In this paper, we analyze the sum of squares hierarchy (SOS) on the ordering principle on $n$ elements. We prove that degree $O(\sqrt{n}log(n))$ SOS can prove the ordering principle. We then show that this upper bound is essentially tight by proving that for any $\epsilon > 0$, SOS requires degree $\Omega(n^{\frac{1}{2} - \epsilon})$ to prove the ordering principle on $n$ elements.

中文翻译：

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排序原则的平方和界限

在本文中，我们根据 $n$ 元素的排序原则分析平方和层次结构 (SOS)。我们证明度$O(\sqrt{n}log(n))$ SOS 可以证明排序原理。然后我们通过证明对于任何 $\epsilon > 0$，SOS 需要度数 $\Omega(n^{\frac{1}{2} - \epsilon})$ 来证明排序$n$ 元素的原则。