Water Jug Problem in AI

Water Jug Problem Using Python

The water jug problem is a classic problem in AI and involves finding a sequence of actions to measure a specific volume of water using two jugs of known capacities. Here's a Python code to solve the water jug problem using a depth-first search:

from collections import deque

# Define the jug capacities
jug1_capacity = 4
jug2_capacity = 3
target_volume = 2

# Initial state with both jugs empty
initial_state = (0, 0)

# Function to check if a state is valid
def is_valid_state(state):
    jug1, jug2 = state
    return 0 <= jug1 <= jug1_capacity and 0 <= jug2 <= jug2_capacity

# Function to perform pour actions
def pour(from_jug, to_jug, state):
    jug1, jug2 = state
    if from_jug == 1:
        if jug1 > 0 and jug2 < jug2_capacity:
            amount_poured = min(jug1, jug2_capacity - jug2)
            new_jug1 = jug1 - amount_poured
            new_jug2 = jug2 + amount_poured
            return (new_jug1, new_jug2)
    else:
        if jug2 > 0 and jug1 < jug1_capacity:
            amount_poured = min(jug2, jug1_capacity - jug1)
            new_jug1 = jug1 + amount_poured
            new_jug2 = jug2 - amount_poured
            return (new_jug1, new_jug2)
    return None

# Function to solve the water jug problem using depth-first search
def water_jug_dfs():
    visited = set()
    stack = deque()
    stack.append((initial_state, []))  # State and actions taken

    while stack:
        current_state, actions = stack.pop()

        if current_state[0] == target_volume or current_state[1] == target_volume:
            return actions

        visited.add(current_state)

        for action in ["Fill Jug 1", "Fill Jug 2", "Empty Jug 1", "Empty Jug 2", "Pour Jug 1 to Jug 2", "Pour Jug 2 to Jug 1"]:
            new_state = None
            if action == "Fill Jug 1":
                new_state = (jug1_capacity, current_state[1])
            elif action == "Fill Jug 2":
                new_state = (current_state[0], jug2_capacity)
            elif action == "Empty Jug 1":
                new_state = (0, current_state[1])
            elif action == "Empty Jug 2":
                new_state = (current_state[0], 0)
            elif action == "Pour Jug 1 to Jug 2":
                new_state = pour(1, 2, current_state)
            elif action == "Pour Jug 2 to Jug 1":
                new_state = pour(2, 1, current_state)

            if new_state is not None and new_state not in visited:
                new_actions = actions + [action]
                stack.append((new_state, new_actions))

    return None

# Solve the water jug problem
solution = water_jug_dfs()
if solution:
    print("Solution Found:")
    for i, action in enumerate(solution, start=1):
        print(f"Step {i}: {action}")
else:
    print("No solution found.")

This code uses depth-first search to find a sequence of actions to reach the target volume in one of the jugs or determine that no solution is possible. You can adjust the **jug1_capacity**, **jug2_capacity**, and **target_volume** variables to solve different instances of the water jug problem.

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