AA Note on John Simplex with Positive Dilation
Zhou Lu ∗ Princeton University [email protected]
December 2020
Abstract
We prove a John’s theorem for simplices in R d with positive dilationfactor d + 2, which improves the previously known d upper bound. Thisbound is tight in view of the d lower bound. Furthermore, we give anexample that d isn’t the optimal lower bound when d = 2. Our resultsanswered both questions regarding John’s theorem for simplices with pos-itive dilation raised by Leme and Schneider [2020]. In the problem of linear function reconstruction (Khachiyan [1995], Summa et al.[2014], Nikolov [2015]), one needs to efficiently reconstruct a linear function f defined on a set X by using a zeroth order oracle. When the cost of oraclequerying grows along with accuracy, a John like simplex serves as a good basisas in Leme and Schneider [2020].A crucial step of their method is to find a simplex T with vertices in X , suchthat X can be contained in some translate of T with dilation, where a smaller(absolute value) dilation factor indicates higher efficiency. When consideringnegative dilation, a translate − dT is able contain X so the upper bound is O ( d )which matches its lower bound.However, if we look at positive dilation, it seems to be less efficient thannegative dilation because only a worse d upper bound was given in Leme andSchneider [2020]. Though a better bound for positive dilation does not immedi-ately help design better algorithms for the reconstruction problem, it is naturalto ask if we can improve it.We prove a John’s theorem for simplices with positive dilation factor d +2 which answers the question affirmatively. Furthermore, we give a counter-example that the d lower bound given in Leme and Schneider [2020] isn’t tightwhen d = 2. ∗ This work is done during the author’s visit at SQZ institution. a r X i v : . [ m a t h . M G ] D ec Background
In the seminal work John [2014], it’s proven that any convex X ⊂ R d canbe sandwiched between two concentric ellipsoids of ratio d . Specifically, thefollowing theorem was given in John [2014]: Theorem 2.1.
Given any convex body X ⊂ R d , let x + T B d denote the minimalvolume ellipsoid containing X , then it holds x + 1 d T B d ⊂ X ⊂ x + T B d (1)The ellipsoid x + T B d is called ’John ellipsoid’. For a more comprehensivereview we refer the readers to Henk [2012]. In this paper we consider general X which is potentially non-convex, and we aim to find a John like simplexwith positive dilation factor O ( d ) so that it can be seen as an analogue to Johnellipsoid. Definition 2.2.
Given d + 1 points v , ..., v d +1 in R d , we call their convex hulla simplex with vertices v , ..., v d +1 . We consider the following: given a compact set X ⊂ R d which is not neces-sarily convex or connected, we want to find d + 1 points in X forming a simplex T such that X is contained in some translate of T with positive dilation.A simple method to do so is by considering the maximum volume simplex(MVS) of X as in Leme and Schneider [2020]. Definition 2.3.
A maximum volume simplex T with vertices in a bounded set X ⊂ R d is one of the simplices whose euclidean measure is no less than anyother simplex of X . Proposition 2.4.
There always exists a MVS as long as X is compact.Proof. Because X is bounded, the volume of its simplex has a least upper bound c ≥
0, therefore we can find a sequence of simplices T j with vertices v ( j )1 , ..., v ( j ) d +1 ,whose volumes converge to c . Due to the compactness of X , any sequence in X has a converging sub-sequence which converges to a point in X , thus we canfind indexes a j such that v ( a j )1 converges to a point ˜ v ∈ X .We consider a new sequence of simplices T a j whose first vertex is replacedby ˜ v while others keep unchanged. It’s easy to see that their volumes stillconverge to c . Again we find a converging sub-sequence of v ( a j )2 which convergesto a point ˜ v ∈ X , and repeat this procedure until we have all ˜ v i . Then thesimplex with vertices ˜ v i is the desired MVS.Leme and Schneider [2020] prove the following lemma by using the ’maxi-mum volume’ property: Lemma 2.5.
Let T be the MVS with vertices in X , then X is contained in atranslate of − dT . d upper bound for positive dilation. We areinterested in proving tighter bounds for the positive dilation factor. Existingupper and lower bounds are d and d respectively as in Leme and Schneider[2020], in their paper two questions regarding John’s theorem for simplices withpositive dilation were raised Question 1:
Can we get a John’s theorem for simplices with positive dila-tion factor O ( d )? Question 2:
Given compact X ⊂ R , can we always find a triangle T withvertices in X so that X is contained in some translate of 2 T ?We give an affirmative answer to question 1 by proving a d + 2 upper bound,and provide a counter-example to question 2. d + 2 upper bound In this section we prove a d + 2 upper bound based on a simple observation:by fully exploiting the ’maximum volume’ property, we can squeeze − dT to amuch smaller set which still contains X .We already know that a simplex T ∈ R d can be covered by a translate of − dT , therefore a naive method (choose T to be MVS and use − d dilation twice)directly leads to a d upper bound for positive dilation.However, the first step of this method is extravagant in that we can finder asmaller set covering X when d >
2. Think about the following example in R :assume T (the MVS of X ) is the regular simplex with vertices v , ..., v and 0as its center, then − T has − v , ..., − v as its vertices. We denote S v,i to bethe hyperplane parallel to hyperplane v , ..., v i − , v i +1 , ..., v , with v lying on it.By the definition of T , X lies between S v i ,i , S − v i ,i for any i = 1 , ...,
4, thereforemuch space is wasted in the ’cones’ of − T .The d bound can be decomposed as 1 + ( d + 1)( d −
1) where for each v i , T extends d − d + 1) = d + 2 upper bound. Theorem 3.1.
For any compact set X ∈ R d , there exists d + 1 points in X forming a simplex T as its vertices, such that X can be covered in a translateof ( d + 2) T .Proof. We choose the MVS of X to be T with vertices v , ..., v d +1 and we assume0 is its center (of gravity) without loss of generality. We denote ˆ v i to be the sym-metric point of v i with respect to hyperplane v , ..., v i − , v i +1 , ..., v d +1 along di-rection v i , and S v,i to be the hyperplane parallel to hyperplane v , ..., v i − , v i +1 , ..., v d +1 ,with v lying on it. By the definition of T , X lies between S v i ,i , S ˆ v i ,i for any i = 1 , ..., d + 1, otherwise the ’maximum volume’ property will be contradicted.We take a close look at the simplex T (cid:48) made up of S ˆ v i ,i . Obviously T (cid:48) is atranslate of T with positive dilation and center (of gravity) 0 unchanged. Definethe intersection point between line v i ˆ v i (we overload the notation of vectorsto denote endpoints of a segment when the context is clear) and hyperplane v , ..., v i − , v i +1 , ..., v d +1 as w i , we have that v i = dd +1 w i v i by the definition of3igure 1: Example in R . The red pyramid is T , the blue one is − T and theyellow ones are the wasted space.center (of gravity). Combing with the fact that w i v i = ˆ v i w i , the dilation factoris (1 + d +1 ) / d +1 = d + 2.Denote the simplex made up of S v i ,i as ˜ T and the region enclosed by S v i ,i , S ˆ v i ,i as ˆ T , we have the following inclusion: X ⊂ ˆ T = T (cid:48) ∩ ˜ T ⊂ T (cid:48) (2)which finishes our proof. > d when d = 2 We give a counter-example to question 2 in this section. The idea behind isintuitive: we construct several discrete points as a hard case, so that thesediscrete points won’t help much when we construct the covering triangle, buthurts a lot when we consider covering them since the whole convex hull needsto be covered implicitly.
Theorem 4.1.
There exists a compact set X ∈ R , such that for any triangle T with vertices in X , X can’t be contained in any translate of T .Proof. We pick 5 points: A = ( − , , B = (1 , , C = ( − (cid:15) − δ, , D = ( (cid:15) + δ, , E = (0 , (cid:15) − X = { A, B, C, D, E } , where constants (TBD) satisfy (cid:15), δ ∈ (0 , T by 6 cases and how they leadto contradiction. Case 1: T = (cid:52) CDE
The intercept along y = 0 of any translate of 2 T has length at most 4( (cid:15) + δ ) <
2, thus AB can’t be covered by any translate of 2 T . Case 2: T = (cid:52) ABE
The intercept along x = 0 of any translate of 2 T has length at most 2 − (cid:15) < − (cid:15) , thus C, E can’t be simultaneously covered by any translate of 2 T . Case 3: T = (cid:52) ACD
The intercept along y = 0 of any translate of 2 T has length at most 4( (cid:15) + δ ) <
2, thus AB can’t be covered by any translate of 2 T . Case 4: T = (cid:52) ABC
The intercept along y = 1 of any translate of 2 T has length at most 2 (cid:15) < (cid:15) +2 δ when its ’bottom’ is below y = (cid:15) −
1, thus CD and E can’t be simultaneouslycovered by any translate of 2 T . Case 5: T = (cid:52) ACE
The intercept along y = 0 of any translate of 2 T has length at most 2 − ( (cid:15) + δ )(1 − (cid:15) )2 − (cid:15) <
2, thus AB can’t be covered by any translate of 2 T . Case 6: T = (cid:52) ADE
We would like to prove that any translate of 2 T can’t cover (cid:52) ABC . Weextend −−→ AD by twice to D (cid:48) = (1 + 2 (cid:15) + 2 δ,
1) and −→ AE by twice to E (cid:48) = (1 , (cid:15) − (cid:52) ABC to fit in (cid:52) A (cid:48) D (cid:48) E (cid:48) where A (cid:48) = A .Because A (cid:48) + −−→ CB = ( (cid:15) + δ, − C to be contained in (cid:52) A (cid:48) D (cid:48) E (cid:48) , B must lie below line y = 11 + (cid:15) + δ ( x − (cid:15) − δ ) − y -coordinate of B is that of the intersectionpoint between line 3 and y = 2 − (cid:15)(cid:15) + δ ( x − − (cid:15) + δ )(1 − (cid:15) )2 − (cid:15) ) (4)By straightforward computation, we have that y = − − (cid:15)(cid:15) + δ + 1 − (cid:15) (5)is the largest possible y -coordinate of B . However, the intercept along line5 of (cid:52) A (cid:48) D (cid:48) E (cid:48) equals(2 − (cid:15) − − (cid:15)(cid:15) + δ + 1 − (cid:15) ) × ( (cid:15) + δ )(1 − (cid:15) )2 − (cid:15) − (cid:15) = 2 + ( (cid:15) + δ )(1 − (cid:15) )2 − (cid:15) − (cid:15) + δ )(1 − (cid:15) )(2 − (cid:15) )= 2 − ( (cid:15) + δ )(1 + 2 (cid:15) − (cid:15) )(1 − (cid:15) )(2 − (cid:15) ) < C, A, B can’t be simultaneously covered by any translate of 2 T .For the choice of constants, any constant pair additionally satisfying (cid:15) + δ < is feasible. In this note, we analyze John’s theorem for simplices with positive dilation andanswer related open questions raised by Leme and Schneider [2020]. We provea tight d + 2 upper bound which matches the d lower bound, improving thepreviously known d bound. We also give a simple counter-example showingthat the d lower bound isn’t optimal. References
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