Mihai Badoiu
MIT

Ken Clarkson
Bell Labs

• Given a set of points S and ε>0,
• find P⊂S so that the smallest ball containing P is within ε of containing S.

• In contrast, the smallest ball (ε=0) is determined by a subset of S of size at most d+1;
• these results are mainly of interest for very large d.
• It is tight because no better can be done with a regular simplex.
• The existence proof is an algorithm, shown below.
• Note that the smallest ball containing P is no larger than the smallest ball containing S.

• The core-set size appears in the exponent in the running times of several algorithms.
• Previous bounds:
• Roughly 128/ε 2 [BI]
• O(1/ε) [KMY]
• 2/ε [BC]

• Find the point a∈S farthest from the center of the smallest ball containing P;
• If that distance is smaller than R(1+ε), we're done; (R is the optimal ball radius.)
• Pick b∈P∪{a}, so that the smallest ball containing P∪{a}∖{b} is as large as possible;
• Set P to P∪{a}∖{b}.

• Let P a ≡P∪{a} ;
• The circumradius of P a must be substantially larger than the circumradius of P
• because a was farthest
• The circumradius of P a∖{b} must be not too much smaller than the circumradius of P a
• because b was not incident to the subset with largest circumradius;
• Combined quantitatively: if not done, the radius of P must increase.

• Any simplex T has a facet F such that r B (F) 2 ≥(1-1/d 2)r B (T) 2.
• Here T has d+1 vertices.
• The smallest ball is alway determined by a simplex;
• The regular (equilateral) simplex is the worst case.
• So the circumradius of P a∖{b} is at least 1-1/⌈1/ε⌉ times the circumradius of P a.

• When farthest point a is added, the circumradius of P a is at least R ^ / r+r/R^2
• Here R^≡R(1+ε), and r was the circumradius of P.
• The basic argument [BI]: if the circumcenter of P a is
  • close to the circumcenter of P, then the new radius must be close to R^;
  • far from the circumcenter of P, then some point of P must be far from that new circumcenter.

• Relation to quadratic programming duality?
• Experimental results?