numbers that can differentiate uniquely between the combinations

[lehloks]

Hi forum users
I need a set of numbers that can differentiate uniquely between the combinations.I choose powers of 3. Starting with 1, that gives multiples of 1, 2 and 3 before I reach the next number, which is 3, with it having multiples of 3, 6, 9, etc. If I write out the combinations you can see what happens when you add them:

1 = 1 (eg. 1 Si)
2 = 1 + 1 (eg. 2 Si)
3 = 1 +
1 + 1 (e.g. 3 Si) or 3 (e.g. 1 Al) alone, and so now I have a problem, I can pick up 1 occurrence of the atom labelled 1, and two occurrences, but then one Al will give the same result as 3 Si. Since the Al atoms with their labels of 3 have multiples of 6, 9, etc, I'll never accidently find 4, 5 and/or 7/8, which I can only reach once I'm adding multiples of 1. With this set I can thus give a unique number to each possible combination of the two (but not three). I can pick any way forward, as long as you make sure that you get unique combinations every time.

Now Im stuck I want extend the assignment to 5 atoms : eg assigning a special number for other atoms like Mg,Ca, O and Na. Any one with a unique idea how can I do this. I want a code to be general so that It can be used for many things. I can assign atom name but my code wont be efficient and it will have many if statements.My assignments are shown in the code below.



[! Assign Si = 3**0 = 1, Al = 3**1 = 3 and unknown metals U = 3**2 = 9
! then we can identify pairs and triads easily without ordering
! [Si,Si] = 3**0 + 3**0 = 1 + 1 = 2
! [Si,Al] = [Al,Si] = 1 + 3 = 4
! [Al,Al] = 3 + 3 = 6
! [U,U] = 9 + 9 = 18
! [Al,U] = 3 + 9 = 12
! [Si,U] = 1 + 9 = 10
! [Si,Si,Si] = 1 + 1 + 1 = 3
! [Si,Si,Al] = 1 + 1 + 3 = 5
! [Si,Al,Al] = 1 + 3 + 3 = 7
! [Al,Al,Al] = 3 + 3 + 3 = 9
! [U,U,U] = 9 + 9 + 9 = 27
! [U,U,Si] = 9 + 9 + 1 = 19
! [U,U,Ai] = 9 + 9 + 3 = 21
! [U,Si,Si] = 9 + 1 + 1 = 11
! [U,Si,Al] = 9 + 1 + 4 = 14
! [U,Al,Al] = 9 + 3 + 3 = 15]

[do i = 1, natms
if (atmname(i) == silicon) then
id(i) = Si

elseif (atmname(i) == aluminium) then

id(i) = Al
else
id(i) = 9
endif
enddo

do i = 1, natms
do k = 1, nmetals
if (atmname(i) == metals(k)) then
do j = i, natms
if (atmname(j) == oxygen) then
bndtab(i,j) = mbonds(k)**2
bndtab(j,i) = bndtab(i,j)
endif
enddo
endif
enddo
enddo

endif
]

Then when I get the assignment correctly I want to use the code as this where I sect a correct or desired case as shown below.

[do i = 1, natms
if ((atmname(i) == oxygen) .and. (cnlist(i) .eq. 2)) then
select case ( id(blist(bindex(i))) + id(blist(bindex(i)+1)) )
case (2)
write(logfile,*)'[Si,Si] bridge'
nbo = nbo + 1
case (4)
write(logfile,*)'[Si,Al] bridge'
nbo = nbo + 1
case (6)
write(logfile,*)'[Al,Al] bridge'
nbo = nbo + 1
case (10)
write(logfile,*)'[Si,Ca/Mg] disconnect'
nnbo = nnbo + 1
case (12)
write(logfile,*)'[Al,Ca/Mg] disconnect'
nnbo = nnbo + 1
case (18)
write(logfile,*)'[Mg/Ca,Ca/Mg] disconnect'
end select
endif
]

 

[salgerman]

For ease, record keeping purposes and uniqueness, I would vote for ditching this idea powers and prime numbers and having a single unique integer as the identifier of a combination...I don't even see the point of it...it is not like you can do math with them like meaningfully subtracting two combinations and get a third one or anything like that.

So, I would simply use arrays or, rather, a single 2D matrix...

The first index of the matrix (row number) would be the i-th combination; the second index (column number) would be the atom...the value in the (i,j) position would be, given the i-th combination, how many times the j-th atom participates in it. As simple as that. So, this matrix would have as many rows as you care to have combination and as many columns as there are elements in the periodic table (118).

With the 2D matrix, you will have at all times the exact composition of each combination, no need to decode anything.