The Water Jugs Problem

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This classic AI problem is described in Artificial Intelligence as follows:

“You are given two jugs, a 4-gallon one and a 3-gallon one. Neither has any measuring markers on it. There is a tap that can be used to fill the jugs with water. How can you get exactly 2 gallons of water into the 4-gallon jug?”.

E. Rich & K. Knight, Artificial Intelligence, 2nd edition, McGraw-Hill, 1991

This program implements an environmentally responsible solution to the water jugs problem. Rather than filling and spilling from an infinite water resource, we conserve a finite initial charge with a third jug: (reservoir).

This approach is simpler than the traditional method because there are only two actions. It is more flexible than the traditional method because it can solve problems that are constrained by a limited supply from the reservoir[1].

To simulate the infinite version, we use a filled reservoir with a capacity greater than the combined capacities of the jugs so that the reservoir can never be emptied.

“Perfection is achieved not when there is nothing more to add, but when there is nothing more to take away.”

Antoine de Saint-Exupéry

Solution

water_jugs

is the entry point. The solution is derived by a simple, breadth-first, state-space search, and translated into a readable format by a DCG.

water_jugs :-
    SmallCapacity = 3,
    LargeCapacity = 4,
    Reservoir is SmallCapacity + LargeCapacity + 1,
    volume( small, Capacities, SmallCapacity ),
    volume( large, Capacities, LargeCapacity ),
    volume( reservoir, Capacities, Reservoir ),
    volume( small, Start, 0 ),
    volume( large, Start, 0 ),
    volume( reservoir, Start, Reservoir ),
    volume( large, End, 2 ),
    water_jugs_solution( Start, Capacities, End, Solution ),
    phrase( narrative(Solution, Capacities, End), Chars ),
    put_chars( Chars ).

water_jugs_solution( +Start, +Capacities, +End, ?Solution )

holds when Solution is the terminal state in a state-space search that begins with an initial state, in which the water-jugs have Capacities and contain the Start volumes, and ends when the water-jugs contain the End volumes.

water_jugs_solution( Start, Capacities, End, Solution ) :-
    solve_jugs( [start(Start)], Capacities, [], End, Solution ).

solve_jugs( +Nodes, +Capacities, +Visited, +End, ?Solution )

holds when Solution is the terminal node in a state-space search, beginning with a first open node in Nodes, and terminating when the water-jugs contain the End volumes. Capacities define the capacities of the water-jugs while Visited is a list of expanded (closed) node states.

The breadth-first operation of solve_jugs is due to the existing Nodes being appended to the new nodes. (If the new nodes were appended to the existing nodes, the operation would be depth-first.)

solve_jugs( [Node|Nodes], Capacities, Visited, End, Solution ) :-
    node_state( Node, State ),
    ( State = End ->
        Solution = Node
    ; otherwise ->
        findall(
            Successor,
            successor(Node, Capacities, Visited, Successor),
            Successors
            ),
        append( Nodes, Successors, NewNodes ),
        solve_jugs( NewNodes, Capacities, [State|Visited], End, Solution )
    ).

successor( +Node, +Capacities, +Visited, ?Successor )

Successor is a successor of Node, for water-jugs with Capacities, if there is a legal transition from Node's state to Successor's state and Successor's state is not a member of the Visited states.

successor( Node, Capacities, Visited, Successor ) :-
    node_state( Node, State ),
    Successor = node(Action,State1,Node),
    jug_transition( State, Capacities, Action, State1 ),
    \+ member( State1, Visited ).

Transition Rules

jug_transition( +State, +Capacities, ?Action, ?SuccessorState )

holds when Action describes a valid transition, from State to SuccessorState, for water-jugs with Capacities.

There are two sorts of Action:

  • empty_into(Source,Target) valid if Source is not already empty and the combined contents from Source and Target, in State, are not greater than the capacity of the Target jug. The Source jug becomes empty in SuccessorState while the Target jug acquires the combined contents of Source and Target in State.
  • fill_from(Source,Target) valid if Source is not already empty, and the combined contents from Source and Target, in State, are greater than the capacity of the Target jug. The Target jug becomes full in SuccessorState, while Source retains the excess of the combined contents of Source and Target in State, over the capacity of the Target jug.

In either case, the contents of the unused jug are unchanged.

jug_transition( State0, Capacities, Action, State1 ) :-
    volume( Source, State0, SourceContents ),
    SourceContents > 0,
    jug_permutation( Source, Target, Unused ),
    volume( Target, State0, TargetContents ),
    volume( Target, Capacities, TargetCapacity ),
    volume( Unused, State0, Unchanged ),
    volume( Unused, State1, Unchanged ),
    CombinedContents is SourceContents + TargetContents,
    ( CombinedContents =< TargetCapacity ->
        Action = empty_into(Source,Target),
        NewSourceContents = 0,
        NewTargetContents = CombinedContents
    ; CombinedContents > TargetCapacity ->
        Action = fill_from(Source,Target),
        NewSourceContents is CombinedContents-TargetCapacity,
        NewTargetContents = TargetCapacity
    ),    
    volume( Source, State1, NewSourceContents ),
    volume( Target, State1, NewTargetContents ).

Data Abstraction

volume( ?Jug, ?State, ?Volume )

holds when Jug (large, small or reservoir) has Volume in State.

volume( small, jugs(Small, _Large, _Reservoir), Small ).
volume( large, jugs(_Small, Large, _Reservoir), Large ).
volume( reservoir, jugs(_Small, _Large, Reservoir), Reservoir ).

jug_permutation( ?Source, ?Target, ?Unused )

holds when Source, Target and Unused are a permutation of small, large and reservoir.

jug_permutation( Source, Target, Unused ) :-
    select( Source, [small, large, reservoir], Residue ),
    select( Target, Residue, [Unused] ).

node_state( ?Node, ?State )

holds when the contents of the water-jugs at Node are described by State.

node_state( start(State), State ).
node_state( node(_Transition, State, _Predecessor), State ).

Definite Clause Grammar

narrative( +Solution, +Capacities, +End )/

is a DCG presenting the water-jugs Solution in a readable format. The grammar is head-recursive because the Solution has the last node outermost.

narrative( start(Start), Capacities, End ) -->
    "Given three jugs with capacities of:", newline,
    literal_volumes( Capacities ),
    "To obtain the result:", newline,
    literal_volumes( End ),
    "Starting with:", newline,
    literal_volumes( Start ),
    "Do the following:", newline.
narrative( node(Transition, Result, Predecessor), Capacities, End ) -->
    narrative( Predecessor, Capacities, End ),
    literal_action( Transition, Result ).
 
literal_volumes( Volumes ) -->
    indent, literal( Volumes ), ";", newline.
 
literal_action( Transition, Result ) -->
    indent, "- ", literal( Transition ), " giving:", newline,
    indent, indent, literal( Result ), newline.
 
literal( empty_into(From,To) ) -->
    "Empty the ", literal( From ), " into the ",
    literal( To ).
literal( fill_from(From,To) ) -->
    "Fill the ", literal( To ), " from the ",
    literal( From ).
literal( jugs(Small,Large,Reservoir) ) -->
    literal_number( Small ), " gallons in the small jug, ",
    literal_number( Large ), " gallons in the large jug and ",
    literal_number( Reservoir ), " gallons in the reservoir".
literal( small ) --> "small jug".
literal( large ) --> "large jug".
literal( reservoir ) --> "reservoir".
 
literal_number( Number, Plus, Minus ) :-
    number( Number ),
    name( Number, Chars ),
    append( Chars, Minus, Plus ).
 
indent --> "  ".
 
newline --> "
 ".

Utility Predicates

Load a small library of Puzzle Utilities.

:- ensure_loaded( misc ).

The code is available as plain text here.

Output

The output of the program is:

?- water_jugs.
 
Given three jugs with capacities of:
   3 gallons in the small jug, 4 gallons in the large jug and 8 gallons in the reservoir;
 To obtain the result:
   0 gallons in the small jug, 2 gallons in the large jug and 6 gallons in the reservoir;
 Starting with:
   0 gallons in the small jug, 0 gallons in the large jug and 8 gallons in the reservoir;
 Do the following:
   - Fill the small jug from the reservoir giving:
     3 gallons in the small jug, 0 gallons in the large jug and 5 gallons in the reservoir
   - Empty the small jug into the large jug giving:
     0 gallons in the small jug, 3 gallons in the large jug and 5 gallons in the reservoir
   - Fill the small jug from the reservoir giving:
     3 gallons in the small jug, 3 gallons in the large jug and 2 gallons in the reservoir
   - Fill the large jug from the small jug giving:
     2 gallons in the small jug, 4 gallons in the large jug and 2 gallons in the reservoir
   - Empty the large jug into the reservoir giving:
     2 gallons in the small jug, 0 gallons in the large jug and 6 gallons in the reservoir
   - Empty the small jug into the large jug giving:
     0 gallons in the small jug, 2 gallons in the large jug and 6 gallons in the reservoir
 
yes
Footnote
  1. For example, the "Three Glass Puzzle" where the jugs have capacities of 8 (filled), 5 and 3 and the goal is to get two equal volumes (4).