@@ -1340,24 +1340,32 @@ def contains_branch_seperately(self, other_transform):
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def __sub__ (self , other ):
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"""
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- Return a transform stack which goes all the way down self's transform
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- stack, and then ascends back up other's stack. If it can, this is
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- optimised::
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-
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- # normally
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- A - B == a + b.inverted()
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-
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- # sometimes, when A contains the tree B there is no need to
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- # descend all the way down to the base of A (via B), instead we
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- # can just stop at B.
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-
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- (A + B) - (B)^-1 == A
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-
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- # similarly, when B contains tree A, we can avoid descending A at
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- # all, basically:
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- A - (A + B) == ((B + A) - A).inverted() or B^-1
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-
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- For clarity, the result of ``(A + B) - B + B == (A + B)``.
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+ Compose *self* with the inverse of *other*, cancelling identical terms
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+ if any::
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+
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+ # In general:
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+ A - B == A + B.inverted()
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+ # (but see note regarding frozen transforms below).
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+
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+ # If A "ends with" B (i.e. A == A' + B for some A') we can cancel
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+ # out B:
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+ (A' + B) - B == A'
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+
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+ # Likewise, if B "starts with" A (B = A + B'), we can cancel out A:
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+ A - (A + B') == B'.inverted() == B'^-1
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+
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+ Cancellation (rather than naively returning ``A + B.inverted()``) is
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+ important for multiple reasons:
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+
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+ - It avoids floating-point inaccuracies when computing the inverse of
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+ B: ``B - B`` is guaranteed to cancel out exactly (resulting in the
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+ identity transform), whereas ``B + B.inverted()`` may differ by a
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+ small epsilon.
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+ - ``B.inverted()`` always returns a frozen transform: if one computes
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+ ``A + B + B.inverted()`` and later mutates ``B``, then
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+ ``B.inverted()`` won't be updated and the last two terms won't cancel
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+ out anymore; on the other hand, ``A + B - B`` will always be equal to
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+ ``A`` even if ``B`` is mutated.
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"""
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# we only know how to do this operation if other is a Transform.
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if not isinstance (other , Transform ):
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