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 α (male) and a (female), β and b, and γ and c., p. 291. This is a shortest solution to the problem, but is not the only shortest solution., p. 291. 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 γ 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: Variations An obvious generalization is to vary the number of jealous couples (or missionaries and cannibals), the capacity of the boat, or both. If the boat holds 2 people, then 2 couples require 5 trips; with 4 or more couples, the problem has no solution. If the boat can hold 3 people, then up to 5 couples can cross; if the boat can hold 4 people, any number of couples can cross., p. 300. A simple graph-theory approach to analyzing and solving these generalizations was given by Fraley, Cooke, and Detrick in 1966. If an