Reconsider equivalence between type.broadcastable and type.shape

[ricardoV94]

Right now TensorType("float64", shape=(None, 5, 1)).broadcastable == (False, False, True)

However, due to enabling dynamic broadcasting, this is not correct as that type can broadcast along the first dimension when that has a size of 1 at runtime. In many rewrites we check for conditions like var1.dtype.broadcastable == var2.dtype.broadcastable, but we shouldn't assume they are the same when they come from None type shape dimensions.

One option would be to change to the following: TensorType("float64", shape=(None, 5, 1)).broadcastable == (np.nan, False, True), or any other type that respects (self == self) is False as np.nan does.

Related to #1089 and #1122

[brandonwillard]

Don't forget that we're trying to phase out use of TensorType.broadcastable altogether. The word "broadcastable" with its obvious interpretation is not correct (and never was) in all cases, and that was a big part of why we wanted to replace it with a quantity that accurately represents the relevant information: i.e. TensorType.shape.

The only reason TensorType.broadcastable is still relevant is due to us not having all the old code refactored yet.

[ricardoV94]

I know that, but many old rewrites relied on the older interpretation of broadcastable which is not the same as the new one.