An Improved Lower Bound for the Traveling Salesman Constant

TL;DR

This paper improves the known lower bound for the Traveling Salesman constant, which describes the asymptotic length of the shortest tour through random points in a unit square, from 0.625 to 0.6277.

Contribution

The authors enhance the lower bound estimate of the Traveling Salesman constant using an approach based on Steinerberger's method, advancing previous results.

Findings

Lower bound for the TSP constant is improved to 0.6277

The approach builds on Steinerberger's method from 2015

The result refines the understanding of TSP tour length asymptotics

Abstract

Let $X_1, X_2, \dots, X_n$ be independent uniform random variables on $[0,1]^2$. Let $L(X_1, \dots, X_n)$ be the length of the shortest Traveling Salesman tour through these points. It is known that there exists a constant $\beta$ such that $$\lim_{n \to \infty} \frac{L(X_1, \dots, X_n)}{\sqrt{n}} = \beta$$ almost surely (Beardwood 1959). The original analysis in (Beardwood 1959) showed that $\beta \geq 0.625$. Building upon an approach proposed in (Steinerberger 2015), we improve the lower bound to $\beta \geq 0.6277$.

Equations

$$\underset{n\to \infty }{\lim }\frac{L(X_1,\dots ,X_n)}{\sqrt{n}}=\beta$$

$$0.625\le \beta \le {\beta }_{+}$$

$$\frac{\mathbb{E}\left[L(X_1,\dots ,X_n)\right]}{\sqrt{n}}\to \beta .$$

$$\frac{\mathbb{E}\left[L({\mathcal{P}}_n)\right]}{\sqrt{n}}\to \beta .$$

$$h(r_1,r_2,r_3)$$

$${\int }_{r_1=0}^{\infty }{\int }_{r_2=r_1}^{\infty }{\int }_{r_3=r_1+2r_2}^{\infty }\left(r_3-\frac{3}{2}{r}_1-\frac{3}{2}{r}_2\right)e^{-n\pi r_3^2}{r}_1{r}_2{r}_{3}d{r}_{3}d{r}_{2}d{r}_1$$

$$\Vert a-c\Vert +\Vert b-c\Vert +\frac{1}{2}\Vert a-d\Vert +\frac{1}{2}\Vert b-e\Vert -\left(\Vert a-c\Vert +\Vert b-c\Vert +\Vert a-b\Vert \right)$$

$$\ge \frac{1}{2}{r}_3+\frac{1}{2}\left(\Vert a-e\Vert -\Vert a-b\Vert \right)-r_1$$

$$\ge \frac{1}{2}{r}_3+\frac{1}{2}\left(r_3-r_1\right)-r_1$$

$$=r_3-\frac{3}{2}{r}_1$$

$$\beta \ge \frac{5}{8}+\frac{\sqrt{n}}{3}{\int }_{r_1=0}^{\infty }{\int }_{r_2=r_1}^{\infty }{\int }_{r_3=r_1+2r_2}^{\infty }\left(r_3-\frac{3}{2}{r}_1-\frac{3}{2}{r}_2\right)e^{-n\pi r_3^2}(2n\pi {)}^3{r}_1{r}_2{r}_{3}d{r}_{3}d{r}_{2}d{r}_1$$

$$\beta \ge \frac{5}{8}+\frac{\sqrt{n}}{3}(2n\pi {)}^3\frac{19}{27648{\pi }^3{n}^{\frac{7}{2}}}=\frac{5}{8}+\frac{19}{10368}\approx \mathrm{0.626833.}$$

$$\beta$$

$$\ge \mathrm{0.6277.}$$

$$\Vert a-c\Vert +\Vert b-c\Vert +\frac{1}{2}\Vert a-d\Vert +\frac{1}{2}\Vert b-e\Vert -\left(\Vert a-c\Vert +\Vert b-c\Vert +\Vert a-b\Vert \right)$$

$$\ge \frac{1}{2}{r}_3+\frac{1}{2}\left(\Vert a-e\Vert -\Vert a-b\Vert \right)-r_1$$

$$\ge \frac{1}{2}{r}_3+\frac{1}{2}\left(r_3-r_1\right)-r_1$$

$$=r_3-\frac{3}{2}{r}_1$$

$$\frac{5}{8}+\frac{\sqrt{n}}{3}{\int }_{r_1=0}^{\infty }{\int }_{r_2=r_1}^{\infty }{\int }_{r_3=r_2+\sqrt{r_1^2+r_2^2}}^{\infty }\left(r_3-\frac{1}{2}{r}_1-\frac{1}{2}{r}_2-\sqrt{r_1^2+r_2^2}\right)e^{-n\pi r_3^2}(2n\pi {)}^3{r}_1{r}_2{r}_{3}d{r}_{3}d{r}_{2}d{r}_1.$$

$${\int }_{r_1=0}^{\infty }{\int }_{r_2=r_1}^{\infty }{\int }_{r_3=r_2+\sqrt{r_1^2+r_2^2}}^{\infty }\left(r_3-\frac{1}{2}{r}_1-\frac{1}{2}{r}_2-\sqrt{r_1^2+r_2^2}\right)e^{-n\pi r_3^2}{r}_1{r}_2{r}_{3}d{r}_{3}d{r}_{2}d{r}_1$$

$$=\left[-\frac{{\left(\frac{1}{1+\sqrt{2}}\right)}^8}{8\cdot 48}-\frac{{\left(\frac{1}{1+\sqrt{2}}\right)}^7}{7\cdot 16}-\frac{{\left(\frac{1}{1+\sqrt{2}}\right)}^6}{6\cdot 16}+\frac{1}{5}\left(\frac{1}{8}+\frac{1}{4}+\frac{1}{6}+\frac{2^{\frac{3}{2}}}{3}\right){\left(\frac{1}{1+\sqrt{2}}\right)}^5-\frac{13{\left(\frac{1}{1+\sqrt{2}}\right)}^4}{64}-\frac{{\left(\frac{1}{1+\sqrt{2}}\right)}^3}{48}+\frac{{\left(\frac{1}{1+\sqrt{2}}\right)}^2}{32}\right]\frac{15}{16{\pi }^3{n}^{\frac{7}{2}}}.$$

$$\frac{5}{8}+\frac{5}{2}\left[-\frac{{\left(\frac{1}{1+\sqrt{2}}\right)}^8}{8\cdot 48}-\frac{{\left(\frac{1}{1+\sqrt{2}}\right)}^7}{7\cdot 16}-\frac{{\left(\frac{1}{1+\sqrt{2}}\right)}^6}{6\cdot 16}+\frac{1}{5}\left(\frac{1}{8}+\frac{1}{4}+\frac{1}{6}+\frac{2^{\frac{3}{2}}}{3}\right){\left(\frac{1}{1+\sqrt{2}}\right)}^5-\frac{13{\left(\frac{r_3}{1+\sqrt{2}}\right)}^4}{64}-\frac{{\left(\frac{1}{1+\sqrt{2}}\right)}^3}{48}+\frac{{\left(\frac{1}{1+\sqrt{2}}\right)}^2}{32}\right]$$

$$=\frac{5}{8}+\frac{3072\sqrt{2}-4325}{5376}$$

$$\approx \frac{5}{8}+0.003621$$

$$\beta \ge \frac{5}{8}+\frac{1}{2}\left(\frac{19}{10368}\right)+\frac{1}{2}\left(\frac{3072\sqrt{2}-4325}{5376}\right)\ge 0.6277$$

$${\int }_{r_1=0}^{\infty }{\int }_{r_2=r_1}^{\infty }{\int }_{r_3=r_1+2r_2}^{\infty }\left(r_3-\frac{3}{2}{r}_1-\frac{3}{2}{r}_2\right)e^{-n\pi r_3^2}{r}_1{r}_2{r}_{3}d{r}_{3}d{r}_{2}d{r}_1$$

$$={\int }_{r_3=0}^{\infty }{\int }_{r_1=0}^{\frac{r_3}{3}}{\int }_{r_2=r_1}^{\frac{r_3-r_1}{2}}\left(r_3-\frac{3}{2}{r}_1-\frac{3}{2}{r}_2\right)e^{-n\pi r_3^2}{r}_1{r}_2{r}_{3}d{r}_{2}d{r}_{1}d{r}_3$$

$$={\int }_{r_3=0}^{\infty }{r}_3{e}^{-n\pi r_3^2}{\int }_{r_1=0}^{\frac{r_3}{3}}{r}_1{\int }_{r_2=r_1}^{\frac{r_3-r_1}{2}}{r}_2\left(r_3-\frac{3}{2}{r}_1-\frac{3}{2}{r}_2\right)d{r}_{2}d{r}_{1}d{r}_3$$

$$\displaystyle=\int_{r_{3}=0}^{\infty}r_{3}e^{-n\pi r_{3}^{2}}\int_{r_{1}=0}^{\frac{r_{3}}{3}}r_{1}\left(\frac{r_{2}^{2}}{2}\left(r_{3}-\frac{3}{2}r_{1}\right)-\frac{1}{2}r_{2}^{3}\right)\Big{|}_{r_{2}=r_{1}}^{\frac{r_{3}-r_{1}}{2}}dr_{1}dr_{3}$$

$$={\int }_{r_3=0}^{\infty }{r}_3{e}^{-n\pi r_3^2}{\int }_{r_1=0}^{\frac{r_3}{3}}{r}_1\left(\frac{{\left(\frac{r_3-r_1}{2}\right)}^2-r_1^2}{2}\left(r_3-\frac{3}{2}{r}_1\right)-\frac{1}{2}\left({\left(\frac{r_3-r_1}{2}\right)}^3-r_1^3\right)\right)d{r}_{1}d{r}_3$$

$$={\int }_{r_3=0}^{\infty }{r}_3{e}^{-n\pi r_3^2}{\int }_{r_1=0}^{\frac{r_3}{3}}\left(\frac{9r_1^4}{8}-\frac{3r_1^3{r}_3}{16}-\frac{r_1^2{r}_3^2}{4}+\frac{r_1{r}_3^3}{16}\right)d{r}_{1}d{r}_3$$

$$\displaystyle=\int_{r_{3}=0}^{\infty}r_{3}e^{-n\pi r_{3}^{2}}\left(\frac{9r_{1}^{5}}{40}-\frac{3r_{1}^{4}r_{3}}{64}-\frac{r_{1}^{3}r_{3}^{2}}{12}+\frac{r_{1}^{2}r_{3}^{3}}{32}\right)\Big{|}_{r_{1}=0}^{\frac{r_{3}}{3}}dr_{3}$$

$$={\int }_{r_3=0}^{\infty }{r}_3{e}^{-n\pi r_3^2}\left(\frac{9{\left(\frac{r_3}{3}\right)}^5}{40}-\frac{3{\left(\frac{r_3}{3}\right)}^4{r}_3}{64}-\frac{{\left(\frac{r_3}{3}\right)}^3{r}_3^2}{12}+\frac{{\left(\frac{r_3}{3}\right)}^2{r}_3^3}{32}\right)d{r}_3$$

$$=\left(\frac{9{\left(\frac{1}{3}\right)}^5}{40}-\frac{3{\left(\frac{1}{3}\right)}^4}{64}-\frac{{\left(\frac{1}{3}\right)}^3}{12}+\frac{{\left(\frac{1}{3}\right)}^2}{32}\right){\int }_{r_3=0}^{\infty }{r}_3^6{e}^{-n\pi r_3^2}d{r}_3$$

$$=\frac{19}{25920}{\int }_{r_3=0}^{\infty }{r}_3^6{e}^{-n\pi r_3^2}d{r}_3$$

$$=\frac{19}{25920}\frac{15}{16{\pi }^3{n}^{\frac{7}{2}}}$$

$$=\frac{19}{27648{\pi }^3{n}^{\frac{7}{2}}}$$

$${\left(r_3-r_2\right)}^2=r_1^2+r_2^2$$

$$r_3^2-2r_2{r}_3+r_2^2=r_1^2+r_2^2$$

$$r_3^2-2r_2{r}_3=r_1^2$$

$$r_2=\frac{r_3^2-r_1^2}{2r_3}$$

Full text

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almost surely ([1]). The original analysis in [1] showed that $\beta\geq 0.625$. Building upon an approach proposed in [2], we improve the lower bound to $\beta\geq 0.6277$.

I Introduction

Let $X_{1},\dots,X_{n}$ be independent uniform random variables on $[0,1]^{2}$. Let $d(x,y)=\|x-y\|_{2}$ be the Euclidean distance. Let $L(X_{1},\dots,X_{n})$ be the distance of the optimal Traveling Salesman tour through these points, under distance $d(\cdot,\cdot)$. In seminal work, Beardwood et al (1959) analyzed the limiting behavior of the value of the optimal Traveling Salesman tour length, under the random Euclidean model.

Theorem 1** ([1]).**

There exists a constant $\beta$ such that

[TABLE]

almost surely.

The authors additionally showed in [1] that

[TABLE]

where $\beta_{+}=2\int_{0}^{\infty}\int_{0}^{\sqrt{3}}\sqrt{z_{1}^{2}+z_{2}^{2}}e^{-\sqrt{3}z_{1}}\left(1-\frac{z_{2}}{\sqrt{3}}\right)dz_{2}dz_{1}.$ This integral is equal to approximately $0.92116$ ([2]). To date, the only improvement to the upper bound was given in [2], showing that $\beta\leq\beta_{+}-\epsilon_{0}$, for an explicit $\epsilon_{0}>\frac{9}{16}10^{-6}$. In [2], the author also claimed to improve the lower bound; however, we have found a fault in the argument.

The rest of this note is structured as follows. In Section II, we present the proof of $\beta\geq 0.625$ by [1]. We then outline the approach of [2] to improve the bound. Section III corrects the result in [2], giving the lower bound $\beta\geq 0.625+\frac{19}{10368}\approx 0.6268$. Finally, Section IV tightens the argument of [2] to derive the improved bound, $\beta\geq 0.6277$.

II Approaches for the Lower Bound

By the following lemma, we can equivalently study the limiting behavior of $\frac{\mathbb{E}\left[L(X_{1},\dots,X_{n})\right]}{\sqrt{n}}$.

Lemma 1** ([1]).**

It holds that

[TABLE]

Further, we can switch to a Poisson process with intensity $n$. Let $\mathcal{P}_{n}$ denote a Poisson process with intensity $n$ on $[0,1]^{2}$.

Lemma 2** ([1]).**

It holds that

[TABLE]

[1] gave the following lower bound on $\beta$.

Theorem 2** ([1]).**

The value $\beta$ is lower bounded by $\frac{5}{8}$.

Proof.

(Sketch) We outline the proof given by [1], giving a lower bound on $\mathbb{E}\left[L(\mathcal{P}_{n})\right]$. Observe that in a valid traveling salesman tour, every point is connected to exactly two other points. To lower bound, we can connect each point to its two closest points. We can further assume that the Poisson process is over all of $\mathbb{R}^{2}$, rather than just $[0,1]^{2}$, in order to remove the boundary effect. The expected distance of a point to its closest neighbor is shown to be $\frac{1}{2\sqrt{n}}$, and the expected distance to the next closes neighbor is shown to be $\frac{3}{4\sqrt{n}}$. Each point contributes half the expected lengths to the closest two other points. Since the number of points is concentrated around $n$, it holds that $\beta\geq\frac{1}{2}\left(\frac{1}{2}+\frac{3}{4}\right)$. ∎

Certainly there is room to improve the lower bound. Observe that short cycles are likely to appear when we connect each point to the two closest other points. In [2], the author gave an approach to identify situations in which $3$-cycles appear, and then lower-bounded the contribution of correcting these $3$-cycles. We outline the approach below.

For point $a$, let $r_{1}$ be the distance of $a$ to the closest point, and let $r_{2}$ be the distance to the next closest point. Let $E_{a}$ be the event that the third closest point is at a distance of $r_{3}\geq r_{1}+2r_{2}$. 2. 2.

The probability that $E_{a}$ occurs is calculated to be $\frac{7}{324}$ for a given point $a$. Therefore, the expected number of points satisfying this geometric property is $\frac{7}{324}n$, and the number of triples involved is at least $\frac{1}{3}\frac{7}{324}n$ in expectation. 3. 3.

Using the relationship $r_{3}\geq r_{1}+2r_{2}$, we can show that if $\{a,b,c,d\}$ satisfy the geometric property with $\|a-b\|=r_{1}$, $\|a-c\|=r_{2}$, and $\|a-d\|=r_{3}$, then the closest two points to $b$ are $a$ and $c$, and the closest two points to $c$ are $a$ and $b$. Therefore, the “count the closest two distances” method would create a triangle in this situation. 4. 4.

To correct for the triangle, subtract the lengths coming from the triangle and add a lower bound on the new lengths. The triangle contribution is calculated to be at most $3(r_{1}+r_{2})$ and the new lengths are calculated to be at least $2r_{3}$. Therefore, whenever the geometric property holds for a triplet of points, the calculated contribution is $2r_{3}-3(r_{1}+r_{2})$. 5. 5.

The final adjustment is calculated to be $\frac{19}{5184}$.

There are two errors in this analysis that are both due to inconsistency with counting edge lengths. If edge lengths are counted from the perspective of vertices, then the right thing to do would be to give each vertex two “stubs.” These stubs are connected to other vertices, and may form edges if there are agreements. A stub from vertex $a$ to vertex $b$ contributes $\frac{1}{2}\|a-b\|$ to the path length. In this way, a triangle comprises 6 stubs, and the contribution to the path length is the sum of the edge lengths. On page 35, the author writes $r_{1}+r_{2}+2\|a-c\|$ as the contribution of the triangle. This is probably a typo and likely $r_{1}+r_{2}+2\|b-c\|$ was meant instead. However, it should be $r_{1}+r_{2}+\|b-c\|\leq 2(r_{1}+r_{2})$. Fixing this error helps the analysis.

The next step is to redirect the six stubs, and determine their length contributions. We break edge $(b,c)$, which means we need to redirect two stubs, while the four stubs that comprise the edges $(a,b)$ and $(a,c)$ remain. The redirected stubs contribute $\frac{1}{2}\|b-d\|+\frac{1}{2}\|c-e\|$. The six stubs therefore yield an overall contribution of $\|a-b\|+\|a-c\|+\frac{1}{2}\|b-d\|+\frac{1}{2}\|c-e\|\geq r_{1}+r_{2}+\frac{1}{2}\left(r_{3}-r_{1}\right)+\frac{1}{2}\left(r_{3}-r_{2}\right)=r_{3}+\frac{1}{2}(r_{1}+r_{2})$. In the analysis above Figure 5 in the paper, the author includes the full lengths $\|b-d\|$ and $\|c-e\|$. The effect of this is to give points $d$ and $e$ a third stub each.

To summarize, the overall contribution for the triangle scenario, after breaking edges $(b,c)$, is $r_{3}+\frac{1}{2}(r_{1}+r_{2})-2(r_{1}+r_{2})=r_{3}-\frac{3}{2}r_{1}-\frac{3}{2}r_{2}$.

III Derivation of the Lower Bound

In this section we use the approach of [2] to derive a lower bound on $\beta$.

Theorem 3**.**

It holds that $\beta\geq\frac{5}{8}+\frac{19}{10368}$.

The proof of Theorem 3 requires Lemmas 3 and 4.

Lemma 3** (Lemma 4 in [2]).**

Let $\mathcal{P}_{n}$ be a Poisson point process on $\mathbb{R}^{2}$ with intensity $n$. Then for any fixed point $p\in\mathbb{R}^{2}$, the probability distribution of the distance between $p$ and the the three closest points to $p$ is given by

[TABLE]

Lemma 4**.**

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Proof of Theorem 3.

First we verify that the lower bound from breaking edge $(b,c)$ is valid. If edge $(a,b)$ is broken instead, the new stub lengths are $\|a-c\|+\|b-c\|+\frac{1}{2}\|a-d\|+\frac{1}{2}\|b-e\|$. The difference after subtracting the original stub lengths is then equal to

[TABLE]

Similarly, if edge $(a,c)$ is broken, the contribution is lower bounded by $r_{3}-\frac{3}{2}r_{2}$. Since $r_{3}-\frac{3}{2}r_{1}-\frac{3}{2}r_{2}\leq r_{3}-\frac{3}{2}r_{2}\leq r_{3}-\frac{3}{2}r_{1}$, we conclude that $r_{3}-\frac{3}{2}r_{1}-\frac{3}{2}r_{1}$ from breaking edge $(b,c)$ is a valid lower bound.

Therefore, from the discussion in Section II and Lemma 3 we adjust the integral in [2] to give

[TABLE]

From Lemma 4,

[TABLE]

∎

IV An Improvement

In this section, we improve upon the bound in Section III by tightening the triangle inequality.

Theorem 4**.**

It holds that

[TABLE]

Proof.

Place a Cartesian grid so that point $a$ is at the origin and point $b$ is at $(r_{1},0)$. Then with probability $\frac{1}{2}$, point $c$ falls into the first or fourth quadrant, and with probability $\frac{1}{2}$, point $c$ falls into the second or third quadrant. Conditioned on point $c$ falling into the first or fourth quadrant, the maximum length of $\|b-c\|$ is $\sqrt{r_{1}^{2}+r_{2}^{2}}$. Conditioned on point $c$ falling into the second or third quadrant, the maximum length of $\|b-c\|$ is $r_{1}+r_{2}$, which corresponds to the computation in Section III.

Conditioned on point $c$ falling into the first or fourth coordinate, the length contribution from breaking edge $(b,c)$ is at least $r_{3}+\frac{1}{2}\left(r_{1}+r_{2}\right)-\left(r_{1}+r_{2}+\sqrt{r_{1}^{2}+r_{2}^{2}}\right)=r_{3}-\frac{1}{2}r_{1}-\frac{1}{2}r_{2}-\sqrt{r_{1}^{2}+r_{2}^{2}}$. If edge $(a,b)$ is broken instead, the new stub lengths are $\|a-c\|+\|b-c\|+\frac{1}{2}\|a-d\|+\frac{1}{2}\|b-e\|$. The difference after subtracting the original stub lengths is then equal to

[TABLE]

Similarly, if edge $(a,c)$ is broken, the contribution is lower bounded by $r_{3}-\frac{3}{2}r_{2}$. Since $r_{3}-\frac{1}{2}r_{1}-\frac{1}{2}r_{2}-\sqrt{r_{1}^{2}+r_{2}^{2}}\leq r_{3}-\frac{3}{2}r_{2}\leq r_{3}-\frac{3}{2}r_{1}$, we conclude that $r_{3}-\frac{1}{2}r_{1}-\frac{1}{2}r_{2}-\sqrt{r_{1}^{2}+r_{2}^{2}}$ from breaking edge $(b,c)$ is a valid lower bound.

We therefore break edge $(b,c)$.

Proposition 1**.**

If $r_{3}\geq r_{2}+\sqrt{r_{1}^{2}+r_{2}^{2}}$, then the closest points to each of $a,b,c$ are the other two points in the set $\{a,b,c\}$, whenever point $b$ is in the first or fourth quadrant.

Proof.

Point $a$: $d(a,b)=r_{1}$, $d(a,c)=r_{2}$, and for any $d\notin\{a,b,c\}$, it holds that $d(a,d)\geq r_{3}\geq r_{2}+\sqrt{r_{1}^{2}+r_{2}^{2}}$. Therefore $d(a,d)\geq d(a,b)$ and $d(a,d)\geq d(a,c)$.

Point $b$: $d(a,b)=r_{1}$, $d(b,c)\leq\sqrt{r_{1}^{2}+r_{2}^{2}}$, and for any $d\notin\{a,b,c\}$, it holds that $d(b,d)\geq d(a,d)-d(a,b)\geq r_{2}+\sqrt{r_{1}^{2}+r_{2}^{2}}-r_{1}$. Therefore $d(b,d)\geq d(a,b)$ and $d(b,d)\geq d(b,c)$.

Point $c$: $d(a,c)=r_{2}$, $d(b,c)\leq\sqrt{r_{1}^{2}+r_{2}^{2}}$, and for any $d\notin\{a,b,c\}$, it holds that $d(c,d)\geq d(a,d)-d(a,c)\geq r_{2}+\sqrt{r_{1}^{2}+r_{2}^{2}}-r_{2}=\sqrt{r_{1}^{2}+r_{2}^{2}}$. Therefore $d(c,d)\geq d(a,c)$ and $d(c,d)\geq d(b,c)$. ∎

The lower bound on $\beta$ is therefore

[TABLE]

Lemma 5**.**

It holds that

[TABLE]

Multiplying the value of the integral in Lemma 5 by $\frac{\sqrt{n}(2n\pi)^{3}}{3}$, we obtain the following lower bound.

[TABLE]

Finally, conditioning on the quadrant, the overall lower bound is

[TABLE]

∎

Appendix

Proof of Lemma 4.

We can change the order of integration to compute the integral more easily.

[TABLE]

∎

Proof of Lemma 5.

Again we change the order of integration to compute the integral more easily.

Given $r_{3}$, the upper bound on $r_{1}$ is derived by setting $r_{3}=r_{1}+\sqrt{2r_{1}^{2}}\iff r_{1}=\frac{r_{3}}{1+\sqrt{2}}$. Given $r_{3}$ and $r_{1}$, set $r_{3}=r_{2}+\sqrt{r_{1}^{2}+r_{2}^{2}}$. We have

[TABLE]

Therefore,

[TABLE]

∎