## Math in Elections Part 7 — The Complexity of Declaring a Victor.

Alright, folks.  Today’s the day.  Today the country votes on who will represent us in government.  But more importantly for the news organizations, today is the day when they get to hype the heck out of every exit poll, every demographically significant neighborhood, every weather report, and, of course, every new set of raw vote data.

And if the polls thus far are any indication, it will be late into the evening or early into tomorrow morning until we learn about the results for the day’s biggest race: Ohio (AKA: the presidency).

So how are votes counted and is that indeed the most efficient way possible?  What is the complexity of determining the winner of each state?

In a presidential election, each state will have many people on the ballot.  Of course Obama and Romney will be there, but Libertarian Gary Johnson will be on the ballot in every state and in DC except possibly Michigan and Oklahoma, where his ballot access has been challenged.  The Green Party’s Jill Stein will be on the ballot in 38 states and in DC.  The Constitution Party and Justice Party will also be on in a handful of states.  And in each state, many more may be on just for that specific state.  But more importantly, most states have a write-in option where you can write in a vote for anyone or anything you like; such as yourself or Daffy Duck or Whateversuitsyourboat or Rex Burkhead or Herman Cain.

Nebraska, for instance, has a write-in option.  And any name is acceptable for submission.  Meaning there is no (finite) bound on the number of possible write-in submissions.

Moreover, let’s assume for the sake of this problem that, although there may be a large collection of write-in submissions for who-knows-whom, that in the end, one of the candidates will get at least 50% of the vote.  (This is not a strong assumption for many elections since oftentimes, unless a candidate gets 50% of the vote, there is a runoff election between the two leading vote-getters anyways.  And hence there would be another election where getting 50% is actually required.)  But for the sake of the algorithm, we cannot and should not narrow down to a finite list who we think will win.  We may not immediately say that Obama or Romney will win, because that would put an immediate bias against Whateversuitsyourboat, which is just not right.

Lastly, for this problem we may even assume more broadly that the tally is “on-line”; meaning the votes are coming in one at a time, and we are viewing them only as they come in.  So we don’t receive them as a pre-arranged set or know how many will eventually be cast.

Ok, so that’s the setup.  Now here’s the question:  How many operations must be done in order to determine which candidate got 50% of the vote?  Can we find a strategy that is fast and is not dependent on the number of candidates submitted?

Say the candidates are enumerated 1 through $k$, but we don’t know how large $k$ is.  One option is to create an array.  Each time a candidate’s name comes in, we search the array to determine if (s)he is already listed.  If so, we increase his/her tally by one.  If not, then we create a new spot for that candidate and set his/her initial tally equal to 1.  And then, at the end of all of this, once every vote has come in, we search through this list and find the largest value.

However, this process is both not nearly as fast as possible and the last step is highly dependent on $k$.  If $k$ is large, then determining the winner at the end could take many operations.

I claim there is an algorithm that will deterministically take the, say, $n$ votes off the line and spit out the winner in at most $n$ operations.

Note that we did not use in our last algorithm that one candidate will get 50% of the vote, so we definitely want to incorporate that in somewhere.  Also note that in the end, we do not care to know the exact vote tallies of all the losers, or even the winner.  So keeping track of them seems like we were doing more work than necessary.  This is a fun problem to think about.  So I encourage you to spend some time mulling it over on your way to the polling booth today, before you come back and read the solution.  But if you wish to know the answer right off the bat, well that’s ok too.  Here it is.

As your votes come in, put them in a list.  Proceed through the list keeping track of two things: one candidate’s name and one number.  Say the sequence started like this:

$A \ \ A \ \ A \ \ C \ \ B \ \ B \ \ B \ \ C \ \ C \ \ C \ \ B \ \ C \ \ C$

So after thirteen votes, 3 candidates have thus far been recorded.  After the first vote, the pair is $(A : 1)$. This is because the first vote was for $A$, but more importantly it means that since the beginning (when no one had a majority), $A$ has had the majority of the votes, and he is in fact 1 vote above the majority (if he lost 1 more vote he would lose the majority).  After step two the pair is $(A : 2)$, and after the third step it is $(A : 3)$.  So, again, since the starting position when no one was winning, $A$ has had the majority and it would take three non-$A$ votes for him to lose it.

The first non-$A$ vote happens next.  After the $C$ the count is at $(A : 2)$, then it goes to $(A : 1)$ after the first $B$.  Note: another way to look at the count at this point is that the number of $A$s at this point is one larger than the sum of the numbers of everything else.  Proceeding from here, after the second $B$ it is $(? : 0)$, meaning at this point no candidate has a majority of the votes.  After the next step, though, we are going to dictate that the the count be $(B : 1)$.  Note that this pair does not mean that $B$ is one vote above the majority in the entire string thus far.  It only means that ever since the $(? : 0)$ point when no candidate had a majority, $B$ is winning.  Continuing in this way, the counts starting from here are

$(B : 1) \to (? : 0) \to (C : 1) \to (C : 2) \to (C : 1) \to (C : 2) \to (C : 3).$

Our conclusion from this is that $C$ must have won.  But why?  We noted that this $(C : 3)$ only means that since the last no-winner point, $C$ has been up by 3 against the majority.  It does not mean, however, that $C$ is 3 above the majority among all votes cast so far.  How do we know that we didn’t run up the tally on $A$ and $B$ early on, causing one of those to be ahead of $C$?

Here’s one way to see it.  Note that certainly $A$ is not the winner, since when he lost the majority, he never gained it back.  $B$ is not the winner for the same reason.  But wait!  We supposed that one of the candidates was going to get at least 50% of the vote.  Therefore, with that supposition and us ruling out $A$ and $B$ as winners, the winner must be $C$.

And how many steps did our algorithm take?  Only one step for each vote.  So with $n$ votes, we require a mere $n$ steps to determine the winner.

Well, that’s it for this series.  I intend to post more often, but you know how that goes.  I stopped mid-series last spring.  Maybe I will pick that up again.  Or maybe I won’t.  Who knows.  Luckily Hotovy is back with some good analysis posts, Zach claims that he’ll post more this year, and Adam is planning on writing on the Noble Prize in Economics soon.  Lots to look forward to!  Of course, even with our joined forces we still can’t even begin to compete with the posting rate over at Andy Soffer’s wonderful blog. But hey, we’re only human.

Until next time!  Now go watch CNN, all you political junkies!