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07Ma6aSol8-2

# 07Ma6aSol8-2 - Math 6a Solution Set 8 Problem 1 Liu 4.2(a...

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Math 6a - Solution Set 8 Problem # 1 Liu 4.2 (a) Three married couples on a journey come to a river where they find a boat which cannot carry more than two persons at a time. The crossing of the river is complicated by the fact that the husbands are all very jealous and will not permit their wives to be left without them in a company where other men are present. Construct a graph to show how the transfer can be made. (b) Prove that the puzzle in part (a) cannot be solved if there are four couples. (c) Prove that the puzzle in part (a) can be solved if there are four couples and the boat holds three persons. Solution. (a) Label the three men as x 1 , x 2 , and x 3 and the three women as y 1 , y 2 and y 3 such that x i is married to y i and let b represent the boat. Then, we create a graph where each node is an ordered pair of the form ( L, R ) where each x i , y i is in L if that person is on the left bank and is in R if that person is on the right bank, and similarly b L if the boat is on the left bank and b R if the boat is on the right bank. So, for example, if x 1 , x 2 , y 1 , y 2 are on the left bank along with the boat, and x 3 and y 3 are on the right bank, we have L = { x 1 , x 2 , y 1 , y 2 , b } and R = { y 1 , y 2 } . Then, there is an edge between vertices ( L, R ) and ( L 0 , R 0 ) if and only if two or less people can take the boat from the side it is in to the other side to get from ( L, R ) to ( L 0 , R 0 ) (note that this is an undirected graph since the same people can go back to the bank they came from to get from ( L 0 , R 0 ) to ( L, R )) . Then, solving the problem is equivalent to finding a path between the configuration where R = to the configuration where L = through the subgraph that consists of only allowable states, ie those states where no y i is in a set containing x j for i 6 = j unless that set also contains x i . Then, note that the following path ( L 1 , R 1 ) ( L 2 , R 2 ) → · · · → ( L 12 , R 12 ) works where L 1 = { x 1 , x 2 , x 3 , y 1 , y 2 , y 3 , b } , L 2 = { x 1 , x 2 , x 3 , y 1 } , L 3 = { x 1 , x 2 , x 3 , y 1 , y 2 , b } , L 4 = { x 1 , x 2 , x 3 } , L 5 = { x 1 , x 2 , x 3 , y 1 , b } , L 6 = { x 1 , y 1 } , L 7 = { x 1 , y 1 , x 2 , y 2 , b } , L 8 = { y 1 , y 2 } , L 9 = { y 1 , y 2 , y 3 , b } , L 10 = { y 1 } , L 11 = { x 1 , y 1 , b } , L 12 = and R i = { x 1 , x 2 , x 3 , y 1 , y 2 , y 3 , b } - L i for 1 i 12 . 1

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2 (b) We define variables the same as above, with the additional x 4 , y 4 for the fourth couple. Assume we have such a path ( L 1 , R 1 ) ( L 2 , R 2 ) → · · · → ( L n , R n ) for some n that represents a solution in this case (so R 1 = , L n = ). Then, note that b L i for each odd i, since the boat must go back and forth in each step. Further, since at most two people can go across, and at least one must come back, if i is odd, | L i +2 | ≥ | L i | - 1 and thus such L i can decrease at most 1 for every two steps. Also, since we must have L n = (so n must be even) it follows that | L n - 1 | ≤ 3 . Since | L i | for odd i can only decrease by 1 at a time, for every m with 4 m 9 (since | L 1 | = 9) there must be some odd i with | L i | = m and with | L i +2 | = m - 1 (otherwise, | L i | ≥ m for all odd i which of course cannot happen).
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