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annotate day12.txt @ 24:19481b061461

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Implement tilting and cycling for Day 14 part 2 Lots of false starts trying to iterate. Eventually looked for "back in same position" to spot a loop. Then took longer to spot that "same position" isn't necessarily "start position" and loop can be offset!
author IBBoard <dev@ibboard.co.uk>
date Sat, 16 Dec 2023 20:39:02 +0000
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1 For each row, the condition records show whether each item is operational (.) or damaged (#). This is the part of the condition records that is itself damaged; for some items, it is simply unknown (?) whether the item is operational or damaged.
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3 However, after each row, the size of each contiguous group of damaged items is listed in the order those groups appear in the row. This list always accounts for every damaged item, and each number is the entire size of its contiguous group (that is, groups are always separated by at least one operational item: #### would always be 4, never 2,2).
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4
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5 So, condition records with no unknown item conditions might look like this:
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7 #.#.### 1,1,3
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8 .#...#....###. 1,1,3
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9 .#.###.#.###### 1,3,1,6
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10 ####.#...#... 4,1,1
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11 #....######..#####. 1,6,5
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12 .###.##....# 3,2,1
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14 However, the condition records are partially damaged; some of the items' conditions are actually unknown (?). For example:
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16 ???.### 1,1,3
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17 .??..??...?##. 1,1,3
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18 ?#?#?#?#?#?#?#? 1,3,1,6
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19 ????.#...#... 4,1,1
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20 ????.######..#####. 1,6,5
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21 ?###???????? 3,2,1
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22
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23 Equipped with this information, it is your job to figure out how many different arrangements of operational and broken items fit the given criteria in each row.
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25 In the first line (???.### 1,1,3), there is exactly one way separate groups of one, one, and three broken items (in that order) can appear in that row: the first three unknown items must be broken, then operational, then broken (#.#), making the whole row #.#.###.
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27 The second line is more interesting: .??..??...?##. 1,1,3 could be a total of four different arrangements. The last ? must always be broken (to satisfy the final contiguous group of three broken items), and each ?? must hide exactly one of the two broken items. (Neither ?? could be both broken items or they would form a single contiguous group of two; if that were true, the numbers afterward would have been 2,3 instead.) Since each ?? can either be #. or .#, there are four possible arrangements of items.
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29 The last line is actually consistent with ten different arrangements! Because the first number is 3, the first and second ? must both be . (if either were #, the first number would have to be 4 or higher). However, the remaining run of unknown item conditions have many different ways they could hold groups of two and one broken items:
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31 ?###???????? 3,2,1
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32 .###.##.#...
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33 .###.##..#..
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34 .###.##...#.
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35 .###.##....#
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36 .###..##.#..
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37 .###..##..#.
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38 .###..##...#
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39 .###...##.#.
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40 .###...##..#
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41 .###....##.#
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42
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43 In this example, the number of possible arrangements for each row is:
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44
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45 ???.### 1,1,3 - 1 arrangement
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46 .??..??...?##. 1,1,3 - 4 arrangements
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47 ?#?#?#?#?#?#?#? 1,3,1,6 - 1 arrangement
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48 ????.#...#... 4,1,1 - 1 arrangement
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49 ????.######..#####. 1,6,5 - 4 arrangements
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50 ?###???????? 3,2,1 - 10 arrangements
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52 Adding all of the possible arrangement counts together produces a total of 21 arrangements.
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53
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54 For each row, count all of the different arrangements of operational and broken items that meet the given criteria. What is the sum of those counts?