Solution: Bird Conundrum

This is a very easy puzzle. It may seem hard, because it's not usually solvable initially; trying to follow the instructions will lead to false statements and general confusion. The key realization is that, as hinted by the flavortext, the puzzle (specifically, the starting bird labels), changes every minute (if the page is refreshed). Then, all you have to do is refresh the page every minute until the puzzle is solvable, and then straightforwardly solve it by following the instructions.

...Ok, not really. It is true that the puzzle isn't usually solvable initially and it changes every minute. However, the intended solve path is not to wait until the puzzle is solvable, but rather to determine the one starting condition from which it's solvable, and follow the instructions from there. (But if you're really lucky and happen on the puzzle when it is solvable, you can just solve it by following the instructions).

There are two parts to determining the starting condition from which the puzzle is solvable. The first part is figuring out how the labels change. The second part is, once you know the possible label configurations, narrowing them down.

Figuring out how the labels change is relatively easy. It's somewhat hinted by "flock" in the flavortext, but the easiest way to realize it is by writing a few sets of labels from consecutive minutes above each other as in the following table, and then reading down each column:

We notice columns seem to spell words, or parts of words. Looking up a few of the shorter/clearer words such as (here) GULP and HOST, we see that these are the collective names for these types of birds (e.g., a GULP is a group of cormerants). (Some types of birds have several collective names, but even a few labels from one of those birds tell us which name to use.) This lets us figure out what the labels will be at any time. For reference, the collective names are (where “Bird number” is the bird’s number in the rest of the solution, for brevity):

The name lengths are 3, 4, 6, 7, 10, and 11 (some repeated). The period after which the labels repeat is thus the lowest common multiple of these, which is 4620. This is 4620 minutes, or 77 hours. This is too long to wait and too many things to manually inspect, but it is possible to brute-force with sheets formulas or programming. It is also possible to solve logically, though, which is what we'll do in the appendix to this solution.

There are many ways to solve it logically; the way given in the appendix doesn't require the rule that "Unless a step says otherwise, there should be only one possible way to do the step, without any need to look ahead to future steps.", but using that rule eliminates a decent number of cases where there are two ways to do a step from some state. (The rule ""The labels of X and Y" always means their labels in that order (first X's label, and then Y's label). When you change labels to turn one element/state abbreviation into another, read the labels in the same order." is very important to keep in mind on any solvepath, though.)

By whatever logic, we now know the starting state is:

We now apply the steps:

Note that we can rule out 4b because NL is not a state abbreviation. The puzzle answer is the final state, THREE AFTER. (And as said in the appendix, this state appears three minutes after the 77-hour cycle starts, in a sense, though teams are not expected to notice this).

Authors' Notes

This puzzle was inspired by two puzzles in CCBC 13: When Will You Come Again? (copied by the author at https://web.mit.edu/yuzhougu/www/20230820/ouroboros.rb) for the "thing repeating after a long period" idea, and a meta (part of another CCBC 13 puzzle, Metaverse) about animal collective names for the bird collective names idea. It was, of course, also inspired by duck conundrums and Cardinality's bird theming.

This puzzle, given those inspirations, was not too hard to come up with. Verification was also not too bad; I just coded it to check for only one solution, and the only requirement was that I had to make sure everything was constrained enough to make it possible to solve manually or get back on track if you did something wrong, and also, like solvers, I had to be careful too. (I think, due to it being hard to debug code, even though this puzzle seems amenable to coding, it may be easier to solve manually because then it's easier to track back what went wrong if something goes wrong. It depends on your coding/manual skill/experience with this type of thing, of course. Definitely during the hunt, many teams solved it manually.)

The hard part was wording. Making sure the wording was as clear and unambiguous as possible was a long process involving lots of testsolver feedback; thanks to all the testsolvers for turning the wording from a long, incomprehensible mess to something that's hopefully readable.

The table of some consecutive starting conditions in this solution was made the Sunday before hunt, PM, Eastern Time. You may notice that it's 29 hours off from what it should be. This is because, during the start of playtesting for this puzzle, I set the "start time" for Bird Conundrum to the UTC epoch, five hours off from when start of hunt ended up being (though we hadn't decided on hunt start time yet) and so five hours off from what it was in hunt. In every testsolve involving Bird Conundrum, I made sure to check that the testsolve wouldn't coincidentally get the one easily solvable state. Usually, it was many hours off. But one day, it seemed like the testsolve might actually hit the one-in-4620 lucky chance (a few hours out, they were within about an hour of doing so based on expected solve times for earlier puzzles), so I set the "start time" one day earlier.

Addendum about the erratum (which at the time of writing I hope remains "the erratum" rather than "one of the many errata"): The erratum resulted from a clarification on states bordering the Atlantic. I originally checked uniqueness without states bordering the Gulf of Mexico, and when, late in editing, a testsolving team said it was unclear whether this included the Gulf of Mexico, I decided to add a clarification to stop teams from having to worry about it. However, at the time I thought the pre-written uniqueness check incorporated states bordering the Gulf of Mexico, and when I looked for states with one letter of their abbreviation changed from IL, I somehow didn't notice AL. In writing this solution, I also didn't go in depth about the IL → FL step enough to notice AL.

In fixing the erratum, we didn't want to say "(not including the Gulf of Mexico)" because whether that includes Florida or not depends on how you read it, so we instead added the vowel-to-consonant part of Step 5.

I'm very sorry that this fell through the cracks. In any case, here's an image of the Bird States of America:

Appendix

Note that there is a time when all the labels are the first letters of the collective name. We'll call this "time 0", and refer to "time x" as x minutes after time 0. This is somewhat motivated by time 0 being on-the-hour, so x:00 for some x (indeed, time 0 can be taken to be the start of hunt). This "time 0" convention is not needed to solve the puzzle, but it will be used in this appendix.

Step 1 is fairly unconstrained at first, so let's start with Step 2. We notice that after Step 1, every bird's label is either a letter from its collective name, or E. This lets us pin down the element in Step 2 quite a bit: it must start with some letter from DESCNT and end with some letter from MOBE. This forces it to be one of the following:

This is a lot of options. However, due to Step 3, the new label for the woodpecker (the first letter of the new element abbreviation) must be the cormorant's label, making it one of GULPE. This restricts us to:

However, the No → Lr case is actually impossible for the following reason: since the cormerant is L (because the woodpecker is L at the start of Step 3), the time must be 2 mod 4 (defining time 0 as when everything uses its first letter, as above, so at time 2 the cormorant uses its third letter L, and every four minutes thereafter). Since the emu is O, the starting time must be 1 mod 3. This means the starting time must be 10 mod 12, and in particular 4 mod 6, meaning the letter used for the crow is E. However, this will make Step 4 impossible, since no state abbreviations start with E.

The Sm → Eu case is also impossible, since it requires the cormorant to have changed to E in Step 1. However, "gulp" is the only collective name with G and "gulp" and "parliament" are the only two with P, so the cormorant couldn't have changed from G or P (remember, only three identical labels can change to E in Step 1). The three with U are "gulp", "murder", and "murmuration", but "murder" is only U when the time is 1 mod 6, but the emu being M (for Sm) means the time is 0 mod 3, which is incompatible, so, there not being enough collective names with U, the cormorant couldn't have changed from U. Finally, the four with L are "gulp", "flamboyance", "exaltation", and "parliament", but for both "exaltation" and "parliament", getting L requires the time to be 3 mod 10, while for "gulp", getting L requires the time to be 2 mod 4, which is incompatible. So the cormorant couldn't have changed from L. Thus the cormorant couldn't have changed to E at all.

We are left with just the Ce → Pr case. This actually tells us a lot. The C tells us the time is 3 mod 7. The Pr means that the cormorant must be P, so the time is 3 mod 4. The E tells us that the emu's label changed to E. Only "mob" and "flamboyance" have B, so it can't have changed from B. Unfortunately, we can't tell at this point if the emu's label was M or O before the change. To summarize what we know, the time is 3 mod 7 and 3 mod 4, and it is 0 or 1 mod 3.

We proceed to Step 4, where we want the labels of the crow and the peacock to form a state abbreviation. The time is 3 mod 4, so if the time is 0 mod 3, it's 3 mod 6 and the crow is D. On the other hand, if the time is 1 mod 3, it's 1 mod 6 and the crow is U. The state abbreviation must then be either Delaware (DE) or Utah (UT). However, this requires the peacock to be E or T, forcing a specific value for the time mod 11. (One might think the peacock could have changed to E. However, this is false, because in the case where the peacock is E, the time is 0 mod 3 and the emu changed from M to E. The peacock's starting label could not have been M.) It's now practical to look at possibilities. Fixing the time mod 3 and 11, we can fully determine the starting labels of everything but the lark and owl, as below (note that this is a table of starting labels, not current labels):

Note that neither the lark or owl can actually be O, because "parliament" has no O and the O in "exaltation" requires a time 8 mod 10, incompatible with 3 mod 4. So neither the second or third rows of this table have three O's. But we know the emu changed to E in Step 1, so we must be using the first row of the table, meaning the time is 0 mod 3 and 3 mod 11. Neither the lark or owl can be M; "exaltation" has no M and "parliament" requires a time 6 mod 10, incompatible with 3 mod 4. But that's fine since we already have three M's.

Looking ahead to Step 7, we see that the emu, flamingo, and starling changed in Step 1, and the woodpecker, cormorant, crow, and peacock change in steps 2, 3, and 4. So the three unchanged labels are those of the lark, owl, and sparrow. We know that two have the same label, which is a consonant, but the sparrow's starting label is E so those two have to be the lark and owl. We see that the lark and owl only have the same label at all when the time is 3 mod 10, and that label is L. This gives us all the starting labels.

As you may have noticed, the time is 3 mod a lot of things. In fact, the time is 3. This means that brute force, if you started when everything was the first letter (time 0), would get you the answer on the fourth try (time 3). However, from testsolving, no one seems to do this.