election_title: "Batch elimination empties the field — the perfect cycle"

scenario_description: |-
  Three voters, three candidates, and the smallest profile in which an
  instant-runoff count has nobody left to elect. Amy, Bruno and Clara each hold
  exactly one first choice, and the three ballots rotate: Amy>Bruno>Clara,
  Bruno>Clara>Amy, Clara>Amy>Bruno. Nobody has a majority (2 of 3), so somebody
  must be eliminated — and every candidate is tied for fewest first choices.
  Under the BATCH convention, which the Stanford Encyclopedia entry and
  pref_voting both use, you remove ALL candidates tied for last in one step. Here
  that is the whole field. The count stops with an empty ballot, and the stated
  answer is that all three candidates TIE for the win.
  That is not a bug in the convention, it is the convention working. This profile
  is perfectly symmetric: rotate the candidate names and you get the same three
  ballots back in a different order. So an ANONYMOUS and NEUTRAL rule — one that
  ignores who cast which ballot and shows no favouritism between names — has
  nothing left to separate Amy from Bruno from Clara, and a three-way tie is the
  only answer it can give. See 07_Concepts/topics/ties/ties_are_forced.md for the
  theorem.
  This engine does not do that. The vendored pyrankvote cuts one candidate,
  transfers, and names a single winner — and WHICH one depends on the order the
  ballots happen to be listed in. All six row orderings of these same three
  ballots were run: the winner is always the first row's first choice. Amy in two
  orderings, Bruno in two, Clara in two. That is an anonymity failure, not a
  neutrality failure, and it is undisclosed. The expected_winners below records
  what this engine prints for THIS row order; it is not the method's answer.
  Lesson: 07_Concepts/topics/ties/batch_elimination.md

video_script: |-
  Three voters. Three candidates. Everybody's first choice is different.
  Nobody has a majority, so instant-runoff has to eliminate somebody. Who?
  Amy has one vote. Bruno has one vote. Clara has one vote. There is no fewest.
  One textbook answer — and it is the one the Stanford Encyclopedia gives — is:
  when several candidates tie for last, drop all of them. Usually that is a
  shortcut. Here it is the whole field. The count ends with nobody left, and the
  honest report is that all three tie.
  And that really is the only answer a fair rule can give. Look at the ballots:
  rename Amy to Bruno, Bruno to Clara, Clara to Amy, and you get the same three
  ballots back. If the rule doesn't care who cast which ballot, and doesn't play
  favourites with names, then it cannot tell these three candidates apart. A
  three-way tie isn't the rule giving up. It is the rule being consistent.
  Now watch what our engine does instead. It picks one. And if you shuffle the
  ballot rows — same votes, same voters, just typed in a different order — it
  picks a different one. Six orderings, three different winners.
  Nothing in the report says so.

voting_method: RCV_IRV
num_winners: 1

ballots: |-
  Amy>Bruno>Clara
  Bruno>Clara>Amy
  Clara>Amy>Bruno

expected_winners:
- Amy