A Polynomial Number of Random Points Does Not Determine the Volume of a Convex Body

Discrete & Computational Geometry, Feb 2011

We show that there is no algorithm which, provided a polynomial number of random points uniformly distributed over a convex body in ℝ n , can approximate the volume of the body up to a constant factor with high probability.

A PDF file should load here. If you do not see its contents the file may be temporarily unavailable at the journal website or you do not have a PDF plug-in installed and enabled in your browser.

Alternatively, you can download the file locally and open with any standalone PDF reader:

https://link.springer.com/content/pdf/10.1007%2Fs00454-011-9328-x.pdf

A Polynomial Number of Random Points Does Not Determine the Volume of a Convex Body

Discrete Comput Geom A Polynomial Number of Random Points Does Not Determine the Volume of a Convex Body Ronen Eldan 0 0 R. Eldan ( ) School of Mathematical Sciences, Tel-Aviv University , Tel-Aviv 69978 , Israel We show that there is no algorithm which, provided a polynomial number of random points uniformly distributed over a convex body in Rn, can approximate the volume of the body up to a constant factor with high probability. Convex bodies; Random point oracle; Volume computing algorithms; Sampling convex bodies; Distribution of mass on convex bodies 1 Introduction Volume-related properties of high-dimensional convex bodies is one of the main topics of convex geometry in research today. Naturally, calculating or approximating the volume of a convex body is an important problem. Starting from the 1980s, several works have been made in the area of finding a fast algorithm for computing the volume of a convex body (see, for example, [1–3, 8, 9], and references therein). These algorithms usually assume that the convex body K ⊂ Rn is given by a certain oracle. An oracle is a “black box” which provides the algorithm some information about the body. One example of an oracle is the membership oracle, which, given a point x ∈ Rn, answers either “x ∈ K ” or “x ∈/ K ”. Another example is the random point oracle, which generates random points uniformly distributed over K . All volume computing algorithms, known to the author, that appear in the literature use the membership oracle. This note deals with a question asked by L. Lovász about Communicated by L. Lovasz. The author is partially supported by the Israel Science Foundation and by a Farajun Foundation Fellowship. the random point oracle. It has been an open problem for a while whether or not it is possible to find a fast algorithm which computes the volume of K , provided that there is access to the random point oracle [4, 7]. We answer this question negatively. In order to formulate our main result, we begin with some definitions. An algorithm which uses the random point oracle is a (possibly randomized) function whose input is a finite sequence of random points generated according to the uniform measure on K and whose output is a number, which is presumed to be an approximation for the volume of K . The complexity of the algorithm will be defined by the length of the sequence of random points. We are interested in the existence of algorithms with complexity which depends polynomially on the dimension n. We say that an algorithm is correct up to C with probability p if for any K ⊂ Rn, given a sequence of random points from K , the output of the algorithm is between Vol(K) and CVol(K), with probability at least p. C We prove the following theorem: Theorem 1 There do not exist constants C, p, κ > 0 such that for any dimension n, there exists an algorithm with complexity O(nκ ) which is correct in estimating the volume of convex bodies in Rn up to C with probability p. It is important to emphasize that this result is not a result in complexity theory. In this note we show that a polynomial number of points actually does not contain enough information to estimate the volume, regardless of the number of calculations, and hence, it is of information-theoretical nature. For convex geometers, the main point in this study may be the additional information on volume distribution in convex bodies it provides. We suggest the reader to look this result in view of the recent results concerning the distribution of mass in convex bodies. In particular, results regarding thin-shell concentration and the Central Limit Theorem for Convex bodies, proved in the general case by B. Klartag, show that essentially all of the mass of an isotropic convex body K is contained in a very thin-shell around the origin and that almost all of the marginals are approximately Gaussian. This may suggest that, in some way, all convex bodies, when neglecting a small portion of the mass, behave more or less the same as a Euclidean ball in many senses. Philosophically, one can also interpret these results as follows: provided a small number of points from a logarithmically concave measure, one cannot distinguish it from a spherically symmetric measure. For definitions and results, see [5]. One of the main stages of our proof is to show that one cannot distinguish between the uniform distribution over certain convex bodies, which are geometrically far from a Euclidean ball, and some spherically symmetric distribution, when the number of sample points is at most polynomially large. Here is a more quantitative formulation of what we prove: Theorem 2 There exist a constant ε > 0 and a number N ∈ N such that for all n > N , there does not exist an algorithm whose input is a sequence of length enε of points generated randomly according to the uniform measure in a convex body K ⊂ Rn, which determines Vol(K) up to enε with probability more than e−nε to be correct. Remark After showing that the volume of a convex body cannot be approximated, one may further ask: what about an algorithm that estimates the volume radius of 1 a convex body, defined by VolRad(K) = Vol(K) n ? A proof which shows that it is also impossible has to be far more delicate than our proof. For example, under the hyperplane conjecture, it is easy to estimate the volume radius of a convex body up to some C > 0. One may also compare this result to the two following related results: a recent result of Goyal and Rademacher [4] shows that in order to learn a convex body, one √ n needs at least 2c ε random points. Learning a convex body roughly means finding a set having at most ε relative symmetric difference with the actual body (see [4]). Klivans, O’Donnel, and Servedio [6] show that any convex body can be agnostically learned with respect to the Gaussian distribution using 2O(√n) labeled Gaussian samples. The general idea of the proof is as follows. Let {Kα}α∈I1 and {Kα}α∈I2 be two families of convex bodies. For i = 1, 2, a probability measure μi on the set of indices Ii induces a random construction of convex bodies, which in turn induces a probability measure Pi on the set of sequences of points in Rn in the following simple way: first generate an index α according to μi , and then generate a sequence of N uniformly distributed random samples from Kα . In the proof we will define two distinct random constructions of convex bodies Ki = ({Kα}α∈Ii , μi ), i = 1, 2, such that: 1. For every α1 ∈ I1 and α2 ∈ I2, the ratio between Vol(Kα1 ) and Vol(Kα2 ) is large. 2. If N is not too large, both distributions P1, P2 are close in total variation distance to some distributions of samples in which the samples are independent and have a spherically symmetric law. 3. The radial profiles (hence the distribution of the Euclidean norm of a random sample) of typical random bodies K1, K2 are very close to each other. In other words, we will define two constructions of random convex bodies for which: 1. The typical volumes of the bodies they produce will be far from equal. 2. They will be both indistinguishable from spherically symmetric constructions for a polynomial number of samples. 3. The radial profiles they produce are indistinguishable from each other for a polynomial number of samples. To go on with the proof, a simple application of Yao’s lemma will help us assume that the algorithm is deterministic. A deterministic algorithm is actually a function F : Rnκ+1 → R which takes a sequence of points and returns the volume of the body. If the total variation distance between the probabilities P1 and P2 defined above is small, then there exists a set A ⊂ Rnκ+1 which has a high probability with respect to both P1 and P2. Obviously, for all x ∈ A, F (x) is wrong in approximating the volume of at least one of the families. In Sect. 2, we will describe how we build these families of bodies, {Kα}, using a random construction which starts from a Euclidean ball, to which deletions which cut out parts of it, generated by some Poisson process, are applied. Then, using elementary properties of the Poisson process and some concentration of measure properties of the ball, we will see that the correlation between different points in polynomially long sequence of random points generated uniformly from the body will be very weak (with respect to the generation of the body itself). Using this fact, we will only have to inspect the distribution of a single random point. The construction will have a spherically symmetric nature, so the density of a single random point will only depend on its distance from the origin, and therefore we will only have to care about the distribution of the distance of a point from the origin in the generated bodies. The role of the following section, which is more technical but fairly delicate, will be to calibrate this construction so that these families have different volumes, yet, approximately the same distribution of distance from the origin. Before we proceed to the proof, let us introduce some notation. In this note the number n will always denote a dimension. For an expression f (n) which depends on n, by f (n) = SE(n) we mean: there exists some n0 ∈ N and > 0 such that for all n > n0, |f (n)| < e−n . Also write f (n) = g(n)(1 + SE(n)) for | fg((nn)) − 1| = SE(n) and f (n) = g(n) + SE(n) for |f (n) − g(n)| = SE(n). The notation f (n) g(n) and f (n) g(n) will be interpreted as f (n) < g(n) and f (n) > g(n) for n large enough. Moreover, we decide that N = N (n) denotes the length of the sequence of random points. All throughout this note we assume that there exists a universal constant ε > 0 such that N (n) < enε . 2 The Deletion Process In this section we will describe the construction of random bodies which will later be used as counterexamples. Our goal, after describing the actual construction, will be to prove, using some simple properties of the Poisson distribution, a weak-correlation property between different points generated from the body. Denote by Dn the n-dimensional Euclidean ball of unit radius, centered at the origin, and by ωn its Lebesgue measure. Recall that for two probability measures P1, P2 on a set Ω , the total-variation distance between the two measures is defined by dTV(P1, P2) = sup P1(A) − P2(A) . A⊆Ω One can easily check that if these measures are absolutely continuous with respect to some third measure Q, then it is also equal half the L1(Q) distance between the two densities. 1 Define r0 = n− 3 and T0(θ ) = Dn ∩ x; x, θ ≤ r0 . Let T be a function from the unit sphere to the set of convex bodies such that for every θ ∈ Sn−1, T (θ ) satisfies T0(θ ) ⊆ T (θ ) ⊆ Dn. (Recall that most of the mass 1 of the Euclidean ball is contained in {x1 ∈ [−1, Cn− 2 ]}. So T (θ ) contains almost all the mass of the Euclidean ball.) Moreover, let m > 0. We will now describe our construction of a random convex body KT ,m. First, suppose that m ∈ N. Let Θ = (θ1, θ2, . . . , θm) be m independent random directions distributed according to the uniform measure on Sn−1. We define KT ,m as KT ,m = Dn T (θi ). i Finally, instead of taking a fixed m ∈ N, we take ζ to be a Poisson random variable with expectation m, independent of the above. We can now define KT ,ζ in the same manner. Let us denote the probability measure on the set of convex bodies induced by the process described above by μ. After generating the body KT ,m, which, from now on will be denoted just by K wherever there is no confusion caused, we consider the following probability space: let Ω = (Dn)N be the set of sequences of length N of points from Dn. Denote by λ the uniform probability measure on Ω , and for a convex body K , denote by λK the uniform probability measure on KN = 1≤i≤N K ⊆ Ω . Finally, define the probability measure P = PT ,m on Ω as follows: for A ⊆ Ω , P (A) = λK (A) dμ(K) = Vol(KN ∩ A) dμ(K). Vol(KN ) (The measure P describes the following process: first, generate the random set K according to construction described above, and then generate N i.i.d. random points, independent of the above, according to the uniform measure on K .) Moreover, for p = (x1, . . . , xN ) ∈ Ω , define πi (p) = xi , the projections onto the ith copy of the Euclidean ball. It easy to check that P is absolutely continuous with respect to λ. We define the following function on Ω : fT ,m(p) = P p ∈ KTN,m = P ∀1 ≤ i ≤ N , πi (p) ∈ KT ,m . As we will see later, the function f is related in a simple way to ddPλ . Namely, we will have dP dλ (p) = 1 + SE(n) f (p) Ω f for all p in some subset of Ω with measure close to 1. For convenience, from now on fT ,m will be denoted by f . We start with some simple geometric observations regarding Ω . Denote by σ the rotation-invariant probability measure on Sn−1. Define, for p ∈ Ω and 1 ≤ i ≤ N , Ai (p) = θ ∈ Sn−1 : πi (p) ∈/ T (θ ) . For 1 ≤ i, j ≤ N , let Fi,j ⊂ ΩN be the event defined by Fi,j = p : σ (Ai (p) ∩ Aj (p)) < e−n0.1 σ (Ai (p)) ( 1 ) ( 2 ) ( 3 ) and let and so, F = Fi,j (which should be understood as “no two points are too close to each other” and, as we will see, will imply that points are weakly correlated). We start with the following simple lemma. Lemma 3 Under the above notation: (i) λ(F ) = 1 + SE(n). (ii) There exists some ε0 > 0 such that if we assume that Pμ Vol(K) < ωne−nε0 < e−nε0 (that is, the volume of K is typically not much smaller than the volume of Dn), then P (F ) = 1 + SE(n). Proof (i) Let p be uniformly distributed in Ω . Denote xi = πi (p), so that x1, x2 are independent points uniformly distributed in Dn. Let us calculate λ(F1,2). First, for a fixed θ ∈ Sn−1, one has P x1 ∈/ T (θ ) ≤ P x1 ∈/ T0(θ ) = P x1, θ ≥ r0 . 1 Recalling that r0 = n− 3 the Euclidean ball, one gets 1 n− 2 , by elementary calculations regarding marginals of ( 4 ) ( 5 ) ( 6 ) ( 7 ) E σ (A1 ∩ A2) x2 = x2 = P(θ ∈ A1) dσ (θ ) = P x1 ∈/ T (θ ) dσ (θ ) A2 σ (A2)e−n0.2 , E σ (A1 ∩ A2) σ (A2) x2 = x2 A2 Now, fix x2 ∈ Dn. Define Ai := Ai (p). One has Now, this is true for every choice of x2, and so integrating over x2 gives Now we use Markov’s inequality to get A union bound completes the proof of (i). E σ (A1 ∩ A2) σ (A2) (ii) First, we can condition on the event {Vol(K) > ωneε0 } (with ε0 to be chosen later). Equation ( 5 ) ensures us that it will happen with probability = 1 − SE(n). Observe that for any event E ⊂ Ω which is measurable by the σ -field generated by π1, π2, we have λK (E) = ωn2λ((K × K × Dn × · · · × Dn) ∩ E) Vol(K)2 ωn2λ(E) ≤ Vol(K)2 . Now, taking E = F1C,2, choosing ε0 to be small enough, and using ( 7 ) and ( 8 ) along with ( 5 ), one gets P (F1,2) = 1 + SE(n). Applying a union bound finishes the proof. We can now turn to the lemma which contains the main ideas of this section: Lemma 4 There exist ε0, ε1 > 0 and n0 such that for every n > n0, the following holds: Whenever m is small enough such that the following condition is satisfied: P {θ ∈ K} > e−nε0 ∀θ ∈ Sn−1 (that is, we are not removing too much volume, in expectation, even from the outer shell), then: P Vol(K) − E Vol(K) > e−nε1 E Vol(K) = SE(n), ( 10 ) (i) We have and also ( 5 ) holds. (ii) For all p ∈ F , we have (iii) then In other words, if we define f˜ : Dn → R as f (p) = 1 + SE(n) P πj (p) ∈ K . N j=1 f˜(x) = P(x ∈ K), f (p) = 1 + SE(n) f˜ πi (p) ∀p ∈ F . i E(Vol(KN ∩ F )) (EVol(K))N − 1 = SE(n). ( 8 ) ( 9 ) (11) (12) Proof We begin by proving (ii). Fix p ∈ F . Define xi = πi (p) and Ai = Ai (p) ⊂ Sn−1 as in ( 2 ). Also define Gj = i≤j {xi ∈ K}. Fix 2 ≤ j ≤ N . Let us try to estimate P (Gj |Gj−1). When conditioning on the event Gj−1, we can consider our Poisson process as a superposition of three “disjoint” Poisson processes: the first one, with intensity λs , only generates deletions that cut xj but leave all the xi ’s for i < j intact. The second one, with intensity λu, deletes xj along with one of the other xi ’s, and the third one is the complement (hence, deletions that do not affect xj ). We have, recalling that the expectation of the number of deletions is m, Moreover, (in the above formula we are including, multiple times, deletions that cut more than two points, and hence it is the inequality rather than equality). Now, using the definition of F , one gets e−(λs (Sn−1)+λu(Sn−1)) ≥ e x −mσ ({θ; |xjj | ∈/T (θ)}) ≥ e−nε0 (the first inequality follows from the fact that T (θ ) are star-shaped). The last two inequalities give λu Sn−1 = SE(n). P (Gj |Gj−1) P ({xj ∈ K}) − 1 = e−(λs(Sn−1)+λu(Sn−1)) − 1 = SE(n). e−λs (Sn−1) λs Sn−1 + λu Sn−1 = mσ (Aj ). λu Sn−1 ≤ m σ (Ai ∩ Aj ) i<j λu(Sn−1) λs (Sn−1) + λu(Sn−1) = SE(n). (13) (14) (15) (16) (17) (18) (19) (20) Note that (9) implies It follows that Moreover, one has This proves (ii). P (GN ) = P (Gj |Gj−1) = j j P (Gj |Gj−1) P {xj ∈ K} . P ({xj ∈ K}) Using (18) and (19), we get f (p) = P (GN ) = 1 + SE(n) P {xj ∈ K} . j (i) Showing that ( 5 ) holds is just a matter of noticing that P(x ∈ K) is decreasing with respect to |x| and taking ε0 small enough. We turn to estimate E(Vol(K)2). We have E Vol(K)2 = P {x1 ∈ K} ∩ {x2 ∈ K} dx1 dx2 = + Dn×Dn (Dn×Dn)∩F1,2 (Dn×Dn)∩F1C,2 P {x1 ∈ K} ∩ {x2 ∈ K} dx1 dx2 P {x1 ∈ K} ∩ {x2 ∈ K} dx1 dx2 (21) (22) (Dn×Dn)∩F1,2 = 1 + SE(n) (we will later see that the second summand is negligible). Now, (20) gives P {x1 ∈ K} ∩ {x2 ∈ K} dx1 dx2 (Dn×Dn)∩F1,2 P {x1 ∈ K} P {x2 ∈ K} dx1 dx2, (23) which also implies that (Dn×Dn)∩F1,2 P {x1 ∈ K} ∩ {x2 ∈ K} dx1 dx2 > 12 e−2nε0 . Recall that λ(F1C,2) = SE(n) (as a result of the previous lemma). Taking ε0 to be small enough, we will get E Vol(K)2 = 1 + SE(n) On the other hand, = 1 + SE(n) E Vol(K) 2 = (Dn×Dn) (Dn×Dn)∩F1,2 (Dn×Dn)∩F1,2 P {x1 ∈ K} ∩ {x2 ∈ K} dx1 dx2 P {x1 ∈ K} P {x2 ∈ K} dx1 dx2. P {x1 ∈ K} P {x2 ∈ K} dx1 dx2. (24) Using the same considerations as above, the part of the integral over F1C,2 can be ignored; hence, E Vol(K) 2 = 1 + SE(n) So, we finally get (Dn×Dn)∩F1,2 P {x1 ∈ K} P {x2 ∈ K} dx1 dx2. (25) E Vol(K)2 = 1 + SE(n) E Vol(K) 2. (26) Recalling that we assume ( 9 ) and using Chebyshev’s inequality, this easily implies (i), which finishes (ii). For the proof of (iii), E Vol KN ∩ F P p ∈ KN dp = 1 + SE(n) P πi (p) ∈ K = F ≤ EVol(K) N . F i Consider the density ddPλ . Our next goal is to find a connection between this density and the function f . Let A ⊆ F ⊂ Ω . Using the concentration properties of Vol(K), we will prove that (where ε1 is the constant from Lemma 4). We have, by the definition of G, G = 1 + SE(n). (27) (28) (29) Now, and so, To prove this, first recall that A ⊆ F . This gives GC Vol(KN ∩ A) dμ(K) (EVol(K))N GC Vol(KN ∩ F ) dμ(K) (EVol(K))N ≤ = EμVol(KN ∩ F ) (EVol(K))N − G Vol(KN ∩ F ) dμ(K) (EVol(K))N . So μ(G) = 1 + SE(n), which gives We will also need that GC (31) (32) (33) (34) (35) (36) G Vol(KN ∩ F ) dμ(K) (EVol(K))N Using part (iii) of Lemma 4, along with (33) and (34), proves (32). Plugging together (28), (30), (31), and (32) implies P (A ) = P (A) + P A ∩ F C . Plugging (35), (36), and (29) proves (27). We would now like to use the result of Lemma 4, to replace f with f˜. Let A ⊆ Ω . Define A = A ∩ F , Part (ii) of Lemma 3 with (27) gives P (A ) = P (A) + SE(n) = A f (p) dp ( Dn f˜(x) dx)N + SE(n). We can now plug in (12) to get So, finally defining P (A ) = A i f˜(πi (p)) dp ( Dn f˜(x))N , we have proved the following lemma: Lemma 5 Suppose that condition ( 9 ) from Lemma 4 holds. Then one has dTV(P , P˜ ) = SE(n). We formulate this accurately in the following lemma. Note that the measure P˜ is not, in general, a probability measure. The lemma, however, ensures that P˜ (Ω) is very close to 1. Recall that our plan is to find two families of convex bodies, achieved by two pairs (T1, m1) and (T2, m2), such that dTV(P1, P2) is small, even though their volumes differ. The above lemma motivates us to try to find such pairs with f˜f1˜1 = f˜f2˜2 + SE(n). Lemma 6 Suppose that there exist two pairs (Ti , mi ) for i = 1, 2 such that ( 9 ) is satisfied and in addition, defining f˜1 and f˜2 as in (11), f˜1 In the next section we deal with how to calibrate Ti and mi so that (37) holds. Our goal in this section is to build convex bodies with a prescribed radial profile. For a measurable body L ⊂ Rn, define gL(r) = 1 − σ This function should be understood as the “profile” of mass of the complement of L, which will eventually be the ratio of mass which a single deletion removes, in expectation, as a function of the distance from the origin. Define gi (r) = gTi (r). Let us try understand exactly what kind of construction we require. Fix x ∈ Dn. Keeping in mind that the function Ti (θ ) commutes with orthogonal transformations, we learn that the probability that x is removed in a single deletion of Ti is exactly gi (|x|). By elementary properties of the Poisson process, this gives P(x ∈ Ki ) = exp −mi gi |x| . In view of (37), we would like the ratio PP((xx∈∈KK21)) to be (approximately) constant. Using (39), we see that the latter follows from m1g1 |x| − m2g2 |x| = C. If we choose to pick m2 = 2m1, this equality will be implied by the following requirements on T1, T2: g1( 1 ) = g2( 1 ) = 0, and g1(r) = 2g2(r), Assuming that ( 5 ) holds and making use of the concentration of the radial profile of Dn, we will actually only be required to make sure that the derivatives are proportional for r ∈ [1 − n−0.99, 1]. Note that when (40) is attained, by picking different values of m1, the ratio between the expected volumes of K1 and K2 can be made arbitrarily large while the expected radial profiles remain about as close. Lemma 6 will then ensure us that this is enough for the distributions to be indistinguishable. The above is established in the main lemma of this section: Lemma 7 For every dimension n, there exist two convex bodies T1, T2 ⊂ Rn satisfying the following: 3 Building the Two Profiles (i) (ii) The radial profiles satisfy, 1 Dn ⊇ Ti ⊇ Dn ∩ x; x, e1 ≤ n− 3 , i = 1, 2. g1( 1 ) = g2( 1 ) = 0, and g1(r) = 2g2(r) (38) (39) (41) and (43) (44) (45) To achieve this, we begin by describing the following construction: Define δ0 = 1 n− 4 and δ1 = n−0.99. For every two constants a, b such that a ∈ [2, 200] and b ∈ [−1000, 1000], let f = fa,b be the linear function with negative slope which satisfies (hence, it is a line of distance aδ0 from the origin which meets the unit circle at x = δ0(1 + bδ1). Note that there exists such a linear function with negative slope since aδ0 δ0(1 + bδ1)). We define the convex body Ta,b by Ta,b = Dn ∩ (x, y) ∈ R × Rn−1 = Rn; |y| ≤ f (x) (an intersection of the ball with a cone defined by a linear equation the coefficients of which depend on a, b). Recall that we require that a > 2 and b > −1000. First of all, from requirement (43) and from the fact that the slope of f is negative it follows directly that Ta,b satisfies (41) (since δ0 n−1/3). Define ga,b(r) = gTa,b (r) as in (38). Let us find an expression for ga,b(r). First, a simple calculation shows that (44) implies that the function fa,b intersects the x axis at x < 2aδ0. This shows that Ta,b ∩ rSn−1 has only one connected component for all r > 12 (hence, the vertex of the cone is inside the sphere). Consider the intersection 1r Ta,b ∩ Sn−1. If r > 12 , it must be a set of the form Sn−1 ∩ {x1 < x(a, b, r)} for some function x(a, b, r). Let us try to find the expression for this function. Equation (44) shows that Ta,b is an intersection of Dn with halfspaces at distance aδ0 from the origin. This implies that x(a, b, r) must satisfy x(a, b, r) = sin arcsin aδ0 r + c aδ0 r for some constant c (draw a picture). To find the value of c, we use (43) to get x(a, b, 1) = δ0(1 + bδ1), and so x(a, b, r) = sin arcsin − arcsin(aδ0) + arcsin δ0(1 + bδ1) . (46) Next, define Ψ (x) = ω1n m1in(x,1) 1 − t 2 n−23 dt, the surface area measure of a cap the base of which has distance x from the origin. We finally have ga,b(r) = σ Sn−1 ∩ x1 ≥ x(a, b, r) = Ψ x(a, b, r) . Given a subset I ⊆ R × R, we define Clearly, KI = Ta,b. (a,b)∈I gI (r) := gKI (r) = sup ga,b(r). (a,b)∈I Our goal is to choose such a subset so that (42) is fulfilled. We will use the following elementary result: Lemma 8 Let c > 0, and let {fα}α∈I be a family of twice-differentiable functions defined on [x1, x2] such that for every triplet (x, y, y ) ∈ [x1, x2] × [y1, y2] × [y1, y2], there exists α ∈ I such that fα(x) = y, fα(x) = y , f (t ) ≤ c ∀t ∈ [x1, x2]. Then for every twice differentiable function g : [x1, x2] → [y1, y2] with there exists a subset I ⊂ I such that g (x) ∈ [y1, y2], g (x) > c, g(x) = sup fα(x). α∈I In view of the above lemma, we would like to show that by choosing appropriate values of a, b, one can attain functions ga,b which, for a fixed r0, have prescribed values ga,b(r0), ga,b(r0), and a small enough second derivative. Define r(u) = 1 − δ1u. Note that substituting r → u, almost all of the mass of the Euclidean ball is contained in u ∈ [0, 1] (the thin shell of the Euclidean ball). We now turn to prove the following lemma: Lemma 9 Suppose that (u, g0, g0) satisfy 0 ≤ u ≤ 1, Ψ (δ0) − 100δ0δ1Ψ (δ0) ≤ g0 ≤ Ψ (δ0) + 100δ0δ1Ψ (δ0), 10δ0δ1Ψ (δ0) ≤ g0 ≤ 100δ0δ1Ψ (δ0). There exist constants a ∈ [2, 200], b ∈ [−1000, 1000] such that ga,b(r(u)) = g0, (ga,b(r(u))) = g0, and ga,b(r(t )) ≤ δ0δ1Ψ (δ0) ∀0 ≤ t ≤ 1. Proof Throughout this proof we always assume u ∈ [0, 1], a ∈ [2, 200], and b ∈ [−1000, 1000]. Let us inspect the function x(a, b, r)) defined in (46). Differentiating it twice, while recalling that aδ0 1 , gives us the following fact: there exists C > 0, inde2 2 pendent of n, such that | ∂∂r2 x(a, b, r)| < C. Consider x(a, b, u) := x(a, b, r(u)). One has xuu(a, b, u) = O δ12 (48) (49) (50) (51) (52) (here and afterwards, by “O” we mean that the term is smaller than some universal constant times the expression inside the brackets, which is valid as long as u, a, b attain values in the intervals defined above). This implies that for all u ∈ [0, 1], xu(a, b, u) = xu(a, b, 0) + O δ12 = aδ0δ1 sin arcsin δ0(1 + bδ1) = aδ0δ1 1 + O(δ0) , 1 + O(δ0) + O δ12 x(a, b, u) = x(a, b, 0) + aδ0δ1u 1 + O(δ0) = δ0 + δ0δ1(au + b) 1 + O(δ0) . Let us now define w(a, b, u) = δ11 ( x(a,bδ,0r(u)) − 1). So, w(a, b, u) = (au + b) 1 + O(δ0) wu(a, b, u) = a 1 + O(δ0) , wuu(a, b, u) = O δ1 δ0 . Next, we consider ga,b(r(u)) = Ψ (x(a, b, u)) = Ψ (δ0(1 + δ1w(a, b, u)). We have 2 2 Ψ x(a, b, u) = Ψ (δ0) + δ0δ1Ψ (δ0)w(a, b, u) + δ0 δ1 Ψ (t )w(a, b, u)2 2 for some t ∈ [δ0, x(a, b, u)]. But note that the following holds: log Ψ (v) Ψ (v) = Ψ (v) = − 2v n −23 (1 − v2) n−25 (1 − v2) n−23 v(n − 3) = − (1 − v2) , and, for all v ∈ [ δ20 , 2δ0], (53) (54) (55) (56) (57) (58) (59) or Combining (56) and (57) gives This finally gives log Ψ (v) = O(nδ0). log Ψ (t ) − log Ψ (δ0) = O nδ02δ1 Ψ (t ) = Ψ (δ0) 1 + O nδ02δ1 . δ0 δ1 Ψ (t ) = O Ψ (δ0)δ12nδ03 = o Ψ (δ0)δ0δ1 . 2 2 Integration of this inequality yields that for t such that t − δ0 = O(δ0δ1), one has ga,b r(u) = Ψ x(a, b, u) = Ψ (δ0) + δ0δ1Ψ (δ0)w(a, b, u) 1 + o( 1 ) = Ψ (δ0) + δ0δ1Ψ (δ0)(au + b) 1 + o( 1 ) . Next we try to estimate the derivative of Ψ (x(a, b, u)). We have ∂ ∂ ∂u ga,b r(u) = ∂u Ψ x(a, b, u) and, using (57), ∂ ∂u = Ψ x(a, b, u) xu(a, b, u) = Ψ x(a, b, u) δ0δ1wu(a, b, u) (60) (61) Ψ x(a, b, u) = Ψ (δ0) 1 + o( 1 ) δ0δ1wu(a, b, u) = aδ0δ1Ψ (δ0) 1 + o( 1 ) . Using the continuity and Ψ and x(a, b, u), we can now conclude the following: for any fixed b ∈ [−1000, 1000] and u ∈ [0, 1], an inspection of (61) teaches us that when a varies in [2, 200], ∂∂u Ψ (x(a, b, u)) can attain all values in the range [3δ0δ1Ψ (δ0), 100δ0δ1Ψ (δ0)]. An inspection of (59) shows that afterwards, by letting b vary in [−1000, 1000], ga,b(r(u)) will attain all values in [Ψ (δ0) − 100δ0δ1Ψ (δ0), Ψ (δ0) + 100δ0δ1Ψ (δ0)]. To estimate the second derivative, ga,b, we write Ψ x(a, b, u) = Ψ x(a, b, u) δ02δ12wu2(a, b, u) + δ0δ1Ψ x(a, b, u) wuu(a, b, u)wu(a, b, u) ∂2 ∂u2 This completes the proof of the lemma. We are now ready to prove the main lemma of the section. Proof of Lemma 7 Define fi (r) = Ψ (δ0) + Ci δ0δ1Ψ (δ0)(u + 1)2 with C1 = 20, C2 = 40. Lemmas ( 9 ) and ( 8 ) show that there exist two subsets I1, I2 of [2, 200] × [−1000, 1000] such that the bodies Ti = TIi that we constructed in (48) satisfy (42). Also, (41) is satisfied, since it is satisfied for Ta,b for all (a, b) ∈ [2, 200] × [−1000, 1000], as we have seen. 4 Tying Up Loose Ends Proof of Theorem 2 Use Lemma 7 two build the two bodies Ti . Let Uθ be an orthogonal transformation which sends e1 to θ . Define Ti (θ ) = Uθ (Ti ) (note that the choice of orthogonal transformation does not matter because Ti are bodies of revolution around e1). Define the functions gi = gTi as in (38). Let m1 = g1n(ε1) with ε > 0 to be chosen later. Define m2 = 2m1. So, (42) implies that m2g2(r) = m1g1(r) + m1g1( 1 ) where ε0 is the constant from ( 9 ). Clearly, this ensures that ( 9 ) holds for both random bodies Ki . Now, ε can be made further smaller, so that concentration properties of the Euclidean ball will give us Dn f˜i = 1 + SE(n) Dn\(1−n−0.99)Dn f˜i for i = 1, 2. Clearly, the above can still be satisfied for some universal constant ε > 0 as long as n is large enough. Next, (64) and (65) imply that (62) (63) (64) (65) (66) dx + SE(n) = SE(n). and so (again, taking ε small enough) one gets f˜1 Dn Dn f˜1 dx = Now use Lemma 6 to get that Denote R = 21 enε . Then E Vol(K1) = 1 + SE(n) 2RE Vol(K2) . Suppose by negation that there exists a classification function F : ω → R that determines the volume of a body K up to a constant enε2 with probability 0.52. Denote L = [ E(VoRl(K1)) , RE(Vol(K1))]. Note that using ( 10 ), the “correctness” of the function implies that Denote A ⊂ Ω as A = {p ∈ Ω : F (p) ∈ L}. Then P1(A) > 0.51, and (66) implies that also P2(A) > 0.51. This means that P1 F (p) ∈ L ≥ 0.51. P2 F (p) ∈ L > 0.5. But clearly, again, ( 10 ) implies that with probability = 1 + SE(n), the volume of K2 is not in L. This contradicts the existence of such a function F . We still have to generalize the above in two aspects: for an even smaller probability of estimating the volume and the possibility that the algorithm is nondeterministic. Upon inspection of the proof above, we notice that it can be easily extended in the following way: instead of taking just two families of random bodies, K1 and K2, one may take d > 2 different families that are all indistinguishable by the algorithm and ε have different volumes. The proof can be stretched as far as d = en 2 . To deal with nondeterministic algorithms, we will use Yao’s lemma (see [10], Lemma 11). Let us generate an index i uniformly distributed in {1, . . . , d }, then a body K from the family Ki , and then a sequence of uniformly distributed random points on K . Following the lines of the above proof, we see that every deterministic algorithm, given this sequence, will be incorrect in estimating the volume of K with probability (at least) 1 = 1 − d + SE(n). It follows from Yao’s lemma that every nondeterministic algorithm will be incorrect with the same probability for at least one of the families Ki . This finishes the proof of the theorem. Acknowledgements I am deeply grateful to my supervisor Prof. Boäz Klartag for very useful discussions and encouragement all along my work on the subject. I would also like to express my thanks to my supervisor, Prof. Vitali Milman for introducing me to this question and encouraging me to work on it, to the referee for providing useful comments and insights, and to Prof. Oded Regev for reviewing a preliminary version of the paper and providing numerous insightful comments and remarks. 1. Bárány , I. , Füredi , Z. : Computing the volume is difficult . Discrete Comput. Geom. 2 , 319 - 326 ( 1987 ) 2. Bollobás , B. : Volume Estimates and Rapid Mixing . Flavors of Geometry , vol. 31 . MSRI Publications, Berkeley ( 1997 ) 3. Dyer , M.E. , Frieze , A.M. , Kannan , R.: A random polynomial time algorithm for approximating the volume of convex bodies . J. ACM 38 ( 1 ), 1 - 17 ( 1991 ) 4. Goyal , N. , Rademacher , L. : Learning convex bodies is hard (submitted manuscript, 2008 ). doi:10.1.1.162.3154 5. Klartag , B. : Power-law estimates for the central limit theorem for convex sets . J. Funct. Anal . 245 , 284 - 310 ( 2007 ) 6. Klivans , A.R. , O 'Donnell , R. , Servedio , R.A. : Learning geometric concepts via Gaussian surface area . In: Proc. 49th IEEE Symposium on Foundations of Computer Science ( 2008 ) 7. Lovász , L. : Private communication 8. Lovász , L. , Simonovits , M. : Random walks in a convex body and an improved volume algorithm . Random Struct. Algorithms 4 , 359 - 412 ( 1993 ) 9. Lovász , L. , Vempala , S. : Simulated annealing in convex bodies and an 0∗(n4) volume algorithm . FOCS 650-672 ( 2003 ) 10. Rademacher , L. , Vempala , S. : Dispersion of mass and the complexity of randomized geometric algorithms . In: Proc. 47th IEEE Symposium on Foundations of Computer Science ( 2006 )


This is a preview of a remote PDF: https://link.springer.com/content/pdf/10.1007%2Fs00454-011-9328-x.pdf

Ronen Eldan. A Polynomial Number of Random Points Does Not Determine the Volume of a Convex Body, Discrete & Computational Geometry, 2011, 29-47, DOI: 10.1007/s00454-011-9328-x