# The smallest enclosing ball problem

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## The smallest enclosing ball problem

 I wanted to solve the following geometric optimization problem, sometimes called the "enclosing ball problem":     Given a set P = {p_1, ..., p_n} of n points in R^d, find a point p_0 such     that max ||p_i - p_0|| is minimized. A known algorithm to solve this as a Qudratic Programming task is as follows:     Define matrix C = (p_1, ..., p_n), i.e. coordinates of points in columns,     and minimize                     x' C' C x - \sum (p_i' p_i) x_i     subject to                     \sum x_1 = 1, x_i >= 0 . Let x0 = (x0_1, ..., x0_n) be an optimal solution, then the linear combination     p0 = \sum x0_i p_i is the center of the smallest enclosing ball, and the negative value of the objective function at x0 is the squared radius of the ball. As an example, find the smallest ball enclosing the points (1,0,0), (0,1,0), and (0.5, 0.5, 0.1) in 3-dim. space. We would expect the center of the ball to be at (0.5, 0.5, 0), and with radius 1/sqrt(2).     C <- matrix( c(1, 0, 0.5,                    0, 1, 0.5,                    0, 0, 0.1), ncol = 3, byrow = TRUE)     # We need to define D = 2    C' C, d = (p_i' p_i), A = c(1,1,1), and b = 1     D <- 2 * t(C) %*% C     d <- apply(C^2, 2, sum)     A <- matrix(1, nrow = 1, ncol = 3)     b <- 1     # Now solve with solve.QP() in package quadprog ...     require(quadprog)     sol1 <- solve.QP(D, d, t(A), b, meq = 1)     # ... and check the result     sum(sol1$solution) # 1 p0 <- c(C %*% sol1$solution)        # 0.50  0.50 -2.45     r0 <- sqrt(-sol1$value) # 2.55 # distance of all points to the center sqrt(colSums((C - p0)^2)) # 2.55 2.55 2.55 As a result we get a ball such that all points lie on the boundary. The same happens for 100 points in 100-dim. space (to keep D positive definite, n = d is required). That is a very nice, even astounding result, but not what I had hoped for! Compare this with another quadratic programming solver in R, for example LowRankQP() in the package of the same name. require(LowRankQP) sol2 <- LowRankQP(D, -d, A, b, u = rep(3, 3), method="LU") p2 <- c(C %*% sol2$alpha)   # 5.000000e-01 5.000000e-01 1.032019e-12     sqrt(colSums((C - p2)^2))   # 0.7071068 0.7071068 0.1000000 The correct result, and we get correct solutions in higher dimensions, too. LowRankQP works also if we apply it with n > d, e.g., 100 points in 2- or 3-dimensional space, when matrix D is not positive definite -- solve.QP( ) does not work in these cases. *** What do I do wrong in calling solve.QP()? *** My apologies for this overly long request. I am sending it through the weekend hoping that someone may have a bit of time to look at it more closely. Hans Werner ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
 Forgot to forward my answer to R-help. Berend On 16-11-2013, at 13:11, Hans W.Borchers <[hidden email]> wrote: > I wanted to solve the following geometric optimization problem, sometimes > called the "enclosing ball problem": > >   Given a set P = {p_1, ..., p_n} of n points in R^d, find a point p_0 such >   that max ||p_i - p_0|| is minimized. > > A known algorithm to solve this as a Qudratic Programming task is as follows: > >   Define matrix C = (p_1, ..., p_n), i.e. coordinates of points in columns, >   and minimize >                   x' C' C x - \sum (p_i' p_i) x_i >   subject to >                   \sum x_1 = 1, x_i >= 0 . > > Let x0 = (x0_1, ..., x0_n) be an optimal solution, then the linear combination > >   p0 = \sum x0_i p_i > > is the center of the smallest enclosing ball, and the negative value of the > objective function at x0 is the squared radius of the ball. > > As an example, find the smallest ball enclosing the points (1,0,0), (0,1,0), > and (0.5, 0.5, 0.1) in 3-dim. space. We would expect the center of the ball to > be at (0.5, 0.5, 0), and with radius 1/sqrt(2). > >   C <- matrix( c(1, 0, 0.5, >                  0, 1, 0.5, >                  0, 0, 0.1), ncol = 3, byrow = TRUE) > >   # We need to define D = 2    C' C, d = (p_i' p_i), A = c(1,1,1), and b = 1 >   D <- 2 * t(C) %*% C >   d <- apply(C^2, 2, sum) >   A <- matrix(1, nrow = 1, ncol = 3) >   b <- 1 > >   # Now solve with solve.QP() in package quadprog ... >   require(quadprog) >   sol1 <- solve.QP(D, d, t(A), b, meq = 1) > >   # ... and check the result >   sum(sol1$solution) # 1 > p0 <- c(C %*% sol1$solution)        # 0.50  0.50 -2.45 >   r0 <- sqrt(-sol1$value) # 2.55 > > # distance of all points to the center > sqrt(colSums((C - p0)^2)) # 2.55 2.55 2.55 > > As a result we get a ball such that all points lie on the boundary. > The same happens for 100 points in 100-dim. space (to keep D positive > definite, n = d is required). > That is a very nice, even astounding result, but not what I had hoped for! > At the risk of making a complete fool of myself. Why the restriction: \sum x_1 = 1, x_i >= 0 ? Isn’t just x_i >= 0 sufficient? Change your code with this A <- diag(3) b <- rep(0,3) sol2 <- solve.QP(D, d, A, b, meq = 0) sol2 resulting in this$solution [1] 0.5 0.5 0.0 $value [1] -0.5$unconstrained.solution [1]  12.75  12.75 -24.50 $iterations [1] 2 0$Lagrangian [1] 0.00 0.00 0.49 $iact [1] 3 And$solution seems to be what you want. And: p0 <- c(C %*% sol2$solution) r0 <- sqrt(-sol2$value) # distance of all points to the center sqrt(colSums((C - p0)^2)) gives the same results as LowRankQP for the last expression. Berend > Compare this with another quadratic programming solver in R, for example > LowRankQP() in the package of the same name. > >   require(LowRankQP) >   sol2 <- LowRankQP(D, -d, A, b, u = rep(3, 3), method="LU") > >   p2 <- c(C %*% sol2$alpha) # 5.000000e-01 5.000000e-01 1.032019e-12 > sqrt(colSums((C - p2)^2)) # 0.7071068 0.7071068 0.1000000 > > The correct result, and we get correct solutions in higher dimensions, too. > LowRankQP works also if we apply it with n > d, e.g., 100 points in 2- > or 3-dimensional space, when matrix D is not positive definite -- solve.QP( ) > does not work in these cases. > > *** What do I do wrong in calling solve.QP()? *** > > My apologies for this overly long request. I am sending it through the weekend > hoping that someone may have a bit of time to look at it more closely. > > Hans Werner > > ______________________________________________ > [hidden email] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code. Reply | Threaded Open this post in threaded view | ## Re: The smallest enclosing ball problem  In reply to this post by Hans W Borchers On 16-11-2013, at 13:11, Hans W.Borchers <[hidden email]> wrote: > I wanted to solve the following geometric optimization problem, sometimes > called the "enclosing ball problem": > > Given a set P = {p_1, ..., p_n} of n points in R^d, find a point p_0 such > that max ||p_i - p_0|| is minimized. > > A known algorithm to solve this as a Qudratic Programming task is as follows: > > Define matrix C = (p_1, ..., p_n), i.e. coordinates of points in columns, > and minimize > x' C' C x - \sum (p_i' p_i) x_i > subject to > \sum x_1 = 1, x_i >= 0 . > > Let x0 = (x0_1, ..., x0_n) be an optimal solution, then the linear combination > > p0 = \sum x0_i p_i > > is the center of the smallest enclosing ball, and the negative value of the > objective function at x0 is the squared radius of the ball. > > As an example, find the smallest ball enclosing the points (1,0,0), (0,1,0), > and (0.5, 0.5, 0.1) in 3-dim. space. We would expect the center of the ball to > be at (0.5, 0.5, 0), and with radius 1/sqrt(2). > > C <- matrix( c(1, 0, 0.5, > 0, 1, 0.5, > 0, 0, 0.1), ncol = 3, byrow = TRUE) > > # We need to define D = 2 C' C, d = (p_i' p_i), A = c(1,1,1), and b = 1 > D <- 2 * t(C) %*% C > d <- apply(C^2, 2, sum) > A <- matrix(1, nrow = 1, ncol = 3) > b <- 1 > > # Now solve with solve.QP() in package quadprog ... > require(quadprog) > sol1 <- solve.QP(D, d, t(A), b, meq = 1) > > # ... and check the result > sum(sol1$solution)                  # 1 >    p0 <- c(C %*% sol1$solution) # 0.50 0.50 -2.45 > r0 <- sqrt(-sol1$value)             # 2.55 > >    # distance of all points to the center >    sqrt(colSums((C - p0)^2))           # 2.55 2.55 2.55 > > As a result we get a ball such that all points lie on the boundary. > The same happens for 100 points in 100-dim. space (to keep D positive > definite, n = d is required). > That is a very nice, even astounding result, but not what I had hoped for! > > Compare this with another quadratic programming solver in R, for example > LowRankQP() in the package of the same name. > >    require(LowRankQP) >    sol2 <- LowRankQP(D, -d, A, b, u = rep(3, 3), method="LU") > >    p2 <- c(C %*% sol2$alpha) # 5.000000e-01 5.000000e-01 1.032019e-12 > sqrt(colSums((C - p2)^2)) # 0.7071068 0.7071068 0.1000000 > > The correct result, and we get correct solutions in higher dimensions, too. > LowRankQP works also if we apply it with n > d, e.g., 100 points in 2- > or 3-dimensional space, when matrix D is not positive definite -- solve.QP( ) > does not work in these cases. > > *** What do I do wrong in calling solve.QP()? *** > After having a closer look at this problem, I believe you did not include the constraint x_i >= 0 in the call to solve.QP. So with this modification of your code A <- matrix(rep(1,3),nrow=4,ncol=3,byrow=TRUE) A[2:4,] <- diag(3) b <- c(1,0,0,0) sol3 <- solve.QP(D, d, t(A), b, meq = 1) # first row of A is an equality sol3 p0 <- c(C %*% sol3$solution) r0 <- sqrt(-sol3$value) p0 r0 sqrt(colSums((C - p0)^2)) one gets the correct answer. BTW LowRankQP seems to postulate x_i >=0 if I read its manual correctly. Berend ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code. Reply | Threaded Open this post in threaded view | ## Re: The smallest enclosing ball problem  In reply to this post by Berend Hasselman Berend Hasselman xs4all.nl> writes: > Forgot to forward my answer to R-help. > > Berend Thanks, Berend, for thinking about it. \sum xi = 1 is a necessary condition to generate a valid geometric solution. The three points in the example are very regular and your apporach works, but imagine some random points: set.seed(8237) C <- matrix(runif(9), 3, 3) D <- 2 * t(C) %*% C d <- apply(C^2, 2, sum) A <- diag(3) b <- rep(0,3) require(quadprog) sol1 <- solve.QP(D, d, A, b, meq = 0) sol1 # now \sum xi = 1is not fulfilled p0 <- c(C %*% sol1$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 > At the risk of making a complete fool of myself. > > Why the restriction:  \sum x_1 = 1, x_i >= 0 ? > > Isn’t just  x_i >= 0 sufficient? > > Change your code with this > > A <- diag(3) > b <- rep(0,3) > sol2 <- solve.QP(D, d, A, b, meq = 0) > sol2 > > resulting in this > > $solution > [1] 0.5 0.5 0.0 >$value > [1] -0.5 > $unconstrained.solution > [1] 12.75 12.75 -24.50 >$iterations > [1] 2 0 > $Lagrangian > [1] 0.00 0.00 0.49 >$iact > [1] 3 > > And $solution seems to be what you want. > > And: > > p0 <- c(C %*% sol2$solution) > r0 <- sqrt(-sol2$value) > > # distance of all points to the center > sqrt(colSums((C - p0)^2)) > > gives the same results as LowRankQP for the last expression. > > Berend > ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code. Reply | Threaded Open this post in threaded view | ## Re: The smallest enclosing ball problem  On 17-11-2013, at 11:32, Hans W.Borchers <[hidden email]> wrote: > Berend Hasselman xs4all.nl> writes: >> Forgot to forward my answer to R-help. >> >> Berend > > Thanks, Berend, for thinking about it. \sum xi = 1 is a necessary condition > to generate a valid geometric solution. The three points in the example are > very regular and your apporach works, but imagine some random points: > > set.seed(8237) > C <- matrix(runif(9), 3, 3) > D <- 2 * t(C) %*% C > d <- apply(C^2, 2, sum) > A <- diag(3) > b <- rep(0,3) > > require(quadprog) > sol1 <- solve.QP(D, d, A, b, meq = 0) > sol1 # now \sum xi = 1is not fulfilled > > p0 <- c(C %*% sol1$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 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 ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
 In reply to this post by Berend Hasselman Berend Hasselman xs4all.nl> writes: > It seems you are absolutely right. I always assumed a quadratic programming solver will -- as all linear programming solvers do -- automatically require the variables to be positive. I checked it for some more examples in 10 and even 100 dimensions, and the results now agree. Still, it's a bit disappointing that 'quadprog' will not solve problems with 10 points in R^3, because the corresponding matrices are not positive definite. Thanks Hans Werner > > After  having a closer look at this problem, I believe you did not include > the constraint x_i >= 0 in the call to solve.QP. > So with this modification of your code > > A <- matrix(rep(1,3),nrow=4,ncol=3,byrow=TRUE) > A[2:4,] <- diag(3)   > b <- c(1,0,0,0) > > sol3 <- solve.QP(D, d, t(A), b, meq = 1) # first row of A is an equality > sol3 > p0 <- c(C %*% sol3$solution) > r0 <- sqrt(-sol3$value) > p0 > r0 > sqrt(colSums((C - p0)^2)) > > one gets the correct answer. > BTW LowRankQP seems to postulate x_i >=0 if I read its manual correctly. > > Berend ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.