Click here to enter your comments. There are four river crossing puzzles. Critters and Carrots Level 1. Scouts Crossing Level 4. Tallbots and Smallbots Level 2. You could discuss why this might be the case. The students can then use either the table suggested in question 2 or a tree diagram as suggested by the curriculum to provide an answer to the question. Even students who understand some probability can have trouble seeing that there are 36 different ways to get the totals 2 to Discuss this, along with the idea that there is more than one way to get a total such as 7.
You could point out that if you get a 1 on the first roll, you can get the desired total 7 by getting a 6 on the second roll. If you get a 2 on the first roll, it could be followed by a 5 on the second. In fact, no matter what you get on the first roll, you have a chance of getting a total of 7 from the two rolls. The fox, the hen and the corn Once upon a time a farmer went to market and bought a fox , a hen and a sack of corn. If left together on the riverbank, the fox would eat the hen, or the hen would eat the corn.
Notes and Background When you try to solve these 'river crossing' puzzles, you are attempting some of the same problems that were set by Alcuin, an ecclesiastic from the 9th century, for the Emperor Charlemagne. Alcuin's version of Problem 1 concerned a wolf, a goat and a cabbage. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.
The third is a finishing move 1,2 , where the fastest two people cross and don't return, completing the solution.
I have been struggling to prove this. I'd like some help with how I might approach it—perspectives, hints, and insights are much more useful to me than full solutions, as I'd enjoy figuring out most of the proof myself. My basic approach has been to prove various necessary properties of optimal strategies such as the fact that in an optimal strategy, the return boat is always piloted by one person, never two and attempt to show that any candidate strategy that satisfies these necessary properties can be converted, by surgery, into a strategy made up of only escorts and ratchets, in a way that can never raise the cost.
In contrast, I am looking for a 1 proof that 2 my particular primitivesratchets and escorts 3 suffice to build an optimal solution no matter what the setup is. These minimal path costs are represented as linear combinations of the individual crossing times. Such minimal path costs are incomparable to one another without knowing the specific values of the crossing times. Given a particular set of transit time values, you can evaluate all of these expressions and whichever evaluates to the lowest number corresponds to the minimal cost solution path.
It was by observing patterns in these cost expressions that I came up with the primitives for this question. Sign up to join this community. The best answers are voted up and rise to the top.
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