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There is a lot of literature and debate on how to rank candidates under preferential voting systems. Two of the methods used to determine winners are those based on some form of Borda count and those based on some form of Condorcet method. Many students of politics and voting systems prefer Condorcet methods to Borda ones for its stronger theoretical qualities. However, while Borda counts, especially in their most recognizable form—average rank, is easy to calculate, Condorecet winners are more difficult, as it requires pairwise comparisons between all candidates. Moreover, the possibility of having a Condorcet cycle grows as the number of candidates grows.

One of the more common methods of solving a Condorcet paradox is the Schulze method. This not only has some strong theoretical qualities, but it also has a relatively simple implementation. However, it is slow. Pairwise ranking, in its straightforward form, is of O(n²) and the Schulze method is O(n³) where n is the number of candidates.

In this post, Rcpp is used to significantly speed up the vote ranking process.

Here is a sample ballot with eight voters ranking five candiates:

  Candidates Vote A Vote B Vote C Vote D Vote E Vote F Vote G Vote H
1     Albert      1      2      1      3      2      1      3      4
2      Bruce      2      4      5      4      5      4      1      2
3    Charles      3      1      3      1      1      2      4      5
4      David      4      5      2      5      4      5      2      1
5     Edward      5      3      4      2      3      3      5      3


Here is some simple code to calculate the average rank of the candidates:

AvgRank <- function(BallotMatrix){
Ballots <- as.matrix(BallotMatrix[, -1], mode = "numeric")
Num_Candidates <- dim(Ballots)
Names <- BallotMatrix[, 1]
Ballots[is.na(Ballots)] <- Num_Candidates + 1 #Treat blanks as one worse than min
MeanRanks <- rowMeans(Ballots)
Rankings <- data.frame(Names, MeanRanks)
Rankings <- Rankings[order(rank(Rankings[, 2], ties.method = "random")), ] #Ties handled through random draw
Rankings <- data.frame(Rankings, seq_along(Rankings[, 1]))
names(Rankings) <- c("Names", "Average Rank", "Position")
return(Rankings)
}

The above ballot would result in the following Borda-based ranking:

    Names Average Rank Position
1  Albert        2.125        1
3 Charles        2.500        2
2   Bruce        3.375        3
4   David        3.500        4
5  Edward        3.500        5


Here is some simplified code to calculate Condorcet and Schulze winners. The ballots have been created to ensure that there is always a unique Schulze winner. In reality, there often is not, and some further form of tiebreaking routine will be necessary:

#This function extracts the matrix of votes from the ballot
VoteExtract <- function(BallotMatrix){
Votes <- as.matrix(BallotMatrix[, -1], mode = "numeric")
Votes[is.na(Votes)] <- Num_Candidates + 1 #Treat blanks as one worse than min
}

#This function performs the pairwise comparison between candidates and results in a square matrix representing the number of wins the candidate in row i has beaten the candidate in column j.
Pairwise <- matrix(nrow = Num_Candidates, ncol = Num_Candidates)
for (CurCand in 1:Num_Candidates) {
for (Pairs in 1:Num_Candidates) {
Pairwise[CurCand, Pairs] <- sum(Pref_Cur_Cand[, Pairs] > 0)
}
}
return(Pairwise)
}

#This function calculates the beatpaths and members of the Schwarz set. A unique member is the Schulze Condorcet winner.
Schulze <- function(PairsMatrix){
size <- dim(PairsMatrix)
p <- matrix(nrow = size, ncol = size)
for (i in 1:size) {
for (j in 1:size){
if (i != j) {
if (PairsMatrix[i, j] > PairsMatrix[j, i]) {
p[i, j] <- PairsMatrix[i, j]
} else {
p[i, j] <- 0
}
}
}
}
for (i in 1:size) {
for (j in 1:size) {
if (i != j) {
for (k in 1:size) {
if (i != k && j != k) {
p[j, k] <- max(p[j, k], min(p[j, i], p[i, k]))
}
}
}
}
}
diag(p) <- 0
return(p)
}

#This function performs the ranking, starting with the full ballot, finding a pure Condorcet or Schulze winner, removing him or her from the ballot, and repeating the process until all candidates are ranked.
CondorcetRank <- function(BallotMatrix)  {
Num_Candidates <- dim(BallotMatrix)
Rankings <- matrix(nrow = Num_Candidates, ncol = 3)
CurrentBallot <- BallotMatrix
CurrentRank <- 1
while (CurrentRank <= Num_Candidates) {
CurrentNames <- as.vector(CurrentBallot[, 1])
CurrentSize <- length(CurrentNames)
Pairwise <- matrix(nrow = CurrentSize, ncol = CurrentSize)
Winner <- vector(length = CurrentSize)

# Check for Condorcet Winner

for (i in 1:CurrentSize) {
Winner[i] <- sum(Pairwise[i, ] > Pairwise[, i]) == (CurrentSize - 1)
}
if (sum(Winner == TRUE) == 1) { #Condorcet Winner Exists
CurrentWinner <- which(Winner == TRUE)
Rankings[CurrentRank, ] <- c(CurrentNames[CurrentWinner], CurrentRank, "Condorcet")
} else {

# Condorcet Winner does not exist, calculate Schulze beatpaths

Pairwise <- Schulze(Pairwise)
for (i in 1:CurrentSize) {
Winner[i] <- sum(Pairwise[i, ] > Pairwise[, i]) == (CurrentSize - 1)
}
if (sum(Winner == TRUE) == 1) { #Schwartz set has unique member
CurrentWinner <- which(Winner == TRUE)
Rankings[CurrentRank, ] <- c(CurrentNames[CurrentWinner], CurrentRank, "Schulze")
}
}
CurrentBallot <- CurrentBallot[-CurrentWinner, ]
CurrentRank = CurrentRank + 1
}
Rankings <- data.frame(Rankings)
names(Rankings) <- c("Name", "Rank", "Method")
return(Rankings)
}

Using the sample ballot, the pairwise matrix is:

     [,1] [,2] [,3] [,4] [,5]
[1,]    0    6    5    6    6
[2,]    2    0    3    5    3
[3,]    3    5    0    5    7
[4,]    2    3    3    0    4
[5,]    2    5    1    4    0


the beatpath matrix for the top ranked candidate (using all ballots) is:

     [,1] [,2] [,3] [,4] [,5]
[1,]    0    6    5    6    6
[2,]    0    0    0    5    0
[3,]    0    5    0    5    7
[4,]    0    0    0    0    0
[5,]    0    5    0    5    0


and the full Condorcet ranking is:

     Name Rank    Method
1  Albert    1 Condorcet
2 Charles    2 Condorcet
3  Edward    3   Schulze
4   Bruce    4 Condorcet
5   David    5 Condorcet


When profiling this code using the actual ballot of 30+ people with multiple Condorcet paradoxes, 81% of the time was spent in the Schulze algorithm, another 12% was spent in the PairCount algorithm, and the remaining 7% was spent on everything else (the actual ranking had multiple steps to handle cases when there was no Schulze winner). To speed up the procedure, I ported the Schulze and PairCount functions to C++:

#include <Rcpp.h>
using namespace Rcpp;

// [[Rcpp::export]]
IntegerMatrix Pairwise(Num_Candidates, Num_Candidates);
for (int CurCand = 0; CurCand < Num_Candidates; CurCand++) {
IntegerMatrix Pref_Cur_Cand(Num_Candidates, Num_Ballots);
for (int i = 0; i < Num_Candidates; i++) {
for (int j = 0; j < Num_Ballots; j++) {
Pref_Cur_Cand(i, j) = Votes(i, j) - CandRank(j);
}
}
for (int i = 0; i < Num_Candidates; i++) {
int G0 = 0;
for (int j = 0; j < Num_Ballots; j++) {
if (Pref_Cur_Cand(i, j) > 0) G0 += 1;
}
Pairwise(CurCand, i) = G0;
}
}
return(Pairwise);
}

// [[Rcpp::export]]
IntegerMatrix Schulze_C(IntegerMatrix Pairs) {
int nrow = Pairs.nrow();
IntegerMatrix Schulze(nrow, nrow);
for (int i = 0; i < nrow; i++) {
for (int j = 0; j < nrow; j++) {
if (i != j) {
if (Pairs(i, j) > Pairs(j, i)) {
Schulze(i, j) = Pairs(i, j);
} else {
Schulze(i, j) = 0;
}
}
}
}
for (int i = 0; i < nrow; i++) {
for (int j = 0; j < nrow; j++) {
if (i != j) {
for (int k = 0; k < nrow; k++) {
if ((i != k) && (j != k)) {
Schulze(j, k) = (std::max)(Schulze(j, k), (std::min)(Schulze(j, i), Schulze(i, k)));
}
}
} else {
if ((i = j)) {
Schulze(i, j) = 0;
}
}
}
}
return(Schulze);
}

It is also interesting to compare these results with those obtained from byte-compiling the functions:

library(compiler)
PairCount_cmp <- cmpfun(PairCount)
PairCount_cmp3 <- cmpfun(PairCount, options=list(optimize = 3))
Schulze_cmp <- cmpfun(Schulze)
Schulze_cmp3 <- cmpfun(Schulze, options=list(optimize = 3))

First, we need to check that the functions return the same values:

all.equal(PairCount(VoteExtract(Ballot)),
PairCount_cmp(VoteExtract(Ballot)),
PairCount_cmp3(VoteExtract(Ballot)),
PairCount_C(VoteExtract(Ballot)))

 TRUE

all.equal(Schulze(PairCount(VoteExtract(Ballot))),
Schulze_cmp(PairCount(VoteExtract(Ballot))),
Schulze_cmp3(PairCount(VoteExtract(Ballot))),
Schulze_C(PairCount_C(VoteExtract(Ballot))))

 TRUE


Now let’s compare the speed:

library(microbenchmark)
V <- VoteExtract(Ballot)
P <- PairCount(V)
microbenchmark(PairCount(V), PairCount_cmp(V), PairCount_cmp3(V), PairCount_C(V), Schulze(P), Schulze_cmp(P), Schulze_cmp3(P), Schulze_C(P), times = 100L)

Unit: microseconds
expr     min      lq  median      uq     max neval
PairCount(V) 375.617 384.634 389.130 400.919 1280.33   100
PairCount_cmp(V) 255.012 263.320 267.079 272.044  361.06   100
PairCount_cmp3(V) 247.473 254.373 259.001 265.401 1120.02   100
PairCount_C(V)   7.334  11.793  12.558  13.214   32.37   100
Schulze(P) 786.678 805.497 812.786 842.798 1677.24   100
Schulze_cmp(P) 230.732 237.673 241.639 251.459 1062.25   100
Schulze_cmp3(P) 195.612 200.953 205.156 214.832 1037.79   100
Schulze_C(P)   3.496   5.712   7.485   8.086   14.26   100


While byte-compiling the PairCount function gives an impressive speedup of ariund 40%, porting it to C++ makes it over 25 times faster (an over 2400% speedup). Results with the Schulze algorithm is even more striking. Byte-compilation gives an increase in speed of between 3 to 3.5 times without any change to the R code, but porting it to C++ is around 100 times as fast (and was over 120 times as fast on a different machine)! Moreover, the PairCount algorithm reads more logically in C++, as the way R handles vectors and matrices, when subtracting the current rank, the resulting matrix ended up transposed with the candidates across the columns.

So with easy-to-read code that results in speed gains of multiple orders of magnitude, what’s not to like?! 