A problem that kept an entire office force puzzled for two days is what is called the cannibal and missionary puzzle. Here it is for you to idle your time away on.
Three cannibals and three missionaries arrive at the bank of a river which they must somehow cross. There is but one boat. This boat will carry but two people. Of the missionary group all three can row, but only one of the cannibals can row. In no case can there be a greater number of cannibals than missionaries left on either bank of the river. The number of missionaries in all cases must equal of exceed the number of cannibals. Try getting them over by using coins of different denominations, or poker chips of different colors to represent the two classes of people, being quite sure to distinguish the cannibal who can row from all the others.
We would not wish to make you lose too much sleep, or unbalance your mind to such an extent as to unfit you for your daily duties, and so publish a diagram and a formula to be followed that will make everybody eventually happy… * For the answer … highlight the “blank” space below!
One, two and three are cannibals – three being the one who can row. Four, five and six are the missionaries, all of whom can row. Three takes one across and returns. Three takes two across and returns; four takes five across and brings back one; four takes three across and brings back two; four takes six across and three returns with the boat; three takes one across and then goes back and gets two, and brings him safely across. Thus the entire party cross the river.