Why is sorting pancakes NP hard?

Question

An article posted yesterday (http://www.i-programmer.info/news/112-theory/3280-pancake-flipping-is-hard-np-hard.html) references a new study released on Arxiv (http://arxiv.org/abs/1111.0434v1) with the following summary:

Pancake Flipping is the problem of sorting a stack of pancakes of different sizes (that is, a permutation), when the only allowed operation is to insert a spatula anywhere in the stack and to flip the pancakes above it (that is, to perform a prefix reversal). In the burnt variant, one side of each pancake is marked as burnt, and it is required to finish with all pancakes having the burnt side down. Computing the optimal scenario for any stack of pancakes and determining the worst-case stack for any stack size have been challenges over more than three decades. Beyond being an intriguing combinatorial problem in itself, it also yields applications, e.g. in parallel computing and computational biology. In this paper, we show that the Pancake Flipping problem, in its original (unburnt) variant, is NP-hard, thus answering the long-standing question of its computational complexity.

The premise is that we don't even know the upper bound maximum for number of flips for a stack any larger than $19$. The instant I read this, I had the solution, and I want you to tell me why I am wrong.

The maximum number of flips for any stack is $2 \times$(number of items $- 1$), and it is a supremely trivial algorithm to implement.

The process is only a single one-or-two-flip step repeated until completion:

• Division point is directly beneath the largest unsorted pancake (unless its on top)
• Flip one brings it to the top (unless its on top)
• Division point #$2$ is directly above the largest properly sorted pancake
• Flip two properly sorts another pancake

From the PDF:

Figure $1$: Examples of efficient flips. Sequence $5$; $2$; $3$; $1$; $4$ is efficiently sortable (in four flips), but $5$; $2$; $3$; $4$; $1$ is not.

I beg to differ...
$52314$. The steps with this approach are: $41325, 23145, 32145, 12345$.
$52341$. The steps with this approach are: $14325, 41325, 23145, 32145, 12345.$
Are they really saying that $4$ is efficient but $5$ is not?

Answer 1

They are saying exactly that - it's easy to see that any stack can be flipped in $2n-1$ flips as you point out, but the question is, what's the precise bound? It does have to be roughly linear by a counting argument ($k$ flips can only bring about $O(n^k)$ configurations and there are $n!\approx n^ne^{-n}$ configurations of $n$ pancakes), but getting the precise constant on the linear term is harder than it appears. The Wikipedia page suggests that the constant is between $15/14$ and $18/11$, better than your naive $2n$ approach.

It's almost more surprising that the problem has been open this long - for instance, many efficiently-solvable puzzles such as the $15$ puzzle have 'easy' solutions but hard bounds on finding optimal solutions. Finding optimal configurations is often hard, and it's not surprising that it's hard here.