The Missionaries and Cannibals Problem.

The Missionaries and Cannibals Problem Statement

Three missionaries and three cannibals find themselves on one side of a river. They have would like to get to the other side. But the missionaries are not sure what else the cannibals agreed to. So the missionaries managed the trip across the river in such a way that the number of missionaries on either side of the river is never less than the number of cannibals who are on the same side. The only boar available holds only two at a time. How can everyone get across the river without the missionaries risking being eaten?

Solution:-

The state for this problem can be defined as

{(i, j)/ i=0, 1, 2, 3, : j=0, 1, 2, 3}

where i represents the number missionaries in one side of a river . j represents the number of cannibals in the same side of river. The initial state is (3,3), that is three missionaries and three cannibals one side of a river , (Bank 1) and ( 0,0) on another side of the river (bank 2) . the goal state is to get (3,3) at bank 2 and (0,0) at bank 1.

To sole this problem we will make the following assumptions:

1. Number of cannibals should lesser than the missionaries on either side.

2. Only one boat is available to travel.

3. Only one or maximum of two people can go in the boat at a time.

4. All the six have to cross the river from bank.

5. There is no restriction on the number of trips that can be made to reach of the goal.

6. Both the missionaries and cannibals can row the boat.

This kind of problem is often solved by a graph search method. Represent the problem as a set of states which are snapshots of the world and operators which transform one state into another state are mapped to nodes of the graph and operators are the edges of the graph. The following procedure shows the graph search algorithm in PROLOG, for Missionaries and Cannibals Problem.

Path finding.

* Write a program to find path from one node to another.

* Must avoid cycles

* A template for the clause is

path(start, Finish, Visited, Path).
Start - the name of the starting node.
Finish - the name of the finishing node.
Visited - the list of nodes aready visited.
Path - the list of nodes on the path, including Start and Finish.

Program for finding a Path.

* The search for a path terminates when we have nowhere t0o go.

path {Node, Node,.., [Node]}

A path from start to Finish, starts with a node, X,connected to start followed by a path from X to Finish.

path (Start, Finish, Visited, {Start [ Path]} :-
edge (Start,X),
not (member( X,Visited)),
path(x,Finish, {X [vISITED],Path)
member (X,[X I_ I).
member (X,I_ IYI):- member(X,Y).

Representing the state.

Now we returen to the peoblem or rrepresenting tyhe missionaries and cannibals problem;
*A State is one "snapshot" in time.

*For this problem the only infoemation we need to fully characterize the state is :
*the number of missionaries on the left bank,
*the number of cannibals on the left bank,
*the side the boat is on.

All other information can be deduced from these thres items.

In PROLOG, the state can be representted by a 3-arity term, state (Missionaries,Cannibals, State).

Representing the solution.

*The solution consists of a list of moves, e.g.
[move( I.I. right), move 2.0, lift)].
*We take this to mean that I missionary and i cannibal moved to the right bank, then 2
missionaries moved to hte left bank.
*Like the graph search problem, we must avoid returing to state we have visited before.
The visited list will have the form:
[Most Recent State I list of previousstates]

Overview of solution.

We follow a simple graph search procedure:

*Start from an initial state
*Find a neighbouring state
*Check that the now stae has not been visited before
*Find a path from the neighbour to the goal.

The search terminates when we have found the state:

state(0, 0, right).

PROLOG Code.

% mandc(CurrentState, Visited, Path)

mandc(start(0, 0, right), -,[]).
mande(Currentstate, visited,[Move I Rest O fMoves]):-
newstate (Current state, NestState, Move).
not (member(NestState,Visited)0,
make., move(Current state, NestState, Move),
mande((Nextstate I visited,[Move I Rest O fMoves])

make_move (state(M1,C1,left) state (M2,C2,right) move(M,C,right)) :-
M is M1 - M2.
C is C1 - C2.

make_move(state(M1,C1,right),state(M2,C2,right),move M,C,left)) :-
M is M2 -M1
C is C2 - C1.

A move is characterized by the nummber of missionar5ies and the number of canniblls taken in the boal at6 one time.

Since the boat can carry no more than two peopel at once, the only possible combinations are:
carry (2,0)
carry (1,0)
carry (1,1)
carry (0,1)
carry (0,2).
Where carry (M,C) means the boat will carry M, missionaries and C.cannibals on one trip.

Feasible Moves

*Once we hafe found a possible move, we have to confirm that it is feasible It is not feasible to move more missionaries or more cannibals than that are present on one bank.
When the stae is stae(M1,C1,left) and we try carry (M,C)then

M ≤ M1 and C ≥ C1

must be true.

*When the state is state(M1, C1, left) and we try carry (M,C) then

M + M1 ≤ AND C + C1 ≤ 3

must be true.

Legal Moves.

* Once we have found a feasible move, we must check that it is legal, i.e no Missionaries must be eaten.

legal(X,X).

legal(3,X).

legal(O,X).

*The only safe combinations are when there are equal numbers of Missionaries and cannibals or all the Missionaries are on one side.

Generating the Next State.

newstate(state(M1,C1,left), state(M2,C2,right)):-

carry(M,C),

M ≤ M1

C ≤ C1

M2 is M1 - M2,

C2 is C1 - C,

legal(M2,C2)

newstate(M1, C1, right), state(M2, C2, left)):-

carry(M,C),

M2 is M1 + M,

C2 is C1 + C,

M2 ≤ 3,

C2 ≤ 3,

legal(M2, C2)