Three Wives & Their Jealous Husbands

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Three jealous husbands with their wives, having to cross a river, find a boat without a boatman.  But the boat is so small that it can contain no more than two of them at a time.

How can these six persons cross the river so that none of the women shall be left in the company of any of the men unless her husband is present?
(Women to row when absolutely necessary to prevent a violation of the above conditions).

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    Let the three husbands be A,B and C and their wives a,b and c respectively.  The following moves are typical of a minimum solution with men rowing where possible.

    Left Bank Crossing Right Bank
    ABCabc
    ACac Bb ®
    ACac ¬ B b
    ABC ac ® b
    ABC ¬ a bc
    Aa BC ® bc
    Aa ¬ Bb Cc
    ab AB ® Cc
    ab ¬ c ABC
    b ac ® ABC
    b ¬ B ACac
    Bb® ACac
    ABCabc

    Note that only a, B, and c need be able to row.

    Detective Expert Answered on 29th September 2015.
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    The missionaries and cannibals problem, and the closely related jealous husbands problem, are classic river-crossing problems

    The missionaries and cannibals problem is a well-known toy problem in artificial intelligence

    Source of solutionWIKIPEDIA

    Solving-
    Amarel devised a system for solving the Missionaries and Cannibals problem whereby the current state is represented by a simple vector <a,b,c>. The vector’s elements represent the number of missionaries on the wrong side, the number of cannibals on the wrong side, and the number of boats on the wrong side, respectively. Since the boat and all of the missionaries and cannibals start on the wrong side, the vector is initialized to <3,3,1>. Actions are represented using vector subtraction/addition to manipulate the state vector. For instance, if a lone cannibal crossed the river, the vector <0,1,1> would be subtracted from the state to yield <3,2,0>. The state would reflect that there are still three missionaries and two cannibals on the wrong side, and that the boat is now on the opposite bank. To fully solve the problem, a simple tree is formed with the initial state as the root. The five possible actions (<1,0,1>, <2,0,1>, <0,1,1>, <0,2,1>, and <1,1,1>) are then subtracted from the initial state, with the result forming children nodes of the root. Any node that has more cannibals than missionaries on either bank is in an invalid state, and is therefore removed from further consideration. The valid children nodes generated would be <3,2,0>, <3,1,0>, and <2,2,0>. For each of these remaining nodes, children nodes are generated by adding each of the possible action vectors. The algorithm continues alternating subtraction and addition for each level of the tree until a node is generated with the vector <0,0,0> as its value. This is the goal state, and the path from the root of the tree to this node represents a sequence of actions that solves the problem.

    SOLUTION
    The earliest solution known to the jealous husbands problem, using 11 one-way trips, is as follows. The married couples are represented as \alpha (male) and a (female), \beta and b, and \gamma and c.

    Trip number Starting bank Travel Ending bank
    (start) \alphaa \betab \gammac
    1 \betab \gammac \alphaa →
    2 \betab \gammac \alpha a
    3 \alpha \beta \gamma bc → a
    4 \alpha \beta \gamma ← a b c
    5 \alphaa \beta\gamma b c
    6 \alphaa \betab \gammac
    7 a b \alpha\beta \gammac
    8 a b ← c \alpha \beta \gamma
    9 b a c → \alpha \beta \gamma
    10 b \beta \alphaa \gammac
    11 \betab → \alphaa \gammac
    (finish) \alphaa \betab \gammac

    This is a shortest solution to the problem, but is not the only shortest solution

    If however, only one man can get out of the boat at a time and husbands must be on the shore to count as with his wife as opposed to just being in the boat at the shore: move 5 to 6 is impossible, for as soon as \gamma has stepped out b on the shore won’t be with her husband, despite him being just in the boat.

    As mentioned previously, this solution to the jealous husbands problem will become a solution to the missionaries and cannibals problem upon replacing men by missionaries and women by cannibals. In this case we may neglect the individual identities of the missionaries and cannibals. The solution just given is still shortest, and is one of four shortest solutions.

    If a woman in the boat at the shore (but not on the shore) counts as being by herself (i.e. not in the presence of any men on the shore), then this puzzle can be solved in 9 one-way trips:

    Trip number Starting bank Travel Ending bank
    (start) \alphaa \betab \gammac
    1 \betab \gammac \alphaa →
    2 \betab \gammac ← a \alpha
    3 \beta \gammac ab → \alpha
    4 \beta \gammac ← b \alphaa
    5 \gammac \betab → \alphaa
    6 \gammac ← b \alphaa \beta
    7 \gamma bc → \alphaa \beta
    8 \gamma ← c \alphaa \betab
    9 \gammac → \alphaa \betab
    (finish) \alphaa \betab \gammac
    Albert Scholar Answered on 29th September 2015.
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