Subfield: Combinatorics


Question: An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:

$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$

Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write

$$
S_{3}[263415]=(132,312,231,213)
$$

More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.

In this power question, you will be asked to analyze some of the properties of labels and signatures.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836)



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825)
For a general $n \geq 2$, compute all $n$-labels that have unique 2 -signatures.

Solution:
The $n$-labels with unique 2 -signatures are $1,2, \ldots, n$ and $n, n-1, \ldots, 1$, and their respective 2-signatures are $(12,12, \ldots, 12)$ and $(21,21, \ldots, 21)$.

Proof: (Not required for credit.) Let $L=a_{1}, a_{2}, \ldots, a_{n}$. The first signature above implies that $a_{1}<a_{2}<\cdots<a_{n}$, which forces $a_{1}=1, a_{2}=2$, and so on. Likewise, the second signature forces $a_{1}=n, a_{2}=n-1$, and so on.

To show that all other $n$-labels fail to have unique 2 -signatures, we will show that in any other $n$-label $L^{\prime}$, there are two numbers $k$ and $k+1$ that are not adjacent. By switching $k$ and $k+1$ we get a label $L^{\prime \prime}$ for which $S_{2}\left[L^{\prime}\right]=S_{2}\left[L^{\prime \prime}\right]$, since the differences between $k$ and its neighbors in $L^{\prime}$ were at least 2 (and likewise for $k+1$ ).

To show that such a $k$ and $k+1$ exist, we proceed by contradiction. Suppose that all such pairs are adjacent in $L^{\prime}$. Then 1 and $n$ must be at the ends of $L^{\prime}$ (or else some intermediate number $k$ will fail to be adjacent to $k+1$ or $k-1$ ). But if $a_{1}=1$, then



that forces $a_{2}=2, a_{3}=3, \ldots, a_{n}=n$. That is, we get the two labels we already covered. (We get the other label if $a_{n}=1$.)

Therefore, none of the $n !-2$ remaining labels has a unique 2 -signature.

Final answer: $1,2, \ldots, n$ and $n, n-1, \ldots, 1$