Lower default tolerance to find optimal matching

[nremenyi]

I noticed that using optmatch that I got very different solution from another package (clue). The reason was the tolerance. I think it would be wise to lower the default tolerance?

### Simulate data set: trt (1) / ctrl (0) indicator and covariate x
N = 10^3
M = 9*10^3
set.seed(124)
data = rbind(
  data.frame(id = 1:N, group = 1, x = rnorm(N, mean = 0, sd = 5)),
  data.frame(id = (N+1):(N+M), group = 0, x = rnorm(M, mean = 0, sd = 5))
)

### Optmatch pairmatch
mhd = optmatch::match_on(group ~ x, data = data)
m1 = optmatch::pairmatch(mhd, data = data)
cat("Total distance for Optmatch pairmatch with default tol:", 
    sum(optmatch::matched.distances(m1, distance = mhd)), "\n")
m2 = optmatch::pairmatch(mhd, data = data, tol = 0.000001)
cat("Total distance for Optmatch pairmatch with tol = 0.000001:", 
    sum(optmatch::matched.distances(m2, distance = mhd)), "\n")

### Clue package (Assignment problem)
sol_h2 = clue::solve_LSAP(x = mhd, maximum = FALSE)
match_mx_h2 = matrix(NA, nrow = min(N,M), ncol = 2)
match_mx_h2[,1] = 1:min(N,M)
match_mx_h2[,2] = sol_h2[]
cat("Total distance for Clue Assignment:", sum(mhd[match_mx_h2]), "\n")

[benthestatistician]

Thanks for reaching out, and thanks for the very nice reprex.

Although I'm relectant to lower the default tolerance alone, I see this as a circumstance we'll be able to address better after a little more work on #164, which will help with flagging solutions that are sub-optimal due only to the tolerance, and #76, which will hep with addressing those problems.