Finally, after reading the articles cited by Warren Smith (listed at the
bottom of this reply) plus some related articles, I can reply to his
insistence that Condorcet-Kemeny calculations take too long to
calculate. Also, this reply addresses the same claim that appears in
Wikipedia both in the "Kemeny-Young method" article and in the
comparison table within the Wikipedia "Voting systems" article (in the
"polynomial time" column that Markus Schulze added).
One source of confusion is that Warren, and perhaps others, regard the
Condorcet-Kemeny problem as a "decision problem" that only has a "yes"
or "no" answer. This view is suggested by Warren's reference (below and
in other messages) to the problem as being NP-complete, which only
applies to decision problems. Although it is possible to formulate a
decision problem based on one or more specified characteristics of the
Condorcet-Kemeny method, that is a different problem than the
Condorcet-Kemeny problem.
In the real world of elections, the Condorcet-Kemeny problem is to
calculate a ranking of all choices (e.g. candidates) that maximizes the
sequence score (or minimizes the "Kemeny score").
Clearly the Condorcet-Kemeny problem is an optimization problem, not a
decision problem (and not a search problem). It is an optimization
problem because we have a way to measure how closely the solution
reaches its goal.
(For contrast, consider the NP-hard "subset sum problem" in which the
goal is to determine whether a specified list of integers contains a
subset that can be added and/or subtracted to yield zero. Any subset
either sums to zero or it doesn't sum to zero. This makes it easy to
formulate the related decision (yes/no) problem that asks whether such a
subset exists for a given set of numbers.)
Because the Condorcet-Kemeny problem is an optimization problem, the
solution to the Condorcet-Kemeny problem can be an approximation. If
this approach is used, it becomes relevant to ask how closely the
approximation reaches the ranking that has the highest sequence score.
Yet even this question -- of "how close?" -- is not a decision problem
(because it goes beyond a yes or no answer).
Keeping in mind that VoteFair popularity ranking calculations are
mathematically equivalent to the Condorcet-Kemeny method, my claim is
that VoteFair popularity ranking calculations yield, at the least, the
same top-ranked choice, and the same few top-ranked choices, as the
solution produced by examining every sequence score -- except (and this
is the important part) in cases where the voter preferences are so
convoluted that any top-ranked choice and any few top-ranked choices
would be controversial. As one academic paper elegantly put it:
"garbage in, garbage out".
More specifically, here is a set of claims that more rigorously state
the above ambiguous claim.
Claim 1: For _some_ _instances_, a polynomial-time calculation can
identify the full ranking that produces the highest Condorcet-Kemeny
sequence score.
Claim 2: For _some_ _instances_, a polynomial-time calculation can rank
the top most-popular candidates/choices and this partial ranking will be
the same as the top portion of the full ranking as determined by
identifying the highest Condorcet-Kemeny sequence score.
Claim 3: For the _remaining_ _instances_ (not covered in Claims 1 and
2), an approximation of the full Condorcet-Kemeny ranking can be
calculated in polynomial time.
Claim 4: For any cases in which the top-ranked candidate/choice
according to the VoteFair popularity ranking algorithm differs from the
top-ranked candidate/choice according to a full calculation of all
sequence scores, the outcome of a runoff election between the two
candidates/choices would be difficult to predict.
As done in the academic literature, I am excluding the cases in which
more than one sequence has the same highest sequence score.
To help clarify the validity of these claims, I'll use an analogy.
Consider a special case of the rigorously studied Traveling Salesman
Problem (TSP), which is NP-hard to solve. (The TSP also can be
expressed as a decision problem, in which case the decision problem is
NP-complete, but that variation is not the problem discussed here.)
The special case -- which I will refer to as the non-returning Traveling
Salesman Problem -- is that we want to know which city the salesman
visits first, and we want to know, with successively less interest,
which city the salesman visits second, third, and so on. Additionally,
for this special case, we specify that the cities to be visited are
roughly located between a beginning point "B" and and ending point "E".
To make this special case mathematically equivalent to the normal
Traveling Salesman Problem in which the salesman returns to the starting
city, we create a path of closely spaced cities (labeled "+" below) that
lead back to the starting city "B".
Here is a diagram of this problem. Remember that the most important
thing we want to know is which city ("*") the salesman visits first.
B = Beginning city
* = City to visit
E = Ending city for main portion
+ = City on path back to beginning
(periods = background; assumes monospace font)
Instance 1:
.................................................B.
.....................................*............+
..................................................+
.....................................*............+
...................................*..............+
..............................*...................+
..................................................+
................................*.................+
.........................*........................+
......................*.....*.....................+
..................................................+
..................*..*.....*......................+
..........*....*..................................+
.......*...............*..........................+
..........*......*................................+
.....*...............*............................+
.........*....*.........*.........................+
..........*........*..............................+
.............*....................................+
E.................................................+
+.................................................+
+.................................................+
+++++++++++++++++++++++++++++++++++++++++++++++++++
In this case it is obvious which city is the first one on the path from
B to E. And it is obvious which are the next four cities on the path.
What we do not know is the sequence of cities after that (for the path
that is shortest).
Now let's consider a different instance of this non-returning Traveling
Salesman Problem.
Instance 2:
.................................................B.
..........................*.......................+
........................*....*....................+
................*.........*...*...................+
.............*.........*....*...*.*...............+
................*...*......*.....*...*............+
.......................*......*...*......*........+
..........*......*.........*......*...*...........+
.............*........*.........*......*..........+
..................*.........*......*..............+
.........*.....*.......*..........................+
.............*.....*..........*....*..............+
..................*..*.....*......................+
..........*....*..................................+
.......*...............*..........................+
..........*......*................................+
.....*...............*............................+
.........*....*.........*.........................+
..........*........*..............................+
.............*....................................+
E.................................................+
+.................................................+
+.................................................+
+++++++++++++++++++++++++++++++++++++++++++++++++++
In this instance we cannot know which city is the first city on the
shortest path until we know the shortest path through all the cities.
Calculating the absolute shortest path in a convoluted case like
Instance 2 might require a calculation time that is super-polynomial
(more than what can be expressed as a polynomial function of the city
count).
However, we can estimate the shortest path.
Such an approximation might identify a first city that is different from
the first city on the absolute shortest path. If the "wrong" city is
identified as the first-visited city, it is understandable that this
occurs because there is not a clearly identifiable first-visit city in
this instance.
This analogy can be extended to the Condorcet-Kemeny problem.
In normal election situations, the most important part of the solution
is the first-ranked winner. In fact, most voting methods are not
_designed_ to identify more than the first-ranked winner.
In contrast, the Condorcet-Kemeny problem is designed to identify a full
ranking. Accordingly, the second-most important part (of solving the
Condorcet-Kemeny problem) is to identify the top few highest-ranked choices.
Both of these important goals can be achieved without fully ranking all
the choices. This is analogous to solving Instance 1 of the
non-returning Traveling Salesman Problem.
The importance of calculating the few top-ranked choices, and the
reduced importance of calculating the lower-ranked choices, is further
demonstrated when the Condorcet-Kemeny method is used to aggregate
(merge/join/etc.) separate rankings from different search engines (to
yield "meta-search" results, which is the intended goal specified by IBM
employees who authored one of the cited articles about Condorcet-Kemeny
calculations). Specifically, a search-engine user is unlikely to look
at the search results beyond the first few pages, which means that
carefully calculating the full meta-search ranking for thousands of
search results is pointless, and therefore the calculation time for a
full ranking is irrelevant.
(As a further contrast, to clarify this point about a partial solution
being useful, the subset-sum problem does not have a partial solution.
All that matters is the existence of at least one solution, or the
absence of any solution.)
Therefore, in some instances we can solve the NP-hard Condorcet-Kemeny
problem "quickly" (in polynomial time) in the same way that we can
"quickly" (in polynomial time) solve some instances -- such as Instance
1 -- of the NP-hard non-returning Traveling Salesman Problem.
In instances where we use an approximate solution for the
Condorcet-Kemeny problem, the approximate solution can be calculated in
polynomial time. Specifically, the algorithm used for VoteFair
popularity ranking, which seeks to maximize the Condorcet-Kemeny
sequence score, always can be solved in polynomial time (as evidenced by
all the programming loops being bounded).
To further clarify these points, consider the following instance of the
non-returning Traveling Salesman Problem.
Instance 3:
.................................................B.
..........................*.......................+
........................*....*....................+
................*.........*...*...................+
.............*.........*....*...*.*...............+
................*...*......*.....*...*............+
.......................*......*...*......*........+
.................*.........*......*...*...........+
.............*........*.........*......*..........+
..................*.........*......*..............+
.......................*..........................+
...................*..............................+
..................*..*............................+
..........*....*..................................+
.......*...............*..........................+
..........*......*................................+
.....*...............*............................+
.........*....*.........*.........................+
..........*........*..............................+
.............*....................................+
E.................................................+
+.................................................+
+.................................................+
+++++++++++++++++++++++++++++++++++++++++++++++++++
For this instance, we can calculate the absolute shortest path through
the group of cities closest to the starting point "B" without also
calculating the absolute shortest path through the group of cities
closest to the ending point "E".
Similarly some instances of the Condorcet-Kemeny problem do not require
calculating the exact order of lower-ranked choices (e.g. candidates) in
order to exactly find the maximum-sequence-score ranking of the
top-ranked choices.
Now that the word "instance" and the concept of a partial order are
clear, I will offer proofs for Claims 1, 2, and 3.
Proof of Claim 1: If an instance has a Condorcet winner and each
successively ranked choice is pairwise preferred over all the other
remaining choices, this instance can be ranked in polynomial time.
Proof of Claim 2: If an instance has a Condorcet winner and the next few
successively ranked choices are each pairwise preferred over all the
remaining choices, the top-ranked choices for this instance can be
ranked in polynomial time.
Proof of Claim 3: There are polynomial-time approximation methods that
can efficiently find a sequence that has a Condorcet-Kemeny sequence
score that is close to the largest sequence score.
(Clarification: I am not claiming that a ranking result based on
approximation will have the same fairness characteristics that are
attributed to the "exact" Condorcet-Kemeny method.)
Using lots of real-life data, plus data that has unusual
calculation-related characteristics, I have tested the VoteFair ranking
algorithm against the full approach that calculates all sequence scores
for up to six choices. In all these cases there are no differences in
the top-ranked choice, nor are there any differences in the full ranking
for the cases that have no ties. (The cases that involve ties involve
multiple sequences that have the same highest score, the resolution of
which is not specified in the Condorcet-Kemeny method.)
Of course Claim 4 would be difficult to prove. (This claim says that if
the two methods do not identify the same winner, the outcome of a runoff
election would be difficult to predict.) The point of Claim 4 is to
clarify the concept of "controversial" and state that if the two methods
identify different winners, neither winner is uncontroversial.
As a reminder (especially for anyone skimming), I am not saying that the
Traveling Salesman Problem is mathematically related to the
Condorcet-Kemeny problem (beyond both being categorized as NP-hard
problems). Instead I am using the well-studied traveling salesman
problem as an analogy to clarify characteristics of the Condorcet-Kemeny
problem that some election-method experts seem to misunderstand.
Perhaps the misunderstanding arises because the Condorcet-Kemeny method
must fully rank all the choices in order to identify the top-ranked
choice. In contrast, other methods do the opposite, namely they
identify the top-ranked choice and then, if a further ranking is needed,
the process is repeated (although for instant-runoff voting and the
Condorcet-Schulze method the process of calculating the winner yields
information that can be used to determine some or all of a full ranking).
If anyone has questions about the calculations done by the open-source
VoteFair popularity ranking software, and especially about its ability
to efficiently identify the highest sequence score based on meaningful
voter preferences, I invite them to look at the clearly commented code.
The code is on GitHub (in the CPSolver account) and on the Perl CPAN
archive (which is mirrored on more than two hundred servers around the
world).
In summary, although the Condorcet-Kemeny method is mathematically
categorized as an NP-hard problem, the instances that are NP-hard to
solve involve either the less-important lower-ranked choices (analogous
to Instance 1 in the non-returning Traveling Salesman Problem), or
involve convoluted top-ranked voter preferences that yield controversial
results (analogous to Instances 2 and 3), or both. For all other
instances -- which include all meaningful election situations --
score-optimized top-ranking results can be calculated in polynomial time.
Clearly, in contrast to what Warren Smith and Markus Schulze and some
other election-method experts claim, the calculation time required by
the Condorcet-Kemeny method is quite practical for use in real-life
elections.
I'll close with a quote from the article by (IBM researchers) Davenport
and Kalananam that Warren cited: "NP-hardness is a only [sic] worst case
complexity result which may not reflect the difficulty of solving
problems which arise in practice."
Richard Fobes
About the citations below: I was not able to read the article by
Bartholdi, Tovey, and Trick because it requires paying a $35 fee. Alas,
it is the article that other articles refer to for the proof of
NP-hardness. However, the other articles, plus related academic
articles, plus Wikipedia articles, provided sufficient perspective.
Again, thank you Warren, for providing the citations.
On 12/24/2011 10:25 AM, Warren Smith wrote:
Rank Aggregation Revisited
Cynthia Dwork, Ravi Kumar, Moni Naor, D. Sivakumar
http://www.eecs.harvard.edu/~michaelm/CS222/rank2.pdf
J. J. Bartholdi, C. A. Tovey, and M. A. Trick: Voting schemes for
which it can be difficult to tell who won the
election, Social Choice and Welfare, 6(2):157–165, 1989.
A Computational Study of the Kemeny Rule for Preference Aggregation
Andrew Davenport and Jayant Kalagnanam
http://www.aaai.org/Papers/AAAI/2004/AAAI04-110.pdf
Cohen, W.; Schapire, R.; and Singer, Y. 1999. Learning to order
things. Journal of Artificial Intelligence Research 10:213-270.
http://www.jair.org/media/587/live-587-1788-jair.ps
etc
and it should be noted that it is NP-complete to find an ordering better than X,
and also NP-hard merely to find the Kemeny winner...
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