Advent Of Code 2021, Day 7: The Treachery of Whales

by Abigail

Challenge

We are given a set $$\mathcal{S}$$ of numbers:

16,1,2,0,4,2,7,1,2,14


We are looking for another number $$N$$ (integer) which minimizes a certain measurement: the sum of the cost between $$N$$ and each element of $$\mathcal{S}$$.

Part One

For Part One, the cost between two numbers is just the absolute value of their difference.

For the given example set, the answer will be 37.

Part Two

For part two, the cost is defined as:

• if the absolute difference is $$1$$, then the cost is $$1$$.
• if the absolute difference is $$2$$, then the cost is $$1 + 2 = 3$$.
• if the absolute difference is $$3$$, then the cost is $$1 + 2 + 3 = 6$$.
• etc.

That is, if the absolute difference is $$k$$, the cost is the $$k^{\text{th}}$$ triangle number.

For the given example set, the answer will be 168.

Solution

We will present two ways of solving the problem, a brute force one, and one which uses statistics.

For triangle numbers, we have:

$1 + 2 + \ldots + n = \sum_{k = 1}^n k = \frac{n * (n + 1)}{2}$

Perl

First, we read in the data:

my @nums = <> =~ /[0-9]+/g;


Next, we define two functions to calculate costs. cost1 calculates the cost for part one, while cost2 calculates the cost for part two. Both functions take two arguments: a number ($target), and a reference to an array with numbers ($nums). What they return is the sum of the costs between $target and each of the elements of $nums:

sub cost1 ($target,$nums) {
say "Solution 2: ", min map {cost2 $_, \@nums} min (@nums) .. max (@nums);  Find the full program on GitHub. Using Statistics The sum of the costs of part one will be minimized for the median of $$\mathcal{S}$$, while the sum of the costs for part two will be minimized for the arithmetic mean of $$\mathcal{S}$$. Now, both the median and the mean may be non-integers. But if we look at the sum of the costs when going from the minimum value of $$\mathcal{S}$$ to the maximum value of $$\mathcal{S}$$, the sum of the costs will strictly decrease from the minimum value of $$\mathcal{S}$$ to the median/mean, and then strictly increase till we have reached the maximum value of $$\mathcal{S}$$. So, we will be calculating the costs for the median/mean rounded up, and rounded down: the minimum of those will be the answer. Which leads to: my$median = median @nums;
my $mean = mean @nums; say "Solution 1: ", min cost1 (floor ($median), \@nums),
cost1 ( ceil ($median), \@nums); say "Solution 2: ", min cost2 (floor ($mean),   \@nums),
cost2 ( ceil (\$mean),   \@nums);


We are importing min and sum from List::Util, median and mean from Statistics::Basic, and floor and ceil from POSIX.

Find the full program on GitHub.

Visualization

Below is a graph show the fuel costs depending on where the crabs converge.