generalise this to the captain and 357 soldiers The boat passes shore to shore a total of

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In this case, you can look at the first few moves.

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Let me give it a try: Let's call the models B1 and B2 since they belong to B and the assistants M1 and M2 and the case F1. B and M B back B and B1 B and M back B and B2 B back M and B M back F and F1 B back B and M M back M and M1 B and M back M and M2 M back finally M and B

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It can be done in a total of as follows.

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They can make it like this: First trip: Second trip: Third trip: Fourth trip: Fifth trip: Sixth trip: Last trip: And they are all across the river with no-one dead or fighting. How I got this:

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For another viewpoint, let's take the "classic" version, of A -> B, B -> C (that is, fox eats goose, goose eats beans) and add a D such that it neither threatens anything nor is threatened by anything (a rock, say), as in the "neutral" element of @Lukasz's answer. Now is it solvable? Well, let's see (\/ = boat): ABCD\/~~~~~~~~~~ A CD ~~~~~~~~~~\/ ...

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UPDATE: UselessInfoMine noticed that Glorfindel’s seemingly optimal solution didn’t actually work. Here’s a working version with the same crossing count. ————— Glorfindel’s approach seems good. Here’s a lateral-thinking solution within the established parameters of the story. First, observe that Abusing those observations, This allows everybody to cross ...

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No lateral thinking tag, but still: The other solution is for them to:

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TL;DR The general formula for fewest trips is: Giving values for (a), (b), (c): The possible combinations for the first trip are: MM, MP, PP, M, P. We can immediately discount a first trip carrying an individual as the return trip would need to be made by that individual, causing no change to the number of beings on the second shore and therefore ...

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xnor posted his answer as I was writing this, so I decided to go ahead and finish it in any case. I think my solution is a bit more elementary, if more convoluted and less general. I will show that the maximum possible value for $n$ is $5$. First, we show five is possible. Let men be $\{A,B,C,D,E\}$ and women $\{a,b,c,d,e\}$, where a man and woman are ...

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While I will defer to puzzledPig's answer as the better one, I would also propose that:

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