# [EM] Unger, wrt tabulation.

robert bristow-johnson rbj at audioimagination.com
Thu Feb 2 11:16:36 PST 2012

```>     2012/2/2 Stephen Unger <unger at cs.columbia.edu
>     <mailto:unger at cs.columbia.edu>>
>
>         A fundamental problem with all these fancy schemes is vote
>         tabulation. All but approval are sufficiently complex to make
>         manual
>         processing messy,
>

hey, listen, it's messy just the way it is.  i participated in a few
manual recounts in the most populous county of this small state.  it
needs lots of (quasi-volunteer) participants from multiple parties or
voting groups, organization, and transparency.  that puts a lid on the
degree of messiness.

>         to the point where even checking the reported
>         results of a small fraction of the precincts becomes a cumbersome,
>         costly operation. (Score/range voting might be workable). Note
>         that,
>         even with plurality voting, manual recounts are rare. With any
>         of the
>         other schemes we would be committed to faith-based elections.
>

naw!  that's hyperbole.

On 2/2/12 11:39 AM, David L Wetzell wrote:
> I wanted to mention that Approval-voting enhanced IRV and STV could be
> tabulated at the precinct level.  You let everyone rank up to 3
> candidates and then you use these rankings to get 3 finalists.  You
> then sort the votes into ten possible ways people could rank the 3
> finalists.

there are 9 possible ways of ranking 3 candidates, unless you're
counting "none of the above".  i guess that would make it 10.

in general, for N candidates and IRV, the number of piles you need to
sort to is

N-1
SUM{ N!/n! }  =  floor( (e-1) N! ) - 1
n=1

for Condorcet you need

N*(N-1)    piles

and, of course, for FPTP it's

N .

Number of necessary subtotals:

FPTP - N:     2      3      4      5       6      7

Condorcet:     2      6     12     20      30     42

IRV:     2      9     40    205    1236   8659

the number of piles grows pretty large for IRV, which is why we normally
call it "not precinct summable".  essentially a physical instrument
(like a thumb drive) that contains the information for each and every
ballot must be (securely) transported from each voting place to the
central tabulation facility (like City Hall).

folks like Kathy Dopp understandably complain about the lack of
transparency about such, while i didn't see it as too bad of a problem
for a small city like Burlington.  however IRV was passed (and vetoed by
the guv back then) in Vermont for the gubernatorial election, and that
centralization of counting would be even more of a problem.  i just
can't see some Town Clerk driving up from Bennington VT to Montpelier to
deliver the opaque physical instrument representing all those votes.

>  But if the third or fourth most often ranked candidates were within a
> small percent of each other then it would not require a manual
> recount.  The IRV cd be done with two sets of 3 candidates so there'd
> be twice as much sorting in the 2nd round and then there'd be a manual
> recount if and only if there's a different outcome in the two sets of
> candidates, which is not likely.

with FPTP, there need be only one team of counters (but more teams will
divide the labor and the results are perfectly summable) and that number
does not grow with N.  a team will normally have 4 people that are known
supporters of the different candidates.  there are two "callers", they
simultaneously examine each ballot, one at a time, and call out the name
of the voted candidate.  there are two "counters" that rack up the
counts.  for every block of 50 or 100 ballots, the counts (between the
two counters) are compared and if there is any discrepancy, that block
is recalled and recounted.

for IRV, this can be done with a single team or multiple teams (to
divide the labor) but the piles (a function of the 1st-choice vote or
the remaining 1st-choice vote) need to be separate so that when a
candidate is eliminated, the votes are "transferred" (as in "STV") at
the end of the pass or round.  then there is retabulation and this
recounting cannot be done in parallel, it must be done sequentially, up
to the final round.

for Condorcet, if the labor is divided, there needs to be a team for
every pair of candidates (essentially the number of piles divided by
two: N*(N-1)/2 ).  each team is concerned only for its assigned pair of
candidates (who is ranked above who) and the ballots are passed from one
team to the adjacent team.  but there is only one pass.  if the number
of teams is not available, it can be done with a single team
sequentially, but would be multiple passes and would be laborious.

--

r b-j                  rbj at audioimagination.com

"Imagination is more important than knowledge."

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