It is not clear in the question what operations are allowed on the input and how to view or access the output. However, since we are dealing with all permutations, the answer is the same whether the input is a stack or queue and whether the output is a stack or a queue. From now on, we will assume both input and output are stacks.
What is the minimum value of $i$ which can generate all permutations of 1 to $n$ at the output?
The options are: 1) 1 $\;$ 2) 2 $\;$ 3) $n-1$ $\;$ 4) $n$
None of the given options is correct.
- All permutations of 1 to 7 can be generated by 3 stacks.
- The permutation 7132465 cannot be generated by 2 stacks.
The above facts can be verified by an exhaustive search by a program easily. So the minimum value of $i$ which can generate all permutations of 1 to 7 is 3. Note that $3\not=2$ and $3\not=7-1$.
Let $s_n$ be the minimum number of stacks (excluding the input stack and the output stack) which can generate all permutations of any given $n$ elements on the input stack.
It is obvious that $s_n$ is non-decreasing.
We have the following recurrence inequality of $s_n$:
$$s_{2n+1} \le 1+s_n, \forall n\ge0$$
Proof: Suppose we are given $\sigma$, an arbitrary permutation of $1$ to $2n$. Here is an procedure that uses $1+s_n$ stacks to produce $\sigma$, i.e., move all numbers from $1$ to $2n$ on the input stack to the output stack so that the output stack contains $\sigma(1), \sigma(2), \cdots, \sigma(2n)$.
Use the first $s_n$ stacks to move the numbers from $1$ to $n$ on the input stack to the last stack, the $s_n$-th stack so that the numbers on the last stack are in the reverse order of $\sigma$.
Similarly, we are able to move the remaining numbers on the input stack, numbers from $n+1$ to $2n$ to the last stack in exactly the same way. However, we will tweak the procedure as follows.
When a number is about to push to the last stack, we will pop out all the numbers on the last stack that should be pushed to the output stack, i.e., those numbers that appears later than that number in the sequence $\sigma(1), \sigma(2), \cdots$. Then we will push that number to the last stack. Also immediately, pop it out from the last stack to the output stack.
Now suppose we are given $\sigma'$, an arbitrary permutation of $1$ to $2n+1$. Now we will change the procedure above in the following way. Whenever we see the number $\sigma'(2n+1)$, we will move it all the way to the output stack immediately. The task remains is to move all other $2n$ numbers to the output stack as wanted, which can be done by the procedure above.
$\checkmark$
Facts:
- $s_1=0$,
- $s_2=1$,
- $s_3=s_6=2$,
- $s_7=s_{13}=3$,
- $s_{27}\le4$,
- $s_{55}\le5$.
Proof. The permutation $\begin{pmatrix} 123\\312\end{pmatrix}$ cannot be generated by 1 stack. $s_6=2$ can be demonstrated by a short brute-force program in Python within 1 second. The permutation $\begin{pmatrix} 1234567\\7132465\end{pmatrix}$ cannot be generated by 2 stacks.
Other facts are either obvious or easily obtained, thanks to the recurrence inequality proved above.
I suspect $s_{14}=3$.
