Water Jug Solver

water_jug.py

"""
https://www.eecis.udel.edu/~mccoy/courses/cisc4-681.10f/lec-materials/handouts/search-water-jug-handout.pdf
Consider the following problem:

A Water Jug Problem:

    You are given two jugs, a 4-gallon one and a 3-gallon
    one, a pump which has unlimited water which you can use
    to fill the jug, and the ground on which water may be
    poured. Neither jug has any measuring markings on it.
    How can you get exactly 2 gallons of water in the 4-gallon jug?

State Representation and Initial State
    we will represent a state of the problem as a tuple (x, y) where
    x represents the amount of water in the 4-gallon jug and
    y represents the amount of water in the 3-gallon jug.
    Note 0 ≤ x ≤ 4, and 0 ≤ y ≤ 3.
    Our initial state: (0,0)

Goal Predicate – state = (2,y) where 0 ≤ y ≤ 3.
"""

class StateMachine:
    """Encode transition verbs from one state to another"""
    def __init__(self, max_capacities):
        self._max = max_capacities

    def start(self):
        return tuple([0 for x in self._max])

    def fill(self, state, bucket):
        newstate = [x if i != bucket else self._max[i] for i, x in enumerate(state)]
        return tuple(newstate)

    def empty(self, state, bucket):
        newstate = [x if i != bucket else 0 for i, x in enumerate(state)]
        return tuple(newstate)

    def transfer(self, state, sourcebucket, destbucket):
        newstate = [x for x in state]
        transfer_amount = min(state[sourcebucket],
                              self._max[destbucket] - state[destbucket])
        newstate[sourcebucket] -= transfer_amount
        newstate[destbucket] += transfer_amount
        return tuple(newstate)

    def breadth_first_search(self, state, steps_from_origin):
        buckets = [i for i in range(len(state))]
        transitions = {}
        for i in buckets:
            result = self.fill(state, i)
            if result != state:
                transitions[tuple([x for x in steps_from_origin] + [tuple(["fill", i])])] = result
            result = self.empty(state, i)
            if result != state:
                transitions[tuple([x for x in steps_from_origin] + [tuple(["empty", i])])] = result
            for j in buckets:
                if j != i:
                    result = self.transfer(state, i, j)
                    if result != state:
                        transitions[tuple([x for x in steps_from_origin] + [tuple(["transfer", i, j])])] = result
        return transitions



if __name__ == "__main__":
    # calculate adjacency lists
    water_jug_state = StateMachine([4, 3])
    nodes_paths = {}
    start = water_jug_state.start()
    nodes_paths[start] = water_jug_state.breadth_first_search(start, [tuple(["start"])])
    all_reached_nodes = {x: k for v in nodes_paths.values() for k, x in v.items()}
    while len(nodes_paths) < len(all_reached_nodes):
        for v, k in all_reached_nodes.items():
            if v not in nodes_paths:
                nodes_paths[v] = water_jug_state.breadth_first_search(v, k)
        all_reached_nodes = {x: k for v in nodes_paths.values() for k, x in v.items()}
    reverse_nodes_paths = {}
    for k, v in nodes_paths.items():
        for k2, v2 in v.items():
            if v2 not in reverse_nodes_paths:
                reverse_nodes_paths[v2] = {k: [k2]}
            elif k not in reverse_nodes_paths[v2]:
                reverse_nodes_paths[v2][k] = [k2]
            else:
                reverse_nodes_paths[v2][k].append(k2)
    print("Valid solutions")
    print([k for k in reverse_nodes_paths if k[0] == 2])
    for solution in [k for k in reverse_nodes_paths if k[0] == 2]:
        print(reverse_nodes_paths[solution])