Ellis Island

Here we go then, the 1st day of 2023’s Advent of Code. 

Day 1 looked deceptively simple, with part 1 completed in pretty quick time. The only real problem was the large amount of data caused memory overflow issues, so I split it into 4 separate parts for the data statements.

I’ll list some code segments with explanation on how I did it below, but I don’t want to post the answers for this current year on here for obvious reasons.

First off, declaring the variables:

I used two subroutines to do the number crunching. It’s not very complicated, this one is used to find the left most number. It’s first set to 100 so we have a benchmark to compare against later on. It then simply loops through each character based on the length of the string and compares it to a digit.

LN (Left Number) is set to a multiple of 10 to satisfy the problem – so if 3 is found first, it needs to be set to 30 as we want to add the single last digit of the string to it. i.e. if the last digit found is 5 then the value for this line would be 35.

Once a left number is found, the loop obviously needs to exit to prevent it finding any more numbers – so we exit as soon as LN is less than 100 (i.e. any of the other values).

SUB FINDNUMSL (TEST$ AS STRING * 96) STATIC
L=LEN(TEST$)
LN = 100  'set to an unfindable value - only required for LN
FOR I=0 TO L-1
    M=MID$(TEST$, I, 1)
    ' Test for left numeric only
    IF M="0" THEN 
    LN=0 
    END IF
    IF M="1" THEN 
    LN=10 
    END IF
    IF M="2" THEN 
    LN=20
    END IF
    IF M="3" THEN 
    LN=30 
    END IF
    IF M="4" THEN 
    LN=40 
    END IF
    IF M="5" THEN 
    LN=50 
    END IF
    IF M="6" THEN 
    LN=60 
    END IF
    IF M="7" THEN 
    LN=70 
    END IF
    IF M="8" THEN 
    LN=80 
    END IF
    IF M="9" THEN 
    LN=90
    END IF
    IF LN < 100 THEN ' We need to exit the loop as soon as a numeric is found
    EXIT FOR
    END IF
NEXT I
    ' Left most digit found
    ' End subroutine
END SUB

The subroutine for the right most number, imaginatively called FINDNUMSR, is almost identical except for 2 differences. We set RN to the single digit as opposed to a multiple of 10, as that will be added to LN, and we don’t exit the loop, meaning we keep going until the end so that RN holds the last value in that string.

With the SUBS setup it’s just a matter of cycling through each input line, calling them both, and totalling up the values as we go.

And as I said, I did this 3 more times for the other 750 lines of data, which yes is a bit cheaty but then XC=BASIC makes me lazy.

Part 2 was trickier as we had to find actual strings of digits that would take precedent over the digits themselves, so if the string looked something like ‘iuskjnineij5hkjlcnd6slkj8klmsstwo‘ the result we want is 92.

To accomplish this I just did more of the same and added extra logic into both subroutines, again exiting the loop early for the left number and letting it complete for the right number.

    'Part 2
    IF M = "o" AND MID$(TEST$,I+1,1) = "n" AND MID$(TEST$,I+2,1) = "e" THEN
    LN=10
    END IF
    IF M = "t" AND MID$(TEST$,I+1,1) = "w" AND MID$(TEST$,I+2,1) = "o" THEN
    LN=20
    END IF
    IF M = "t" AND MID$(TEST$,I+1,1) = "h" AND MID$(TEST$,I+2,1) = "r" AND MID$(TEST$,I+3,1) = "e" AND MID$(TEST$,I+4,1) = "e" THEN
    LN=30
    END IF
    IF M = "f" AND MID$(TEST$,I+1,1) = "o" AND MID$(TEST$,I+2,1) = "u" AND MID$(TEST$,I+3,1) = "r" THEN
    LN=40
    END IF
    IF M = "f" AND MID$(TEST$,I+1,1) = "i" AND MID$(TEST$,I+2,1) = "v" AND MID$(TEST$,I+3,1) = "e" THEN
    LN=50
    END IF
    IF M = "s" AND MID$(TEST$,I+1,1) = "i" AND MID$(TEST$,I+2,1) = "x" THEN
    LN=60
    END IF
    IF M = "s" AND MID$(TEST$,I+1,1) = "e" AND MID$(TEST$,I+2,1) = "v" AND MID$(TEST$,I+3,1) = "e" AND MID$(TEST$,I+4,1) = "n" THEN
    LN=70
    END IF
    IF M = "e" AND MID$(TEST$,I+1,1) = "i" AND MID$(TEST$,I+2,1) = "g" AND MID$(TEST$,I+3,1) = "h" AND MID$(TEST$,I+4,1) = "t" THEN
    LN=80
    END IF
    IF M = "n" AND MID$(TEST$,I+1,1) = "i" AND MID$(TEST$,I+2,1) = "n" AND MID$(TEST$,I+3,1) = "e" THEN
    LN=90
    END IF
    ' End of part 2

The total run time, for both part 1 and part 2 in the same script was 12 minutes 56 seconds.

This screen shows just 1 part completing – there were 3 more after this and I totalled up the output for my answers.

I like some direction and order in my life, or maybe I need it rather than like it, but either way Advent of Code is something that very much appeals to me as it gives us daily problems to solve and actually gets me coding and thinking.

This year, actually my first year, I will be attempting to solve each problem in XC=BASIC. The main advantage of using XC=BASIC is, of course, the modern IDE and the slightly more flexible way I can get data in. The way AoC works is they give you data inputs for the problems, and sometimes these are quite large amounts of data. XC=BASIC helps overcome some of the limitations of CBM Basic while still allowing it to compile down and be run on the Commodore 64.

So, to get me started and by way of a bit of a ‘toe dipping’ exercise, I am going back to AOC 2022 and completing the first few days. I’m not interested in leader boards or competing online. I will be taking my time with each problem and solving them in my own, no doubt incorrect and inefficient way.

This is day 1.

You can find the full text of the problem here and, obviously, if you want to try this out for yourself, stop reading now as the answer is given below!

The first part was relatively simple, define an array, add the data, then cycle through the values adding them up and comparing them to the last highest value, which gives us the elf carrying the most calories compared to its pals.

I replaced all the blank lines in the large dataset with a zero to indicate the end of one elf’s total calories and the start of the next.

So you can see each time a zero is encountered we can total up the calories and compare to the last elf.

You’ll see there I also set a LOW counter which is needed for part 2 of this challenge. It basically sets the first load of calories, in this case 53308, to LOW and uses that to compare against later, along with the HIGH value.

So, a relatively gentle one to start with but very glad I did it as I now have my XC=BASIC development flow set up again properly and I have reminded myself of the little foibles that the language has – I’m looking forward to starting 2023’s calendar!

Oh, and for what it’s worth, I got 2 gold stars for this problem 🙂

This is probably going to keep me busy until well into the new year, after which time I will be returning to adventuring! I’ll try and squeeze another 2022 problem in before the 1st….but no guarantees!

Thanks for reading.