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 $(12,12,12,21,12,21)$ :

<img_3144>



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:

<img_3777>
Show that for each $k \geq 2$, the number of unique $2^{k-1}$-signatures on the set of $2^{k}$-labels is at least $2^{2^{k}-3}$.

Solution:
Let $s_{k}$ denote the number of such unique signatures. We proceed by induction with base case $k=2$. From $8(\mathrm{c})$, a 2-signature for a label $L$ is unique if and only if consecutive numbers in $L$ appear together in some window. Because $k=2$, the consecutive numbers must be adjacent


in the label. The $2^{2}$-labels 1234 and 4321 satisfy this condition, ${ }^{1}$ so their $2^{1}$-signatures are unique. Thus we have shown that $s_{2} \geq 2=2^{2^{2}-3}$, and the base case is established.

Now suppose $s_{k} \geq 2^{2^{k}-3}$ for some $k \geq 2$. Let $L_{k}$ be a $2^{k}$-label with a unique $2^{k-1}$-signature. Write $L_{k}=\left(a_{1}, a_{2}, \ldots, a_{2^{k}}\right)$. We will expand $L_{k}$ to form a $2^{k+1}$ label by replacing each $a_{i}$ above with the numbers $2 a_{i}-1$ and $2 a_{i}$ (in some order). This process produces a valid $2^{k+1}$-label, because the numbers produced are all the integers from 1 to $2^{k+1}$. Furthermore, different $L_{k}$ 's will produce different labels: if the starting labels differ at place $i$, then the new labels will differ at places $2 i-1$ and $2 i$. Therefore, each starting label produces $2^{2^{k}}$ distinct $2^{k+1}$ labels through this process. Summarizing, each valid $2^{k}$-label can be expanded to produce $2^{2^{k}}$ distinct $2^{k+1}$-labels, none of which could be obtained by expanding any other $2^{k}$-label.

It remains to be shown that the new label has a unique $2^{k-1}$-signature. Because $L_{k}$ has a unique $2^{k-1}$-signature, for all $i \leq 2^{k}-1$, both $i$ and $i+1$ appeared in some $2^{k-1}$-window. Therefore, there were fewer than $2^{k-1}-1$ numbers between $i$ and $i+1$. When the label is expanded, $2 i$ and $2 i-1$ are adjacent, $2 i+1$ and $2 i+2$ are adjacent, and $2 i$ and $2 i+1$ are fewer than $2 \cdot\left(2^{k-1}-1\right)+2=2^{k}$ places apart. Thus, every pair of adjacent integers is within some $2^{k}$-window.

Since each pair of consecutive integers in our new $2^{k+1}$-label coexists in some $2^{k}$-window for every possible such expansion of $L_{k}$, that means all $2^{2^{k}}$ ways of expanding $L_{k}$ to a $2^{k+1}$-label result in labels with unique $2^{k}$-signatures. We then get

$$
\begin{aligned}
s_{k+1} & \geq 2^{2^{k}} \cdot s_{k} \\
& \geq 2^{2^{k}} \cdot 2^{2^{k}-3} \\
& =2^{2^{k+1}-3}
\end{aligned}
$$

which completes the induction.