Detailed explanation of the following Python code:

from collections import deque

def BFS(a, b, target):

m = {}

isSolvable = False

path = []

# Queue to maintain states

q = deque()

# Initializing with initial state

q.append((0, 0))

while (len(q) > 0):

# Current state

u = q.popleft()

# q.pop() #pop off used state

# If this state is already visited

if ((u[0], u[1]) in m):

continue

# Doesn't met jug constraints

if ((u[0] > a or u[1] > b or

u[0] < 0 or u[1] < 0)):

continue

# Filling the vector for constructing

# the solution path

path.append([u[0], u[1]])

# Marking current state as visited

m[(u[0], u[1])] = 1

# If we reach solution state, put ans=1

if (u[0] == target or u[1] == target):

isSolvable = True

if (u[0] == target):

if (u[1] != 0):

# Fill final state

path.append([u[0], 0])

else:

if (u[0] != 0):

# Fill final state

path.append([0, u[1]])

# Print the solution path

sz = len(path)

for i in range(sz):

print("(", path[i][0], ",",

path[i][1], ")")

break

# If we have not reached final state

# then, start developing intermediate

# states to reach solution state

q.append([u[0], b]) # Fill Jug2

q.append([a, u[1]]) # Fill Jug1

for ap in range(max(a, b) + 1):

# Pour amount ap from Jug2 to Jug1

c = u[0] + ap

d = u[1] - ap

# Check if this state is possible or not

if (c == a or (d == 0 and d >= 0)):

q.append([c, d])

# Pour amount ap from Jug 1 to Jug2

c = u[0] - ap

d = u[1] + ap

# Check if this state is possible or not

if ((c == 0 and c >= 0) or d == b):

q.append([c, d])

# Empty Jug2

q.append([a, 0])

# Empty Jug1

q.append([0, b])

# No, solution exists if ans=0

if (not isSolvable):

print("No solution")

# Driver code

if __name__ == '__main__':

Jug1, Jug2, target = 4, 3, 2

print("Path from initial state "

"to solution state :")

BFS(Jug1, Jug2, target)