The simultaneous semi-random model for TSP

Abstract

Worst-case analysis is a performance measure that is often too pessimistic to indicate which algorithms we should use in practice. A classical example is in the context of the Euclidean Traveling Salesman Problem (TSP) in the plane, where local search performs very well in practice even though it only achieves an $\Omega (\frac{\log n}{\log \log n})$ worst-case approximation ratio. In such cases, a natural alternative approach to worst-case analysis is to analyze the performance of algorithms in semi-random models. In this paper, we propose and investigate a novel semi-random model for the Euclidean TSP. In this model, called the simultaneous semi-random model, an instance over n points consists of the union of an adversarial instance over $(1-\alpha )n$ points and a random instance over $\alpha n$ points, for some $\alpha \in [0, 1]$. As with smoothed analysis, the semi-random model interpolates between distributional (random) analysis when $\alpha = 1$ and worst-case analysis when $\alpha = 0$. In contrast to smoothed analysis, this model trades off allowing some completely random points in order to have other points that exhibit a fully arbitrary structure. We show that with only an $\alpha = \frac{1}{\log n}$ fraction of the points being random, local search achieves an $\mathcal {O}(\log \log n)$ approximation in the simultaneous semi-random model for Euclidean TSP in fixed dimensions. On the other hand, we show that at least a polynomial number of random points are required to obtain an asymptotic improvement in the approximation ratio of local search compared to its worst-case approximation, even in two dimensions.

Appendices

Appendix 1: Proof of Lemma 17

We start with an auxiliary lemma.

Lemma 23

Let $\varepsilon >0$. Consider the distinct points $v,v',x,x'\subseteq \mathbb {R}^2$ and suppose that $\left\Vert v-v' \right\Vert =\varepsilon \le 1$, $\left\Vert x-x' \right\Vert =\ell \le 2$, and

$$\begin{aligned} d(v, [x,x'])^2>\max \{3\varepsilon ,\varepsilon \ell +\varepsilon ^2\} \end{aligned}$$

where $[x,x']$ denotes the segment from x to $x'$. Then the pair of edges $\{(v,v'), (x,x')\}$ is unordered 2-optimal.

Proof

Without loss of generality, assume $x=(0,0)$, $x'=(\ell ,0)$ and v lies in the upper half-plane. Hence, letting $a=\left\Vert x-v \right\Vert$, $b=\left\Vert v'-x' \right\Vert$, $h = d(v, [x,x'])$, and $\phi (v,v') = a+b-\ell -\varepsilon$, we want to show that

$$\begin{aligned} \phi (v,v')>0 \end{aligned}$$

for all choices of $v,v'$ ($\phi (v',v) > 0$ follows symmetrically). Note that when $v, x',$ and $\epsilon$ are fixed, b is minimized when $v'$ is on the line joining v and $x'$, so we may assume it is the case. Note moreover that we can assume that $v' \in [v,x']$ since otherwise $\varepsilon ^2> b^2 > h^2$, contradicting the hypothesis of the lemma. Hence, $b+\varepsilon = \left\Vert x'-v \right\Vert$.

Let t be the horizontal coordinate of v. We first consider the case $t<0$. If $t<0$, we have $h = \left\Vert v-x \right\Vert =a$. Consider the right triangle with vertices $x,(0,h),x'$. Its hypotenuse has length $\left\Vert x'-(0,h) \right\Vert \le \left\Vert x'-v \right\Vert = b+\varepsilon$, for $t<0$. Its catheti are of length $\left\Vert x-(0,h) \right\Vert =h$ and $\left\Vert x-x' \right\Vert =\ell$. By Pythagoras’ theorem, we have $\left\Vert x'-(0,h) \right\Vert ^2 = h^2 +\ell ^2$. Thus

$$\begin{aligned} (b + \varepsilon )^2 \ge \left\Vert x'-(0,h) \right\Vert ^2 = h^2 + \ell ^2>3\varepsilon + \ell ^2 \, \Rightarrow \, b^2 > 3 \varepsilon + \ell ^2 - \varepsilon ^2 -2b \varepsilon , \end{aligned}$$

where we used the hypothesis of the lemma.

We have

$$\begin{aligned} (a+b)^2 - (\ell +\varepsilon )^2= & {} (h+b)^2 - (\ell + \varepsilon )^2 \\ {}= & {} h^2 + b^2 + 2hb - \ell ^2 - \varepsilon ^2 - 2\varepsilon \ell \\ {}> & {} 3\varepsilon + b^2 + 2hb - \ell ^2 - \varepsilon ^2 - 2\varepsilon \ell \\> & {} 3\varepsilon + (3 \varepsilon + \ell ^2 - \varepsilon ^2 -2b \varepsilon ) + 2hb - \ell ^2 - \varepsilon ^2 - 2\varepsilon \ell \\= & {} 6\varepsilon - 2 \varepsilon ^2 -2b\varepsilon +2hb - 2\varepsilon \ell \\= & {} 2\varepsilon (1-\varepsilon ) + 2b (h - \varepsilon ) + \varepsilon (4 - 2 \ell ) \\> & {} 0. \end{aligned}$$

where the last inequality is since $\varepsilon \in [0, 1]$, $h\ge \varepsilon$, and $\ell \le 2$. We get that $a+b > \ell +\varepsilon$, which implies $\phi (v,v')>0$.

Now consider the case $t\ge \ell$, and let $v=(t,q)$. Note that in this case $h=\left\Vert x'-v \right\Vert =b+\varepsilon$. Thus by the hypothesis of the lemma

$$\begin{aligned} (b + \varepsilon )^2> 3\varepsilon \, \Rightarrow \, b^2 > 3 \varepsilon - \varepsilon ^2 -2b \varepsilon . \end{aligned}$$

Consider the right triangle with vertices x, v, (0, t). Its hypotenuse has length $\left\Vert x-v \right\Vert =a$, while its catheti have length t and q. In turn, the hypotenuse of the right triangle with vertices $x',(0,t),v$ has length $\left\Vert v-x' \right\Vert =b+\varepsilon$ while its catheti have length $t-\ell$ and q. Hence:

$$\begin{aligned} a^2 = t^2 + q^2 = (\ell + (t-\ell ))^2 + q^2> \ell ^2 + (t-\ell )^2 + q^2 = \ell ^2 + (b+\varepsilon )^2 > \ell ^2 + 3\varepsilon . \end{aligned}$$

Thus,

$$\begin{aligned} (a+b)^2 - (\ell +\varepsilon )^2> & {} \ell ^2 + 3\varepsilon + b^2 + 2ab - \ell ^2 -\varepsilon ^2 - 2 \ell \varepsilon \\> & {} \ell ^2 + 3\varepsilon + (3 \varepsilon - \varepsilon ^2 -2b \varepsilon ) + 2ab - \ell ^2 -\varepsilon ^2 - 2 \ell \varepsilon \\= & {} 2\varepsilon (1-\varepsilon ) + 2b (a-\varepsilon ) + 2\varepsilon (2 - \ell ), \end{aligned}$$

and $\phi (v,v')>0$ follows since $\varepsilon \in [0, 1]$, $a > \varepsilon$, and $\ell \le 2$.

Last, assume $t\in [0,\ell ]$. Hence, $v=(t,h)$ and we want to find the minimum of

$$\begin{aligned} \phi (v,v')=a+b-\ell -\varepsilon = \sqrt{t^2+h^2}+\sqrt{(\ell -t)^2+h^2}-\ell -2\varepsilon . \end{aligned}$$

By studying the variations of $\phi$ as t changes, one finds that it has a minimum at $t=\frac{\ell }{2}$ with value

$$\begin{aligned} \phi (\left( \frac{\ell }{2},h\right) ,v') = 2\sqrt{\frac{\ell ^2}{4}+h^2}-\ell -2\varepsilon \end{aligned}$$

and we have $\phi \left( \left( \frac{\ell }{2},h\right) ,v'\right) >0$ if and only if $h^2>\varepsilon \ell +\varepsilon ^2$, which we assumed. $\square$

We can now prove Lemma 17.

Proof of Lemma 17

Let $x,y\in K$ and $(v,v')\in P$. Let $\varepsilon = \left\Vert v-v' \right\Vert \le \delta$ and $\ell =\left\Vert x-x' \right\Vert \le \sqrt{2}\le \frac{3}{2}$. We have

$$\begin{aligned} d(v,[x,x'])^2\ge d(v,K)^2\ge d(V,K)^2 \ge 3\delta \ge 3\varepsilon \ge 3 \varepsilon ^2\ge \varepsilon ^2+\sqrt{2}\varepsilon \ge \varepsilon ^2+ \varepsilon \ell , \end{aligned}$$

where the first inequality holds by the convexity of K and the fourth, fifth, and sixth because $\varepsilon \le \delta \le 1$. Thus by Lemma 23, the swap is impossible. $\square$

Appendix 2: Proof of Lemma 18

Proof of Lemma 18

Let $a = \left\Vert e \right\Vert$ and assume without loss of generality that $e = (e_1,e_2)=((0,a),(0,2a))$ and $f=(x,x') = ((x_1,x_2),(x'_1,x'_2))$ (with $x\ne x'$, $x_1,x'_1\ge 0$ and $x_2,x'_2\le 0$).

Without loss of generality, assume $x'_1\ge x_1$.

Case 1: $x_2'\ge x_2$. Assume we have a possible 2-swap between e and f, and rename $e_1$ and $e_2$ to $e_1'$, $e_2'$ so that, after the swap, $e_1'$ is linked to x and $e_2'$ to $x'$. For any vector u denote $u_1,u_2$ its horizontal and vertical coordinates. Then,

$$\begin{aligned} \left\Vert e_2'-x' \right\Vert \ge (e_2'-x')_1 = x'_1 = |x'_1| \ge |x'_1-x_1|, \end{aligned}$$

where last inequality follows from $x_1'\ge x_1 \ge 0$. Moreover,

$$\begin{aligned} \left\Vert e_1'-x \right\Vert \ge (e_1'-x)_2 \ge a-x_2 = \left\Vert e_1-e_2 \right\Vert + |x_2| \ge \left\Vert e_1-e_2 \right\Vert + |x_2-x'_2|, \end{aligned}$$

where last inequality follows from $0\ge x'_2 \ge x_2$. Notice that at least one of these inequalities is strict, because either $x_2'>x_2$ or $x'_1>x_1$ (otherwise we would have $x=x'$). Thus,

$$\begin{aligned} \left\Vert e_2'-x' \right\Vert + \left\Vert e_1'-x \right\Vert >|x'_1-x_1| + \left\Vert e_1-e_2 \right\Vert + |x_2-x'_2|\ge \left\Vert e_1-e_2 \right\Vert + \left\Vert x-x' \right\Vert , \end{aligned}$$

so the 2-swap is not possible. This finishes the first case.

Case 2: $x'_2<x_2$. As before assume we have a possible 2-swap and rename $e_1,e_2$ to $e_1'$, $e_2'$ according to the swap. We then have

$$\begin{aligned} \left\Vert e_1'-x \right\Vert \ge (e_1'-x)_2\ge a-x_2 \ge a = \left\Vert e_1-e_2 \right\Vert , \end{aligned}$$

where we used $x_2<0$. Moreover, since

$$\begin{aligned} x_1'\ge x_1 \ge 0 \ \Rightarrow \ x_1' \ge (x_1'-x_1)\ge 0 \end{aligned}$$

and

$$\begin{aligned} a> 0\ge x_2> x'_2 \ \Rightarrow a-x_2'> x_2 - x_2' > 0, \end{aligned}$$

we have

$$\begin{aligned} \left\Vert e_2'-x' \right\Vert \ge \sqrt{(a-x'_2)^2+(x'_1)^2} > \sqrt{(x_2-x'_2)^2+(x'_1-x_1)^2}=\left\Vert x-x' \right\Vert , \end{aligned}$$

so that

$$\begin{aligned} \left\Vert e_1'-x \right\Vert + \left\Vert e_2'-x' \right\Vert >\left\Vert e_1-e_2 \right\Vert +\left\Vert x-x' \right\Vert , \end{aligned}$$

which concludes the proof. $\square$