# [R] The smallest enclosing ball problem

Berend Hasselman bhh at xs4all.nl
Sun Nov 17 12:05:16 CET 2013

On 17-11-2013, at 11:32, Hans W.Borchers <hwborchers at googlemail.com> wrote:

> 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)
>
>    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$alpha) # 0.5783628 0.5372570 0.2017087 > sqrt(colSums((C - p2)^2)) # 0.3736386 0.3736386 0.3736386 > > But the strangest thing is that with \sum xi =1 solve.QP positions all points > on the boundary, though (in my opinion) no constraint requests it. So the > question remains: > *** What do I do wrong in calling solve.QP()? *** > > Hans Werner See my second reply to your original post. Modify your code with A <- matrix(rep(1,3),nrow=4,ncol=3,byrow=TRUE) A[2:4,] <- diag(3) b <- c(1,0,0,0) to include constraints x_i>=0 (which LowRankQP includes automatically!) and run solve.QP as follows sol2 <- solve.QP(D, d, t(A), b, meq = 1) sol2 p0 <- c(C %*% sol2$solution)
r0 <- sqrt(-sol2\$value)
p0
r0
# distance of all points to the center
sqrt(colSums((C - p0)^2))

and the answers now agree with LowRankQP.

Berend