Globe-hopping

01/17/2020 ∙ by Dmitry Chistikov, et al. ∙ University of Cambridge Boise State University University of Warwick 0

We consider versions of the grasshopper problem (Goulko and Kent, 2017) on the circle and the sphere, which are relevant to Bell inequalities. For a circle of circumference 2π, we show that for unconstrained lawns of any length and arbitrary jump lengths, the supremum of the probability for the grasshopper's jump to stay on the lawn is one. For antipodal lawns, which by definition contain precisely one of each pair of opposite points and have length π, we show this is true except when the jump length ϕ is of the form πp/q with p,q coprime and p odd. For these jump lengths we show the optimal probability is 1 - 1/q and construct optimal lawns. For a pair of antipodal lawns, we show that the optimal probability of jumping from one onto the other is 1 - 1/q for p,q coprime, p odd and q even, and one in all other cases. For an antipodal lawn on the sphere, it is known (Kent and Pitalúa-García, 2014) that if ϕ = π/q, where q ∈N, then the optimal retention probability of 1-1/q for the grasshopper's jump is provided by a hemispherical lawn. We show that in all other cases where 0<ϕ < π/2, hemispherical lawns are not optimal, disproving the hemispherical colouring maximality hypotheses (Kent and Pitalúa-García, 2014). We discuss the implications for Bell experiments and related cryptographic tests.

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I Introduction

The grasshopper problem was first introduced in Ref. Kent and Pitalúa-García (2014), which analysed Bell inequalities for the case where two parties carry out spin measurements about randomly chosen axes and obtain the spin correlations for pairs of axes separated by angle . It was noted Kent and Pitalúa-García (2014) that tighter bounds could be obtained by a version of the following problem on the Bloch sphere. Half the area of a sphere is covered by a lawn, with the property that exactly one of every pair of antipodal points belongs to the lawn. A grasshopper lands at a random point on the lawn, and then jumps in a random direction through spherical angle . What lawn shape maximises the probability that the grasshopper remains on the lawn after jumping, and what is this maximum probability (as a function of )?

Ref. Goulko and Kent (2017) studied the planar version of the grasshopper problem, giving a combination of analytic and numerical results, and also discussed several interesting variants. As these discussions illustrate, the grasshopper problem is an appealing problem in geometric combinatorics which is of intrinsic interest, independent of its original motivations.

In this paper we consider versions of the grasshopper problem on the circle and the sphere. One can similarly motivate this work as an exploration of geometric combinatorics in simple manifolds with non-trivial topologies. Although the circle defines the simplest non-trivial version of the problem, its solution still has some interesting features. Results for antipodal lawns on the circle also imply some optimality results for antipodal lawns on the sphere, since a spherical antipodal lawn defines a circular one for every great circle. The spherical version of the grasshopper problem has some features in common with the planar version, but the non-trivial topology and compactness mean that planar results do not always have direct parallels, and one would not expect all regimes of the planar “phase diagrams” of Ref. Goulko and Kent (2017) to have qualitatively similar parallels in the spherical case.

Another strong motivation is to develop further the analysis of Bell inequalities initiated in Ref. Kent and Pitalúa-García (2014)

, which considered the average anti-correlations attainable by local hidden variable models for random pairs of measurement axes separated by a given fixed angle. Both the circle and the sphere are relevant here, since the circle parametrizes the simplest class of projective polarization measurements commonly used in Bell experiments on photons, while the sphere parametrizes all possible projective polarization measurements, or more generally all possible projective measurements on any physical system defining a qubit. We show that, perhaps surprisingly, local hidden variable models can produce perfect anticorrelations for random pairs of axes separated by angle

on the circle, unless with odd, even and , i.e.  and are coprime. This means that any imprecision in the axis separation, however slight, allows classical simulation of quantum correlations. For the sphere, our results show that hemispherical lawns are not optimal unless , where is odd and . This means that local hidden variable models can achieve stronger anticorrelations than previously realised for generic . We discuss these results and their implications further below.

Ii The grasshopper on a circle

We first consider what seems to be the simplest non-trivial version of the problem, in which the grasshopper is constrained to jump through a known fixed angle around a circle of circumference . This also allows us to study the significance of an antipodal condition, by considering lawns of length that contain precisely one of every pair of antipodes.

Then we consider the case in which there are two potentially independent (maybe overlapping) antipodal lawns, in which the grasshopper starts at a random point on one. As before, it jumps through a known fixed angle in a random direction. In this case, the question is how to configure the lawns to maximize the probability that it lands on the second.

Ref. Pitalúa-García (2013) previously discussed the grasshopper on the circle, with two antipodal lawns, and noted the optimality of the lawn for the case of rational jumps with even numerator (see below).

These versions of the problem are physically motivated as follows. A great circle on the Bloch sphere defines linear polarization measurements, which are easily implemented and commonly used in Bell and other cryptographic tests. “Classical” hidden variable models for these measurement outcomes are defined by antipodal colourings of the circle. A simple hidden variable model for the quantum singlet state, in which the outcomes of the same measurements on both subsystem are perfectly anticorrelated, is defined by a single antipodal colouring. The most general model, in which measurement outcomes may be independent, is defined by a pair of antipodal colourings. The retention probabilities define the anticorrelations predicted by these models, which can be compared to those predicted by quantum theory via Bell inequalities. As we discuss below, our results have interesting implications for testing and simulating quantum entanglement.

ii.1 Statement of the problem

General lawns:   Following Ref. Goulko and Kent (2017)

, the most general version of the grasshopper problem on the circle allows lawns of variable density, defined by a measurable probability density function

on the circle satisfying for all and

(1)

Here is the lawn length. We take the circle to have length , so the non-trivial cases have .

It will suffice for most of our discussion to consider indicator functions with . We represent such lawns by measurable subsets , where has measure .

The functional is defined by

(2)

Here and below all angles are taken modulo . We refer to the expression in Equation (2) as the retention probability: it defines the probability that a grasshopper starting at a randomly chosen point on the lawn remains on the lawn after jumping through angle in a random direction.

The grasshopper problem is then to answer the following. What is the supremum of over all such functions , for each value of ? Which , if any, attain the supremum? Or if none, which sequences approach the supremum value?

We will show below that the supremum value is for all .

Antipodal lawns:   For the case , we also consider these questions restricted to antipodal lawns, for which and for all .

We will show below that the supremum value is for all except of the form where and are coprime and is odd, when it is .

Two independent antipodal lawns:   We can extend the grasshopper problem to the case of two lawns given by antipodal measurable subsets of the circle, defined by suitable indicator functions , where . Here the relevant functional for jump is defined by

As we explain below, this is the version of the problem relevant to analysing general local hidden variable models for bipartite quantum states, where the measurements chosen on the two subsystems are independent and each parametrised by the circle. We will show below that the supremum value is for all except of the form where and are coprime with odd and even, when it is .

This illustrates that the two-lawn version of the problem is a non-trivial extension of the one-lawn version, even in the case of the circle. In particular, for jump values of the form with and coprime and both odd, the optimal one-lawn jump probability is strictly less than , while the optimal two-lawn jump probability is .

One-directional grasshopper:   A natural and apparently simpler version of the grasshopper problem in one dimension is given by assuming that the grasshopper always jumps in the same direction. For a grasshopper on the real line Goulko and Kent (2017), this restriction makes no essential difference: the supremum probability is still for any jump length, and the same limiting construction works for the one-directional and standard bidirectional case.

Similarly, one can assume that the grasshopper on the circle only jumps clockwise (or anticlockwise). For one lawn, our results and constructions for the bidirectional case carry over straightforwardly to this case, so we will not discuss it separately. With two lawns, the construction , obviously gives retention probability for clockwise jump .

For this construction defines antipodal lawns. The one-directional grasshopper problem on the circle thus turns out to be of no independent interest, and we will not discuss it further.

ii.2 Solution for general lawns

Lemma 1 (rational case).

The optimal retention probability is for a lawn of length with jump , where the highest common factor .

Proof.

Define the lawn to be

(3)

An example of such a lawn in depicted in Figure 1. Clearly has length and retention probability . ∎

This solves the problem for rational jumps.

Figure 1: Sketch of an optimal general lawn with , for a rational jump with and .
Lemma 2 (irrational case).

The supremum retention probability is for a lawn of length when the jump is an irrational multiple of .

Proof.

Define the lawn to be

(4)

where (following the usual definition of set union) any overlapping intervals are counted only once. Let be the largest value of such that has length smaller than . Weyl’s equidistribution theorem Weyl (1910) guarantees that such a exists for any in the range . Now define the lawn by

(5)

where is the smallest value in that range such that has length . Unless the grasshopper starts in the first interval and jumps anticlockwise, or the last interval and jumps clockwise, it remains on the lawn, which thus has retention probability . A sequence with gives us the supremum value of . ∎

Thus we obtain

Theorem 3.

For unconstrained lawns of length the supremum retention probability is 1. The supremum is attained for rational jumps .

ii.3 Solution for antipodal lawns

Lemma 4 (rational case, even numerator).

The optimal retention probability is for an antipodal lawn with jump , where and is even.

Proof.

An antipodal lawn has length . Since and is even, is odd. From Eqn. (3), we have

(6)

Clearly has length , is antipodal, and has retention probability . ∎

An example for such a lawn in depicted in Figure 2.

Figure 2: Sketch of an optimal antipodal lawn for a rational jump with even numerator and .

Now consider a lawn of length with jump , where and is odd. We take here; we consider the special case of (which for odd implies a jump through angle ) separately below.

Lemma 5 (rational case, odd numerator).

The optimal retention probability is for an antipodal lawn with jump , where , is odd and .

Proof.

First, consider lawns defined by indicator functions with . Consider any starting point and the set of points . These points are all distinct, and the points are antipodal, where the sum is taken modulo . Hence any antipodal lawn will contain precisely of these points. A sequence of jumps in either direction takes a point on the lawn to the antipodal point, which must be off the lawn. Thus, whichever points are on the lawn, a clockwise jump from at least one of them takes the grasshopper off the lawn, and similarly for anti-clockwise jumps. Because this is true for all such sets of lawn points, the probability of leaving the lawn is at least , i.e. the retention probability is at most .

For a general indicator function, consider again the discrete set of points , and suppose , with and . Suppose the grasshopper first lands at a point in . It then jumps to another point in . In what follows we take all probabilities conditioned on the first landing being in .

Let be the probability of first landing at , so we have since the lawn is antipodal. The retention probability is

The retention probability as a function of is linear in each variable, therefore its maximum value is attained for some choice of the extreme values or for each variable. This consideration returns us to the case with indicator functions previously discussed.

In the case , the semi-circular lawn

(7)

which has retention probability , is thus optimal.

For general odd , the lawn

(8)

is antipodal and has retention probability , and is thus optimal. ∎

There are also other optimal lawns that are not rotated versions of these. For general , we can construct a lawn with the maximal retention probability as follows.

(9)

Here the two unions define disjoint “demi-lawns” of length . Several examples are plotted in Figure 3.

Figure 3: Sketch of an optimal antipodal lawn for a rational jump with odd numerator (left), (middle), (right), and (in all panels) . The two demi-lawns are marked by different line styles (black solid lines and red dashed lines).

The special case of a jump through angle takes any point to its antipodal point. In the case when the indicator function is restricted to , all antipodal lawns have probability 0. However, a higher probability can be attained for general indicator functions.

Lemma 6 (jump through angle ).

The optimal retention probability is for an antipodal lawn with jump .111We thank Carlo Piosevan for noting this special case.

Proof.

Since antipodal points have lawn density and for some , the optimal density for each pair is given by , which maximizes . The optimal lawn in this case thus has uniform density and the retention probability is . ∎

Lemma 7 (irrational case).

The supremum retention probability is for antipodal lawns with jump angle , when is irrational.

Proof.

By Hurwitz’s theorem Hurwitz (1891) there are infinitely many rationals such that

(10)

We could apply this result directly, considering separately the possibilities that is odd or even. A slightly shorter argument follows from an extension of Hurwitz’s theorem due to Uchiyama Uchiyama (1980), elaborated by Elsner Elsner (1996). This implies that for any irrational there are infinitely many rationals with even and odd such that

(11)

Let be a rational approximation to of this type. For jump , the lawn has leaving probability

(12)

By considering an infinite sequence of approximations with increasing , we can make this arbitrarily small. Hence the supremum retention probability is . ∎

Combining the lemmas for antipodal lawns we obtain

Theorem 8.

For antipodal lawns with general indicator functions, the supremum retention probability is 1, except for jumps of the form , where and is odd. In the latter case, it is for and otherwise.

ii.4 Two antipodal lawns

We now consider the case of two lawns, and , where the grasshopper starts on one and we are interested in optimizing the probability that it lands on the other. In versions of the problem where the supremum probability is for a single lawn, we can obviously obtain this supremum by taking . The only case that remains of interest is thus two antipodal (thus length ) lawns with jump , where and is odd.

Lemma 9.

The optimal retention probability for two antipodal lawns is when the jump , where , is odd and is even.

Proof.

We have that , and . Again, we first consider the case in which the lawn densities take only values or . Any such lawn is defined by the set . Define the complementary lawn

If is antipodal, then so is , since is the set of points antipodal to points in .

Consider any starting point and the sets of points

These points are all distinct. The points are antipodal, as are the points , where the subscript sums are taken modulo . Hence the antipodal lawns contain precisely of the points respectively. Starting at a point on , a sequence of jumps clockwise through reaches the antipodal point, which must be off the lawn. Thus at least one of these jumps links a point on to a point off or a point on to a point off . In the first case, there is at least one clockwise jump from a point in to . In the second case, there is at least one clockwise jump from to . As the lawns are antipodal, this implies at least one anticlockwise jump from to . Thus, whichever points lie in , there is at least one jump from one of them in one direction that takes the grasshopper to , Because this is true for all such sets of lawn points, the probability of leaving the lawn is at least , i.e. the retention probability is at most .

For the general case with unrestricted density functions, we can argue as before. If lawn has a point such that , where , then at least one of the choices and does not decrease the retention probability. The same argument holds for lawn . We can thus restrict without loss of generality to lawns with density or everywhere.

In the case , where , we can attain this retention probability by taking (i.e. identical semicircular lawns). A jump from clockwise remains on unless it begins in the segment ; similarly an anticlockwise jump remains on unless it begins in the segment .

For general odd, we can construct a lawn with the same retention probability by taking , where is the antipodal lawn defined above, which has leaving probability . ∎

Lemma 10.

The optimal retention probability for two antipodal lawns is when the jump , where , is odd and is odd.

Proof.

We have . Define the lawns

and

These lawns are antipodal, and the jump probability from to for a jump of angle is , so this lawn configuration is optimal. ∎

Combining the lemmas for a pair of antipodal lawns we obtain

Theorem 11.

For antipodal lawns the supremum retention probability is 1, except for jumps of the form , where , is odd and is even. In the latter case, it is .

Iii The grasshopper on a sphere

iii.1 Statement of the problem

We now consider the spherical (one-lawn, antipodal) version of the problem. The lawn is now a subset of the sphere, , in three-dimensional space, that is antipodal: every point of the sphere belongs to if and only if the opposite point does not belong to . As before, the grasshopper starts at a point chosen uniformly at random in the lawn, and jumps a fixed distance in a direction chosen uniformly at random. The goal is to pick the shape of that maximizes the probability of a successful jump, i.e., the probability of staying on the lawn. Put differently, the goal is to maximize the integral

(13)

where the point corresponds to the surface element on the sphere , the point is taken from the circle  of radius  around the point , the angle is the corresponding position on the circle, and the function is defined by if and only if .

Theorems 12 and 22 below show that the retention probability, which is proportional to (13), is maximised by the hemispherical lawn if and only if , with integer.

iii.2 Corollaries of results on the circle

Theorem 12.

The optimal retention probability is for an antipodal lawn on the sphere with jump , where is a positive integer.

Proof.

For a single antipodal lawn on the circle, with jump for positive integer , we have shown that the optimal retention probability is and is attained by a semi-circular lawn.

Now consider an antipodal lawn on the sphere with the same jump . Any starting point on the lawn and any jump direction together define a great circle. The start point is equally likely to be in any arc of given length on the circle. Hence our argument for the upper bound for antipodal lawns on the circle also applies to antipodal lawns on the sphere. ∎

Since a hemispherical lawn attains this bound, we see that hemispherical lawns are optimal for jumps , where is a positive integer. We show below that hemispherical lawns are not optimal for any other jump values.

For a pair of antipodal lawns on the circle, with jump for even positive integer , we have shown that the optimal retention probability is and is attained by a pair of identical semi-circular lawns. Again, our arguments extend to a pair of antipodal lawns on the sphere: the optimal retention probability is and is attained by a pair of identical hemi-spherical lawns. This result was proved in Theorem 1 of Ref. Kent and Pitalúa-García (2014).

iii.3 Construction of lawns with greater retention probability than the hemisphere for

We now consider jumps and show that hemispherical lawns are not optimal in these cases, by constructing lawns that have higher retention probability. Each of the lawns we construct will be a hemisphere with a “cogged” boundary akin to the construction in Goulko and Kent (2017). The number, size and layout of the cogs varies according to the size  of the grasshopper’s jump. We consider three cases according as  is irrational, rational with even numerator, or rational with odd numerator. These cases are resolved in Lemmas 20, 19 and 21, respectively.

Lemma 19 will deal with the simplest case: where and are coprime and is even. This construction is illustrated in Figure 4 (left) for . Starting from any point on the equator, take the set of points visited by a sequence of consecutive jumps of size  travelling round and round the equator times before returning to the starting point. Take a second sequence with an antipodal starting point and continuing with the sequence of points antipodal to the first sequence. Note that these points are distinct and separated from each other by a distance of at least . The points are spaced evenly, distance apart, round the equator. Draw circles of sufficiently small radius around these points such that there are no overlaps between them. Our lawn consists of the southern hemisphere with these circles added for the first sequence and removed for the second sequence. In other words, we modify the hemisphere to fill semi-circular caps above the equator and remove semi-circular cups below the equator, creating our cogged hemisphere.

Figure 4: Construction of modified lawns for rational and even, with and (left panel, Lemma 19) and for irrational with (right panel, Lemma 20) The caps and cups around the equator are shown as black and white dots respectively.

For comparison, the right panel of Figure 4 looks ahead to the corresponding construction when is irrational. This case will be considered in Lemma 20.

Our construction ensures that a jump of length  cannot connect any point in a cap to any point in a cup, or vice versa. For Lemma 19

, most of our analysis involves estimating the probability of a jump between caps or between cups. For Lemmas 

20 and 21 we need also to consider jumps from caps or cups to the whole hemisphere ; we describe the modified constructions in the proofs of these lemmas. Lemma 21 requires the further complication of having caps and cups of varying sizes.

iii.4 Preliminaries

Figure 5: Spherical triangle (left) and spherical coordinates (right).

We recall some basic formulae in spherical geometry. Using the notation in Figure 5, where are (great-circle) lengths and are angles, we have the sine rule:

(14)

the cosine rule:

(15)

and, in particular,

(16)

The next formula follows easily:

(17)

For points on the sphere, it is convenient to use pairs of angles based on (longitude, latitude), where is the azimuthal angle and is the elevation or copolar angle, see Figure 5. We take the radius of the sphere to be unity, so a typical point on the surface with spherical coordinates has cartesian coordinates , where the -axis passes through the poles.

For and points on the sphere, the dot-product of their cartesian coordinates gives their angular distance. In Figure 5,

(18)

We recall the notation for the cosecant function: .

In Section III.8 we make use of the following trigonometric identities and corollaries.

Proposition 13.

For all integers ,

Proof.

In each case the summands are the real parts of a set of complex numbers regularly spaced around the origin in the Argand diagram. The sets are and . These symmetric sets and therefore their sums are each invariant under a rotation by about the origin. Hence each sum is zero. ∎

Corollary 14.

(i) ;
(ii) .

Proof.

For (i),

For (ii),

Finally, in the analysis of our construction the following proposition will be useful.

Proposition 15.

Let and satisfy , , and for some as . Then

Proof.

Assume without loss of generality that ; then

and therefore

iii.5 Analysis of the construction

In our constructions for any fixed jump distance , we will take the cap and cup radius to be sufficiently small. We can assume that is much smaller than and formally we will take .

For each of our lemmas we need to analyse the difference between jumps on a hemisphere and jumps on our lawn . Denote by , and , the caps, the cups and the southern hemisphere respectively. Note that consists of with added and taken away. The set of successful jumps from to

can be classified as jumps from

to plus jumps from to and vice versa, plus jumps from to , but minus jumps to or from . These latter are jumps from to and vice versa since our construction ensures that no jump is possible between and . We account for jumps involving as jumps to plus jumps to minus jumps to , since these last were counted twice. In symbols we may express this as:

(19)

For subsets of the spherical surface, we denote by the probability that the grasshopper starts at a point in and ends up at a point in . For a sequence of caps corresponding to distance and a corresponding antipodal sequence, we define the following quantities: is the probability of a jump from one particular cap to another particular cap distance from the first, for the corresponding probability for cups, for the jump probability between one cap and , for the corresponding probability for a cup, for the probability of a jump from to , and finally for a jump from a cap to the northern hemisphere. We define and similarly. For Lemmas 19 and 20, each cap and cup is the same size. Later, for Lemma 21, this will not be the case and we will revise our notation accordingly.

It is easy to show that, for all ,

(20)

By symmetry we find that

(21)
Figure 6: Jump geometry

In Figure 6, we show two successive caps of radius , whose centres , are at angular distance from each other, and a sample point in . For jumps from towards , and are the angles between the latitude through and the direction of jumps to the equator at and to the circumference of at , respectively. We see that always, but it is possible that , for example, if is close enough to the point in Figure 6.

Proposition 16.

The circle of possible jump destinations from intersects the upper semicircle of radius centred at exactly once, so is well-defined.

Proof.

We first note that the point always lies within the diameter of . Let be the longitude of relative to , so . By Equation (16) applied to the right-angled spherical triangle , is the solution to

Since is a decreasing function of and , showing that will establish that , and so lies within the diameter of .

Since and by Equation (16),

The right-hand side of this equation is an increasing function of and takes the value at . Since , we have as desired.

We use without proof the geometrically obvious fact that the jump circle around intersects the equator exactly once to the right of . This is true as long as is fixed and . Combined with the argument above, this fact implies that the jump circle intersects the circumference of , as well as the circumference of the mirror image of below the equator.

It remains to see why the intersection point with the upper semicircle (the circumference of ) is unique. Notice that each of the two curves, the jump circle and the upper semicircle, lies in a single plane, and these two planes are different. Therefore, each possible location for the point belongs to the line of intersection of these planes. This line passes through at most two points of the circle around . We already know that the jump circle intersects not only the circumference of but also its mirror image; therefore, there is exactly one intersection point with the circumference itself. ∎

We see that

where is a surface element of a cap .

To prepare estimates for these integrals we first find probabilities for a sample point in a cap. In the following proposition we use the notation given by Figure 6, i.e., and are the azimuth and elevation (longitude and latitude) of  relative to the centre of and is the radius of the cogs.

In the next proposition and everywhere below, the constants in our notation depend on , but not on , and we let .

Proposition 17.

For ,

Proof.

Let us first prove that and are both . Observe that are . Notice that is an angle in the right-angled spherical triangle . Hence by equation (17), , and so . Next, by the sine rule (14) for , we have , and so , since the points and are both inside a circle of radius . Therefore, and .

We now prove parts  and of the proposition. We take the vertex in Figure 6 as the point with spherical coordinates . Then has spherical coordinates and cartesian coordinates . Let be the point with spherical coordinates on the equator inside at distance from . From equation (17) we derive

which implies part  of the proposition.

Let be the point with spherical coordinates on the circumference of at distance from . Then

(22)
(23)

from equations (18) and (16).

We observe that ; note that for this is by Proposition 16. Equation (23) gives

and we will now focus on equation (22). We have

Equation (22) thus becomes

So

giving . By our choice of ,

and so and . From