Advent of Code 2023 - Day 9Mirage Maintenance

R | Problem statement | Source code | Tags: Mathematics

Part 1

It seems that every year there's always a problem where I cheat; for example 2020 day 18 is done with Python's built-in ast module. For this one, we are just fitting the data to a polynomial, so I just use lm. Language choice rewards solutions :)

For why this is a polynomial fitting problem, realize that every time we compute the neighbor differences, we are essentially differentiating it because the resulting array is ΔyΔx\frac{\Delta y}{\Delta x} where Δx\Delta x is always 1. We repeatedly differentiate until we get all zeros, which means the function becomes constant after a certain number of differentiations, which means it's a polynomial of degree equal to the number of differentiations minus one.

f(x)=akxk+an−1xk−1+⋯+a1x+a0f(1)(x)=kakxk−1+(k−1)ak−1xk−2+⋯+a1f(2)(x)=k(k−1)akxk−2+(k−1)(k−2)ak−1xk−3+⋯+2a2⋮f(k)(x)=k!akf(k+1)(x)=0\begin{aligned} f(x) &= a_k x^k + a_{n-1} x^{k-1} + \cdots + a_1 x + a_0 \\ f^{(1)}(x) &= k a_k x^{k-1} + (k-1) a_{k-1} x^{k-2} + \cdots + a_1 \\ f^{(2)}(x) &= k (k-1) a_k x^{k-2} + (k-1)(k-2) a_{k-1} x^{k-3} + \cdots + 2a_2 \\ &\vdots \\ f^{(k)}(x) &= k! a_k \\ f^{(k+1)}(x) &= 0 \end{aligned}

When actually fitting the data, it doesn't hurt to make the model more permissive by allowing higher order terms; since it's guaranteed to be a perfect fit, the extra terms will just be zero. Each differentiation reduces the length by 1, and we need at least 1 number left (which is zero) to assert that the differences are all zero, so we can at most differentiate length(data) - 1 times, which means the polynomial can at most have degree length(data) - 2. After obtaining the model, the next value is just plugging in length(data) + 1 into the model.

predict_next <- function(seq) {
  n <- length(seq)
  x <- 1:n
  fit <- lm(seq ~ poly(x, n - 2))
  predict(fit, newdata = data.frame(x = n + 1))[[1]]
}
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Part 2

Predicting the previous value just requires plugging in x = 0 instead of x = n + 1. I added a round because a lot of these values are close to 0 and have some floating point error, and the problem guarantees that the answer is an integer.

round(predict(fit, newdata = data.frame(x = 0))[[1]])
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