% % Missionaries and Cannibals Puzzle -- % % There are 3 missionaries, 3 cannibals, and a boat on the west bank of % a river. All wish to cross, but the boat holds at most 2 people. % If the missionaries ever outnumber the cannibals on either bank of the % river or in the boat, the outnumbered cannibals will be converted. % Can they all safely cross the river? If so, how? % (The boat cannot cross empty.) % % state(X,Y,Z) means that X missionaries, Y cannibals, and the boat % are on the Z side of the river. % set(binary_res). set(prolog_style_variables). op(900, xfx, ||). op(900, xfx, &&). make_evaluable(_-_, \$DIFF(_,_)). make_evaluable(_+_, \$SUM(_,_)). make_evaluable(_<_, \$LT(_,_)). make_evaluable(_<=_, \$LE(_,_)). make_evaluable(_>_, \$GT(_,_)). make_evaluable(_>=_, \$GE(_,_)). make_evaluable(_==_, \$EQ(_,_)). make_evaluable(_&&_, \$AND(_,_)). make_evaluable(_||_, \$OR(_,_)). list(usable). % MBS - missionaries on same side as the boat % CBS - cannibals on same side as the boat % BP - position of boat (either west or east) % M - missionaries to cross % C - cannibals to cross -state(MBS,CBS,BP) % If we have an achievable state, | -pick(M) % missionaries to cross | -pick(C) % cannibals to cross | -(M<=MBS) | -(C<=CBS) | -(M+C>0) % if number in boat > 0, | -(M+C<=2) % if number in boat <= 2, | -(C>=M||C==0) % if no conversions in the boat, % if no conversions after the boat leaves current side, | -(CBS-C>=MBS-M || CBS-C==0) % if no conversions when the boat arrives at the other side, | -((3-CBS)+C >= (3-MBS)+M || (3-CBS)+C==0) % then a crossing can occur | state((3-MBS)+M, (3-CBS)+C, otherside(BP)). % The following give the number of each that can cross. pick(0). pick(1). pick(2). -state(3,3,east). % goal state end_of_list. list(sos). state(3,3,west). % initial state end_of_list. list(demodulators). otherside(west) = east. otherside(east) = west. end_of_list.