I liked the twist in the Day 13 puzzle. Implementing a literal solution to part 1, where I had the scanners walk back and forth, was straightforward. And Part 2 looked easy enough. Then I peeked at how slowly my algorithm was proceeding and realized I would need to refactor.

### Part 1

This function has been modified in two places for Part 2. I added the first line in the function, using the modulo operator `%%`

to trim the time down to a single back-and-forth pass. This made the commented-out direction change line obsolete.

I worked out the `time %% ((size-1)*2)`

bit by counting examples on my fingers 🙂

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# Time corresponds to "depth", size = "range" move <- function(size, time){ time <- time %% ((size-1)*2) # added this line for part 2 to be O(n) not O(n^2) pos <- 1 increment <- 1 for(i in seq_len(time)){ # with the modulo line this could be done w/o loop # if(pos == 1) { increment <- 1 } # Don't need this b/c of adding the modulo if(pos == size) { increment <- -1} pos <- pos + increment } pos } library(testthat) expect_equal(move(3, 0), 1) expect_equal(move(2, 1), 2) expect_equal(move(4, 4), 3) expect_equal(move(4, 6), 1) dat <- read.csv("13_dat.txt", sep = ":", header = FALSE) %>% setNames(c("depth", "size")) sum((dat$size * dat$depth)[with(dat, mapply(move, size, depth)) == 1]) # 1728 |

### Part 2

Without the modulo line above, my loop will walk the scanner back and forth for `time`

steps. As the delay increases, so does the time. The answer to this problem is a delay of approximately 4 million time units; at that point, each scanner is walked 4 million times… it’s been a long time since CS 201 but I think that makes the algorithm O(N^{2}). In practice, I realized I had a problem when progress slowed dramatically.

Realizing this, I eliminated the unnecessary walking. Now, this could still be much more efficient. Because I’m recycling code from part 1, I test all scanner depths for collisions at each iteration; a faster solution would move on to the next delay value after a single scanner collision is found. But hey, that is a mere ~20x slowdown or so, and is still O(N).

Having started with R in 2014, I rarely feel like an old-timer. I learned dplyr from the beginning, not `tapply`

. But I had that feeling here… being somewhat comfortable with `mapply`

gets in the way of sitting down to learn the `map2`

functions from purrr, which I suspect will be useful to know.

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# Part 2 system.time({ delay <- 0 hits <- sum(with(dat, mapply(move, size, depth)) == 1) while( hits != 0 ){ dat$depth <- dat$depth + 1 delay <- delay + 1 hits <- sum(with(dat, mapply(move, size, depth)) == 1) } } ) user system elapsed 906.512 2.804 909.319 > delay [1] 3946838 |

My runtime was about 15 minutes.

Looking at another solution, https://github.com/exunckly/Advent2017/blob/master/day13.R, I realize how literal and inefficient my approach is.

I calculate the scanner’s position, but that’s not important. All that matters is whether the scanner’s position is 1, which can be determined by the period of the scanner’s oscillation and returned as a logical.

Would you mind showing the process of working out this bit time %% ((size-1)*2)?

Thanks

If the range of the scanner (“size”) above is X, it will take (x-1)*2 time steps to go from the first position down to the end and then back to the first position. It needs to go down X and back X, which would be 2X, except that at each end it doesn’t pause for a beat, it bounces back. Those end positions get touched only once. So 2X-2, which simplifies to (x-1)*2.

Try it out with your hand, where size = 5. To count from your pinky to your thumb and back is 8 … (5-1)*2. Do it again and you’re at 16, then 24, etc. Using %% to get the remainder avoids the time of walking the scanner back through all of these full periods. It only walks the final period of (x-1)*2 or fewer steps.