Fiddler: The Helical Ball?

Instead of requiring a vertical line down the cylinder’s surface, now any helix down the surface is allowed. An example of such a helix passing through all three gaps is shown below.

What is the probability that there exists at least one such helix that can pass through each ring’s gap?

Again, let's assume that we have cylinder with radius one, that the broken rings are embedded within the planes $z=2,$ $z=1$ and $z=0$ and finally that the gap on the ring at $z=2$ is from $\theta \in [0, \pi / 4],$ while the gaps at $z=1$ and $z=0$ are from $\theta \in [\alpha, \alpha+\pi/4]$ and $\theta \in [\beta, \beta + \pi / 4]$ for some $\alpha, \beta \sim U(0,2\pi),$ respectively. Here again we will play a little fast and loose with notation and say that $[\psi_1, \psi_2] \subseteq [0,2\pi]$ should mean the set $[0, \psi_2 - 2\pi] \cup [\psi_1, 2\pi]$ in the case that $\psi_1 \lt 2\pi \lt \psi_2.$

We also see that the helical path from any point $(2, \theta_2)$ to $(1, \theta_1)$ for any $\theta_1, \theta_2 \in [0,2\pi)$ is given by $z(t) = 2-t,$ $\theta(t) = \theta_2 + t(\theta_1 - \theta_2).$ All we need to do is understand where this helical path will hit the plane $z = 0,$ in this case at $\theta(2) = 2 \theta_1 - \theta_2.$ Since any helical path that gets through the upper and middle gaps will have $\theta_2 \in [0, \pi/4]$ and $\theta_1 \in [\alpha, \alpha + \pi/4],$ we see that $\theta_0 = \theta(2) = 2 \theta_1 - \theta_2 \in [2 \alpha - \pi/4, 2 \alpha + \pi/2].$ See the figure below, for instance for the choice of $\alpha = 4.613846199712232.$

So in order to fit through all of the gaps, then we would need to have $$[\beta, \beta + \pi/4] \cap [2 \alpha - \pi/4, 2 \alpha + \pi/2 ] \ne \emptyset,$$ that is, we would need to have $\beta \in [ 2 \alpha - \pi/2, 2\alpha + \pi/2].$ Therefore, the conditional probability of there being a helical path conditional on choice of $\alpha$ is $$p(\alpha) = \mathbb{P} \{ \beta \in [ 2 \alpha - \pi/2, 2 \alpha + \pi/2 ] \mid \alpha \} = \frac{\pi}{2\pi} \equiv \frac{1}{2}.$$ Therefore, the probability of there being at least one such helix that can pass through each of the gaps is $p = \int_0^{2\pi} p(\alpha) \,d\alpha = \frac{1}{2}.$

Ring around the cylinder

A long vertical cylinder has three narrow open rings, each of which wraps around seven-eighths of the cylinder (leaving a one-eighth “gap”). The rings are evenly spaced vertically, but are otherwise randomly rotated about the cylinder’s central axis. For some orientations of the rings, there exists at least one vertical line down the cylinder’s surface that passes through each ring’s gap, as illustrated below.

What is the probability that at least one such vertical line exists?

Let's define some terms. Let's assume we have a unit cylinder and that the broken rings are embedded within the planes $z=2,$ $z=1$ and $z=0.$ Let's, without loss of generality, assume that the opening at the topmost ring is one-eighth of the way around and oriented such that it is from $\theta \in [0, \pi/4].$ Let's further assume that for $\alpha, \beta \sim U(0,2\pi)$ that the middle and bottom rings at z=1 and z=0 are given by $\theta \in [\alpha, \alpha+\pi/4]$ and $\theta \in [\beta, \beta+\pi/4],$ respectively. Here we shall abuse notation slightly and assume that if $\psi_1 \lt 2\pi \lt \psi_2$ then by $[\psi_1, \psi_2] \subseteq [0,2\pi]$ we mean $[0, \psi_2 - 2\pi] \cup [ \psi_2, 2\pi].$

Any vertical line on this cylinder can be written in cylindrical coordinates as $\theta = \theta_0$ for some $\theta_0 \in [0,2\pi).$ If a vertical line can pass through the top and middle gaps, then we must have $[0, \pi/4] \cap [\alpha, \alpha+\pi/4] \ne \emptyset,$ or equivalently, $\alpha \leq \pi/4$ and $\alpha + \pi/4 \geq 0,$ that is $\alpha \in [-\pi/4, \pi/4].$ In order for this line to pass through all three gaps, we must have $$[ \beta, \beta + \pi/4 ] \cap [ \alpha, \alpha + \pi/4 ] \cap [0, \pi/4] \ne \emptyset,$$ or equivalently $\beta \in [ \max \{ 0, \alpha \} - \pi/4, \min \{0, \alpha\} + \pi/4 ].$

Therefore, the probability that a vertical line can pass through all gaps is \begin{align*}p &= \int_{-\pi/4}^{\pi/4} \int_{\max \{0, \alpha\} - \pi/4}^{\min \{0, \alpha\} + \pi/4} \frac{d\beta}{2\pi} \, \frac{d\alpha}{2\pi} \\ &= \int_{-\pi/4}^{\pi/4} \left( \frac{\pi}{2} - |\alpha| \right) \,\frac{d\alpha}{4\pi^2}\\ &= \frac{1}{2\pi^2} \int_0^{\pi/4} \left( \frac{\pi}{2} - \alpha \right), d\alpha\\ &= \frac{1}{2\pi^2} \int_{\pi/4}^{\pi/2} u \,du = \frac{1}{4\pi^2} \left[ \left(\frac{\pi}{2}\right)^2 - \left(\frac{\pi}{4}\right)^2 \right] \\ &= \frac{1}{4\pi^2} \frac{3\pi^2}{16} = \frac{3}{64} = 4.6875\%\end{align*}

Fiddling for the win

Congratulations to Fiddler Nation for making it to the semifinals of the World Cup! All four teams that made it this far are equally matched in that they each possess the same total amount of “energy.” In advance of each semifinal game, teams must independently decide how much of their energy to allocate to the match; all remaining energy goes toward the finals. The team that spends more energy in any given game will win. The semifinals and finals occur so close in time that teams can’t recuperate any of their energy in between.

You’ve heard that the managers for the other three teams are abysmal and have no idea how to allocate their teams’ energy. Each of the other managers will independently pick a random percentage between 0 and 100 and allocate that portion of their team’s energy to the semifinal game; the rest of that team’s energy will go toward the final. Since you’re the cleverest manager of the bunch, you can choose an optimal strategy that will maximize Fiddler Nation’s probability of winning the World Cup. What is this optimal probability?

Let's define some terms, let's let $V$, $W$ and $X \sim U(0,1)$ be the i.i.d. random energy levels selected by the other three coaches to the first match. Without loss of generality, let's assume Fiddler Nation's first draw is against the team that devotes $V$ energy to the first match. Therefore, if we devote some level of energy $E \in [0,1]$ to the first match, then the probability of making the final is $\mathbb{P} \{ V \leq E \} = E.$

On the other hand since the other two coaches devoted $W$ and $X$ energies respectively to the first match, the coach that devoted $\max \{ W, X \}$ will have won the match, and the amount of energy remaining for the final is only $1 - \max \{W, X \}.$ Again, if our strategy was to devote $E$ to the first match that Fiddler Nation's fierce fiddling team would only have $1-E$ energy left for the final, which means that we will win the final only if $1 - E \geq 1 - \max \{W,X\},$ or equivalently, $\max \{W,X \} \geq E.$ So the probability of winning the final if you only have $1-E$ energy remaining is $$\mathbb{P} \{ \max \{W, X \} \geq E \} = 1 - \mathbb{P} \{ \max \{ W, X \} \leq E \} = 1 - E^2.$$

Putting these together we see that the probability of winning the final by devoting energy $E$ to the semifinal is $$f(E) = \mathbb{P} \{ V \leq E \} \mathbb{P} \{ \max \{W, X \} \geq E \} = E (1-E^2) = E - E^3.$$ In this case, we want to maximize this probability, which will occur when $f^\prime(E) = 1-3E^2 = 0,$ or at $E^* = \frac{1}{\sqrt{3}}.$ In this case, we see that the maximum probability of winning the World Cup is $$f(E^*) = \frac{1}{\sqrt{3}} - \left(\frac{1}{\sqrt{3}}\right)^3 = \frac{2\sqrt{3}}{9} \approx 0.38490017946\dots.$$

The one where they end up at the same place

The climber and the sprinter are racing up a perfectly sinusoidal hill. They go from the base, where the gradient is 0 percent, to the peak, where the gradient is again 0 percent. For them to reach the top at the same time, what should the maximum gradient of the hill be?

Ok, so a few things are left unstated here, so let's insert them. Let's assume that the horizontal distance of this hill section is 1, since units don't seem to matter in the slightest at Tour de Fiddler. Since we know that the hill is perfectly sinusoidal, we see that the height at horizontal distance $x \in [0,1]$ from the start is given by $$h(x) = \frac{h}{2} + \frac{h}{2} \sin \left( \pi x - \frac{\pi}{2} \right) = \frac{h}{2} \left(1 - \cos (\pi x) \right),$$ where $h$ is the total height of the hill. In particular, we additionally see that the gradient at a vertical distance of $x$ from the start is given by $$g(x) = h^\prime (x) = \frac{\pi h}{2} \sin (\pi x).$$ So like the Classic problem we will solve the problem with respect to one variable, in this case the height $h,$ and then transform it into the desired variable, in this case $$g_{max} = \max_{x \in [0,1]} \frac{\pi h}{2} \sin(\pi x) = \frac{\pi h}{2}.$$

In order to figure out the time elapsed for a rider, we might first want to chop the horizontal distance into say $N$ small pieces, say with $x_n = \frac{n}{N},$ for $n = 0, 1, \dots, N.$ In this case, we can determine for each short segment of horizontal distance from say $x_{n-1}$ to $x_n,$ what the horizontal inclination $\theta_n$ is, in order to determine the speed of the rider over that small interval $v_n = v(\theta_n).$ Then we see that since $v_n$ represents the total speed, that the horizontal speed is only $v_n \cos \theta_n,$ so that the time it would take for the rider to traverse the small segment would be $$t_n = \frac{x_{n+1} - x_n}{ v_n \cos \theta_n} = \frac{1}{N v_n \cos \theta_n}.$$ Then the total time elapsed is \begin{align*}T = \sum_{n=1}^{N} t_n = \sum_{n=1}^N \frac{1}{N (v_n \cos \theta_n)} &\to \int_0^1 \frac{dx}{ v(\theta(x)) \cos \theta(x) } \\ &= \int_0^1 \frac{dx}{ \frac{P}{m \sin \theta(x) + 10} \cos \theta(x) } \\ &= \frac{1}{P} \int_0^1 \left(m \tan \theta(x) + 10 \sec \theta(x) \right) \,dx \end{align*} as $N \to \infty.$

Since $$\theta(x) = \tan^{-1} g(x) = \tan^{-1} h^\prime(x) = \tan^{-1} \left( \frac{\pi h}{2} \sin (\pi x) \right),$$ we have $$T(h) = \frac{1}{P} \int_0^1 \left( m \tan \left( \tan^{-1} \left( \frac{\pi h}{2} \sin (\pi x) \right) \right) + 10 \sec \left( \tan^{-1} \left( \frac{\pi h}{2} \sin (\pi x) \right) \right)\right) \,dx.$$ Because $\tan( \tan^{-1} u) = $u and $\sec (\tan^{-1} u) = \sqrt{1 + u^2}$ for all $u \in \mathbb{R},$ we have \begin{align*} T(h) &= \frac{m}{P} \int_0^1 \frac{\pi h}{2} \sin(\pi x) \,dx + \frac{10}{P} \int_0^1 \sqrt{ 1 + \left( \frac{\pi h}{2} \sin(\pi x)\right)^2} \,dx \\ &= \frac{mh}{P} + \frac{10}{P} \int_0^1 \sqrt{1 + \left( \frac{\pi h}{2} \sin(\pi x)\right)^2} \,dx \\ &= \frac{mh}{P} + \frac{20}{P \pi} E\left(-\frac{\pi^2 h^2}{4}\right),\end{align*} where $E(k) = \int_0^{\pi/2} \sqrt{1 - k \sin^2 t} \,dt$ is the complete elliptical integral of the second kind.

In this case, in particular, if we wanted to see when does $T_c(h) = T_s(h),$ then we have \begin{align*}0 = T_c(h) - T_s(h) &= \left( \frac{m_c h}{P_c} + \frac{20}{P_c \pi} E\left( - \frac{\pi^2 h^2}{4} \right) \right) - \left( \frac{m_s h}{P_s} + \frac{20}{P_s \pi} E \left( - \frac{\pi^2 h^2}{4} \right) \right) \\ &= \frac{1}{P_cP_s} \left( h (P_s m_c - P_c m_s) + \frac{20 (P_s-P_c)}{\pi} E\left( - \frac{\pi^2 h^2}{4} \right) \right),\end{align*} so the two times will be equal when we have $$h \left( \frac{P_c m_s - P_s m_c}{10 (P_s - P_c)} \right) = \frac{2}{\pi} E \left( - \frac{\pi^2 h^2}{4} \right).$$ For our particular climber and sprinter, we know from the Classic problem that $\frac{P_c m_s - P_s m_c}{10 (P_s - P_c)} = 18,$ so we have the implicit equation $$18h = \frac{2}{\pi} E \left(-\frac{\pi^2h^2}{4} \right).$$

To properly root-solve for this implicit equation, we can use a fixed-point iteration method to solve for the point where $f(x) = \frac{1}{9\pi} E(-\frac{\pi^2 x^2}{4} )$ has a fixed point where $f(x)=x.$ In particular, we can define the small Python snippet, using the built-in special function $\textsf{scipy.special.ellipe}$:

import numpy as np

import scipy as sp

def find_root(init_guess, maxiter=100, tol=1e-16):

    xlast = init_guess

    i = 0

    dx = 1

    while (i < maxiter) & (dx > tol):

        yiter = (2 / np.pi) * sp.special.ellipe(-(0.5 * np.pi * xlast)**2)

        xiter = yiter / 18

        dx = np.abs(xiter - xlast)

        xlast = xiter

        i += 1

    return xiter, yiter, i, dx

Since $\frac{2}{\pi} E(z) \approx 1$ for $z \approx 0,$ let's start with an initial guess of $h_0 = \frac{1}{18}.$ Doing so, quickly retrieves, after 6 iterations the approximate solution of $$h^* \approx 0.05566157787953384\dots,$$ which means that for them to reach the top of the hill at the same time that maximum gradient should be $g_{max} = \frac{\pi h^*}{2} = 8.743300207677981\dots \%.$

The one where they go the same speed

As it’s now July, the Tour de Fiddler is back!

This time, we’ll be looking at a model for a cyclist’s speed $v$ as a function of their pedaling power $P,$ their mass $m,$ and the ground’s angle of inclination $\theta$: $$v = \frac{P}{m \sin \theta + 10}.$$

In cycling, roads are marked with a gradient $g,$ which is a hill’s slope, typically expressed as a percentage. Thus, an incredibly steep 45-degree incline has a gradient of 1, or “100 percent.”

Consider the following two riders:

• A “climber,” who has a power of 300 and a mass of 60
• A “sprinter,” who has a power of 325 and a mass of 80

At what gradient will the climber and sprinter cycle at the same speed?

In this case, since the formula involves the ground's angle of inclination $\theta,$ but we then want an answer in terms of gradient, $g,$ we need to understand how these two quantities relate to one another. In particular, we see that $g = \tan \theta,$ or equivalently $\theta = \tan^{-1} g,$ but more on this later.

In general if $v_c$ and $v_s$ are the climber's and sprinter's velocities, then they will be equal when $$\frac{P_c}{m_c \sin \theta + 10} = \frac{P_s}{m_s \sin \theta + 10},$$ or equivalently when $$P_c(m_s \sin \theta + 10) = P_s(m_c \sin \theta + 10)$$ which in turn is equivalent to when $$(P_cm_s - P_sm_c) \sin \theta = 10 (P_s - P_c),$$ or when $$\sin \theta^* = \frac{ 10 (P_s - P_c) }{ P_cm_s - P_s m_c }.$$ In particular, when we have $P_c = 300,$ $m_c = 60,$ $P_s = 325,$ and $m_c = 80,$ then we have $$\sin \theta^* = \frac{10 \cdot ( 325 - 300 )}{ 300 \cdot 80 - 325 \cdot 60 } = \frac{250}{4500} = \frac{1}{18},$$ or equivalently $\theta^* = \sin^{-1} \frac{1}{18}.$

Since we have $$\tan (\sin^{-1} u) = \frac{u}{\sqrt{1-u^2}},$$ for any $u \in [-1,1],$ we see that the gradient at which the climber and sprinter will have the same speed is $$g^* = \tan \left( \sin^{-1} \frac{1}{18} \right) = \frac{ \frac{1}{18} }{ \sqrt{ 1 - \left( \frac{1}{18} \right)^2 } } = \frac{1}{\sqrt{18^2 -1}} = \frac{\sqrt{323}}{323} \approx 5.56414884\dots\%.$$

Retsy Boss VIII's Star Fitting Homage

After 250 years, the nation has commissioned Retsy Boss VIII to design a new flag with one star for each of the nation’s current $58$ states. As an homage to the original flag design, Retsy wants to select $58$ stars from the square grid that are all at most some distance $R$ from a point on the plane. What is the minimum distance $R$ that Retsy can use?

Thankfully for Retsy Boss VIII, her namesake left her the Python code to run the optimization to find the number of stars given $h,$ $k$ and $R.$ Also, fortunately for Retsy Boss VIII she knew to look at OEIS to look up the integer sequence A000328, which captures the case of $h=k=0$ and $R \in \mathbb{N},$ to give her a decent idea of where to start. In particular, we see that $N(0,0,4) = 49$ and $N(0,0,5)=81,$ so let's start looking for $R \in [4,5].$ In particular, she wants to find $$R^* = \inf \left\{ R \in [4,5] \mid N^*(R) = \sup_{(h,k) \in [0,1]^2} N(h,k,R) \geq 58 \right\}.$$

Retsy didn't want to exhaustively search the entire phase space, so first she went about implementing a secant method update to arrive at some point $R$ for which $N^*(R) = 58.$ In particular, let's let $R_0 = 4,$ $R_1 = 5$ and $$R_{n+1} = \frac{R_{n-1} N^*(R_n) - R_n N^*(R_{n-1}) + 58 (R_n - R_{n-1})}{N^*(R_n) - N^*(R_{n-1})}.$$ Note that relatively quickly we arrive at some value with $N^*(R) = 58,$ that is,

$n$	$R_n$	$N^*(R_n)$
$0$	$4$	$49$
$1$	$5$	$81$
$2$	$4.20689655\dots$	$59$
$3$	$4.17084639\dots$	$58$

That is all well and good, but Retsy VIII wants the minimal such value of $R$ such that $N^*(R) = 58.$ Now she turns to a binary search of the interval $[a_0,b_0] = [4, 4.17084639],$ to within a tolerance of $\varepsilon = 10^{-6}.$ For each $n,$ she defines $c_n = \frac{a_n+b_n}{2}.$ If $N^*(c_n) \lt 58,$ then she defines $a_{n+1} = c_n$ and $b_{n+1} = b_n;$ however if $N^*(c_n) = 58,$ then she defines $a_{n+1} = a_n$ and $b_{n+1} = c_n.$ We can stop as soon as

$n$	$a_n$	$b_n$	$c_n$	$N^*(c_n)$
$0$	$4$	$4.17084639$	$4.08542320$	$56$
$1$	$4.08542320$	$4.17084639$	$4.12813480$	$57$
$2$	$4.12813480$	$4.17084639$	$4.14949060$	$57$
$3$	$4.14949060$	$4.17084639$	$4.16016850$	$58$
$4$	$4.14949060$	$4.16016850$	$4.15482955$	$57$
$5$	$4.15482955$	$4.16016850$	$4.15749902$	$58$
$6$	$4.15482955$	$4.15749902$	$4.15616428$	$58$
$7$	$4.15482955$	$4.15616428$	$4.15549691$	$58$
$8$	$4.15482955$	$4.15549691$	$4.15516323$	$57$
$9$	$4.15516323$	$4.15549691$	$4.15533007$	$58$
$10$	$4.15516323$	$4.15533007$	$4.15524665$	$58$
$11$	$4.15516323$	$4.15524665$	$4.15520494$	$58$
$12$	$4.15516323$	$4.15520494$	$4.15518409$	$57$
$13$	$4.15518409$	$4.15520494$	$4.15519451$	$58$
$14$	$4.15518409$	$4.15519451$	$4.15518930$	$57$
$15$	$4.15518930$	$4.15519451$	$4.15519191$	$57$
$16$	$4.15519191$	$4.15519451$	$4.15519321$	$58$
$17$	$4.15519191$	$4.15519321$	$4.15519256$	$58$

So we see that the optimal value of $R^*$ is about $4.15519256$ to within a tolerance of $\varepsilon = 10^{-6}.$ At this point, we can use map a contour plot, shown below, similar to what we did for the Classic problem. It is hard to discern, but the only place that the maximum is attained (up to the various symmetries of the $N$ function) when $R = 4.15519256$ is in a small neighborhood of the point $(h,k) = (1/2, 1/8)$ (seem familiar ... wink, wink?). If we were to graph the circle of radius $R = 4.15519256$ centered at $(1/2, 1/8),$ we see that it ever so slightly includes the lattice points $(-1,4),$ $(2,4),$ $(0,-4)$ and $(1,-4).$ In particular, since we see that the distance from $(1/2, 1/8)$ to any of these four lattices points is exactly the same, we can reduce the value of $R$ to precise that distance and maintain a value of $N^*(R) = 58,$ and hence the minimum distance that Retsy Boss VIII can use to honor her forebear's design is exactly $$R^* = \sqrt{0.5^2 + 4.125^2} = \sqrt{1.5^2 + 3.875^2} = \frac{\sqrt{1105}}{8} \approx 4.155192534648665\dots.$$

Move Over Ed McMahon, It's Retsy Boss's Star Search

When designing her new nation’s flag, Retsy Boss wanted to compactly arrange some stars. These stars were positioned along a square grid, but she only wanted to include stars whose centers were at most two units away from some point on the plane.

For example, if she had centered the circle on a star itself, then she could have placed a total of 13 stars on the flag, as shown below:

What is the greatest number of stars Retsy could have placed on the flag?

Let's define the function $N(h,k,R)$ to be the number of integer lattice points within distance $R$ of the point $(h,k) \in \mathbb{R}^2,$ in particular, we have $$N(h,k,R) = \# \left\{ (m,n) \in \mathbb{Z}^2 \mid (m-h)^2 + (n-k)^2 \leq R^2 \right\}.$$ If we have a fixed value of $k,$ let's say, then we see will have $$\eta_n(h,k,R) = \left\lfloor \sqrt{R^2 - (n-k)^2} + h \right\rfloor + \left\lfloor \sqrt{R^2 - (n-k)^2} - h \right\rfloor + 1,$$ where for clarity $\lfloor x \rfloor = \max \{ n \in \mathbb{Z} \mid n \leq x \}$ and in particular $\lfloor x \rfloor = -1$ for $x \in [-1,0).$ So in particular, we have $$N(h,k,R) = \sum_{n=-\lfloor R - k \rfloor}^{\lfloor R + k \rfloor} \eta_n (h,k,R).$$ We see for instance, that when $h=k=0$ and $R=2,$ that we indeed recover $\eta_{-2}(0,0,2) = \eta_2(0,0,2) = 1, \eta_{-1}(0,0,2) = \eta_1(0,0,2) = 3,$ and $\eta_0 (0,0,2)= 5,$ so we have $$N(0,0,2) = \sum_{n=-2}^2 \eta_n(0,0,2) = 1 + 3 + 5 + 3 + 1 = 13.$$

Now, since we also see that everything is periodic and symmetric, in order to solve Retsy's problem we only need to find $$N^* = \max \left\{ N(h,k,2) \mid 0 \leq k \leq h \leq \frac{1}{2} \right\},$$ since everything else in the unit square will can be recovered from this region and everything else in the plane is periodic so $N(h,k,2) = N([h], [k],2),$ for all $(h,k) \in \mathbb{R}^2,$ where $[x] = x - \lfloor x \rfloor$ is the fractional part function.

We can code this up in the following Python code to explore the phase space and empirically determine $N^*$:

 

import math

import numpy as np

​

def _stars(h, k, R):

    num = 0

    lst_y = list(range(-math.floor(R-k), math.floor(R+k)+1))

    for y in lst_y:

        root = math.sqrt(R*R - (y-k)*(y-k))

        num += math.floor(root + h) + math.floor(root - h) + 1

    return num

​

def stars(h, k, R):

    tmph = h.copy()

    tmpk = k.copy()

    shape = h.shape

    if len(shape) == 1:

        shape = (shape[0],1)

    out = np.zeros(shape=shape)

    tmph.resize(shape)

    tmpk.resize(shape)

    for i in range(shape[0]):

        for j in range(shape[1]):

            out[i,j] = _stars(tmph[i,j], tmpk[i,j], R)

    return out

Using this Python code, we can plot the following contour plot, which shows that anywhere in the dark blue region produces the the largest possible number of stars that Retsy could place on the flag is $N^*=14.$ In particular, we can take the point $(h,k) = (0.5, 0.125)$ for foreshadowing purposes and explicitly compute that \begin{align*} \eta_{-1}(0.5,0.125,2) &= \lfloor \sqrt{ 4 - (-1-0.125)^2 } + 0.5 \rfloor + \lfloor \sqrt{ 4 - (-1-0.125)^2 } - 0.5 \rfloor + 1 \\ &\quad = \lfloor 1.654\dots + 0.5 \rfloor + \lfloor 1.654\dots - 0.5 \rfloor + 1 = 4\\ \eta_{0}(0.5,0.125,2) &= \lfloor \sqrt{ 4 - (-0.125)^2 } + 0.5 \rfloor + \lfloor \sqrt{ 4 - (-0.125)^2 } - 0.5 \rfloor + 1\\ &\quad =\lfloor 1.996\dots + 0.5 \rfloor + \lfloor 1.996\dots - 0.5 \rfloor + 1 = 4\\ \eta_{1}(0.5,0.125,2) &= \lfloor \sqrt{ 4 - (1-0.125)^2 } + 0.5 \rfloor + \lfloor \sqrt{ 4 - (1-0.125)^2 } - 0.5 \rfloor + 1\\ &\quad =\lfloor 1.798\dots + 0.5 \rfloor + \lfloor 1.798\dots - 0.5 \rfloor + 1 = 4\\ \eta_{2}(0.5,0.125,2) &= \lfloor \sqrt{ 4 - (2-0.125)^2 } + 0.5 \rfloor + \lfloor \sqrt{ 4 - (2-0.125)^2 } - 0.5 \rfloor + 1\\ &\quad =\lfloor 0.696\dots + 0.5 \rfloor + \lfloor 0.696\dots - 0.5 \rfloor + 1 = 2\end{align*} so that $$N(0.5,0.125,2)=N^*=4+4+4+2=14.$$