mit 6.0001 recursion - towers of hanoi

towers_rec.py

def towers_rec(n, home, destination, temp):
  if n == 1:
    moves.append((home+' to '+destination))
  else:
    towers_rec(n-1, home, temp, destination)
    towers_rec(1, home, destination, temp)
    towers_rec(n-1, temp, destination, home)

if __name__ == "__main__":
  print('Elements are stacked at position A, which is \'home\'.\n  \
  Move them to position B, which is \'destination\'.\nImportant!   \
  The home stack is ranked as:\nbottom biggest, top smallest...\n  \
  and the destination stack must be ranked the same way.\nThere\'s \
  one spare position you can use called \'temp\' which starts at   \
  position C.\nTwo rules:\n1. move one element at a time, and\n2.  \
  never place an element on top of a smaller element.')
  moves=[]
  n = int(input('How many stacked rings?: '))
  towers_rec(n, 'A', 'B', 'C')
  print(f'The winning set of moves: {moves}\nIt took (2^n)-1 = {len(moves)} moves!')