Yes, well, there is a question about which test cases you need to verify
that you will always get the correct results at runtime. Will any
variable go out of the bounds defined by comparison tolerance and the
precision of the datatypes anywhere in the calculations? Each verb doing
calculations should only be defined for arguments within these bounds?
This would mean J and most code written in J needs to be rewritten? Or
you have to accept that your programs sometimes give incorrect results?
/Erling
Den 2017-09-08 kl. 19:50, skrev Jimmy Gauvin:
Hi,
I was curious how J's cousin, APL, deals with this.
So I used Dyalog APL (Version: 15.0.27982.0 64 Unicode) and got a
surprising result:
⍪(¯4 ¯3 ¯2 ¯1 0 1 2 3 4+14*2) | 5729082486784839
0
0
0
0
0
104
0
113
0
Changing from IEE-754 64-bit to IEE-754-2008 128-bit floating point
representation corrects the problem
⎕FR
645
⎕FR←1287
⍪(¯4 ¯3 ¯2 ¯1 0 1 2 3 4+14*2) | 5729082486784839
135
146
23
39
147
104
171
113
39
Jimmy
PS
I guess a manual for beginners should have some sort of warning
paragraph/blurb about the perils of numerical computations (and the good
habits of verifying our results just like we were taught in elementary
school).
PPS for history buffs
Back in the older days, Burroughs computer had modulo-3 checking of
arithmetic operations
http://bitsavers.trailing-edge.com/pdf/burroughs/BSP/
BSP_Floating_Point_Processor.pdf
http://www.acsel-lab.com/arithmetic/arith4/papers/ARITH4_Gajski.pdf
On Fri, Sep 8, 2017 at 9:21 AM, Erling Hellenäs <erl...@erlinghellenas.se>
wrote:
Hi all !
It seems an IEEE double can hold this value without lost precision. I
don't know what might happen in the calculations. Where the precision is
lost. Maybe something with comparison tolerance.
5729082486784839 - 5729082486784839 - 0.999
1
JWithATwist does not lose precision. Here with integer and float right
argument.
{ ( 14 ^ 2 ) {! ] - [ * |<. ] % [ } 1.0 * 5729082486784839 }
147
{ ( 14 ^ 2 ) {! ] - [ * |<. ] % [ } 5729082486784839 }
147
Cheers,
Erling Hellenäs
Den 2017-09-08 kl. 13:19, skrev Erling Hellenäs:
Case 3 seems to be working in the latest Beta(8.06.03). The problem seems
to be solved. /Erling
3!:0 (x: n^2)
128
(x: n^2) | 5729082486784839
147
3!:0 (x: n^2)
128
Den 2017-09-07 kl. 20:40, skrev Erling Hellenäs:
Hi all !
Case 1:
3!:0 [ (n^2)
8
(n^2) | 5729082486784839
0
It is non-intuitive that an integer raised to an integer is a float, but
I think it is normal. It would be possible with a performance penalty to
get an integer result. One problem with that is that it would easily
overflow. It would also be possible to have a special operation for this
case.
When the left argument is a float the right argument has to be converted
to a float. It must be assumed that this conversion is intentional, even
though it is implicit.
Case 2:
3!:0 [ (n^2)
8
(n^2) | 5729082486784839x
0
3!:0 (n^2) | 5729082486784839x
8
Here the rational seems to be converted to a float and the result is
float. Shouldn't we have an error instead of converting rationals to float?
Case 3:
3!:0 (x: n^2)
128
(x: n^2) | 5729082486784839
0
3!:0 (x: n^2) | 5729082486784839
128
I have a hard time understanding what happens here. This result seems
very peculiar. Is the left and right argument converted to float and the
float result converted to rational?!
Shouldn't we have an error instead of converting rationals to float?
We could not have floats auto-converted to rationals?
In this case the integer should be converted to rational and we should
get a rational result?
It is non-intuitive that (*: n) does not give the same result as (n^2).
Maybe once this was decided because of performance reasons.
Cheers,
Erling Hellenäs
On 2017-09-05 18:41, Rob B wrote:
Could someone explain this please?
n=.14
n
14
(*: n) | 5729082486784839
147
196 | 5729082486784839
147
(n^2) | 5729082486784839
0
(n^2) | 5729082486784839x
0
(x: n^2) | 5729082486784839
0
(x: n^2) | 5729082486784839x
147
Regards, Rob Burns
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