# How to Solve the Tower of Hanoi Puzzle Using a For Loop in Python?

The Tower of Hanoi puzzle is a classic problem in computer science and mathematics that involves moving a stack of disks from one peg to another, using a third peg as a temporary buffer, such that no larger disk is ever placed on top of a smaller disk.

Here is an example of solving the Tower of Hanoi puzzle using a for loop in Python:

``````def tower_of_hanoi(n, source, dest, temp):
for i in range(1, 2**n):
if i & i-1:
if i % 3 == 1:
move_disk(source, dest)
elif i % 3 == 2:
move_disk(source, temp)
else:
move_disk(temp, dest)
else:
if n % 2 == 0:
move_disk(source, temp)
else:
move_disk(source, dest)``````

Here, `n` represents the number of disks, `source` represents the starting peg, `dest` represents the destination peg, and `temp` represents the temporary buffer peg. The `move_disk` function is not defined in this example, but it can be a simple function that takes two arguments representing the source and destination pegs and prints a message indicating that a disk has been moved from the source peg to the destination peg.

The for loop iterates `2^n - 1` times, where `n` is the number of disks. The `if i & i-1` condition checks if the current iteration number is a power of 2 or not. If it is, then we are at the start of a new round of moves, and we need to determine whether to move the top disk from the `source` peg to the `dest` peg, the `source` peg to the `temp` peg, or the `temp` peg to the `dest` peg. If it’s not a power of 2, then we are in the middle of a round of moves, and we simply move the top disk from the `source` peg to the `temp` peg if `n` is even, or from the `source` peg to the `dest` peg if `n` is odd.

Note that this implementation assumes that the number of disks is less than or equal to the maximum integer representable in Python, which is `sys.maxsize` on most systems.