Optimality of the coordinate-wise median mechanism for strategyproof facility location in two dimensions

Abstract

We consider the facility location problem in two dimensions. In particular, we consider a setting where agents have Euclidean preferences, defined by their ideal points, for a facility to be located in $\mathbb {R}^2$. We show that for the p-norm ($p \ge 1$) objective, the coordinate-wise median mechanism (CM) has the lowest worst-case approximation ratio in the class of deterministic, anonymous, and strategyproof mechanisms. For the minisum objective and an odd number of agents n, we show that CM has a worst-case approximation ratio (AR) of $\sqrt{2}\frac{\sqrt{n^2+1}}{n+1}$. For the p-norm social cost objective ($p\ge 2$), we find that the AR for CM is bounded above by $2^{\frac{3}{2}-\frac{2}{p}}$. We conjecture that the AR of CM actually equals the lower bound $2^{1-\frac{1}{p}}$ (as is the case for $p=2$ and $p=\infty$) for any $p\ge 2$.

Appendices

A Proofs for Section 4 (The minisum objective)

Theorem 2

For n odd, $X=\mathbb {R}^2$, and $sc(z,\mathbf {p})=\sum _{i=1}^n \Vert z-p_i \Vert$,

$$\begin{aligned} AR(CM)=\sqrt{2}\dfrac{\sqrt{n^2+1}}{n+1}. \end{aligned}$$

Proof

Define Centered Perpendicular (CP) profiles as all profiles $\mathbf {p}\in (\mathbb {R}^2)^n$ such that

• $c(\mathbf {p})=(0,0)$

• for all i, either $x_i=0$ or $y_i=0$ or $p_i=g(\mathbf {p})$

• if $p_i' \in (p_i, g(\mathbf {p}))$, then $c(p_i', p_{-i}) \ne (0,0)$

Lemma 10

(CP) For any profile $\mathbf {p}\in (\mathbb {R}^2)^n$, there exists a profile $\mathbf {\chi } \in CP$ such that $AR(\mathbf {\chi }) \ge AR (\mathbf {p})$.

Proof

Let $\mathbf {p}\in (\mathbb {R}^2)^n$ be a profile. Let $\mathbf {p}'$ be the profile where $p_i' = p_i - c(\mathbf {p})$. Then $\mathbf {p}'$ has the same approximation ratio and $c(\mathbf {p}') = (0,0)$. Denote $A = \{i: x_i=0\}$ and $B = \{i: y_i=0\}$. Note that since $c(\mathbf {p}') = (0, 0)$, it follows from the definition of $c(\mathbf {p}')$ that $A \ne \emptyset$ and $B \ne \emptyset$. Let $\Gamma = \{(x, y) \, : \, x = 0 \text { or } y = 0\} \cup g(\mathbf {p}')$. Starting from $i=1$ and going until n, define $p_i''$ to be the point in $[p_i', g(\mathbf {p}')] \cap \Gamma$ that is closest to $g(\mathbf {p}')$ under the constraint that $c(p_1'', p_2'', \dots , p_i'', p_{i+1}, \dots , p_n)=(0,0)$. Then $\mathbf {p}'' \in CP$. Further, by lemma 7$AR(\mathbf {p}'') \ge AR(\mathbf {p}') = AR(\mathbf {p})$; hence, taking $\mathbf {\chi } = \mathbf {\mathbf {p}''}$ completes the proof.

Define Isosceles-Centered Perpendicular (I-CP) profiles as all $\mathbf {p}\in CP$ for which there exists $t \ge 0$ such that

• $p_1= \dots =p_m=(t,0)$

• $p_{m+1}=(-t, 0)$

• $p_{m+2}= \dots = p_{2m+1}=(0,1)$

• $g(\mathbf {p})=(0,1)$.

Next, we prove some lemmas that will be useful in reducing the search space for the worst-case profile from CP to $I-CP$.

First, we show that we can reduce the number of half-axes that the points lie on from (at most) four to (at most) three.

Lemma 11

(Reduce axes) Suppose $\mathbf {p}$ and $\mathbf {p}'$ are profiles which differ only at i where for some $a > 0$, $p_i = (0, -a)$ and $p_i' = (-a, 0)$, and for which $c(\mathbf {p}) = c(\mathbf {p}') = (0, 0)$ and $y_g(\mathbf {p}) \ge x_g(\mathbf {p}) \ge 0$. Then $AR(\mathbf {p}') \ge AR(\mathbf {p})$.

Proof

Again $c(\mathbf {p}')=c(\mathbf {p})$ and $sc(c(\mathbf {p}'), \mathbf {p}')=sc(c(\mathbf {p}), \mathbf {p})$. Thus, it is sufficient to show that $sc(g(\mathbf {p}'), \mathbf {p}') \le sc(g(\mathbf {p}), \mathbf {p})$. For this, we just need to show that $d(p_i', g(\mathbf {p})) \le d(p_i, g(\mathbf {p}))$. This follows from the following simple calculation:

$$\begin{aligned} d\left( p_i', g(\mathbf {p})\right) ^2&= \left( x_g(\mathbf {p}) + a \right) ^2 + y_g(\mathbf {p})^2\\&= x_g(\mathbf {p})^2 + 2 x_g(\mathbf {p})a + a^2 + y_g(\mathbf {p})^2\\&\le x_g(\mathbf {p})^2 + y_g(\mathbf {p})^2 + 2a y_g(\mathbf {p}) + a^2\\&= x_g(\mathbf {p})^2 + \left( y_g(\mathbf {p}) + a \right) ^2\\&= d\left( p_i, g(\mathbf {p})\right) ^2. \end{aligned}$$

Next, we show that we can combine points on each of the three half-axes while weakly increasing the approximation ratio.

Lemma 12

(Convexity) Let $\mathbf {p}\in CP$ and let $S \subseteq N$ be such that for all $i \in S$, $x_i > 0$ and $y_i = 0$. Let $p_S$ be the mean of the $p_i$ across $i \in S$. Let $\mathbf {p}'$ be the profile where

• 1. $p_j' = p_j$ for $j \notin S$ and

• 2. $p_j' = p_S$ for $j \in S$.

Then $AR(\mathbf {p}') \ge AR(\mathbf {p})$.

Proof

It is immediate that $c(\mathbf {p}') = c(\mathbf {p})$. Further, $sc(c(\mathbf {p}'),\mathbf {p}')=sc(c(\mathbf {p}),\mathbf {p}')=sc(c(\mathbf {p}),\mathbf {p}))$ and $sc(g(\mathbf {p}'),\mathbf {p}')<sc(g(\mathbf {p}),\mathbf {p}') <sc(g(\mathbf {p}),\mathbf {p})$ where the last inequality follows from convexity of the distance function.

The same argument applies for any of the other strict half axes.

Next, we show that we can move all the points that are on the geometric median to the axis in a way that weakly increases the approximation ratio.

Lemma 13

(Double Rotation) Let $\mathbf {p}$ and $\mathbf {p}'$ be profiles that differ only at $i_1$ and $i_2$, such that for some $a \ge 0$

• $c(\mathbf {p}) = (0, 0)$,

• $y_g(\mathbf {p}) \ge x_g(\mathbf {p}) > 0$,

• $p_{i_1} = (-a, 0)$,

• $p_{i_1}' = (a + 2 x_g(\mathbf {p}), 0)$,

• $p_{i_2} = g(\mathbf {p})$, and

• $p_{i_2}' = (0 , d(g(\mathbf {p}), (0, 0)))$.

Then $c(\mathbf {p}') = (0, 0)$ and $AR(\mathbf {p}') \ge AR(\mathbf {p})$.

Proof

The first claim is immediate.

For the second claim, let

$$\begin{aligned} A&= \sum _{i \ne i_1}{d\left( p_i, c(\mathbf {p})\right) }\\ B&= \sum _{i \ne i_2}{d\left( p_i, g(\mathbf {p})\right) }. \end{aligned}$$

By Lemma 3,

$$\begin{aligned} A + d\left( p_{i_1}, c(\mathbf {p})\right) \le \sqrt{2} B. \end{aligned}$$

Hence, it follows that

$$\begin{aligned} \left[ A + d\left( p_{i_1}, c(\mathbf {p})\right) \right] d\left( p_{i_2}', g(\mathbf {p})\right) \le \sqrt{2} B d\left( p_{i_2}', g(\mathbf {p})\right) . \end{aligned}$$

But since $y_g(\mathbf {p}) \ge x_g(\mathbf {p})$, it follows that $d(p_{i_2}', g(\mathbf {p})) \le \sqrt{2} x_g(\mathbf {p})$. Hence,

$$\begin{aligned} \left[ A + d\left( p_{i_1}, c(\mathbf {p})\right) \right] d\left( p_{i_2}', g(\mathbf {p})\right)&\le 2 B x_g(\mathbf {p})\\&= B \left( d\left( p_{i_1}', c(\mathbf {p})\right) - d\left( p_{i_2}', c(\mathbf {p})\right) \right) . \end{aligned}$$

From this it follows that

$$\begin{aligned} \left( A + d\left( p_{i_1}, c(\mathbf {p})\right) \right) \left( B + d\left( p_{i_2}', g(\mathbf {p})\right) \right)&= AB + B d\left( p_{i_1}, c(\mathbf {p})\right) + \left[ A + d\left( p_{i_1}, c(\mathbf {p})\right) \right] d\left( p_{i_2}', g(\mathbf {p})\right) \\&\le AB + B d\left( p_{i_1}', c(\mathbf {p})\right) \\&= \left( A + d\left( p_{i_1}', c(\mathbf {p})\right) \right) B \end{aligned}$$

and hence

$$\begin{aligned} AR(\mathbf {p})&= \frac{A + d\left( p_{i_1}, c(\mathbf {p})\right) }{B}\\&\le \frac{A + d\left( p_{i_1}', c(\mathbf {p})\right) }{B + d\left( p_{i_2}', g(\mathbf {p})\right) }\\&= \frac{A + d\left( p_{i_1}', c(\mathbf {p}')\right) }{B + d\left( p_{i_2}', g(\mathbf {p})\right) }\\&\le AR(\mathbf {p}'). \end{aligned}$$

Lemma 14

(Geometric to axis) Suppose that $\mathbf {p}$ is a profile such that there are $a \ge 0$ and $b, c > 0$ and subsets $L, R, U \subseteq N$ with $L \cap R = L \cap U = R \cap U = \emptyset$, $L \cup R \cup U = N$, $|L| = 1$, $|U| = |R| = m$, and

• $p_i = (-a, 0)$ for $i \in L$

• $p_i = (0, b)$ for $i \in U$

• $p_i = (c, 0)$ for $i \in R$

and so that $x_g(\mathbf {p}) > 0$.

Then, there exists another profile $\mathbf {z}$ such that $AR(\mathbf {z}) > AR(\mathbf {p})$.

Proof

We’ll consider two separate cases.

First assume that $a>0$. Let $\mathbf {p}(t)$ be the profile which is the same as $\mathbf {p}$ for $i \notin U$ and which has $\mathbf {p}_i(t) = g(\mathbf {p})+t\left( -x_g(\mathbf {p}),b-y_g(\mathbf {p})\right)$ for $i \in U$. Then there exists $\epsilon > 0$ such that for $t \in [0, 1+\epsilon ]$,

$$\begin{aligned} AR(\mathbf {p}(t))&= \frac{(a + (1-t)x_g(\mathbf {p})) + m(c - (1-t)x_g(\mathbf {p})) + mb_g(\mathbf {p})+mt(b-y_g(\mathbf {p}))}{d((-a, 0), g(\mathbf {p})) + m d((c,0), g(\mathbf {p})) + mt d((0, b), g(\mathbf {p}))}\\&=\dfrac{t\left( (m-1) x_g(\mathbf {p})+m(b-y_g(\mathbf {p}))\right) +a+x_{g}(\mathbf {p})+m(c-x_{g}(\mathbf {p})+y_g(\mathbf {p}))}{tmd((0, b), g(\mathbf {p}))+d((-a, 0), g(\mathbf {p})) + m d((c,0), g(\mathbf {p}))} \end{aligned}$$

Note that $\mathbf {p}(1)=\mathbf {p}$. Now since the denominator of $AR(\mathbf {p}(t))$ is strictly positive for $t \ge 0$ and since both the numerator and the denominator are linear in t, $AR(\mathbf {p}(t))$ is monotonic on $[0, 1 + \epsilon ]$. There are three possibilities.

If $AR(\mathbf {p}(t))$ is strictly increasing, then $AR(\mathbf {p}(1+\epsilon )) > AR(\mathbf {p})$.

If $AR(\mathbf {p}(t))$ is strictly decreasing, then $AR(\mathbf {p}(0)) > AR(\mathbf {p})$.

If $AR(\mathbf {p}(t))$ is constant, consider the profile $\mathbf {z}'$ obtained by putting t such that $-a=(1-t)x_g(\mathbf {p})$. Then, under $\mathbf {z}'$, we have 1 agent at $(-a,0)$, m agents at $(-a,bt+(1-t)y_g(\mathbf {p}))$ and m agents at (c, 0). Also, $AR(\mathbf {p})=AR(\mathbf {z}')$. We can translate this profile by a to the right and get a profile of points in the case where essentially we have $a=0$. We’ll deal with this case now.

Now let’s consider the case where $\mathbf {p}$ is such that $a=0$. To begin, note that since $g(\mathbf {p})$ must be in the convex hull of (0, 0), (0, b) and (c, 0), if $x_g(\mathbf {p}) \ge \frac{c}{2}$ and $y_g(\mathbf {p}) \ge \frac{b}{2}$, then $g(\mathbf {p})=(\frac{c}{2}, \frac{b}{2})$. But then $\sum _{i=1}^n \dfrac{p_i-g(\mathbf {p})}{\Vert p_i-g(\mathbf {p}) \Vert }=\dfrac{(c,b)}{\sqrt{c^2+b^2}} \ne 0$, contradicting the characterization of the geometric median in Lemma 2. Hence, it must be that either $x_g(\mathbf {p})<\frac{c}{2}$ or $y_g(\mathbf {p}) < \frac{b}{2}$.

Suppose that $x_g(\mathbf {p})<\frac{c}{2}$. Let $\mathbf {z}$ be the profile obtained from $\mathbf {p}$ by moving the point at (0, 0) to $(x_g(\mathbf {p}) - \frac{c}{2}, 0)$ and moving one of the points at (c, 0) to $(\frac{c}{2} + x_g(\mathbf {p}), 0)$. This transformation leaves coordinate-wise median unchanged, as well as leaving the sum of distances to the coordinate-wise median unchanged. However, the sum of distances to $g(\mathbf {p})$ strictly decreases, since for the unaltered points the distance to $g(\mathbf {p})$ remains the same, and the sum of the distances from the altered points to $g(\mathbf {p})$ is

$$\begin{aligned} d((0, 0), g(\mathbf {p})) + d((c, 0), g(\mathbf {p}))&= d\left( \left( 2x_g(\mathbf {p}), 0\right) , g(\mathbf {p})\right) + d((c, 0), g(\mathbf {p}))\\&> 2d\left( \left( x_g(\mathbf {p}) + \frac{c}{2}, 0 \right) , g(\mathbf {p})\right) \\&= d\left( \left( x_g(\mathbf {p}) - \frac{c}{2}, 0\right) , g(\mathbf {p})\right) + d\left( \left( x_g(\mathbf {p}) + \frac{c}{2}, 0\right) , g(\mathbf {p})\right) , \end{aligned}$$

where the inequality follows from convexity of $d(\cdot , g(\mathbf {p}))$. Since $AR(\mathbf {z})$ is bounded below by the ratio of the sum of distances to the coordinate-wise median to the sum of distances to $g(\mathbf {p})$, it follows that $AR(\mathbf {z}) > AR(\mathbf {p})$.

Next, suppose that $y_g(\mathbf {p}) < \frac{b}{2}$. Let $\mathbf {z}$ be the profile obtained from $\mathbf {p}$ by moving the point at (0, 0) to $(0, y_g(\mathbf {p}) - \frac{b}{2})$ and moving one of the points at (0, b) to $(0, \frac{b}{2} + y_g(\mathbf {p}))$. By essentially the same argument just given, $AR(\mathbf {z}) > AR(\mathbf {p})$.

Finally, the following lemma shows that we can use convexity to make the triangle formed by the three groups of points isosceles.

Lemma 15

(Isosceles) Let $\mathbf {p}$ be a profile such for which are m points at (a, 0), 1 point at $(-b,0)$ and m points at (0, c), and for which $g(\mathbf {p})=(0,c)$ and $c(\mathbf {p})=(0,0)$. Let $\mathbf {p}'$ be the profile where there are m points at $\left( \dfrac{ma+b}{m+1},0\right)$, 1 point at $\left( -\dfrac{ma+b}{m+1},0\right)$, and m points at (0, c). Then, $AR(\mathbf {p}') \ge AR(\mathbf {p})$.

Proof

Note that $c(\mathbf {p})=c(\mathbf {p}')=(0,0)$. Since $ma+b=m\frac{(ma+b)}{m+1}+\frac{ma+b}{m+1}$, we get that the numerator in $AR(\mathbf {p})$ and $AR(\mathbf {p}')$ remains the same. Thus, we only need to argue that the denominator goes down as we go from $AR(\mathbf {p})$ to $AR(\mathbf {p}')$.

Even though $g(\mathbf {p}')$ may not be equal to $g(\mathbf {p})$ we have that $sc(g(\mathbf {p}),\mathbf {p}') \le sc(g(\mathbf {p}),\mathbf {p})$ by the convexity of the distance function which would imply $sc(g(\mathbf {p}'),\mathbf {p}') \le sc(g(\mathbf {p}),\mathbf {p})$ by definition of $g(\mathbf {p})$. Thus, we have that $AR(\mathbf {p}') \ge AR(\mathbf {p})$.

Now, we use above lemmas to reduce the search space to I-CP.

Lemma 16

(ICP) For every $\mathbf {p}\in CP$, there exists $\chi \in I-CP$ such that $AR(\chi ) \ge AR(\mathbf {p})$.

Proof

Without loss of generality, consider any profile $\mathbf {p}\in CP$ such that $y_g(\mathbf {p}) \ge x_{g}(\mathbf {p}) \ge 0$. Applying Lemma 11 to all points on the $-y$-axis gives a profile $\mathbf {p}'$ with a weakly higher approximation ratio. In $\mathbf {p}'$, we have all points on $+x$-axis, $-x$-axis, $+y$-axis and the geometric median. Using lemma 12, we can combine the points on $+x$-axis, $-x$-axis, $+y$-axis to some $(a,0), (0,b), (-c, 0)$ while weakly increasing AR. Let this profile be $\mathbf {p}''$. Now, we use lemma 13 to move points on the geometric median to $+y$-axis. Using 12 again, we get a profile $\mathbf {p}'''$ with m points on some (a, 0), 1 point on $(-c, 0)$ and m points on (0, b).

So we know that there must be a worst-case profile that takes this form. From Lemma 14, we can say that if the geometric median of such a profile is not on the y-axis, it cannot be a worst-case profile. Thus, there must be a worst-case profile $\mathbf {z}$ with m points on some (a, 0), 1 point on $(-c, 0)$ and m points on (0, b) and $x_g(\mathbf {z})=0$. Further, such a profile must have $y_g(\mathbf {z}) = b$, since otherwise, the profile with m points on (a, 0), 1 point on $(-c, 0)$, and m points on $(0, y_g(\mathbf {z}))$ would have a strictly higher approximation ratio than $\mathbf {z}$. By Lemma 15, since $\mathbf {z}$ is a worst-case profile, it must be that $c = a$.

Now, since $\mathbf {z}$ is a worst-case profile, the profile $\chi$ with $\chi _i = \frac{1}{b}z_i$ is also a worst-case profile, and since $\chi \in I-CP$, the result follows.

Using Lemma 16, we can now restrict attention to profiles in $I-CP$. Define

$$\begin{aligned} \mathbf {\eta }_t = \left( p_1^t, \dots , p_{2m+1}^t\right) , \end{aligned}$$

where

$$\begin{aligned} p_i^t = {\left\{ \begin{array}{ll} (t, 0) &{} i = 1, \dots , m \\ (-t, 0) &{} i = m+1 \\ (0, 1) &{} i = m+2, \dots , 2m+1 \end{array}\right. } \end{aligned}$$

Then, $I-CP = \left\{ \mathbf {\eta }_t \, : \, t \ge \sqrt{\frac{2m + 1}{2m - 1}} \right\}$. Defining $\alpha (t) = \frac{(m+1)t + m}{(m+1) \sqrt{t^2 + 1}}$, we get that for $t \ge \sqrt{\frac{2m + 1}{2m - 1}}$, $AR(\mathbf {\eta }_t) = \alpha (t)$, and that $\alpha (t)$ is maximized at $t^* = \frac{m+1}{m} > \sqrt{\frac{2m + 1}{2m - 1}}$, from which it follows that

$$\begin{aligned} \text {Approximation ratio of CM} = \alpha \left( \frac{m+1}{m} \right) = \sqrt{2}\dfrac{\sqrt{(2m+1)^2+1}}{(2m+1)+1} = \sqrt{2}\dfrac{\sqrt{n^2+1}}{n+1}. \end{aligned}$$

Thus, we get that the worst case approximation ratio is $\sqrt{2}\dfrac{\sqrt{n^2+1}}{n+1}$ as required.

B Proofs for Section 5 (p-norm objective)

Theorem 3

For $X=\mathbb {R}^2$ and the p-norm objective with $p \ge 2$,

$$\begin{aligned} 2^{1-\frac{1}{p}} \le \sup _{n \in \mathbb {N}} AR(CM) \le 2^{\frac{3}{2}-\frac{2}{p}} \end{aligned}$$

Proof

We’ll prove that the lower bound actually holds for any deterministic, strategyproof mechanism (defined for all $n \in \mathbb {N}$) and hence, it holds for the coordinate-wise median mechanism. So suppose f is any deterministic, strategyproof mechanism. With $n=2m+1$ agents,⁵ for any profile $\mathbf {p}$ such that m agents have ideal point $\alpha$, $m+1$ agents have ideal point $\beta \ne \alpha$, and $f(\mathbf {p}) = \beta$,

$$\begin{aligned} AR_f(\mathbf {p}')&=\dfrac{sc(f(\mathbf {p}),\mathbf {p})}{sc(OPT(\mathbf {p}),\mathbf {p})}\\&\ge \dfrac{\left( m*\Vert \alpha - \beta \Vert ^p\right) ^\frac{1}{p}}{\left( (2m+1) (\Vert \alpha - \beta \Vert /2)^p\right) ^\frac{1}{p}}\\&= 2^{1-\frac{1}{p}} \cdot \left( \dfrac{m}{m + 1/2}\right) ^\frac{1}{p} \cdot \end{aligned}$$

To see that such a profile always exists, consider the profile $\mathbf {p}$ where agents 1 through m have ideal point $(-1, 0)$ and agents $m+1$ through $2m+1$ have ideal point (1, 0). If $f(\mathbf {p}) = (1, 0)$, then $\mathbf {p}$ is such a profile; if not, let $\mathbf {p}'$ be the profile where agents 1 through $m+1$ have ideal point $f(\mathbf {p})$ and agents $m+2$ through $2m+1$ have ideal point at (1, 0). Since $\mathbf {p}'$, every agent’s ideal point is either the same as under $\mathbf {p}$ or equal to $f(\mathbf {p})$, it follows from strategyproofness that $f(\mathbf {p}') = f(\mathbf {p})$, and so $\mathbf {p}'$ is such a profile.

Thus, for any $n = 2m+1$,

$$\begin{aligned} AR(f) \ge 2^{1-\frac{1}{p}} \cdot \left( \dfrac{n-1}{n}\right) ^\frac{1}{p}, \end{aligned}$$

so

$$\begin{aligned} \sup _n{AR(f)} \ge 2^{1-\frac{1}{p}}. \end{aligned}$$

Now we show that the asymptotic AR of coordinate-wise median mechanism is bounded above by $2^{\frac{3}{2}-\frac{2}{p}}$. Consider any profile $\mathbf {p}\in (\mathbb {R}^2)^n$. Let $g(\mathbf {p})=(x_g(\mathbf {p}), y_g(\mathbf {p}))$ and $c(\mathbf {p})=(x_c(\mathbf {p}),y_c(\mathbf {p}))$. Then, we have that

$$\begin{aligned} sc(g(\mathbf {p}), \mathbf {p})^p&=\sum _{i=1}^n \Vert g(\mathbf {p})-p_i \Vert ^p\\&\ge \left( \sum _{i=1}^n \Vert x_g(\mathbf {p})-x_i \Vert ^p+\sum _{i=1}^n \Vert y_g(\mathbf {p})-y_i \Vert ^p\right) \\&\ge \left( \sum _{i=1}^n \Vert OPT(\mathbf {x})-x_i \Vert ^p+\sum _{i=1}^n \Vert OPT(\mathbf {y})-y_i \Vert ^p\right) \\&\ge \dfrac{1}{2^{p-1}}\left( \sum _{i=1}^n \Vert x_c(\mathbf {p})-x_i \Vert ^p+\sum _{i=1}^n \Vert y_c(\mathbf {p})-y_i \Vert ^p\right) \\&\ge \dfrac{2^{1-\frac{p}{2}}}{2^{p-1}}\sum _{i=1}^n \Vert c(\mathbf {p})-p_i \Vert ^p\\&=2^{2-\frac{3p}{2}}sc(c(\mathbf {p}),\mathbf {p})^p \end{aligned}$$

Thus, we get $AR(CM) \le 2^{\frac{3}{2}-\frac{2}{p}}$ for $p \ge 2$ as required.