I am studying the Shortest Vector Problem and I have some troubles understanding the actual Lovàsz condition used in the LLL algorithm.

On the one hand, the original LLL article, the Springer book "The LLL Algorithm" and A Complete Analysis of the BKZ Lattice Reduction Algorithm state that the LLL algorithm outputs a reduced basis meaning that:

  • $|\mu_{i,j}| \le 0.5\quad\forall 1\le j < i \le n$ and
  • $\delta\|b_k^*\|^2 \le \|b^*_{k+1} + \mu_{k+1,k}b^*_k\|^2$

On the other hand, other sources, like International Symposium on Mathematics, Quantum Theory, and Cryptography, page 198, Wikipedia and these course notes state that the LLL algorithm outputs a reduced basis meaning that:

  • $|\mu_{i,j}| \le 0.5\quad\forall 1\le j < i \le n$ and
  • $\delta\|b_k^*\|^2 \le \|b^*_{k+1}\|^2 + \mu_{k+1,k}^2\|b^*_k\|^2$

I am a bit confused as I cannot even apply the triangle inequality to infer the latter from the former.

Any hint is appreciated.

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    $\begingroup$ Little search Prove Euclidean norm satisfies the triangle inequality $\endgroup$
    – kelalaka
    Commented Nov 3, 2023 at 20:05
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    $\begingroup$ Maybe the closest answer is about the definition of a norm by Mark On the spectral norm in lattice-based cryptography $\endgroup$
    – kelalaka
    Commented Nov 3, 2023 at 20:16
  • $\begingroup$ I thought of triangle inequality but while $\|b^*_{k+1} + \mu_{k+1,k} b^*_k\| \le \|b^*_{k+1}\| + \mu_{k+1,k}\|b^*_k\|$ is valid, if we square it we obtain $\|b^*_{k+1} + \mu_{k+1,k} b^*_k\|^2 \le \|b^*_{k+1}\|^2 + \mu_{k+1,k}^2\|b^*_k\|^2 \mathbf{+ 2\mu_{k+1,k}\|b^*_{k+1}\|\|b^*_k\|}$. I know no ways to check if $\|b^*_{k+1} + \mu_{k+1,k} b^*_k\|^2 \le \|b^*_{k+1}\|^2 + \mu_{k+1,k}^2\|b^*_k\|^2$ is also true (which is what we want). Or am I missing something? $\endgroup$ Commented Nov 3, 2023 at 20:57
  • $\begingroup$ And even if the triangle inequality explained the validity of the second definition, they are not equivalent: if LLL returns a basis $B$ that followed the first definition, then, assuming the inequality can be applied, it follows also the second one, but the opposite is not true (if $B$ follows the second "version" of the Lovász condition, we have no ways to say it also follows the first). PS: a small note, I'm unable to edit my former comment, but there is a typo: $\|b^*_{k+1}+\mu_{k+1,k}b^*_k\|\le\|b^*_{k+1}\|+\mathbf{|\mu_{k+1,k}|}\|b^*_k\|$ $\endgroup$ Commented Nov 3, 2023 at 21:06
  • $\begingroup$ It is worth mentioning that often when working with the squared euclidean norm $\lVert x+y\rVert_2^2$, one may apply convexity of $x\mapsto \lVert x\rVert_2^2$ as a substitute for the triangle inequality. Working on this briefly I it hasn't sufficed (for me) to show equivalence, but it might plausibly be useful anyway, hence this comment. $\endgroup$
    – Mark Schultz-Wu
    Commented Nov 3, 2023 at 21:29

1 Answer 1


Actual answer to your question:

For the Euclidean norm, both conditions are equivalent because $b^*_{k}$ and $b^*_{k+1}$ are orthogonal.

Remember that they are obtained via Gram-Schmidt orthogonalization, thus, $b^*_{k} \cdot b^*_{k+1} = 0$.

Also, remember that for any vectors $u$ and $v$, because the Euclidean norm is $\|u\| = \sqrt{u \cdot u}$, we have

$\|u + v\|^2 = (u + v) \cdot (u + v) = u\cdot u + v \cdot v + 2\cdot u\cdot v = \|u\|^2+\|v\|^2 + 2\cdot u\cdot v$

Then, if $u$ and $v$ are orthogonal, we obtain $\|u + v\|^2 = (u + v) \cdot (u + v) = u\cdot u + v \cdot v + 2\cdot u\cdot v = \|u\|^2+\|v\|^2$.

A comment about your comments:

From your comments, it seems that you are getting confused because you are using the "squared triangular inequality", which says that

$ \|u + v\|^2 \le \|v\|^2 + \|v\|^2 + 2\|u\|\|v\| $

This is true, of course, but this inequality is often obtained using the Cauchy–Schwarz inequality, i.e., $u\cdot v \le \|u\|\|v\|$... That is, we write

$\|u + v\|^2 = (u + v) \cdot (u + v) = u\cdot u + v \cdot v + 2\cdot u\cdot v = \|u\|^2+\|v\|^2 + 2\cdot u\cdot v ~~ (1)$

then use Cauchy–Schwarz to obtain

$\|u + v\|^2 \le \|u\|^2+\|v\|^2 + 2\cdot \|u\|\cdot \|v\|$.

But of course, in your case, instead of using this upper bound, you can just set $u \cdot v = 0$ in Equality (1)

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    $\begingroup$ @kelala thanks! $\endgroup$ Commented Nov 6, 2023 at 21:02
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    $\begingroup$ So just to complete the trivial passages: $\delta\|b^*_k\|^2 \le \|b^*_{k+1} + \mu_{k+1,k}b^*_k\|^2 = \|b^*_{k+1}\|^2 + \mu_{k+1,k}^2\|b^*_k\|^2$ therefore $(\delta - \mu_{k+1,k}^2)\|b^*_k\| \le \|b^*_{k+1}\|^2$ which is the common other "version" of the Lovász condition. Thanks a lot for your answer! :) $\endgroup$ Commented Nov 7, 2023 at 12:39

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