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I have been using cif2cell for converting CIF files into quantum ESPRESSO input files. But cif2cell always keeps the ibrav value to be zero and indicates the CELL_PARAMETERS. As per the quantum ESPRESSO website, the following is provided.

with ibrav=0 lattice vectors must be given with a sufficiently large number of digits and with the correct symmetry, or else symmetry detection may fail and strange problems may arise in symmetrization. The link

Plus there is this excerpt in their GitLab repo :

There is no sure way to distinguish between true symmetry breaking and sloppy data. If it works for you (it should if you provide enough significant digits for crystal axis and atomic positions), no problem, but nobody who uses ibrav=0 is allowed to complain any longer about symmetry. The link

So does it mean that when considering systems with a larger ntyp value {number of unique elements}, that I should resort to finding out the symmetry using an explicit ibrav value, rather than using ibrav = 0 and specifying the cell parameters.

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There is no guaranteed increase in performance if you specify a non-zero value for ibrav. There is an additional layer to this - Whether the atomic positions are specified in cartesian or crystal coordinates (sometimes some symmetries can be missed out on QE). On both points, you would need to run a test calculation and see how many symmetries QE detects. This is mentioned in the output file after some data about the system and the K points. If all the symmetries you expect to be detected are found to be QE, it will greatly reduce the computational cost of the calculation.

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    $\begingroup$ Thank you for your answer!. So technically, if the reduction in computational cost boils down to symmetry detection, then if the symmetry is specified explicitly via ibrav wouldn't it improve performance. $\endgroup$ Dec 5 '20 at 1:24
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    $\begingroup$ 'ibrav' alone is not sufficient to detect symmetries. It also depends to some extent on the position of the atoms that you specify. If the position of atoms perfectly follow the positions generated by say the symmetry group, it shouldn't matter if you specify ibrav =0 or not equal to zero. QE will detect symmetries in either case. $\endgroup$
    – Xivi76
    Dec 5 '20 at 3:42
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    $\begingroup$ Okay :). so ibrav is just for scenarios in which the user is too lazy to convert the lattice length and lattice angles to specify the lattice vectors in CELL PARAMETERS Name card. right? but then what's with this excerpt form a developer of QE " but nobody who uses ibrav=0 is allowed to complain any longer about symmetry. ". That's what throws me off from the previous "ibrav doesn't influence performance" assumption $\endgroup$ Dec 5 '20 at 6:32
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    $\begingroup$ @AnoopANair In the original answer, I state that it isn't guaranteed to improve performance. The point is that, you need to see if QE detects the symmetries in your system. $\endgroup$
    – Xivi76
    Dec 5 '20 at 7:29
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    $\begingroup$ Yes! I do understand that! QE could in fact reverse calculate using the cell parameters in order to do the symmetry identification. $\endgroup$ Dec 5 '20 at 9:34

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