"Problematic" Ballot Exhaustion examples  RCV IRV

We need three small, illustrative elections to demonstrate each ‘problematic’ box separately (avoid ‘Less Problematic’ Exhausted Ballots).
Assumptions:
 ranking limit 3 (or 5 if easier to recreate)
 number of candidates 5 (or less if easier)
 all voters correctly rank all the ballots (no spoilers, no bullet voting)
 all voters rank all ranks within Ranking Limit (3 or 5)
Context:

article https://docs.google.com/document/d/1ASC5BS10rCfAYZWGeCyS7dKdKc4p5wwI6DHs4F7ScGc/edit

presentation https://docs.google.com/presentation/d/1ipof9WSSy0GenVKWfKLu_jlYwUIT0TU0WVSAap0VZ5Q/edit#slide=id.g27f6bb33467_0_1
Comments, feedback  very much appreciated!

I believe you mean "spoiled" as in "voided", not "spoiler".
With your assumptions, some of the "middling" problems can't happen because every voter fully filled out their ballot correctly. Basically the red and yellow boxes become the same.
Regardless, this exercise is fairly trivial once you wrap you head around the basic approach, which I'm happy to walkthrough
First off, all ranked ballots have an imaginary "last rank" that isn't shown on the ballot. For example, if I can only rank 3 candidates explicitly, then in practice I can explicitly rank all candidates if there are up to 4 of them because unranked candidates are considered ranked 4th. So, given your assumptions, if we want to create these kinds of problems, we need at least 2 more candidates than the number of ranks on the ballot. 3 ranks and 5 candidates is perfect.
Next up, we just make a ballot set that produces RCV results where no candidate is within a couple votes of beating another candidate. Here's one with little thought:
40: A>B>C
30: D>E>B
20: C>A>D
5: E>B>ANow we run through the tally.
B, E, and C are all eliminated in the first round. All 25 votes transfer to A. A beats D 65 to 30. Looking back, I guess we need to make sure it comes down to a number of finalists no more than the difference between the number of candidates and the number of available ranks (so 53=2), which happened in this case.
Now, we add a voter who ranked all 3 of those eliminated candidates.
40: A>B>C
30: D>E>B
20: C>A>D
5: E>B>A
1: B>E>CWe also need a voter whose ballot stops transferring partway through their ranks but ultimately goes to a losing finalist. This means a losing finalist needs to be in the middle of their marked rankings. The only losing finalist is D and there are only 3 explicit ranks available, so the ballot needs to rank first a candidate who gets eliminated, then rank second D, and then rank third ideally a candidate who also lost, but the third rank could really be any candidate because RCV is nonmonotonic.
40: A>B>C
30: D>E>B
20: C>A>D
5: E>B>A
1: B>E>C
1: B>D>CBoom. Those last two ballots show the problems you're looking for.
Of course, another voter intent issue in RCV is nonmontonicity, which this ballot set doesn't show. Again, it'd be fairly easy to come up with a ballot set that shows all of the problems (my TikTok series does that), but nonmontonicity is probably best presented in isolation because of how screwy it is. The following is the simplest ballot set demonstrating nonmontonicity in RCV:
8: A>B>C
7: C>B>A
6: B>A>CIf 2 C voters strategically rank A first, then A would lose.
And that's it.

@masiarek hm I feel like a small demonstration of the issue under consideration might be helpful. To me, I already think IRV is questionable because it fails Condorcet and participation (these are mutually exclusive criteria), is not monotonic, and uses a very strict noncompensatory method to screen out candidates.