def calculateBondSymmetryNumber(molecule, atom1, atom2):
"""
Return the symmetry number centered at `bond` in the structure.
"""
bond = atom1.edges[atom2]
symmetryNumber = 1
if bond.isSingle() or bond.isDouble() or bond.isTriple():
if atom1.equivalent(atom2):
# An O-O bond is considered to be an "optical isomer" and so no
# symmetry correction will be applied
if atom1.atomType.label == 'Os' and atom2.atomType.label == 'Os' and atom1.radicalElectrons == atom2.radicalElectrons == 0:
return symmetryNumber
# If the molecule is diatomic, then we don't have to check the
# ligands on the two atoms in this bond (since we know there
# aren't any)
elif len(molecule.vertices) == 2:
symmetryNumber = 2
else:
molecule.removeBond(bond)
structure = molecule.copy(True)
molecule.addBond(bond)
atom1 = structure.atoms[molecule.atoms.index(atom1)]
atom2 = structure.atoms[molecule.atoms.index(atom2)]
fragments = structure.split()
if len(fragments) != 2: return symmetryNumber
fragment1, fragment2 = fragments
if atom1 in fragment1.atoms: fragment1.removeAtom(atom1)
if atom2 in fragment1.atoms: fragment1.removeAtom(atom2)
if atom1 in fragment2.atoms: fragment2.removeAtom(atom1)
if atom2 in fragment2.atoms: fragment2.removeAtom(atom2)
groups1 = fragment1.split()
groups2 = fragment2.split()
# Test functional groups for symmetry
if len(groups1) == len(groups2) == 1:
if groups1[0].isIsomorphic(groups2[0]): symmetryNumber *= 2
elif len(groups1) == len(groups2) == 2:
if groups1[0].isIsomorphic(groups2[0]) and groups1[1].isIsomorphic(groups2[1]): symmetryNumber *= 2
elif groups1[1].isIsomorphic(groups2[0]) and groups1[0].isIsomorphic(groups2[1]): symmetryNumber *= 2
elif len(groups1) == len(groups2) == 3:
if groups1[0].isIsomorphic(groups2[0]) and groups1[1].isIsomorphic(groups2[1]) and groups1[2].isIsomorphic(groups2[2]): symmetryNumber *= 2
elif groups1[0].isIsomorphic(groups2[0]) and groups1[1].isIsomorphic(groups2[2]) and groups1[2].isIsomorphic(groups2[1]): symmetryNumber *= 2
elif groups1[0].isIsomorphic(groups2[1]) and groups1[1].isIsomorphic(groups2[2]) and groups1[2].isIsomorphic(groups2[0]): symmetryNumber *= 2
elif groups1[0].isIsomorphic(groups2[1]) and groups1[1].isIsomorphic(groups2[0]) and groups1[2].isIsomorphic(groups2[2]): symmetryNumber *= 2
elif groups1[0].isIsomorphic(groups2[2]) and groups1[1].isIsomorphic(groups2[0]) and groups1[2].isIsomorphic(groups2[1]): symmetryNumber *= 2
elif groups1[0].isIsomorphic(groups2[2]) and groups1[1].isIsomorphic(groups2[1]) and groups1[2].isIsomorphic(groups2[0]): symmetryNumber *= 2
return symmetryNumber
def calculateBondSymmetryNumber(molecule, atom1, atom2):
"""
Return the symmetry number centered at `bond` in the structure.
"""
bond = atom1.edges[atom2]
symmetryNumber = 1
if atom1.equivalent(atom2):
# An O-O bond is considered to be an "optical isomer" and so no
# symmetry correction will be applied
if atom1.atomType.label == 'O2s' and atom2.atomType.label == 'O2s' and atom1.radicalElectrons == atom2.radicalElectrons == 0:
return symmetryNumber
# If the molecule is diatomic, then we don't have to check the
# ligands on the two atoms in this bond (since we know there
# aren't any)
elif len(molecule.vertices) == 2:
symmetryNumber = 2
else:
molecule.removeBond(bond)
structure = molecule.copy(True)
molecule.addBond(bond)
atom1 = structure.atoms[molecule.atoms.index(atom1)]
atom2 = structure.atoms[molecule.atoms.index(atom2)]
fragments = structure.split()
if len(fragments) != 2: return symmetryNumber
fragment1, fragment2 = fragments
if atom1 in fragment1.atoms: fragment1.removeAtom(atom1)
if atom2 in fragment1.atoms: fragment1.removeAtom(atom2)
if atom1 in fragment2.atoms: fragment2.removeAtom(atom1)
if atom2 in fragment2.atoms: fragment2.removeAtom(atom2)
groups1 = fragment1.split()
groups2 = fragment2.split()
# Test functional groups for symmetry
if len(groups1) == len(groups2) == 1:
if groups1[0].isIsomorphic(groups2[0]): symmetryNumber *= 2
elif len(groups1) == len(groups2) == 2:
if groups1[0].isIsomorphic(groups2[0]) and groups1[1].isIsomorphic(groups2[1]): symmetryNumber *= 2
elif groups1[1].isIsomorphic(groups2[0]) and groups1[0].isIsomorphic(groups2[1]): symmetryNumber *= 2
elif len(groups1) == len(groups2) == 3:
if groups1[0].isIsomorphic(groups2[0]) and groups1[1].isIsomorphic(groups2[1]) and groups1[2].isIsomorphic(groups2[2]): symmetryNumber *= 2
elif groups1[0].isIsomorphic(groups2[0]) and groups1[1].isIsomorphic(groups2[2]) and groups1[2].isIsomorphic(groups2[1]): symmetryNumber *= 2
elif groups1[0].isIsomorphic(groups2[1]) and groups1[1].isIsomorphic(groups2[2]) and groups1[2].isIsomorphic(groups2[0]): symmetryNumber *= 2
elif groups1[0].isIsomorphic(groups2[1]) and groups1[1].isIsomorphic(groups2[0]) and groups1[2].isIsomorphic(groups2[2]): symmetryNumber *= 2
elif groups1[0].isIsomorphic(groups2[2]) and groups1[1].isIsomorphic(groups2[0]) and groups1[2].isIsomorphic(groups2[1]): symmetryNumber *= 2
elif groups1[0].isIsomorphic(groups2[2]) and groups1[1].isIsomorphic(groups2[1]) and groups1[2].isIsomorphic(groups2[0]): symmetryNumber *= 2
return symmetryNumber
def calculateAxisSymmetryNumber(molecule):
"""
Get the axis symmetry number correction. The "axis" refers to a series
of two or more cumulated double bonds (e.g. C=C=C, etc.). Corrections
for single C=C bonds are handled in getBondSymmetryNumber().
Each axis (C=C=C) has the potential to double the symmetry number.
If an end has 0 or 1 groups (eg. =C=CJJ or =C=C-R) then it cannot
alter the axis symmetry and is disregarded::
A=C=C=C.. A-C=C=C=C-A
s=1 s=1
If an end has 2 groups that are different then it breaks the symmetry
and the symmetry for that axis is 1, no matter what's at the other end::
A\ A\ /A
T=C=C=C=C-A T=C=C=C=T
B/ A/ \B
s=1 s=1
If you have one or more ends with 2 groups, and neither end breaks the
symmetry, then you have an axis symmetry number of 2::
A\ /B A\
C=C=C=C=C C=C=C=C-B
A/ \B A/
s=2 s=2
"""
symmetryNumber = 1
# List all double bonds in the structure
doubleBonds = []
for atom1 in molecule.vertices:
for atom2 in atom1.edges:
if atom1.edges[atom2].isDouble() and molecule.vertices.index(atom1) < molecule.vertices.index(atom2):
doubleBonds.append((atom1, atom2))
# Search for adjacent double bonds
cumulatedBonds = []
for i, bond1 in enumerate(doubleBonds):
atom11, atom12 = bond1
for bond2 in doubleBonds[i + 1 :]:
atom21, atom22 = bond2
if atom11 is atom21 or atom11 is atom22 or atom12 is atom21 or atom12 is atom22:
listToAddTo = None
for cumBonds in cumulatedBonds:
if (atom11, atom12) in cumBonds or (atom21, atom22) in cumBonds:
listToAddTo = cumBonds
if listToAddTo is not None:
if (atom11, atom12) not in listToAddTo:
listToAddTo.append((atom11, atom12))
if (atom21, atom22) not in listToAddTo:
listToAddTo.append((atom21, atom22))
else:
cumulatedBonds.append([(atom11, atom12), (atom21, atom22)])
# Also keep isolated double bonds
for bond1 in doubleBonds:
for bonds in cumulatedBonds:
if bond1 in bonds:
break
else:
cumulatedBonds.append([bond1])
# For each set of adjacent double bonds, check for axis symmetry
for bonds in cumulatedBonds:
# Do nothing if less than two cumulated bonds
if len(bonds) < 1:
continue
# Do nothing if axis is in cycle
found = False
for atom1, atom2 in bonds:
if molecule.isBondInCycle(atom1.edges[atom2]):
found = True
if found:
continue
# Find terminal atoms in axis
# Terminal atoms labelled T: T=C=C=C=T
axis = []
for bond in bonds:
axis.extend(bond)
terminalAtoms = []
for atom in axis:
if axis.count(atom) == 1:
terminalAtoms.append(atom)
if len(terminalAtoms) != 2:
continue
# Remove axis from (copy of) structure
bondlist = []
for atom1, atom2 in bonds:
bond = atom1.edges[atom2]
bondlist.append(bond)
molecule.removeBond(bond)
structure = molecule.copy(True)
terminalAtoms = [structure.vertices[molecule.vertices.index(atom)] for atom in terminalAtoms]
for bond in bondlist:
molecule.addBond(bond)
atomsToRemove = []
for atom in structure.vertices:
if len(atom.edges) == 0 and atom not in terminalAtoms: # it's not bonded to anything
atomsToRemove.append(atom)
for atom in atomsToRemove:
structure.removeAtom(atom)
# Split remaining fragments of structure
end_fragments = structure.split()
#
# there can be two groups at each end A\ /B
# T=C=C=C=T
# A/ \B
# to start with nothing has broken symmetry about the axis
symmetry_broken = False
end_fragments_to_remove = []
for fragment in end_fragments: # a fragment is one end of the axis
# remove the atom that was at the end of the axis and split what's left into groups
terminalAtom = None
for atom in terminalAtoms:
if atom in fragment.atoms:
terminalAtom = atom
fragment.removeAtom(atom)
break
else:
continue
groups = []
if len(fragment.atoms) > 0:
groups = fragment.split()
# If end has only one group then it can't contribute to (nor break) axial symmetry
# Eg. this has no axis symmetry: A-T=C=C=C=T-A
# so we remove this end from the list of interesting end fragments
if len(groups) == 0:
end_fragments_to_remove.append(fragment)
continue # next end fragment
elif len(groups) == 1 and terminalAtom.radicalElectrons == 0:
if terminalAtom.atomType.label == "N3d":
symmetry_broken = True
else:
end_fragments_to_remove.append(fragment)
continue # next end fragment
elif len(groups) == 1 and terminalAtom.radicalElectrons != 0:
symmetry_broken = True
elif len(groups) == 2:
if not groups[0].isIsomorphic(groups[1]):
# this end has broken the symmetry of the axis
symmetry_broken = True
for fragment in end_fragments_to_remove:
end_fragments.remove(fragment)
# If there are end fragments left that can contribute to symmetry,
# and none of them broke it, then double the symmetry number
# NB>> This assumes coordination number of 4 (eg. Carbon).
# And would be wrong if we had /B
# =C=C=C=C=T-B
# \B
# (for some T with coordination number 5).
if end_fragments and not symmetry_broken:
symmetryNumber *= 2
return symmetryNumber
def calculateAxisSymmetryNumber(molecule):
"""
Get the axis symmetry number correction. The "axis" refers to a series
of two or more cumulated double bonds (e.g. C=C=C, etc.). Corrections
for single C=C bonds are handled in getBondSymmetryNumber().
Each axis (C=C=C) has the potential to double the symmetry number.
If an end has 0 or 1 groups (eg. =C=CJJ or =C=C-R) then it cannot
alter the axis symmetry and is disregarded::
A=C=C=C.. A-C=C=C=C-A
s=1 s=1
If an end has 2 groups that are different then it breaks the symmetry
and the symmetry for that axis is 1, no matter what's at the other end::
A\ A\ /A
T=C=C=C=C-A T=C=C=C=T
B/ A/ \B
s=1 s=1
If you have one or more ends with 2 groups, and neither end breaks the
symmetry, then you have an axis symmetry number of 2::
A\ /B A\
C=C=C=C=C C=C=C=C-B
A/ \B A/
s=2 s=2
"""
symmetryNumber = 1
# List all double bonds in the structure
doubleBonds = []
for atom1 in molecule.vertices:
for atom2 in atom1.edges:
if (atom1.edges[atom2].isDouble() or atom1.edges[atom2].order > 2) \
and molecule.vertices.index(atom1) < molecule.vertices.index(atom2):
doubleBonds.append((atom1, atom2))
# Search for adjacent double bonds
cumulatedBonds = []
for i, bond1 in enumerate(doubleBonds):
atom11, atom12 = bond1
for bond2 in doubleBonds[i + 1:]:
atom21, atom22 = bond2
if atom11 is atom21 or atom11 is atom22 or atom12 is atom21 or atom12 is atom22:
listToAddTo = None
for cumBonds in cumulatedBonds:
if (atom11, atom12) in cumBonds or (atom21,
atom22) in cumBonds:
listToAddTo = cumBonds
if listToAddTo is not None:
if (atom11, atom12) not in listToAddTo:
listToAddTo.append((atom11, atom12))
if (atom21, atom22) not in listToAddTo:
listToAddTo.append((atom21, atom22))
else:
cumulatedBonds.append([(atom11, atom12), (atom21, atom22)])
# Also keep isolated double bonds
for bond1 in doubleBonds:
for bonds in cumulatedBonds:
if bond1 in bonds:
break
else:
cumulatedBonds.append([bond1])
# For each set of adjacent double bonds, check for axis symmetry
for bonds in cumulatedBonds:
# Do nothing if less than two cumulated bonds
if len(bonds) < 1: continue
# Do nothing if axis is in cycle
found = False
for atom1, atom2 in bonds:
if molecule.isBondInCycle(atom1.edges[atom2]): found = True
if found: continue
# Find terminal atoms in axis
# Terminal atoms labelled T: T=C=C=C=T
axis = []
for bond in bonds:
axis.extend(bond)
terminalAtoms = []
for atom in axis:
if axis.count(atom) == 1: terminalAtoms.append(atom)
if len(terminalAtoms) != 2: continue
# Remove axis from (copy of) structure
bondlist = []
for atom1, atom2 in bonds:
bond = atom1.edges[atom2]
bondlist.append(bond)
molecule.removeBond(bond)
structure = molecule.copy(True)
terminalAtoms = [
structure.vertices[molecule.vertices.index(atom)]
for atom in terminalAtoms
]
for bond in bondlist:
molecule.addBond(bond)
atomsToRemove = []
for atom in structure.vertices:
if len(
atom.edges
) == 0 and atom not in terminalAtoms: # it's not bonded to anything
atomsToRemove.append(atom)
for atom in atomsToRemove:
structure.removeAtom(atom)
# Split remaining fragments of structure
end_fragments = structure.split()
#
# there can be two groups at each end A\ /B
# T=C=C=C=T
# A/ \B
# to start with nothing has broken symmetry about the axis
symmetry_broken = False
end_fragments_to_remove = []
for fragment in end_fragments: # a fragment is one end of the axis
# remove the atom that was at the end of the axis and split what's left into groups
terminalAtom = None
for atom in terminalAtoms:
if atom in fragment.atoms:
terminalAtom = atom
fragment.removeAtom(atom)
break
else:
continue
groups = []
if len(fragment.atoms) > 0:
groups = fragment.split()
# If end has only one group then it can't contribute to (nor break) axial symmetry
# Eg. this has no axis symmetry: A-T=C=C=C=T-A
# so we remove this end from the list of interesting end fragments
if len(groups) == 0:
end_fragments_to_remove.append(fragment)
continue # next end fragment
elif len(groups) == 1 and terminalAtom.radicalElectrons == 0:
if terminalAtom.atomType.label == 'N3d':
symmetry_broken = True
else:
end_fragments_to_remove.append(fragment)
continue # next end fragment
elif len(groups) == 1 and terminalAtom.radicalElectrons != 0:
symmetry_broken = True
elif len(groups) == 2:
if not groups[0].isIsomorphic(groups[1]):
# this end has broken the symmetry of the axis
symmetry_broken = True
for fragment in end_fragments_to_remove:
end_fragments.remove(fragment)
# If there are end fragments left that can contribute to symmetry,
# and none of them broke it, then double the symmetry number
# NB>> This assumes coordination number of 4 (eg. Carbon).
# And would be wrong if we had /B
# =C=C=C=C=T-B
# \B
# (for some T with coordination number 5).
if end_fragments and not symmetry_broken:
symmetryNumber *= 2
return symmetryNumber