A tight $\sqrt{2}$-approximation for linear 3-cut

Abstract

We investigate the approximability of the linear 3-cut problem in directed graphs. The input here is a directed graph $D=(V,E)$ with node weights and three specified terminal nodes $s,r,t\in V$, and the goal is to find a minimum weight subset of non-terminal nodes whose removal ensures that s cannot reach r and t, and r cannot reach t. The precise approximability of linear 3-cut has been wide open until now: the best known lower bound under the unique games conjecture (UGC) was 4 / 3, while the best known upper bound was 2 using a trivial algorithm. In this work we completely close this gap: we present a $\sqrt{2}$-approximation algorithm and show that this factor is tight under UGC. Our contributions are twofold: (1) we analyze a natural two-step deterministic rounding scheme through the lens of a single-step randomized rounding scheme with non-trivial distributions, and (2) we construct integrality gap instances that meet the upper bound of $\sqrt{2}$. Our gap instances can be viewed as a weighted graph sequence converging to a “graph limit structure”. We complement our results by showing connections between the linear 3-cut problem and other fundamental cut problems in directed graphs.

Appendix A. Necessary conditions for (2)

Claim

Condition (2) is satisfied by $\xi$ if and only if the following hold for $\zeta$:

$$\begin{aligned} \zeta \left( \frac{\sqrt{2}-1}{\sqrt{2}}\right)&=0, \end{aligned}$$

(13)

$$\begin{aligned} (1-y) \zeta (y)+\int _y^{1-y}\zeta (z)\, \mathrm {d} z&=\frac{1}{2}\quad \text {if }\frac{\sqrt{2}-1}{\sqrt{2}} \le y \le \frac{1}{2}, \end{aligned}$$

(14)

$$\begin{aligned} (1-y) \zeta (y)-\int _{1-y}^y \zeta (z)\, \mathrm {d} z&=\frac{1}{2}\quad \text {if }\frac{1}{2} \le y \le \frac{1}{\sqrt{2}}. \end{aligned}$$

(15)

Proof

The direction showing that (13)–(15) imply (2) was already shown in Claim 4. We now argue the necessity of (13), (14) and (15).

To see the necessity of (14), we consider $a=b=y$ for $\frac{\sqrt{2}-1}{\sqrt{2}} \le y \le \frac{1}{2}$. For this choice of a and b, condition (2) necessitates that

$$\begin{aligned} 1&= 2\int _0^1 \xi (y,z)\, \mathrm {d} z + 2 \int _y^1 \xi (z,y)\, \mathrm {d} z\\&= 2\int _0^{1-y} \xi (y,z)\, \mathrm {d} z + 2 \int _y^{1-y} \xi (z,y)\, \mathrm {d} z\\&= 2\int _0^{1-y} \zeta (y)\, \mathrm {d} z + 2 \int _{y}^{1-y} \zeta (z)\, \mathrm {d} z\\&= 2 (1-y) \zeta (y)+ 2 \int _y^{1-y}\zeta (z)\, \mathrm {d} z, \end{aligned}$$

which shows the necessity of (14). The second equation above is because $\xi (y,z)=\xi (z,y)=0$ for $z > 1-y$ since $y\le 1/2$.

To see the necessity of (15), we consider $a=y$, $b=1-y$ for some y such that $\frac{1}{2} \le y \le \frac{1}{\sqrt{2}}$. For this choice of a and b, condition (2) necessitates that

$$\begin{aligned} 1&=\int _0^1 \left( \xi (y,z)+\xi (1-y,z)\right) \, \mathrm {d} z + \int _y^1 \xi (z,1-y)\, \mathrm {d} z + \int _{1-y}^1 \xi (z,y)\, \mathrm {d} z \nonumber \\&= \int _0^1 \xi (y,z)\, \mathrm {d} z + \int _{0}^1 \xi (1-y,z)\, \mathrm {d} z + 0 + 0 \nonumber \\&= \int _0^{1-y} \xi (y,z)\, \mathrm {d} z + \int _{0}^y \xi (1-y,z)\, \mathrm {d} z \nonumber \\&= \int _0^{1-y} \zeta (y)\, \mathrm {d} z + \int _{0}^{y}\zeta (1-y)\, \mathrm {d} z \nonumber \\&= (1-y) \zeta (y) + y\zeta (1-y). \end{aligned}$$

(16)

We note that the bounds on y imply that $\frac{\sqrt{2}-1}{\sqrt{2}} \le 1-y \le \frac{1}{2}$. Hence, by (14) applied to $y':=1-y$, we obtain that

$$\begin{aligned} y \zeta (1-y) = \frac{1}{2} - \int _{1-y}^{y}\zeta (z)\, \mathrm {d} z. \end{aligned}$$

(17)

Substituting (17) in (16) and rewriting in the required form shows the necessity of (15).

To see the necessity of (13), we consider $a=b<\frac{\sqrt{2}-1}{\sqrt{2}}$. For this choice of a and b, condition (2) necessitates that

$$\begin{aligned} 1&=2 \int _0^1 \xi (a,z)\, \mathrm {d} z + 2 \int _a^1 \xi (z,a)\, \mathrm {d} z =0 + 2 \int _a^1 \xi (z,a)\, \mathrm {d} z = 2\int _{\frac{\sqrt{2}-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\zeta (z)\, \mathrm {d} z. \end{aligned}$$

(18)

Now, by (14) applied to $y=\frac{\sqrt{2}-1}{\sqrt{2}}$, we obtain that

$$\begin{aligned} \frac{1}{2}=\frac{1}{\sqrt{2}}\zeta \left( \frac{\sqrt{2}-1}{\sqrt{2}}\right) + \int _{\frac{\sqrt{2}-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \zeta (z)dz = \frac{1}{\sqrt{2}}\zeta \left( \frac{\sqrt{2}-1}{\sqrt{2}}\right) + \frac{1}{2} \end{aligned}$$

where the second equation is obtained by substituting (18). Hence, $\zeta (\frac{\sqrt{2}-1}{\sqrt{2}})=0$, showing the necessity of (13). $\square$