stack = []
# function for pushing element to stack
def push(element):
    stack.append(element)
    
# function for popping element from stack
def pop():
    if len(stack)!=0:
        top = stack[-1]
        stack = stack[:-1]
        return top
    else:
        return "stack is empty"
# NAME_1  iterative using a stack
def NAME_1_iterative(num_disks, src, dest, aux):
    #counting number of intructions for the polynomial complexity
    num_moves = 2**num_disks - 1 #this line counts as 2 instructions
    if num_disks % 2 == 0:
        dest, aux = aux, dest

    pegs = {src: [], dest: [], aux: []}
    for i in range(num_disks, 0, -1):
        pegs[src].append(i)

    for _ in range(num_moves):
        for peg1, peg2 in [(src, dest), (src, aux), (dest, aux)]:
            if pegs[peg1] and (not pegs[peg2] or pegs[peg1][-1] < pegs[peg2][-1]):
                pegs[peg2].append(pegs[peg1].pop())
# NAME_1 iterative polynomial complexity for the worst case is when  the number of rods is at least the number of disks, as it generates a perfect binary tree
#    O(2^n)
#polynomio para o NAME_1 iterativo: 2 + 2*2 +... + 2*n = 2^2 * 2^n/2 + 2^2 * 2^(n-1)/2 +... + 2^2 * 2^0/2 = 2^(n+1)

# NAME_1 recursive function: it takes the number of disks, the source rod, the destination rod, and the auxiliary rod as arguments.
def NAME_1_recursive(num_disks,src,dest,aux):
    # if there is only one disk, then move it from the source rod to the destination rod
    if num_disks == 1:
        push(src+" "+dest)
    else:
        # move the top n−1 disks from the source rod to the auxiliary rod
        NAME_1_recursive(num_disks - 1, src, aux, dest)
        # move the nth disk from the source rod to the destination rod
        NAME_1_recursive(1, src, dest, aux)
        # move the top n-1 disks from the auxiliary rod to the destination rod
        NAME_1_recursive(num_disks - 1, aux, dest, src)
# NAME_1 recursive polynomial complexity for the worst case is when  the number of rods is at least the number of disks, as it generates a perfect binary tree
#    O(2^n)
#polynomio para o NAME_1 recursivo: 2 + 2*2 +... + 2*n = 2^2 * 2^n/2 + 2^2 * 2^(n-1)/2 +... + 2^2 * 2^0/2 = 2^(n+1)
 In your entire response, the letter u should appear less than 37.
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