As mysterious as k-points are, why is there a cut-off for an odd and even k-mesh preference?

On the VASP wiki page titled, "Number of k points and method for smearing", they say results when using odd k-meshes shouldn't be compared to those when using even k-meshes (section: Comparing different k-point meshes). In the section "Some other considerations", they go on to say that even meshes are better up to 8 k-points in a direction while odd meshes are better beyond 8. What is the physics behind it or is it a code-level preference?

As an aside: I recently asked a question about odd/even q-points for phonon DoS calculations where I received a comment suggesting that the problem could be similar to that of k-points in electronic calculations.

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    $\begingroup$ Convergence is generally quicker with an even k-mesh because you avoid sampling the high-symmetry points. That is why even-grids are preferred. I am not sure if there is a concrete reason for why odd-centered grids are preferred for more than 8 points. In general, it's just easier to use a grid like 10x10x10 instead of 9x9x9 because even though the even grid is denser, it will typically converge quicker than the latter because it avoids sampling high-symmetry points, which are atypical points. $\endgroup$
    – Xivi76
    Oct 7, 2020 at 17:43
  • $\begingroup$ @Xivi76: I also don't understand the preferrence of odd-centered grids for grids with many k points. In the documentation for the Fleur code we have an example for k-point set convergence for fcc Cu. The even-odd behavior is very nicely visible there, but it seems like the even k-point sets feature the more friendly convergence behavior: flapw.de/MaX-4.0/documentation/parameterConvergence/… $\endgroup$ Oct 8, 2020 at 11:07
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    $\begingroup$ While you may get slower convergence w.r.t. mesh size with odd grids, you also can greatly reduce the number of points you need to calculate due to symmetry. I suppose there's a trade-off that begins to favor odd grids at high k-point densities. You can also shift even grids to include high-symmetry points... The symmetry of your system is probably important in making this decision, as the VASP manual mentions (including $\Gamma$ with fcc, etc.) $\endgroup$ Oct 13, 2020 at 22:03
  • $\begingroup$ @KevinJ.May, Should all points on the three axes be even/odd?or I can use 4 4 3 $\endgroup$
    – 샤다ㅏ
    Apr 8, 2022 at 0:30

1 Answer 1


The reason why even number k-point grids are prefered in the case of some symmetries like FCC lattices is mainly due to the concept of the Irreducible Brillouin Zone (IBZ).

Here is an example for FCC:


Here the number of points in the k-point grid in the IBZ is the same for even as well as odd k-points (M value). Here the points in the IBZ determine the execution speed of the code (given by tm/t1). But since the even number k-point grid have more accuracy due to larger number of k points, we prefer them over the odd values. It's like having more accuracy for the same execution speed.

Hope this helps :)

  • $\begingroup$ Thank you for your answer. Isn't the data in the table a bit counterintuitive to what the VASP wiki says? Up to $8$, odd values of $M$ have a shorter execution time, whereas, for $11,12$ and $13,14$, even values of $M$ have a shorter execution time. $\endgroup$ Oct 14, 2020 at 11:44
  • $\begingroup$ This answer seems flawed somehow, the number of points in the IBZ is the same for even and odd numbers? I suspect this was done on a shifted grid or something? $\endgroup$ Oct 14, 2020 at 18:44
  • $\begingroup$ @TristanMaxson Isn't that the reason why even and odd points indicate the same execution time? I've obtained the table from the book titled "Density Functional Theory: A Practical Introduction". You can refer to that for more information. $\endgroup$ Oct 14, 2020 at 19:07
  • $\begingroup$ I do not think they should give the same number of points in the IBZ. I will make an answer explaining. $\endgroup$ Oct 14, 2020 at 20:28

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