Sharper bounds on four lattice constants

Abstract

The Korkine–Zolotareff (KZ) reduction and its generalisations, are widely used lattice reduction strategies in communications and cryptography. The KZ constant and Schnorr’s constant were defined by Schnorr in 1987. The KZ constant can be used to quantify some useful properties of KZ reduced matrices. Schnorr’s constant can be used to characterize the output quality of his block 2k-reduction and is used to define his semi block 2k-reduction, which was also developed in 1987. Hermite’s constant, which is a fundamental lattice constant, has many applications, such as bounding the length of the shortest nonzero lattice vector and the orthogonality defect of lattices. Rankin’s constant was introduced by Rankin in 1953 as a generalization of Hermite’s constant. It plays an important role in characterizing the output quality of block-Rankin reduction, proposed by Gama et al. in 2006. In this paper, we first develop a linear upper bound on Hermite’s constant and then use it to develop an upper bound on the KZ constant. These upper bounds are sharper than those obtained recently by the authors, and the ratio of the new linear upper bound to the nonlinear upper bound, developed by Blichfeldt in 1929, on Hermite’s constant is asymptotically 1.0047. Furthermore, we develop lower and upper bounds on Schnorr’s constant. The improvement to the lower bound over the sharpest existing one developed by Gama et al. is around 1.7 times asymptotically, and the improvement to the upper bound over the sharpest existing one which was also developed by Gama et al. is around 4 times asymptotically. Finally, we develop lower and upper bounds on Rankin’s constant. The improvements of the bounds over the sharpest existing ones, also developed by Gama et al., are exponential in the parameter defining the constant.

Appendices

Appendix A Proof of Lemma 1

Proof

The left hand side of (14) is referred to as the midpoint rule for approximating the integral on the right hand side in numerical analysis. It is well known that

$$\begin{aligned} \int _a^b f(s) d s - (b-a) f\left( \frac{a+b}{2}\right) = \frac{1}{24}(b-a)^3 f''(z) \end{aligned}$$

(A.1)

for some $z \in (a,b)$. This formula can be easily proved as follows. By Taylor’s theorem,

$$\begin{aligned} f(s) =&f\left( \frac{a+b}{2}\right) + f'\left( \frac{a+b}{2}\right) \Big (s- \frac{a+b}{2}\Big ) \\&+ \frac{1}{2} f''(\zeta (s)) \Big (s- \frac{a+b}{2}\Big )^2, \end{aligned}$$

where $\zeta (s)$ depends on $s\in (a, b)$. Integrating both sides of the above equality over [a, b] and using the Mean-Value-Theorem for integrals immediately lead to (A.1). Then using the condition that $f''(t)\ge 0$ for $t\in [a,b]$, we obtain (14). $\square$

Appendix B Proof of Lemma 2

Proof

By some direct calculations, we have

$$\begin{aligned} f'(t)&\!=\frac{1}{t}\left[ \frac{1}{t+18}-\frac{1}{t} \ln \left( \frac{ t+18}{8.5} \right) \right] . \end{aligned}$$

Then,

$$\begin{aligned} f''(t)=\!\frac{g(t)}{t^3(t+18)^2}, \end{aligned}$$

where

$$\begin{aligned} g(t)= 2(t+18)^2\ln \left( \frac{t+18}{8.5}\right) -(3t^2+36t),\,\; t\ge 0. \end{aligned}$$

Hence, to show $f''(t)> 0$, it sufficient to show that $g(t)>0$ for $t>0$.

By some direct calculations, we have

$$\begin{aligned} g'(t)=\&4(t+18)\ln \left( \frac{t+18}{8.5}\right) -4t,\\ g''(t)=\&4\ln \left( \frac{t+18}{8.5}\right) . \end{aligned}$$

Clearly, $g''(t)>0$ when $t\ge 0$, hence $g'(t)$ is increasing with t. Furthermore, $g'(0)>0$, therefore $g'(t)>0$ for $t\ge 0$. By some direct calculations, $g(0)>0$. Therefore, $g(t)>0$ for $t> 0$. $\square$

Appendix C Proof of Corollary 1

Proof

Since

$$\begin{aligned} \frac{2^{3\ln 2}}{17^{2\ln 2}}<0.083216, \end{aligned}$$

(C.2)

by (26), to show (30), we show

$$\begin{aligned} \left( \frac{4x-1}{17} \right) ^\frac{1}{2x-1}<1.04521<\frac{0.08698}{0.083216}, \text{ for } \text{ integer } x\ge 5. \end{aligned}$$

(C.3)

Let

$$\begin{aligned} f(x)=\left( \frac{4x-1}{17} \right) ^\frac{1}{2x-1},\,\; x\ge 5, \end{aligned}$$

then

$$\begin{aligned} f(12)=1.045206\ldots <1.04521. \end{aligned}$$

Hence, to show Corollary 1, we only need to show that

$$\begin{aligned} f(x)\le f(12), \text{ for } \text{ any } \text{ integer } x\ge 5. \end{aligned}$$

(C.4)

To this end, let $g(x)=\ln (f(x))$, then

$$\begin{aligned} g'(x)=\frac{(1-4x)\ln \frac{4x-1}{17}+4x-2}{(2x-1)^2(4x-1)}. \end{aligned}$$

Let

$$\begin{aligned} h(x)=(1-4x)\ln \frac{4x-1}{17}+4x-2, \,\; x\ge 5 \end{aligned}$$

then

$$\begin{aligned} h'(x)=-4\ln \frac{4x-1}{17}<0. \end{aligned}$$

Hence, h(x) is monotonically decreasing. Furthermore, by some direct calculations, one can show that $h(11)>0, h(12)<0$, therefore, $h(x)>0$ for $x\le 11$ and $h(x)<0$ for $x\ge 12$. Hence $g'(x)>0$ for $x\le 11$ and $g'(x)<0$ for $x\ge 12$. Since $g(x)=\ln (f(x))$, f(x) is increasing for $x\le 11$ and decreasing for $x\ge 12$, therefore (C.4) holds. $\square$

Appendix D Proof of Lemma 4

Proof

By the definition of $\zeta (s)$, we have

$$\begin{aligned} \zeta (2)< 1+\sum _{n=2}^{\infty }\frac{1}{n(n-1)} =1+\sum _{n=2}^{\infty }\left( \frac{1}{n-1}-\frac{1}{n}\right) <2. \end{aligned}$$

Furthermore, for $i\ge 3$, we have

$$\begin{aligned} \zeta (i)\le \zeta (3)<\frac{\pi ^2}{8}, \end{aligned}$$

where the second inequality follows from [6]. Then, we have

$$\begin{aligned} \prod _{i=2}^{k}\zeta (i)\le 2 \Big (\frac{\pi ^2}{8}\Big )^{k-2}=\pi ^{2(k-2)}2^{7-3k}. \end{aligned}$$

Since $\zeta (i)>1$ for $i\ge k+1$, we have

$$\begin{aligned} \frac{\prod _{i=k+1}^{2k}\zeta (i)}{\prod _{i=2}^{k}\zeta (i)} > \frac{1}{\prod _{i=2}^{k}\zeta (i)} \ge \pi ^{4-2k}2^{3k-7}. \end{aligned}$$

$\square$

Appendix E Proof of Lemma 5

Proof

By [3, Theorem 1.6], for $x\ge 1$, we have

$$\begin{aligned} (x/e)^x\sqrt{2\pi x}<\Gamma (x+1)<(x/e)^x\sqrt{2\pi (x+1/2)}. \end{aligned}$$

Therefore,

$$\begin{aligned}&\frac{\prod _{i=k+1}^{2k}\Gamma (\frac{i}{2}+1)}{\prod _{i=2}^{k}\Gamma (\frac{i}{2}+1)}\nonumber \\&\quad > \frac{\prod _{i=k+1}^{2k}(i/2e)^{i/2}\sqrt{\pi i}}{\prod _{i=2}^{k}(i/2e)^{i/2}\sqrt{\pi (i+1)}}\nonumber \\&\quad = \sqrt{\pi }\sqrt{\frac{2(2k)!}{k!(k+1)!}}(2e)^{\frac{1}{2}(\sum _{i=2}^ki-\sum _{i=k+1}^{2k}i)} \sqrt{\frac{\prod _{i=k+1}^{2k}i^{i}}{\prod _{i=2}^{k}i^{i}}}\nonumber \\&\quad = \sqrt{\pi }\sqrt{\frac{2(2k)!}{k!(k+1)!}}(2e)^{-\frac{k^2+1}{2}} \sqrt{\frac{\prod _{i=k+1}^{2k}i^{i}}{\prod _{i=2}^{k}i^{i}}}\nonumber \\&\quad = \sqrt{\pi }2^{-\frac{k^2}{2}}e^{-\frac{k^2+1}{2}}\sqrt{\frac{(2k)!}{k!(k+1)!}} \sqrt{\frac{\prod _{i=k+1}^{2k}i^{i}}{\prod _{i=2}^{k}i^{i}}}. \end{aligned}$$

(E.5)

In the following, we give a lower bound on the last term of the right-hand side of (E.5). Since $x\ln x$ is an increasing function for $x\ge \frac{1}{e}$, we have

$$\begin{aligned}&\ln \left( \frac{\prod _{i=k+1}^{2k}i^{i}}{\prod _{i=2}^{k}i^{i}}\right) \\&\quad = \sum _{i=k+1}^{2k}i\ln i-\sum _{i=2}^ki\ln i\\&\quad \ge \sum _{i=k+1}^{2k}\int _{i-1}^i x\ln xdx-\sum _{i=2}^k\int _{i}^{i+1}x\ln xdx\\&\quad = \int _{k}^{2k} x\ln xdx-\int _{2}^{k+1}x\ln x dx\\&\quad = \Big [2k^2\ln (2k)-k^2-\frac{k^2\ln k}{2}+\frac{k^2}{4}\Big ]\\&\qquad -\Big [\frac{(k+1)^2\ln (k+1)}{2}-\frac{(k+1)^2}{4}-2\ln 2+1\Big ]\\&\quad \overset{(a)}{>} \Big (2k^2\ln k+2k^2\ln 2-\frac{k^2\ln k}{2}-\frac{3k^2}{4}\Big )\\&\qquad -\Big (\frac{(k^2+2k+1)\ln k}{2}+1-\frac{k^2}{4}\Big )\\&\quad =\frac{(2k^2-2k-1)}{2}\ln k+2k^2\ln 2-\frac{k^2+2}{2}, \end{aligned}$$

where (a) follows from the following inequalities,

$$\begin{aligned}&\frac{(k+1)^2\ln (k+1)}{2}-\frac{(k+1)^2}{4}-2\ln 2+1\\&\quad< \frac{(k+1)^2\ln k}{2}+\frac{(k+1)^2\ln (1+\frac{1}{k})}{2}-\frac{(k+1)^2}{4}\\&\quad < \frac{(k^2+2k+1)\ln k}{2}+\frac{(k^2+2k+1)\frac{1}{k}}{2}-\frac{(k+1)^2}{4}\\&\quad = \frac{(k^2+2k+1)\ln k}{2}+\frac{k}{2}+1+\frac{1}{2k}-\frac{k^2+2k+1}{4}\\&\quad \le \frac{(k^2+2k+1)\ln k}{2}+1-\frac{k^2}{4}, \end{aligned}$$

where the last inequality is from $k\ge 2$. Hence, we have

$$\begin{aligned} \sqrt{\frac{\prod _{i=k+1}^{2n}i^{i}}{\prod _{i=2}^{k}i^{i}}} \ge 2^{k^2}e^{-\frac{k^2+2}{4}}k^{\frac{2k^2-2k-1}{4}}, \end{aligned}$$

which combines (E.5) imply (33). $\square$