Another year has come around quickly and so another chance to have some fun solving the Advent Of Code puzzles for 2023. Once again, I will be attempting to solve these puzzles using Jactl. The version I am using is the latest version (1.3.1 as of time of writing).

Day 1 - Trebuchet?!

See Day 1 for a detailed description of the problem.

To run the Jactl shown here save the code into a file (e.g. advent01.jactl) and run it like this (where advent01.txt is your input file from the Advent of Code site for Day 1):

$ cat advent01.txt | java -jar jactl-2.1.0.jar advent01.jactl

Part 1

For Part 1 we are given input where each line contains set of characters and digits. The idea is to find the first and last digits and combine them into a two-digit number and sum these numbers across all the lines of input. If there is only one digit in the line then that digit should be used twice.

So input like:

abc2def3x
xyz4ynn

Would produce 23 for the first line and 44 for the second line.

A nice simple problem for day 1 as they usually are:

stream(nextLine).map{ /^.*?(\d)(.*(\d))?/n; 10*$1+($3?:$1) }.sum()

I just used a regex to extract both the digits and then multiply the first digit by 10 before adding the second digit. The only tricky part was getting the optional matching for the last digit to work.

The first part of the pattern ^.*?(\d) is pretty straightforward. It says to match any character from the beginning of the line until a digit is found. The digit will be in the first capture group so $1 will end up having the value of this digit. The non-greedy match .*? means that it will stop on at the first digit.

The next part of the pattern is (.*(\d))?. This says to optionally do a greedy match on any character followed by a digit. Since we use a greedy match we will only match a digit when it is the last digit on the line. There are two groups here and the inner group will capture the value of the last digit (if it exists) as $3. If there is no match for a second digit then $3 will be null so we use $3 ?: $1 to get the value of $3 if it exists or return $1 when $3 is null.

Part 2

For Part 2 we need to treat lines that have digit spelled out as letters as well as digits themselves. So assuming we have a line like this:

xoney3zfour2z

We then need to convert that to 12 rather than 32.

One of the problems I encountered is that making it work for examples like the one above was not too hard but if the line was instead something like this:

x3zatwoneyz

we have to get 31 but my initial naive approach would return 32 because the two was matched even though one was actually the match we wanted.

The other issue is support input like:

threeight

In theory this should return 38 as the value. It is not clear from the problem description whether this is possible or not but I decided to make the solution cater for this type of input anyway. While it probably possible to make a single regex which can support overlapping digits like this, it was easier to split it up into one regex for the first digit and one for the second.

In the end this is the solution I came up with:

def vals = [zero:0, one:1, two:2, three:3, four:4, five:5, six:6, seven:7, eight:8, nine:9] + (10.map{ ["$it",it] } as Map)
def n = vals.map{it[0]}.join('|')      // number pattern
stream(nextLine).map{ [/^.*?($n)/r, vals[$1], /.*($n).*$/r, vals[$1]] }
                .map{ 10 * it[1] + it[3] }
                .sum()

The approach I took was to construct a map of words to numeric values, including the digits themselves as keys in the map. Then I create a pattern string n for matching the words or digits. It ends up being:

zero|one|two|three|four|five|six|seven|eight|nine|0|1|2|3|4|5|6|7|8|9

We then use two patterns. One for the first digit /^.*?($n)/ which uses a non-greedy match for characters at the beginning of the line to therefore find the first set of characters that matches our digits pattern stored in n. The $1 value then ends up being the matching value which we look up in vals to get the numeric value.

The second regex is /.*($n).*$/ where we use a greedy match to grab everything possible at the start of the line and then find the characters matching $n. Since we used a greedy match we know that there are no matches earlier in the line. Again $1 will have the result.

The map method converts each line to a list of four values being:

  1. Regex match value for first digit (always true)
  2. The numeric value of the matched digit characters,
  3. The regex match value for the second digit (always true), and
  4. The numeric value of the matched second digit characters.

Then we calculate the number by multiplying the first digit by 10 and adding the second one and finally, sum all of these numbers together.