bitwXor64 <- function(x, y) {
+ x <- as.bitstring(as.integer64(x))
+ y <- as.bitstring(as.integer64(y))
+
+ base::xor(
+ as.integer(str_split_1(x, "")),
+ as.integer(str_split_1(y, ""))
+ ) |>
+ as.integer() |>
+ str_c(collapse = "") |>
+ structure(class = "bitstring") |>
+ as.integer64() |>
+ as.numeric()
+}
+
+
+
+
+
+
+
+
+Day 22
+Advent of Code: Worked Solutions
+Setup
+
+
+
+# Libraries
+library(tidyverse)
+library(bit64)
+library(memoise)
+
+# Read input from file
+input <- read_lines("../input/day22.txt", skip_empty_rows = TRUE) |>
+ as.numeric()Part 1
+Define custom bitwise XOR function, needed to handle large integers:
+
+
+
+bitwXor64 <- function(x, y) {
+ x <- as.bitstring(as.integer64(x))
+ y <- as.bitstring(as.integer64(y))
+
+ map2_chr(
+ x |> str_split("") |> map(as.integer),
+ y |> str_split("") |> map(as.integer),
+ ~ base::xor(.x, .y) |>
+ as.integer() |>
+ str_c(collapse = "")
+ ) |>
+ structure(class = "bitstring") |>
+ as.integer64() |>
+ as.numeric()
+}Define the algorithm for producing a sequence of “secret” numbers:
+
+
+
+mix <- memoise::memoise(\(a, b) bitwXor64(a, b))
+prune <- \(x) x %% 16777216
+
+secret_alg <- function(x) {
+ x1 <- prune(mix(x, x * 64))
+ x2 <- prune(mix(x1, floor(x1 / 32)))
+ x3 <- prune(mix(x2, x2 * 2048))
+ return(x3)
+}
+
+secret_seq <- function(init, len) {
+ out <- list(init)
+ for (i in 2:len) {
+ out[[i]] <- secret_alg(pluck(out, i - 1))
+ }
+ out
+}Run puzzle input:
+
+
+
+secret_nums <- secret_seq(input, len = 2001)
+
+secret_nums |>
+ tail(n = 1) |>
+ unlist() |>
+ sum()
+
+[1] 20401393616Part 2
+
+
+
+
+
+# Convert sequences to a data frame by buyer and time
+diffs <- secret_nums |>
+ imap_dfr(\(x, idx) tibble(time = idx, secret_number = x)) |>
+ mutate(
+ buyer_id = row_number(),
+ .by = time
+ ) |>
+ mutate(
+ # Get the price at each time by taking the ones digit of each secret number
+ price = secret_number %% 10L,
+ # Compute the difference in price at the current time vs the previous time
+ diff = price - lag(price),
+ # Compute the sequence of 4 price changes preceeding the current price
+ lag1 = lag(diff, n = 1L),
+ lag2 = lag(diff, n = 2L),
+ lag3 = lag(diff, n = 3L),
+ diff_seq = str_c(lag3, lag2, lag1, diff, sep = ","),
+ .by = buyer_id
+ ) |>
+ arrange(buyer_id, time)
+
+# For each price change seq, compute the bananas you will get from each buyer:
+bananas_by_seq <- diffs |>
+ filter(!is.na(diff_seq)) |>
+ summarize(
+ bananas = head(price, 1),
+ .by = c(buyer_id, diff_seq)
+ )
+
+# Find the most advantageous sequence:
+bananas_by_seq |>
+ summarize(bananas = sum(bananas), .by = diff_seq) |>
+ slice_max(bananas) |>
+ pull(bananas)
+
+[1] 2272