Mischievous SMARTS Queries

Last year I extended the CDK SMARTS implementation to match component groupings and stereochemistry. Specifying stereochemistry presents some interesting logical predicate that might be tricky to handle.

Here are some examples that I came up with for testing the correctness of query handling. They start simple before getting a little mischievous. First, recursion and component grouping.

query	targets	nmatch	Comment
Component grouping (fragment)
(O).(O)	O=O	0	Example from Daylight
	OCCO	0
	O.CCO	2
Component grouping (connected)
(O.O)	O=O	2	Example from Daylight
	OCCO	2
	O.CCO	0
Recursion, ad infinitum
[$(CC[$(CCO),$(CCN)])]	CCCCO	1
	CCCCN	1
	CCCCC	0
Recursive component grouping
[O;D1;$(([a,A]).([A,a]))][CH]=O	OC=O.c1ccccc1	1	Feature/Bug #1312
	OC=O	0

These next ones are concerned with logic and stereochemistry.

query	targets	nmatch	Comment
Ensure local stereo matching
*[@](*)(*)(*)	O[C@](N)(C)CC	12	tetrahedrons have 12 rotation symmetries
	O[C@@](N)(C)CC	12
	O[C](N)(C)CC	0
Implicit (hydrogen or lone-pair) neighbour
CC[S@](C)=O	CC[S@](C)=O	1
	CC[S@@](C)=O	0
	CC[S](C)=O	0
Either (tetrahedral)
CC[@,@@](C)O	CC[C@H](C)O	1
	CC[C@@H](C)O	1
	CCC(C)O	0
Both (tetrahedral)
CC[@&@@](C)O	CC[C@H](C)O	0
	CC[C@@H](C)O	0
	CCC(C)O	0
Respect logical precedence 1
CC[@,Si@@](C)O	CC[C@H](C)O	1
	CC[C@@H](C)O	0
	CCC(C)=O	0
Respect logical precedence 2
CC[C@,Si@@](C)O	CC[C@H](C)O	1
	CC[C@@H](C)O	0
	CCC(C)O	0
	CC[Si@H](C)O	0
	CC[Si@@H](C)O	1
	CC[Si](C)O	0
Unspecified
CC[@@?](C)O	CC[C@H](C)O	0
	CC[C@@H](C)O	1
	CCC(C)O	1
Negation
CC[!@](C)O	CC[C@H](C)O	0	!@@ is also equivalent to @?
	CC[C@@H](C)O	1
	CCC(C)O	1
Neither (tetrahedral) using 'or unspecified'
CC[@?@@?](C)O	CC[C@H](C)O	0
	CC[C@@H](C)O	0
	CCC(C)O	1
Neither (tetrahedral) using negation
CC[!@!@@](C)O	CC[C@H](C)O	0
	CC[C@@H](C)O	0
	CCC(C)O	1
Either (geomeric)
C/C=C/,\C	C/C=C/C	1
	C/C=C\C	1
	CC=CC	0
Neither (geomeric)
C/C=C!/!\C	C/C=C/C	0
	C/C=C\C	0
	CC=CC	1

The last two are quite tricky (and not currently implemented) but once the atom-centric handling is correct it's a simple reduction. It's quite fun to work out so i'll leaf that up to the reader.