If I add (1/2, 1) to (1, 2), then I should get (3/2,3), which is not representable with any of the above values.
So what is the result and in what sense is it correct?
As I understand it, it's just "floating" interval arithmetic, so like in interval arithmetic you should understand a representation [a,b] as 'The number is contained in the interval [a,b]'. In other words, interval arithmetic with carefully assigned floating point limits such those don't violate the interval arithmetic and are concise.
Not sure how it would go, I think it depends if you maintain the precision or increase it after this operation. I guess 2 bit: (1/2,1)+(1,2)=(1,inf) (?) and 4 bit (2 exp and 2 mantissa?): (1/2,1)+(1,2)=(3/2,3). Maybe there's a built in check to compare how short your interval can get and stop at a reasonable precision (in this case there's no point going for more than 4 bits).
Honestly I find it quite elegant and is at least trying to solve a big issue with bandwidth limitations. I do wish the exposition was more clear/straightforward.
Ah, that makes sense. The 4-bit representations do not suffice by themselves, but imply the support of some or all of the smaller representations as well, and I guess you always need a representation for [-inf,inf].
Not sure how it would go, I think it depends if you maintain the precision or increase it after this operation. I guess 2 bit: (1/2,1)+(1,2)=(1,inf) (?) and 4 bit (2 exp and 2 mantissa?): (1/2,1)+(1,2)=(3/2,3). Maybe there's a built in check to compare how short your interval can get and stop at a reasonable precision (in this case there's no point going for more than 4 bits).
Honestly I find it quite elegant and is at least trying to solve a big issue with bandwidth limitations. I do wish the exposition was more clear/straightforward.