A General Reduction Theorem with Applications to Pathwidth and the Complexity of MAX 2-CSP

Abstract

We prove a general reduction theorem which allows us to extend bounds for certain graph parameters on cubic graphs to bounds for general graphs taking into account the individual vertex degrees. As applications, we give an algorithm for Max $2$-CSP whose complexity matches the algorithm of Scott and Sorkin in the case of $d$-regular graphs, $d \le 5$, but is otherwise faster. It also improves on the previously fastest known algorithm in terms of the average degree, given by Golovnev and Kutzkov. Also from the general theorem, we derive a bound for the pathwidth of a general graph which equals that of Fomin et al. and Gaspers for graphs of degree at most $6$, but is smaller otherwise, and use this to give an improved exponential-space algorithm for Max $2$-CSP. Finally we use the general result to give a faster algorithm for Max $2$-CSP on claw-free graphs.

Appendix

In this section we prove the properties of the functions $g_{\alpha }$ and $g_{\alpha }'$ which are listed in Sect. 3.1. First recall the definition of the functions concerned:

For any $n \ge 2$, and $\alpha$ with $0 \le \alpha \le 1$, define the function $g_{\alpha }(n)$ by setting $g_{\alpha }(2)=0$, $g_{\alpha }(3)=\alpha$, and for any $n \ge 4$,

$$\begin{aligned} n g_{\alpha }(n) = (n-2) g_{\alpha }(n-1) + g_{\alpha }(n-2) + 1. \end{aligned}$$

(6)

We extend $g_{\alpha }$ to all real numbers at least $2$ by linear interpolation, i.e., if $r = n + x$, where $n \ge 2$ is an integer, and $0 \le x \le 1$, then we set $g_{\alpha }(r) = (1-x)g_{\alpha }(n) + xg_{\alpha }(n+1)$.

Also, for any $n \ge 3$, define $g_{\alpha }'(n) = g_{\alpha }(n) - g_{\alpha }(n-1)$.

Lemma 7.1

For any integer $n \ge 2$,

$$\begin{aligned} g_{\alpha }(n) = (4-3\alpha ) \frac{A(n)}{n!} + (2-3\alpha )\frac{(-1)^n}{n!} - (3-3\alpha ) \end{aligned}$$

where $A(n)$ is the alternating factorial function given by

$$\begin{aligned} A(n) = n! - (n-1)! + \cdots - (-1)^n \cdot 1!. \end{aligned}$$

Proof

Let $h(n) = 1-g_{\alpha }(n)$. Then from the recurrence (6) above we obtain, for $n \ge 4$,

$$\begin{aligned} n(1-h(n)) = (n-2)(1-h(n-1)) + 1 - h(n-2) + 1 \end{aligned}$$

or

$$\begin{aligned} nh(n) = (n-2)h(n-1) + h(n-2), \end{aligned}$$

from which

$$\begin{aligned} nh(n) + h(n-1) = (n-1)h(n-1) + h(n-2) \end{aligned}$$

Thus $nh(n) + h(n-1)$ is a constant, $K$ say, for all $n \ge 3$. Multiplying through by $(n-1)!$, we obtain

$$\begin{aligned} n!h(n) + (n-1)!h(n-1) = K(n-1)! \end{aligned}$$

or (replacing $n$ by $n+1$)

$$\begin{aligned} (n+1)!h(n+1) + n!h(n) = Kn! \end{aligned}$$

for $n \ge 2$. Setting $j(n) = (n+1)!h(n+1)/K$ for $n \ge 1$, we obtain

$$\begin{aligned} j(n) + j(n-1) = n! \end{aligned}$$

for all $n \ge 2$. The alternating factorial function $A(n)$ satisfies

$$\begin{aligned} A(n) + A(n-1) = n! \end{aligned}$$

so that

$$\begin{aligned} j(n) - A(n) = -(j(n-1) - A(n-1)). \end{aligned}$$

Hence $j(n) - A(n) = C(-1)^n$ for some constant $C$. Thus

$$\begin{aligned} h(n) = (K/n!)j(n-1) = (K/n!)A(n-1) - (KC/n!)(-1)^n \end{aligned}$$

or

$$\begin{aligned} g_{\alpha }(n) = 1 - (K/n!)(n! - A(n)) + (KC/n!)(-1)^n. \end{aligned}$$

Thus

$$\begin{aligned} g_{\alpha }(n) = (K/n!)A(n) + (KC/n!)(-1)^n - (K-1). \end{aligned}$$

Recall that $K = (n+1)h(n+1) + h(n)$, so setting $n=2$ gives $K = 3(1-\alpha )+1 = 4 - 3\alpha$. Also $C = A(1) - j(1) = 1 - 2h(2)/K$, so $KC = K - 2h(2) = 4 - 3\alpha -2 = 2 - 3\alpha$. So finally,

$$\begin{aligned} g_{\alpha }(n) = (4-3\alpha ) \frac{A(n)}{n!} + (2-3\alpha )\frac{(-1)^n}{n!} - (3-3\alpha ) \end{aligned}$$

as required. $\square$

Lemma 7.2

For all real $d \ge 2$,

$$\begin{aligned} g_{\alpha }(d) = 1 - \frac{4-3\alpha }{d+1} + O(1/d^3). \end{aligned}$$

Proof

We have, for $n \ge 2$,

$$\begin{aligned} g_{\alpha }(n)&= (4-3\alpha ) \frac{A(n)}{n!} + (2-3\alpha )\frac{(-1)^n}{n!} -(3-3\alpha )\\&= (4\!-\!3\alpha ) \frac{n!-(n-1)!+(n-2)!-A(n-3)}{n!} \!+\! (2-3\alpha )\frac{(-1)^n}{n!} -(3-3\alpha )\\&= (4-3\alpha ) \frac{n!-(n-1)!+(n-2)!}{n!} - (3-3\alpha ) + O(1/n^3)\\&= 1 - (4-3\alpha ) \left( \frac{1}{n} - \frac{1}{n(n-1)} \right) + O(1/n^3)\\&= 1 - \frac{(4-3\alpha )}{n+1} + (4-3\alpha ) \left( \frac{1}{n+1} - \frac{1}{n} + \frac{1}{n(n-1)} \right) + O(1/n^3)\\&= 1 - \frac{(4-3\alpha )}{n+1} + (4-3\alpha ) \left( \frac{2}{n(n-1)(n+1)} \right) + O(1/n^3)\\&= 1 - \frac{(4-3\alpha )}{n+1} + O(1/n^3). \end{aligned}$$

The extension to non-integer arguments is straightforward. $\square$

Lemma 7.3

If $\alpha \le 1/2$, then $g_{\alpha }$ is non-decreasing, i.e.,

$$\begin{aligned} g_{\alpha }(n+1) \ge g_{\alpha }(n) \end{aligned}$$

for all integers $n \ge 2$.

Proof

This is proved by induction, exactly as for the function $g = g_{1/4}$ in [8]. First a very easy induction shows that $g_{\alpha } < 1$ for all $n$. Then we show by induction that for all $n \ge 3$,

$$\begin{aligned} g_{\alpha }(n-1) \le g_{\alpha }(n) \le (1 + g_{\alpha }(n-1))/2 \end{aligned}$$

This is true for $n = 3$, so suppose it is true for some $n$. Then firstly,

$$\begin{aligned} (n+1)g_{\alpha }(n+1)&= (n-1)g_{\alpha }(n) + g_{\alpha }(n-1) + 1 \ge (n-1)g_{\alpha }(n) + 2g_{\alpha }(n) \\&= (n+1)g_{\alpha }(n), \end{aligned}$$

and secondly

$$\begin{aligned} (n+1)g_{\alpha }(n+1) = (n-1)g_{\alpha }(n) + g_{\alpha }(n-1) + 1 \le n g_{\alpha }(n) + 1 \end{aligned}$$

so that

$$\begin{aligned} g_{\alpha }(n+1) \le \frac{n}{n+1} g_{\alpha }(n) + \frac{1}{n+1} \le \frac{1}{2} g_{\alpha }(n) + \frac{1}{2} \end{aligned}$$

since $n \ge 1$ and $g_{\alpha }(n) < 1$. $\square$

Note that if $\alpha > 1/2, g_{\alpha }$ is not non-decreasing, since $g_{\alpha }(4) = (1 + 2\alpha )/4 < \alpha = g_{\alpha }(3)$.

Lemma 7.4

If $1/6 \le \alpha \le 3/10$, then $g_{\alpha }'$ is non-increasing, i.e.,

$$\begin{aligned} g_{\alpha }'(n+1) \le g_{\alpha }'(n) \end{aligned}$$

for all integers $n \ge 3$.

Proof

We have

$$\begin{aligned} n g_{\alpha }(n)&= (n-2) g_{\alpha }(n-1) + g_{\alpha }(n-2) + 1, \\ (n-1) g_{\alpha }(n-1)&= (n-3) g_{\alpha }(n-2) + g_{\alpha }(n-3) + 1 \end{aligned}$$

from which we obtain

$$\begin{aligned} n g_{\alpha }'(n) = (n-3)g_{\alpha }'(n-1) + g_{\alpha }'(n-2). \end{aligned}$$

for $n \ge 5$. Then it is easily shown by an induction similar to the one above that

$$\begin{aligned} g_{\alpha }'(n-1) \ge g_{\alpha }'(n) \ge g_{\alpha }'(n-1)/3. \end{aligned}$$

for all $n \ge 4$. The base case requires that $g_{\alpha }'(3) \ge g_{\alpha }'(4) \ge g_{\alpha }'(3)/3$, or $\alpha \ge (1-2\alpha )/4 \ge \alpha /3$, which is true since $1/6 \le \alpha \le 3/10$. $\square$

Lemma 7.5

If $1/6 \le \alpha \le 3/10, g_{\alpha }(n)/n$ is strictly decreasing for $n \ge 5$.

Proof

First note that $g_{\alpha }(6) = (50\alpha +53)/120 > 1/2$. Now $g_{\alpha }(n+1)/(n+1) < g_{\alpha }(n)/n$ if and only if $ng_{\alpha }(n+1) < (n+1)g_{\alpha }(n)$. Since

$$\begin{aligned} (n+1)g_{\alpha }(n+1) = (n-1)g_{\alpha }(n) + g_{\alpha }(n-1) + 1, \end{aligned}$$

we have

$$\begin{aligned} ng_{\alpha }(n+1) = (n-1)g_{\alpha }(n) + g_{\alpha }(n-1) - g_{\alpha }(n+1) + 1, \end{aligned}$$

so $ng_{\alpha }(n+1) < (n+1)g_{\alpha }(n)$ if and only if

$$\begin{aligned} g_{\alpha }(n-1) - g_{\alpha }(n+1) + 1 < 2g_{\alpha }(n), \end{aligned}$$

or

$$\begin{aligned} 2g_{\alpha }(n+1) - g_{\alpha }'(n+1) + g_{\alpha }'(n) > 1, \end{aligned}$$

which is true since $g_{\alpha }(n+1) \ge g_{\alpha }(6) > 1/2$ and $g_{\alpha }'$ is non-increasing. In fact $g_{\alpha }(n)/n$ attains its maximum at $n=5$ for all $\alpha$ in this range. $\square$