Yes, the test is very close to the edge:
T=: 2^_34
2 =!.T 2+T+T
1
2 =!.T 2+T+T+2^_50
0
Henry Rich
On 2/19/2014 8:33 PM, Roger Hui wrote:
You have to consider the definition of tolerant equality carefully. It is
the _relative_ difference that matters, not the absolute difference. The
absolute difference between 2 and 2+T+T is 2*T, but the relative difference
is (2*T)%2+T+T, and this relative difference is less than T.
On Wed, Feb 19, 2014 at 5:24 PM, Ian Clark <earthspo...@gmail.com> wrote:
Okay, I was reading that as *less* than 2^_34
All is not as it seems however...
] T=: 2^_34
5.82077e_11
2 =!.T 2+T
1
2 =!.T 2+T+T NB. I'd expect 0 here
1
2 =!.T 2+T+T+T
0
NB. at last!
On Thu, Feb 20, 2014 at 12:37 AM, Henry Rich <henryhr...@nc.rr.com> wrote:
Comparison tolerance cannot be more than 2^_34.
(So that integer comparisons on 32-bit machines do not have to worry
about
tolerance)
Henry Rich
On 2/19/2014 7:03 PM, Ian Clark wrote:
I'm trying to construct an example using Fit (!.) to customize a given
primitive (e.g. =) to set its tolerance.
Try as I might I cannot get variants of this to work:
T=: 0.001
2 =!.T 2.0001
|domain error
| 2 =!.T 2.0001
What am I doing wrong?
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