# The Sequence

The Look & Say sequence was originally introduced and analyzed by John Conway. The sequence begins as follows: $$ \begin{aligned} 1, 11, 21, 1211, 111221, 312211, \dots \end{aligned} $$ To generate a member of the sequence from the previous member, read the digits of the previous member, counting the number of digits in groups of the same digit. For example:

- $1$ is read as "one $1$" or $11$.
- $11$ is read as "two $1$s" or $21$.
- $21$ is read as "one $2$, one $1$" or $1211$.
- $1211$ is read as "one $1$, one $2$, two $1$s" or $111221$.

Notice you can start with a different "seed" or starting number. This leads to a different sequence. Interestingly enough, $22$ is the only seed that leads to a sequence that loops. All other seeds generate a sequence that grows indefinitely.

# The Function Definition

How would you make *this* into a function using only the constructs we've
introduced? This post will be longer than usual because of the many parts
required to complete this task. It's not that daunting, but it's going to
take a bit of explaining on my part.

The function we want to create will take in an integer $x$ and output the next term in the sequence. For example, using the values we computed in the section above, we would have the following input/output pairs: $$ \begin{aligned} L(1) &= 11 \\ L(11) &= 21 \\ L(21) &= 1211 \\ L(1211) &= 111221 \\ &\vdots \end{aligned} $$ The function could also take any positive integer, and not just those from the most common sequence. For example, $$ \begin{aligned} L(6) &= 16 \\ L(73) &= 1713 \\ L(4444) &= 44 \\ &\vdots \end{aligned} $$ This is the function we want to write out. Let's start off with some intuition first to get our feet wet.

# Implementation

## Grouping

When visually computing the next Look & Say number, we group consecutive
similar digits together. For example, we would group the digits of the
number $11244$ as $[11][2][44]$. Then we use the contents of each group
to generate the next term. This "grouping" action can be done numerically
by computing a number that encodes where these groups start and end. One
way to do this is to create a number that has $1$s on the boundaries of
groups and $0$s otherwise. For example, using the number $11244$ again,
we would have
$$
\begin{aligned}
11244 \to {_11_01_12_14_04_1} \to 101101
\end{aligned}
$$
This is equivalent to placing $1$s between digits that differ and
$0$s between digits that are the same. Finally pad this number with $1$s
on the left and right side of the entire integer. Here's how we can do this:
$$
\begin{aligned}
G_{inner}(x) &=
\sum_{i = 0}^{\text{len}(x) - 2} [\text{at}(x, i) \ne \text{at}(x, i + 1)] \cdot 10^i \\
G(x) &= 1 + 10^{\text{len}(x)} + G_{inner}(x) \cdot 10
\end{aligned}
$$
*
Note: I've omitted the base $b$ subscript found in $\text{len}_b(x)$ as
it's assumed to be $10$. This will be done throughout the rest of this post.
*

The keen eye will notice that we haven't defined $\text{neq}$ yet, and therefore should not be able to use $\neq$, but, it's easily defined as $$ \begin{aligned} \text{neq}(x, y) = 1 - \text{eq}(x, y) \end{aligned} $$ The way $G$ works is $G_{inner}$ goes through each adjacent pair of digits in $x$ and "concatenates" a $1$ if they're different and a $0$ otherwise. Notice this doesn't take into account the $1$s that should be at the beginning and the end of this grouping number. To take care of this, $G$ shifts $G_{inner}$ over by one digit, and adds a $1$ at the start and end. These manipulations are easier to think about in terms of string operations, instead of literally adding numerals.

## Counting Groups

### Intuition

The next big step is to iterate over these groups and build up the next Look & Say number. A sentence description of what needs to happen would be something like: For each $1$ in $G(x)$ (from right to left) at index $i$, let $j$ be the index of the next $1$ in $G(x)$ after the one at index $i$, then $at(x, i)$ is the digit that occured $j - i$ times in $x$. If we concatenate these instances of $j - i$ and $at(x, i)$ for each group in $G(x)$, we'll have the next number.

If we remember from previous post, concatenation is hard
because when concatenating $x$ to $y$, we need to know the length of $y$. This
is difficult if you're building up a number via concatenation. In the sorting
case this was easy because each iteration of the loop we would add only one
digit. Here we may add a variable number of digits since $j - i$ could be *any*
positive integer. Thus, we need a function to count the length of the Look &
Say number we have built *so far*, up until group $i$, so that we can
concatenate properly.

Some useful things to notice are:

- The number of $1$s in $G(x)$ is one greater than the number of groups of digits in $x$.
- The number of $0$s in between two $1$s in $G(x)$ is one less than the number of digits in that group.
- In a group bounded by $1$s at indices $i, j$ where $i < j$, the length that group contributes to the next Look & Say number is $\text{len}(j - i) + 1$.

## Formalizing

When iterating through $G(x)$, we only really care about performing operations
where the $1$s are. Thus, we can start off with
$$
\begin{aligned}
L(x) = \sum_{i = 0}^{\text{len}(G(x)) - 2} at(G(x), i) \cdot [\dots]
\end{aligned}
$$
*
Note: The upper bound of the sum is subtracted by $2$ because the last
$1$ in $G(x)$ is not the start boundary of a group, but instead and the end.
*

The $[\dots]$ represents stuff we still need to write. The $\text{at}(G(x), i)$ serves as discarding the computations in $[\dots]$ when the digit at index $i$ is $0$.

Now, as we said in the previous section, we need to find the index of the next $1$, use that to compute the length of the group, and then concatenate that length and the corresponding digit in $x$ to our ongoing computation. If $G_{len}(x, i)$ is the length of the group at index $i$ and $\sigma(x, i)$ is new index of the group length & digit pair, then we have $$ \begin{aligned} \sigma(x, i) &= [\dots] \\ G_{len}(x, i) &= [\dots] \\ L_{pair}(x, i) &= G_{len}(x, i) \cdot 10 + \text{at}(x, i) \\ L(x) &= \sum_{i = 0}^{\text{len}(G(x)) - 1} \text{at}(G(x), i) \cdot L_{pair}(x, i) \cdot 10^{\sigma(x, i)} \end{aligned} $$

## Tackling $\sigma$

I think $\sigma(x, i)$ is pretty easy assuming we already have $G_{len}$. It's
essentially the number of digits that have been concatenated already. Thus,
it's the sum of all of the *lengths* of $L_{pair}$s for indices less than $i$,
since that's what we're concatenating together. Thus, we would write this as
$$
\begin{aligned}
\sigma(x, i) = \sum_{j = 0}^{i - 1} at(G(x), j) \cdot (\text{len}(L_{pair}(x, j)))
\end{aligned}
$$
*Note: again, we don't care about indices where $G(x)$ is $0$.*

## Tackling $G_{len}$

$G_{len}(x, i)$ is simply the length of the group at index $i$ in $x$. So we know the first $1$ is at index $i$. Then we need to find the index of the next $1$. Here's an example $G(x)$ with $i, j$ indicated when $i = 2$: $$ \begin{aligned} G(22344455) &\to {_12_02_13_14_04_04_15_05_1} \to 101100101 \\ G(x) &= 101\underset{j}{1}00\underset{i}{1}01 \\ \end{aligned} $$ The number we care about is $j - i$, since that is equal to the length of the group. We would expect, in this case, that $G_{len}(22344455, 2) = 3$, since the group of $4$s is of length $3$. One way to compute $j - i$ is to truncate the digits at indices before and including $i$ in $G(x)$, then take the length of the reverse of the resulting number and subtract it from the length of the truncation before the reversal and add $1$. Here's a diagram that's hopefully more understandable: $$ \begin{aligned} [10&1\underset{j}{1}00]\underset{i}{1}01 \\ &\downarrow \text{truncate} \\ a = 10&1100 \\ &\downarrow \text{reverse} \\ b = 11&01 \\ &\downarrow \text{compute} \\ 1 + \text{len}(a) &- \text{len}(b) = 3 \end{aligned} $$ We would write that like this: $$ \begin{aligned} G_{tr}(x, i) &= \left\lfloor \frac{G(x)}{10^{i + 1}} \right\rfloor \\ G_{len}(x, i) &= 1 + \text{len}(G_{tr}(x, i)) - \text{len}(\text{rev}(G_{tr}(x, i))) \end{aligned} $$

## Recap

Let's just rewrite all of the functions here for convenience $$ \begin{aligned} G_{inner}(x) &= \sum_{i = 0}^{\text{len}(x) - 2} [\text{at}(x, i) \ne \text{at}(x, i + 1)] \cdot 10^i \\ G(x) &= 1 + 10^{\text{len}(x)} + G_{inner}(x) \cdot 10 \\ \sigma(x, i) &= \sum_{j = 0}^{i - 1} at(G(x), j) \cdot (G_{len}(x, j) + 1) \\ G_{tr}(x, i) &= \left\lfloor \frac{G(x)}{10^{i + 1}} \right\rfloor \\ G_{len}(x, i) &= 1 + \text{len}(G_{tr}(x, i)) - \text{len}(\text{rev}(G_{tr}(x, i))) \\ L_{pair}(x, i) &= G_{len}(x, i) \cdot 10 + \text{at}(x, i) \\ L(x) &= \sum_{i = 0}^{\text{len}(G(x)) - 1} \text{at}(G(x), i) \cdot L_{pair}(x, i) \cdot 10^{\sigma(x, i)} \end{aligned} $$

# Codified Functions

Let's keep with the tradition of writing everything out as python functions.

Does it work?

Heck yea it works! Isn't that crazy? What's even cooler is looking at some $x$ values that we don't usually consider, and see if we can find meaning in those.

Hmmm... Well, I guess that makes sense right? We could interpret this as there being no $0$s, hence we'd expect $L(0) = 00 = 0$. Fair enough.

Ah, this is because $\text{len}$ relies on $\log$, whose domain is only the positive reals. $\text{len}(0) = 0$, but is undefined for any smaller integer. No real way around that I think.

Notice this handles groups of length greater than $9$ just fine. As we'd hopefully expect, $L(21111111113)$ results in reading "one $2$, ten $1$s, and one $3$", which is the output we get. (This is just me showing off).

# Conclusion

This was a pretty fun "adventure" to go on as a high schooler. I previously saw the Look & Say numberphile video and was so intrigued by the sequence. When I came up with this I thought I could be on the show, but of course I was (and still am) just a kid. It took me about 2-3 weeks to come up with this and make sure it actually worked.

I'm not sure how much further we can go with this. I was thinking if it's possible to make a function $L_{idx}(s, i)$ that can take a seed $s$ and an index $i$ without using the super-function notation. That is, this could be written as $$ \begin{aligned} L_{idx}(s, i) = L^{(i)}(s) \end{aligned} $$ but I was wondering if there was a way to do this without introducing any new notation. I haven't come up with anything yet though.

On the next (and likely last) functions post, we'll talk about stuff I haven't figured out yet and maybe gloss over the other interesting functions I haven't yet mentioned.