dt2 <- as.numeric(data.table::fread("input.txt", header = FALSE)[1])
t <- table(dt2)
v <- c(0, 0, t, rep(0, 8 - length(t)))
for(i in 1:80)
v <- c(v[2:7], v[8] + v[1], v[9:10], v[2])
sum(v)[1] 3631012021-06: Lanternfish
Part 1
Part 2
Speed Edition
This was a classic Advent Of Code misdirection. Once I figured it out, my solution was quick to write and execute.
Fastest solution
# part 1 and 2
dt2 <- c(1,3,3,4,5,1,1,1,1,1,1,2,1,4,1,1,1,5,2,2,4,3,1,1,2,5,4,2,2,3,1,2,3,2,1,
1,4,4,2,4,4,1,2,4,3,3,3,1,1,3,4,5,2,5,1,2,5,1,1,1,3,2,3,3,1,4,1,1,4,1,
4,1,1,1,1,5,4,2,1,2,2,5,5,1,1,1,1,2,1,1,1,1,3,2,3,1,4,3,1,1,3,1,1,1,1,
3,3,4,5,1,1,5,4,4,4,4,2,5,1,1,2,5,1,3,4,4,1,4,1,5,5,2,4,5,1,1,3,1,3,1,
4,1,3,1,2,2,1,5,1,5,1,3,1,3,1,4,1,4,5,1,4,5,1,1,5,2,2,4,5,1,3,2,4,2,1,
1,1,2,1,2,1,3,4,4,2,2,4,2,1,4,1,3,1,3,5,3,1,1,2,2,1,5,2,1,1,1,1,1,5,4,
3,5,3,3,1,5,5,4,4,2,1,1,1,2,5,3,3,2,1,1,1,5,5,3,1,4,4,2,4,2,1,1,1,5,1,
2,4,1,3,4,4,2,1,4,2,1,3,4,3,3,2,3,1,5,3,1,1,5,1,2,2,4,4,1,2,3,1,2,1,1,
2,1,1,1,2,3,5,5,1,2,3,1,3,5,4,2,1,3,3,4)
fishTable <- c(0, 0, table(dt2), rep(0, 8 - max(dt2)))
fish <- function(v, d) {
if (d == 0) {
return(sum(v, digits = 999))
} else if (d == 176) {
print(sum(v))
}
fish(c(v[2:7], v[8] + v[1], v[9:10], v[2]), d - 1)
}
fish(fishTable, 256)[1] 363101[1] 1644286075023Benchmark
bench <- rbenchmark::benchmark(
"First try" = {
# part 1
dt <- as.numeric(unlist(stringr::str_split(readLines("input.txt"), ",")))
dt2 <- dt
for(d in 1:80) {
dt2 <- dt2 - 1
if(sum(dt2 == -1) > 0) {
dt2 <- append(dt2, rep(8, sum(dt2 == -1)))
dt2[dt2 == -1] <- 6
}
}
s <- length(dt2)
# part 2
dt <- as.numeric(unlist(stringr::str_split(readLines("input2.txt"), ",")))
dt2 <- dt
dt <- data.frame("m" = 0, "zero" = 0, "one" = 0, "two" = 0,
"three" = 0, "four" = 0, "five" = 0, "six" = 0,
"seven" = 0, "eight" = 0)
for(i in 1:length(dt2)) {
if(dt2[i] == 0) {dt$zero[1] <- dt$zero[1] + 1}
if(dt2[i] == 1) {dt$one[1] <- dt$one[1] + 1}
if(dt2[i] == 2) {dt$two[1] <- dt$two[1] + 1}
if(dt2[i] == 3) {dt$three[1] <- dt$three[1] + 1}
if(dt2[i] == 4) {dt$four[1] <- dt$four[1] + 1}
if(dt2[i] == 5) {dt$five[1] <- dt$five[1] + 1}
if(dt2[i] == 6) {dt$six[1] <- dt$six[1] + 1}
if(dt2[i] == 7) {dt$seven[1] <- dt$seven[1] + 1}
if(dt2[i] == 8) {dt$eight[1] <- dt$eight[1] + 1}
}
for (i in 1:255) {
dt$zero[1] <- dt$one[1]
dt$one[1] <- dt$two[1]
dt$two[1] <- dt$three[1]
dt$three[1] <- dt$four[1]
dt$four[1] <- dt$five[1]
dt$five[1] <- dt$six[1]
dt$six[1] <- dt$seven[1] + dt$m[1]
dt$seven[1] <- dt$eight[1]
dt$eight[1] <- dt$m[1]
dt$m[1] <- dt$zero[1]
}
options(scipen = 999)
s <- sum(dt[1, ])
},
"Third try" = {
# part 1 and 2
dt2 <- as.numeric(data.table::fread("input.txt", header = FALSE)[1])
t <- table(dt2)
v <- c(0, 0, t, rep(0, 8 - length(t)))
for (i in 1:256) {
v <- c(v[2:7], v[8] + v[1], v[9:10], v[2])
if(i == 80)
s <- sum(v)
}
options(scipen = 999)
s <- sum(v)
},
"With recursion!" = {
# part 1 and 2
dt2 <- c(1,3,3,4,5,1,1,1,1,1,1,2,1,4,1,1,1,5,2,2,4,3,1,1,2,5,4,2,2,3,1,2,3,
2,1,1,4,4,2,4,4,1,2,4,3,3,3,1,1,3,4,5,2,5,1,2,5,1,1,1,3,2,3,3,1,4,
1,1,4,1,4,1,1,1,1,5,4,2,1,2,2,5,5,1,1,1,1,2,1,1,1,1,3,2,3,1,4,3,1,
1,3,1,1,1,1,3,3,4,5,1,1,5,4,4,4,4,2,5,1,1,2,5,1,3,4,4,1,4,1,5,5,2,
4,5,1,1,3,1,3,1,4,1,3,1,2,2,1,5,1,5,1,3,1,3,1,4,1,4,5,1,4,5,1,1,5,
2,2,4,5,1,3,2,4,2,1,1,1,2,1,2,1,3,4,4,2,2,4,2,1,4,1,3,1,3,5,3,1,1,
2,2,1,5,2,1,1,1,1,1,5,4,3,5,3,3,1,5,5,4,4,2,1,1,1,2,5,3,3,2,1,1,1,
5,5,3,1,4,4,2,4,2,1,1,1,5,1,2,4,1,3,4,4,2,1,4,2,1,3,4,3,3,2,3,1,5,
3,1,1,5,1,2,2,4,4,1,2,3,1,2,1,1,2,1,1,1,2,3,5,5,1,2,3,1,3,5,4,2,1,
3,3,4)
fishTable <- c(0, 0, table(dt2), rep(0, 8 - max(dt2)))
fish <- function(v, d) {
if (d == 0) {
return(sum(v, digits = 999))
} else if (d == 176) {
s <- sum(v)
}
fish(c(v[2:7], v[8] + v[1], v[9:10], v[2]), d - 1)
}
s <- fish(fishTable, 256)
},
replications = 100, columns = c(1:5), order = "user.self")
bench$per <- bench$user.self / bench$replications
bench test replications user.self sys.self elapsed per
3 With recursion! 100 0.079 0.001 0.079 0.00079
2 Third try 100 0.437 0.012 0.450 0.00437
1 First try 100 7.977 1.149 9.155 0.07977