4 ms·
With respect to the domain it isn't.
by UncleMeat 5y ago
With respect to the domain it isn't.
- ncmncm 5y agoThe domain is, exactly, lossy.
- UncleMeat 5y agoIf you aren't considering an abstraction with respect to a specific domain then there every single abstraction is lossy, not just the useful ones. You cannot transform state in any way without loss because, given enough effort, all properties of state are observable. This makes "lossy" a worthless descriptor using your approach. Instead, I propose that we consider abstractions with respect to some problem domain. If you have states {null, notnull} then this is lossless but {null, notnull, top, bottom} is lossy.
- ncmncm 5y agoWhere you put your lossiness is a matter of choice and taste.