diff options
| -rw-r--r-- | src/day7.rs | 177 | ||||
| -rw-r--r-- | src/lib.rs | 1 |
2 files changed, 178 insertions, 0 deletions
diff --git a/src/day7.rs b/src/day7.rs new file mode 100644 index 0000000..9b04ead --- /dev/null +++ b/src/day7.rs @@ -0,0 +1,177 @@ +use aoc_runner_derive::*; + +//This problem has an interesting property that we can exploit. Between the positions of the crabs +//the function we check is linear. +//This means that we can immediately disregard all positions where no crab is at the start of the +//computation. Those points might be a solution, but only if the next crab's location is a solution +//as well. +// +//Another property that we can exploit is that the fuel costs of each crab are linear (except +//directly at its starting position) and extend infinitely. This means that the slope on both +//sides of the global minimum is always strictly positive. In other words: there are no local +//minima. +// +//We could now start searching for the minimum, for instance with a bisection solver. +//(Side note: Newton's method, the first that comes to mind, is not suitable here, as it assumes +//that the function behaves quadratic at the minimum.) +// +//However looking at the above constraints, and at the form of our problem, which basically is +//f(x) = sum(abs(x-pos(crab)), crabs) +// +//we can see another interesting detail: The slope of the derivative of this function is simply +//given by the number of crabs to the right minus the number of crabs to the left. +// +//f'(x) = sum(sign(x-pos(crab)), crabs) +// +//Now that's something, isn't it? +// +//Long story short: we need the position of the first crab, at which there are more crabs left of and at +//the position than there are crabs right of it. +//There is a word for that condition: Median. + +fn compute_fuel_costs_for_position_part_1(input : &Vec<usize>, position : usize) -> usize { + input.iter().map(|c| { + if *c < position { + position - c + } + else { + c - position + } + }).sum() +} + +#[aoc_generator(day7)] +pub fn input_generator(input : &str) -> Result<Vec<usize>, std::num::ParseIntError> { + input.split(",").try_fold(Vec::new(), |mut v,string| { + v.push(string.trim().parse::<usize>()?); + Ok(v) + }) +} + +#[aoc(day7, part1, Median)] +pub fn solve_day7_part1_median(input : &Vec<usize>) -> usize { + let mut input = input.clone(); + if input.len() == 0 { + 0 + } + else { + let midpoint = input.len()/2; + let (_, &mut optimum_position, _) = input.select_nth_unstable(midpoint); + compute_fuel_costs_for_position_part_1(&input, optimum_position) + } +} + +//Part 2 is more "difficult". Here the fuel function is no longer linear, but rather quadratic. +//(see https://de.wikipedia.org/wiki/Gaußsche_Summenformel - no, there's no English wiki page +//on this) +// +//This means all the nice properties we found in the first part are not applicable. On the plus +//side, now we have quadratic behaviour around the minimum, so Newton's method is on the table +//again. +//But before jumping to conclusions, let's analyze the maths. +//The fuel costs for a single crab are (|Δx|+1)*(0.5*|Δx|). Multiplied out we have +//f(Δx) = 0.5 * (Δx²+|Δx|) +//f'(Δx) = Δx + 0.5*sign(Δx) +//The 0.5*sign(Δx) "offsets" all crabs parabolas by 0.5 to the left if viewed from the right, +//or by 0.5 to the right, if viewed from the left. +// +//Still, put in words, the global minimum is there, where the sum of all distances + half the count +//is the same on both sides. +// +//Finding the exact point is not straightforward, but we can get a reasonably good estimate and +//an upper bound for the error. +// +//First, let's write the full function that has to be fulfilled: +//sum((x-pos[i]), i) + 0.5*sum(sign(x-pos[i]),i) = 0 +//sum((x-pos[i]), i) + 0.5*sum(1, i where pos[i] < x) - 0.5*sum(1, i where pos[i] > x) = 0 +//sum((x-pos[i]), i) = 0.5*(n_right - n_left) +// +//Now let's look at how those two values change when we change x by 1. +//The left hand side will change by the total number of crabs, as each term will change by 1. +//That is already the total range the right side can ever cover, as it can, at maximum change by +//the total number of crabs if all crabs were within a single unit. It is also bounded, meaning +//that it can never be larger than half the number of crabs (if all crabs are at the same side). +// +//In other words, if we find a solution to +//sum((x-pos[i]),i) = 0 +//we are already at most 1 unit away from the exact solution, and the exact solution is always in +//the direction that makes the imbalance 0.5*(n_right - n_left) smaller, or, put differently, +//towards the median. +// +//This information about the direction allows us to further reduce our error estimate. The latest +//after we have passed half the number of crabs, the sign of the imbalance 0.5*(n_right - n_left) +//must flip, meaning that the solution to sum((x-pos[i]),i) = 0 actually can never be more than 0.5 +//units away from the true solution. (Assuming otherwise would need us to correct our result in +//both directions at the same time, what is absurd -> point proven.) +// +//We could have gotten the same reduction in error by exploiting the quantization of the problem. +//With full integer steps it is not possible to move "over" a crab by moving one step, at best we +//can move "on" or "off" a crab. This already limits the maximum change of the imbalance to half +//the number of crabs. +// +//Now knowing that the maximum error from solving +//sum((x-pos[i]),i) = 0 +//is 0.5 steps, we can limit our search range to two integer position values, namely the rounded +//solution, and the rounded solution ±1 in direction towards the median. +// +//The condition that fulfills +//sum((x-pos[i]),i) = 0 +//again is a well known quantity: The arithmetic mean. +// +//So, long story short: If we test the fuel consumption at the rounded arithmetic mean of the +//crab positions, the two neighbouring positions, and then pick the result with the lowest value, +//we have a solution. + +fn compute_fuel_costs_for_position_part2(input : &Vec<usize>, position : usize) -> usize { + input.iter() + .map(|&x| if x > position { x-position } else { position - x }) + .map(|dx| dx*dx+dx).sum::<usize>() + /2 +} + +fn rounded_arithmetic_mean(v : &Vec<usize>) -> usize { + ((2*v.iter().sum::<usize>()) / v.len() + 1) / 2 +} + +#[derive(Debug)] +pub struct NoInputError; +impl std::fmt::Display for NoInputError { + fn fmt(&self, f : &mut std::fmt::Formatter) -> Result<(), std::fmt::Error> { + write!(f, "No input from generator. Could not compute fuel consumption.") + } +} +impl std::error::Error for NoInputError {} + +#[aoc(day7,part2,Mean)] +pub fn solve_part2_mean(input :&Vec<usize>) -> Result<usize, NoInputError> { + //computing the fuel consumption a third time is probably cheaper than finding the median. + //Let's just try three values, take the best one. + let mean = rounded_arithmetic_mean(input); + (mean-1..=mean+1).map(|position| compute_fuel_costs_for_position_part2(input,position)).min() + .ok_or(NoInputError) +} + +#[cfg(test)] +mod tests { + use super::*; + fn get_day7_test_string() -> &'static str { + "16,1,2,0,4,2,7,1,2,14" + } + #[test] + fn test_day7_generator() { + let input = input_generator(get_day7_test_string()); + assert_eq!(input, Ok(vec![16,1,2,0,4,2,7,1,2,14])) + } + #[test] + fn test_day7_part1_median() { + let input = input_generator(get_day7_test_string()).unwrap(); + let result = solve_day7_part1_median(&input); + assert_eq!(result, 37); + } + #[test] + fn test_day7_part2_average() { + let input = input_generator(get_day7_test_string()).unwrap(); + let result = solve_part2_mean(&input).unwrap(); + assert_eq!(result,168) + } +} @@ -6,5 +6,6 @@ pub mod day3; pub mod day4; pub mod day5; pub mod day6; +pub mod day7; aoc_lib!{ year = 2021 } |