The smallest enclosing ball problem

Berend Hasselman <bhh <at> xs4all.nl> writes:

Forgot to forward my answer to R-help.

Berend

Thanks, Berend, for thinking about it. \sum xi = 1  is a necessary condition 
to generate a valid geometric solution. The three points in the example are 
very regular and your apporach works, but imagine some random points:

   set.seed(8237)
   C <- matrix(runif(9), 3, 3)
   D <- 2 * t(C) %*% C
   d <- apply(C^2, 2, sum)
   A <- diag(3)
   b <- rep(0,3)

   require(quadprog)
   sol1 <- solve.QP(D, d, A, b, meq = 0)
   sol1                                # now \sum xi = 1is not fulfilled

   p0 <- c(C %*% sol1$solution)        # 0.3707410 0.3305265 0.2352084
   r0 <- sqrt(-sol1$value)             # 0.5495631

   # distance of all points to the center
   sqrt(colSums((C - p0)^2))           # 0.5495631 0.3119314 0.5495631

Unfortunately, this is not the smallest enclosing ball.
LowRankQP will find the true solution with radius 0.3736386 !

   require(LowRankQP)
   A <- matrix(1, nrow = 1, ncol = 3)
   b <- 1

   sol2 <- LowRankQP(D, -d, A, b, u = rep(1, 3), method="LU")

   p2 <- c(C %*% sol2$alpha)   # 0.5783628 0.5372570 0.2017087
   sqrt(colSums((C - p2)^2))   # 0.3736386 0.3736386 0.3736386

But the strangest thing is that with \sum xi =1 solve.QP positions all points
on the boundary, though (in my opinion) no constraint requests it. So the
question remains:
                 *** What do I do wrong in calling solve.QP()? ***

Hans Werner