The famous 10-7-4 water jug problem

water_10_7_4.py

# 10-7-4 Water Jug Problem

# Get 2 units of water in the 7-container or the 4-container

# Starting state is (0,7,4)
# Operations are F10, F7, F4, T107, T104, T710, T74, T410, T47
# Goal states are (x,2,z) and (x,y,2)

operations = {
    'F10'  : lambda x, y, z: [10, y, z],
    'F7'   : lambda x, y, z: [x, 7, z],
    'F4'   : lambda x, y, z: [x, y, 4],
    'T107' : lambda x, y, z: [0, y+x, z] if y+x<=7 else [x-(7-y), 7, z],
    'T104' : lambda x, y, z: [0, y, z+x] if z+x<=4 else [x-(4-z), y, 4],
    'T710' : lambda x, y, z: [x+y, 0, z] if x+y<=10 else [10, y-(10-x), z],
    'T74'  : lambda x, y, z: [x, 0, z+y] if z+y<=4 else [x, y-(4-z), 4],
    'T410' : lambda x, y, z: [x+z, y, 0] if x+z<=10 else [10, y, z-(10-x)],
    'T47'  : lambda x, y, z: [x, y+z, 0] if y+z<=7 else [x, 7, z-(7-y)],
}

def dfs(state, visited, solution):
    solution.append(state)
    if state[1] == 2 or state[2] == 2:
        return True
    for (op, transition) in operations.items():
        next = transition(*state)
        if str(next) not in visited and dfs(next, visited|{str(next)}, solution):
            return True
    solution.pop()
    return False


solution = []
if dfs([0,7,4], set(), solution):
    print(solution)