4 ms·

Why interval arithmetic is so useful (2007) [pdf]

29 points by maho 11y ago

paperwork 11y ago
Could such intervals be represented by random variables of uniform distributions?
- grayclhn 11y ago
  No, the rules for adding and subtracting (etc...) random variables are pretty different than what's at the link.
- Someone 11y ago
  No. You cannot do that because adding/subtracting/multiplying/... two uniformly distributed independent variables does not produce a uniformly distributed variable (see for example http://en.m.wikipedia.org/wiki/Irwin–Hall_distribution http://en.m.wikipedia.org/wiki/Irwin–Hall_distribution) Thats true for other distributions, too. You could use the obvious choice of a normal distribution, if all you did is add and subtract, but that's (about?) as good as it gets.
imh 11y ago
I think this kind of thing is great, but it runs into some of the same considerations that make probabilistic graphical models hard. If I have two variables defined as x = [0,1]; y = [0,1]; z = x - y; Then we see that z should be the interval [-1,1]. But for a similar problem, a = [0,1]; b = a - a; x = a; y = a; z = x - y; b should clearly be the interval [0,0]. z is computed the same way as the first example, as the difference between two [0,1] intervals. But this time we want it to be [0,0]. That means we want to keep around some notion similar to the joint distribution of x and y. That makes arithmetic immensely harder than simple rules. Not to say it's not worth it, just that it's hard.
- anon4 11y ago
  You get the same behaviour with regular floats: >>> a=10.0**10 >>> a 10000000000.0 >>> b = 0.1 >>> b+a 10000000000.1 >>> c=b+a >>> d=c-a >>> d 0.10000038146972656 >>> d == b False >>> a + b - b == b False >>> You could build a variable-tracking system on top of the numbers, or you could just use a symbolic computation package, then simplify the final equations so each variable figures only once.

29 points by maho 11y ago

paperwork 11y ago

Could such intervals be represented by random variables of uniform distributions?

grayclhn 11y ago
No, the rules for adding and subtracting (etc...) random variables are pretty different than what's at the link.
Someone 11y ago
No. You cannot do that because adding/subtracting/multiplying/... two uniformly distributed independent variables does not produce a uniformly distributed variable (see for example http://en.m.wikipedia.org/wiki/Irwin–Hall_distribution http://en.m.wikipedia.org/wiki/Irwin–Hall_distribution) Thats true for other distributions, too. You could use the obvious choice of a normal distribution, if all you did is add and subtract, but that's (about?) as good as it gets.

imh 11y ago

I think this kind of thing is great, but it runs into some of the same considerations that make probabilistic graphical models hard. If I have two variables defined as x = [0,1]; y = [0,1]; z = x - y; Then we see that z should be the interval [-1,1]. But for a similar problem, a = [0,1]; b = a - a; x = a; y = a; z = x - y; b should clearly be the interval [0,0]. z is computed the same way as the first example, as the difference between two [0,1] intervals. But this time we want it to be [0,0]. That means we want to keep around some notion similar to the joint distribution of x and y. That makes arithmetic immensely harder than simple rules. Not to say it's not worth it, just that it's hard.

anon4 11y ago
You get the same behaviour with regular floats: >>> a=10.0**10 >>> a 10000000000.0 >>> b = 0.1 >>> b+a 10000000000.1 >>> c=b+a >>> d=c-a >>> d 0.10000038146972656 >>> d == b False >>> a + b - b == b False >>> You could build a variable-tracking system on top of the numbers, or you could just use a symbolic computation package, then simplify the final equations so each variable figures only once.