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423         // OPTIMIZATION: if there are three or more identical Ts, then
510 int more = 1;
511 while (fTs.end() - &span[more - 1] > 1 && AlmostEqualUlps(span[more].fPt, span->fPt)) {
512 if (span[more - 1].fDone) {
515 double tEndInterval = span[more].fT - newT;
517 if ((span->fTiny = span[more].fTiny)) {
530 span[more - 1].fSmall = true;
531 bool tiny = span[more].fPt == span->fPt;
532 span[more - 1].fTiny = tiny;
533 span[more - 1].fDone = true;
534 if (approximately_negative(span[more].fT - newT) && tiny) {
535 if (approximately_greater_than_one(span[more].fT)) {
536 span[more + 1].fUnsortableStart = true;
537 span[more].fUnsortableEnd = true;
540 span[more].fUnsortableStart = true;
541 span[more - 1].fUnsortableEnd = true;
545 ++more;
560 // have three or more coincident edges. Shouldn't this subtract the difference
1319 // OPTIMIZATION: this may fix indices more than once. Build an array of unique segments to
1522 // more than one viable candidate -- measure angles to find best
1653 // more than one viable candidate -- measure angles to find best
1875 // and use more concise logic like the old edge walker code?
1903 // if the topmost T is not on end, or is three-way or more, find left
2254 // wastes time, it shouldn't do any more than spin through the T spans.
2515 // return span if when chasing, two or more radiating spans are not done
2519 // or this code could be more complicated in detecting this case. Worth it?