2015-05-15 23:57 GMT+02:00 Richard Wordingham < [email protected]>:
> On Fri, 15 May 2015 22:09:13 +0200 > Philippe Verdy <[email protected]> wrote: > > > 2015-05-15 9:10 GMT+02:00 Richard Wordingham < > > [email protected]>: > > > This is because you don't understand the issue ! > > > > Now, a program to check whether a trace matching > > > {\u0323|\u0302)* matches (\u0323\u0302)* is very simple. It just > > > counts the number of times \u0323 occurs and the number of times > > > \u0302 occurs, and returns whether they are equal. > > > This is wrong. \0323\0323\0302\0302 and \0323\0302\0323\0302 would > > pass your counting test (which does not work in a FSA) but they are > > NOT canonically equivalent because the identical combining characters > > are blocking each other (so arbitrary ordering is not possible). > > TUS7.0: D108 Reorderable pair: > Two adjacent characters A and B in a coded character sequence > <A, B> are a Reorderable Pair if and only if ccc(A) > ccc(B) > 0. > > Now, ccc(U+0302) = 230 > 220 = ccc(U+0323) > 0, so (U+0302, U+0303) is > a reorderable pair. > I do NOT contest that U+0323 and U+0302 can reorder, but the fact that U+0323 blocks another occurence of U+0323 because it has the **same** combining class.
