Subject: Re: CCL:gaussian and hollerith string Summury

From: Bernd Schubert <[email-masked]>
Date: Tue, 17 Apr 2001 23:18:00 +0200

On Monday 9 April 2001 I asked how to define the maximum allowed symmetry in 
Gaussian.

I got two replies to my answer and the problem is completely solved (thanks 
both for their reply).

The maximum allowed symmetry might be specified by using the undocumented 
keyword symm(pg=pointgroup). 
For example: symm(pg=cs)

If someone wants to use iop(2/19) for defining then the required number can 
be determinated by using the following procedure:

1.) The hollerith string is a 4 byte string
2.) the string must be written from the right to the left, unused characters 
are replaced by space
for example for the cs group: <SPACE>  <SPACE>  <S> <C>
3.) The characters are replaced by hexadecimal numbers of the ascii table
for my example giving: 20 20  53 43
4.) The resulting hex. number (20205343) is translated into a decimal number 
giving: 538989379
5.) This decimal number can be used for iop(2/19)
for my example: iop(2/19=538989379)

By using a lower symmetry than the molecule has there might be several equal 
symmetry planes possible to use as main symmetry plane. 
IOP(2/20) is needed to define which one is the main symmetry plane.

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Here are the 2 answers I got:

From: Stefan Fau <[email-masked]>

Hi Bernd,

I ran across the problem of Hollerith strings
in connection with symmetry in Gaussian some
years ago. Unfortunately I lost access to the
handbook in which I wrote down the solution ...

As far as I remember, the Hollerith string is a
hexadecimal representation of the characters
in the symmetry label. I think it went like this:
" C2v", characters 32, 67, 50 and 118 of the
ASCII character table. In hexadecimal this
corresponds to 20, 43, 32, 76 giving the
Hollerith string "20433276".

This string depends on which character set is
default on your computer. You can find out
using the FORTRAN command char('C'). I'm
not sure anymore if the Hollerith string expects 8
hexadecimal numbers and about small or capital
characters.

If it doesn't work at first, better look into the
source code ...

Stefan

P.S.: please post the final solution to CCL when
you find it, so it gets into the archives.
______________________________________________________________________
Dr. Stefan Fau                    |      [email-masked]
Quantum Theory Project     |     (352) 392-6714
University of Florida
Gainesville, FL 32611-8435

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From: "Cust. Service Jim" <[email-masked]>

Dear Bernd,
   There is an undocumented feature that will get around your
iop(2/19=2HCS) problem, and that is to use something like "symm=(pg=cs)". 
This will read in the point group you requested. As for opt(2/20), this is
necessary when it is ambiguous how the symmetry will be reduced. For
example, if your initial structure is C4V and you request a reduction of
symmetry to CS, the program wants to know which of the symmetry planes in
the C4V point group you'd like as the CS plane. 

Regards,

Jim Hess
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