THATS MY CODE. IMPROVEMENTS?

class Stack:
 def __init__(self):
 self.items = []
 d

I am attaching my code to solve the Hanoi problem. What do you think? Would you change something? I am attaching the instructions for solving the problem, does the code solve the question? In the role of the 3 towers, 3 cartridges will be used: s1,s2,s3 which initially the numbers 1,2,3,...,n (the big number at the bottom of the stack) are found in s1 and the other two stacks are empty.

It is required to develop a solution characterized by the following distinctions: Arrange the 3 towers in a circle. At every odd step, move the small ring clockwise. In every even step, make the possible move that does not involve the small ring

Requirements for implementing the program: Define n as the number of rings. You will produce 3 cartridges and enter values ​​for the first time. Activate the Hanoi function which accepts 3 cartridges. Requirements for the algorithm: The algorithm will transfer the rings from S1 to one of the other towers. After each transfer of a ring, the contents of the 3 cartridges, and the step number, must be printed. (Each ring transfer is a step).

class Stack:
 def __init__(self):
 self.items = []
 def is_empty(self):
 return self.items == []
 def push(self, item):
 self.items.append(item)
 def pop(self):
 return self.items.pop()
 def getLen(self):
 return len(self.items)
 
def print_stacks(S1, S2, S3):
 print("S1:", S1.items)
 print("S2:", S2.items)
 print("S3:", S3.items)
def hanoi(n, source, helper, target, step_count):
 if n > 0:
 hanoi(n - 1, source, target, helper, step_count)
 target.push(source.pop())
 step_count[0] += 1
 print("Step:", step_count[0])
 print_stacks(S1, S2, S3)
 print()
 hanoi(n - 1, helper, source, target, step_count)
N = 3
S1 = Stack()
S2 = Stack()
S3 = Stack()
for i in range(N, 0, -1):
 S1.push(i)
print("Initial state of the stacks:")
print_stacks(S1, S2, S3)
print()
step_count = [0]
hanoi(N, S1, S2, S3, step_count)
print("Final state of the stacks:")
print_stacks(S1, S2, S3)
print()
print("Length of S1:", S1.getLen())
print("Length of S2:", S2.getLen())
print("Length of S3:", S3.getLen())
print("Total steps:", step_count[0])

I am attaching my code to solve the Hanoi problem. What do you think? Would you change something? I am attaching the instructions for solving the problem, does the code solve the question? In the role of the 3 towers, 3 cartridges will be used: s1,s2,s3 which initially the numbers 1,2,3,...,n (the big number at the bottom of the stack) are found in s1 and the other two stacks are empty.

It is required to develop a solution characterized by the following distinctions: Arrange the 3 towers in a circle. At every odd step, move the small ring clockwise. In every even step, make the possible move that does not involve the small ring

Requirements for implementing the program: Define n as the number of rings. You will produce 3 cartridges and enter values ​​for the first time. Activate the Hanoi function which accepts 3 cartridges. Requirements for the algorithm: The algorithm will transfer the rings from S1 to one of the other towers. After each transfer of a ring, the contents of the 3 cartridges, and the step number, must be printed. (Each ring transfer is a step).

class Stack:
 def __init__(self):
 self.items = []
 def is_empty(self):
 return self.items == []
 def push(self, item):
 self.items.append(item)
 def pop(self):
 return self.items.pop()
 def getLen(self):
 return len(self.items)
 
def print_stacks(S1, S2, S3):
 print("S1:", S1.items)
 print("S2:", S2.items)
 print("S3:", S3.items)
def hanoi(n, source, helper, target, step_count):
 if n > 0:
 hanoi(n - 1, source, target, helper, step_count)
 target.push(source.pop())
 step_count[0] += 1
 print("Step:", step_count[0])
 print_stacks(S1, S2, S3)
 print()
 hanoi(n - 1, helper, source, target, step_count)
N = 3
S1 = Stack()
S2 = Stack()
S3 = Stack()
for i in range(N, 0, -1):
 S1.push(i)
print("Initial state of the stacks:")
print_stacks(S1, S2, S3)
print()
step_count = [0]
hanoi(N, S1, S2, S3, step_count)
print("Final state of the stacks:")
print_stacks(S1, S2, S3)
print()
print("Length of S1:", S1.getLen())
print("Length of S2:", S2.getLen())
print("Length of S3:", S3.getLen())
print("Total steps:", step_count[0])

THATS MY CODE. IMPROVEMENTS?

class Stack:
 def __init__(self):
 self.items = []
 d

I am attaching my code to solve the Hanoi problem.What do yot think? would you chancge something? I am attaching the instructions for solving the problem, does the code solve the question? In the role of the 3 towers, 3 cartridges will be used: s1,s2,s3 which initially the numbers 1,2,3,...,n (the big number at the bottom of the stack) are found in s1 and the other two stacks are empty.

It is required to develop a solution characterized by the following distinctions: Arrange the 3 towers in a circle At every odd step move the small ring clockwise. In every even step, make the possible move that does not involve the small ring

class Stack:
 def __init__(self):
 self.items = []
 def is_empty(self):
 return self.items == []
 def push(self, item):
 self.items.append(item)
 def pop(self):
 return self.items.pop()
 def getLen(self):
 return len(self.items)
 
def print_stacks(S1, S2, S3):
 print("S1:", S1.items)
 print("S2:", S2.items)
 print("S3:", S3.items)
def hanoi(n, source, helper, target, step_count):
 if n > 0:
 hanoi(n - 1, source, target, helper, step_count)
 target.push(source.pop())
 step_count[0] += 1
 print("Step:", step_count[0])
 print_stacks(S1, S2, S3)
 print()
 hanoi(n - 1, helper, source, target, step_count)
N = 3
S1 = Stack()
S2 = Stack()
S3 = Stack()
for i in range(N, 0, -1):
 S1.push(i)
print("Initial state of the stacks:")
print_stacks(S1, S2, S3)
print()
step_count = [0]
hanoi(N, S1, S2, S3, step_count)
print("Final state of the stacks:")
print_stacks(S1, S2, S3)
print()
print("Length of S1:", S1.getLen())
print("Length of S2:", S2.getLen())
print("Length of S3:", S3.getLen())
print("Total steps:", step_count[0])```

I am attaching my code to solve the Hanoi problem. What do you think? Would you change something? I am attaching the instructions for solving the problem, does the code solve the question? In the role of the 3 towers, 3 cartridges will be used: s1,s2,s3 which initially the numbers 1,2,3,...,n (the big number at the bottom of the stack) are found in s1 and the other two stacks are empty.

It is required to develop a solution characterized by the following distinctions: Arrange the 3 towers in a circle. At every odd step, move the small ring clockwise. In every even step, make the possible move that does not involve the small ring

class Stack:
 def __init__(self):
 self.items = []
 def is_empty(self):
 return self.items == []
 def push(self, item):
 self.items.append(item)
 def pop(self):
 return self.items.pop()
 def getLen(self):
 return len(self.items)
 
def print_stacks(S1, S2, S3):
 print("S1:", S1.items)
 print("S2:", S2.items)
 print("S3:", S3.items)
def hanoi(n, source, helper, target, step_count):
 if n > 0:
 hanoi(n - 1, source, target, helper, step_count)
 target.push(source.pop())
 step_count[0] += 1
 print("Step:", step_count[0])
 print_stacks(S1, S2, S3)
 print()
 hanoi(n - 1, helper, source, target, step_count)
N = 3
S1 = Stack()
S2 = Stack()
S3 = Stack()
for i in range(N, 0, -1):
 S1.push(i)
print("Initial state of the stacks:")
print_stacks(S1, S2, S3)
print()
step_count = [0]
hanoi(N, S1, S2, S3, step_count)
print("Final state of the stacks:")
print_stacks(S1, S2, S3)
print()
print("Length of S1:", S1.getLen())
print("Length of S2:", S2.getLen())
print("Length of S3:", S3.getLen())
print("Total steps:", step_count[0])