Wolf, sheep, cabbage is a logic puzzle, in which the player (boatman) has to transport a wolf, a sheep and a cabbage from one river bank to the other. In the game the player must obey these rules:
- The player can use a boat to transport the objects, but he may take at maximum one thing with him every time.
- If the sheep remains unguarded on the same bank as the cabbage, than the sheep will eat the cabbage.
- If the wolf remains unguarded on the same bank as the sheep, than the wolf will eat the sheep.
Solution
From the algorithmic point of view the puzzle can be easily solved by a backtracking algorithm. The backtracking algorithm in every step tries to transport one thing to the other bank and then it checks the consistency of the partial solution. If the solution is consistent, than it repeats the step (transports another thing). If not, than the backtracking algorithm returns to the preceding decision and changes it (transports something else, or returns to the previous decision, if all possible objects were already tried out). Using this technique of organized trials and failures the algorithm finds the solution.
Code
01.
/**
02.
* Solves the sheep-cabbage-wolf riddle and prints out its solution
03.
*/
04.
public
static
void
solveSheepCabbageWolf(){
05.
sheepCabbageWolf(
false
,
false
,
false
,
false
,
new
LinkedList<String>());
06.
}
07.
/**
08.
* Solves the sheep-cabbage-wolf riddle and prints out its solution (the actual worker method)
09.
* @param sheep true if the sheep is on the right bank, false otherwise
10.
* @param cabbage true if the cabbage is on the right bank, false otherwise
11.
* @param wolf true if the wolf is on the right bank, false otherwise
12.
* @param farmer true if the farmer (boat) is on the right bank, false otherwise
13.
* @param solution partial solution
14.
* @return false if the partial solution is invalid
15.
*/
16.
private
static
boolean
sheepCabbageWolf(
boolean
sheep,
boolean
cabbage,
boolean
wolf,
boolean
farmer, Deque<String> solution) {
17.
if
(sheep && cabbage && wolf && farmer) {
18.
printSolution(solution);
19.
return
true
;
20.
}
21.
if
(!checkConsistency(sheep, cabbage, wolf, farmer)) {
22.
return
false
;
23.
}
24.
if
(solution.isEmpty() || !solution.peek().equals(
"boatman"
)) {
25.
solution.addFirst(
"boatman"
);
26.
if
(sheepCabbageWolf(sheep, cabbage, wolf, !farmer, solution)) {
27.
return
true
;
28.
}
29.
solution.pop();
//backtrack
30.
}
31.
if
(sheep == farmer && (solution.isEmpty() || !solution.peek().equals(
"sheep"
))) {
32.
solution.addFirst(
"sheep"
);
33.
if
(sheepCabbageWolf(!sheep, cabbage, wolf, !farmer, solution)) {
34.
return
true
;
35.
}
36.
solution.pop();
//backtrack
37.
}
38.
if
(cabbage == farmer && (solution.isEmpty() || !solution.peek().equals(
"cabbage"
))) {
39.
solution.addFirst(
"cabbage"
);
40.
if
(sheepCabbageWolf(sheep, !cabbage, wolf, !farmer, solution)) {
41.
return
true
;
42.
}
43.
solution.pop();
//backtrack
44.
}
45.
if
(wolf == farmer && (solution.isEmpty() || !solution.peek().equals(
"wolf"
))) {
46.
solution.addFirst(
"wolf"
);
47.
if
(sheepCabbageWolf(sheep, cabbage, !wolf, !farmer, solution)) {
48.
return
true
;
49.
}
50.
solution.pop();
//backtrack
51.
}
52.
return
false
;
53.
}
54.
55.
/**
56.
* Check consistency of the partial solution
57.
* @param sheep if the sheep is on the right bank, false otherwise
58.
* @param cabbage if the cabbage is on the right bank, false otherwise
59.
* @param wolf if the wolf is on the right bank, false otherwise
60.
* @param farmer if the farmer is on the right bank, false otherwise
61.
* @return true if the solution is consistent, false otherwise
62.
*/
63.
private
static
boolean
checkConsistency(
boolean
sheep,
boolean
cabbage,
boolean
wolf,
boolean
farmer) {
64.
if
(sheep == cabbage && sheep != farmer) {
65.
return
false
;
66.
}
else
if
(sheep == wolf && sheep != farmer) {
67.
return
false
;
68.
}
69.
return
true
;
70.
}
71.
72.
/**
73.
* Prints out the solution of the sheep-cabbage-wolf problem
74.
* @param solution
75.
*/
76.
private
static
void
printSolution(Deque<String> solution) {
77.
while
(!solution.isEmpty()) {
78.
System.out.print(solution.pollLast() +
" "
);
79.
}
80.
System.out.println();
81.
}
01.
% farmer, sheep, cabbage, wolf
02.
start :- transport(left, left, left, left, empty,
X
).
03.
04.
opposite(left, right).
05.
opposite(right, left).
06.
07.
safe(
X
,
X
,
X
).
08.
safe(
X
,
Y
,
Z
) :-
X
\\=
Y
.
09.
10.
transport(right, right, right, right, _, []).
11.
12.
transport(
F
,
K
,
Z
,
V
,
B
,
X
) :-
13.
B
\\= empty,
14.
opposite(
F
,
F
2),
15.
safe(
K
,
Z
,
F
2),
16.
safe(
K
,
V
,
F
2),
17.
transport(
F
2,
K
,
Z
,
V
, empty,
X
1),
18.
append
([farmer],
X
1,
X
).
19.
20.
transport(
F
,
K
,
Z
,
V
,
B
,
X
) :-
21.
F
==
V
,
22.
B
\\= wolf,
23.
opposite(
F
,
F
2),
24.
opposite(
V
,
V
2),
25.
safe(
K
,
V
2,
F
2),
26.
safe(
K
,
Z
,
F
2),
27.
transport(
F
2,
K
,
Z
,
V
2, wolf,
X
1),
28.
append
([wolf],
X
1,
X
).
29.
30.
transport(
F
,
K
,
Z
,
V
,
B
,
X
) :-
31.
F
==
K
,
32.
B
\\= sheep,
33.
opposite(
F
,
F
2),
34.
opposite(
K
,
K
2),
35.
safe(
K
2,
Z
,
F
2),
36.
safe(
K
2,
V
,
F
2),
37.
transport(
F
2,
K
2,
Z
,
V
, sheep,
X
1),
38.
append
([sheep],
X
1,
X
).
39.
40.
transport(
F
,
K
,
Z
,
V
,
B
,
X
) :-
41.
F
==
Z
,
42.
B
\\= cabbage,
43.
opposite(
F
,
F
2),
44.
opposite(
Z
,
Z
2),
45.
safe(
K
,
Z
2,
F
2),
46.
safe(
K
,
V
,
F
2),
47.
transport(
F
2,
K
,
Z
2,
V
, cabbage,
X
1),
48.
append
([cabbage],
X
1,
X
).