Recursive QAOA outperforms the original QAOA for the MAX-CUT problem on complete graphs

Abstract

Quantum approximate optimization algorithms are hybrid quantum-classical variational algorithms designed to approximately solve combinatorial optimization problems such as the MAX-CUT problem. In spite of its potential for near-term quantum applications, it has been known that quantum approximate optimization algorithms have limitations for certain instances to solve the MAX-CUT problem, at any constant level p. Recently, the recursive quantum approximate optimization algorithm, which is a non-local version of quantum approximate optimization algorithm, has been proposed to overcome these limitations. However, it has been shown by mostly numerical evidences that the recursive quantum approximate optimization algorithm outperforms the original quantum approximate optimization algorithm for specific instances. In this paper, we analytically prove that the recursive quantum approximate optimization algorithm is more competitive than the original one to solve the MAX-CUT problem for complete graphs with respect to the approximation ratio.

Appendix: The positivity of the function g(t)

In this appendix, we want to show that for all $t\in [0,1]$,

$$\begin{aligned} g(t):=\frac{4}{(4n-1)^2}+\frac{1}{4n-1}\left( 1-\left( 2t-1\right) ^{2n-2} \right) - (1-t)t^{2n-2} > 0. \end{aligned}$$

To find the minimum of g(t), we consider the equation

$$\begin{aligned} g'(t) = -\frac{4n-4}{4n-1}(2t-1)^{2n-3}+t^{2n-3}\left( -(2n-2)+(2n-1)t\right) =0. \end{aligned}$$

(9)

Since g is continuous, we need to see that $g(0)>0$, $g(1) >0$, and $g(t^*)>0$ for all critical points $t^* \in \left[ 0,1\right]$.

Observe that

$$\begin{aligned} g(0)=g(1)=\frac{1}{(4n-1)^2}>0. \end{aligned}$$

For any critical point $t^* \in (0,1)$, it is clear that

$$\begin{aligned} g'(t^*)=-\frac{4n-4}{4n-1}(2t^*-1)^{2n-3} + {t^*}^{2n-3}\left( -(2n-2)+(2n-1)t^*\right) = 0, \end{aligned}$$

(10)

that is,

$$\begin{aligned} \frac{4n-4}{4n-1}(2t^*-1)^{2n-3} = {t^*}^{2n-3}\left( -(2n-2)+(2n-1)t^*\right) . \end{aligned}$$

(11)

Now, by imposing the condition in Eq. (11) on the function g, we have

$$\begin{aligned} g(t^*)= & {} \frac{4}{(4n-1)^2}+\frac{1}{4n-1}\left( 1-\left( 2t^*-1\right) ^{2n-2} \right) - (1-t^*){t^*}^{2n-2} \nonumber \\= & {} \frac{4}{(4n-1)^2}+\frac{1}{4n-1} \left[ 1-\frac{4n-1}{4n-4}(2t^*-1){t^*}^{2n-3}(-(2n-2)+(2n-1)t^*) \right] \nonumber \\{} & {} \quad -(1-t^*){t^*}^{2n-2} \nonumber \\= & {} \frac{4}{(4n-1)^2}+\frac{1}{4n-1} +{t^*}^{2n-3} \left( -\frac{1}{2n-2}{t^*}^2 + \frac{2n-1}{4n-4}t^*-\frac{1}{2} \right) \nonumber \\= & {} \frac{4}{(4n-1)^2}+\frac{1}{4n-1} +\frac{{t^*}^{2n-3}}{2n-2} \left( -\left( t^*-\frac{2n-1}{4} \right) ^2 + \frac{(2n-1)^2}{16}-(n-1) \right) . \nonumber \\ \end{aligned}$$

(12)

If we regard the third term in Eq. (12) as a function of $t^*$, we can easily see that it is decreasing on (0, 1). Therefore,

$$\begin{aligned} g(t^*)= & {} \frac{4}{(4n-1)^2}+\frac{1}{4n-1} +\frac{{t^*}^{2n-3}}{2n-2} \left( -\left( t^*-\frac{2n-1}{4} \right) ^2 + \frac{(2n-1)^2}{16}-(n-1) \right) \\> & {} \frac{4}{(4n-1)^2}+\frac{1}{4n-1} +\frac{1}{2n-2} \left( -\left( 1-\frac{2n-1}{4} \right) ^2 + \frac{(2n-1)^2}{16}-(n-1) \right) \\= & {} \frac{4}{(4n-1)^2}+\frac{1}{4n-1} +\frac{1}{2n-2} \left( -\frac{1}{2} \right) \\= & {} \frac{4n-13}{4(n-1)(4n-1)^2} \\> & {} 0 \end{aligned}$$

for all $n \ge 4$. Hence $g(t)>0$ for all $t \in [0,1]$ when $n \ge 4$.

For $n<4$, we can prove that $g(t)>0$ on [0, 1] from direct calculations as follows. When $n=2$, g(t) becomes

$$\begin{aligned} g(t)= & {} \frac{4}{49} + \frac{1}{7}\left( 1-(2t-1)^2\right) - (1-t)t^2 \\= & {} \frac{1}{49} \left( 49t^3-77t+28t+4\right) >0 \end{aligned}$$

since it can easily be shown that g(t) is increasing on [0, 1] and $g(0)=4>0$. When $n=3$, g(t) becomes

$$\begin{aligned} g(t)=\frac{1}{121} \left( t^5 - 177t^4 +352t^3-264t^2 +88t +4\right) \end{aligned}$$

and we can show that it is concave on [0, 1] and $g(0)=g(1)=4>0$. Thus, $g(t)>0$ on the interval [0, 1].