Re: A problem with range objects and floating point numbers

[s.jo]

Daniel Sebald wrote
> On 09/29/2014 11:36 AM, s.jo wrote:
>> Well, I don't think that way on the printf format:
>>
>> octave:18>  sprintf('%.43g',0.1)
>> ans = 0.1000000000000000055511151231257827021181583
>>
>> Do you think that the above ans is rounded around 12th position?
>> Conversely, the decimation at 12th position is the way how Matlab shows
>> the
>> number in display
>> only when the explicit precision number is not specified such as '%12f'
>> or
>> '%.43g'.
>> Even though Matlab shows that way on display, any raw bits are NOT
>> discarded.
>> You can check it by using num2hex command in matlab.
> 
> I understand what you are looking at.
> 
> 
>> Anyhow, it is fun to check the detail behaviors:
>> Here are the test result up-to test 8:
>>
>> % generate ranges :
>> %  1. range object, 2. matrix object, 3. linspace, 4. perfect decimation
>> N=1000;
>> range_test1=(-N:0.1:N);
>> range_test2=[-N:0.1:N];
>> range_test3=linspace(-N,N,20*N+1);
>> range_test4=(str2num(sprintf('%.12f ',-N:0.1:N)));
>> range_test5=(-N/0.1:1:N/0.1)*0.1;
>> range_test6=[-N/0.1:1:N/0.1]*0.1;
>> % same as range_test6
>> NN=20*N+1;
>> range_test7=zeros(1,NN);
>> for i=(1:1:NN)
>>     range_test7(i)=(i-1-N/0.1)*0.1;
>> endfor
>> % the linear spacing (interpolation) algorithm
>> range_test8=zeros(1,NN); A=-N; B=N;
>> for i=(1:1:NN)
>>     range_test8(i)=(A*(NN-i) + B*(i-1))/(NN-1);
>> endfor
>>
>> + sum (abs (sin (range_test1 * pi))<  eps (range_test1 * pi))
>> ans =       1055
>> + sum (abs (sin (range_test2 * pi))<  eps (range_test2 * pi))
>> ans =       1168
>> + sum (abs (sin (range_test3 * pi))<  eps (range_test3 * pi))
>> ans =       2001
>> + sum (abs (sin (range_test4 * pi))<  eps (range_test4 * pi))
>> ans =       2001
>> + sum (abs (sin (range_test5 * pi))<  eps (range_test5 * pi))
>> ans =       1055
>> + sum (abs (sin (range_test6 * pi))<  eps (range_test6 * pi))
>> ans =       2001
>> + sum (abs (sin (range_test7 * pi))<  eps (range_test7 * pi))
>> ans =       2001
>> + sum (abs (sin (range_test8 * pi))<  eps (range_test8 * pi))
>> ans =       2001
>>
>> The results are as you expected.
> 
> Great!  So this may be the way to go with linspace():
> 
> result(i) = (A*(N-i) + B*i)/N;
> 
> And if so, perhaps the range operator can be changed to utilize the 
> linear-spacing algorithm.  That is
> 
> Step 1) First compute what the final B value is for the range, i.e.,
> 
>    N = floor((limit-base)/increment);
>    B = base + N * increment;
> 
> Step 2) Apply the linear spacing routine, i.e., the equivalent of 
> linspace(base, B, N).
> 
> Now that first step for computing B could be fraught with it's own 
> numerical peculiarity, so would need some thought in that as well.
> 
> Dan

Dan,

It will a good routine for NEW range operator.
* A traditional incremental range operator (say, op1)
* A new interpolated range operator (say, op2)
The incremental range should be supported anyway.

My idea is that colon operator can select a routine between implicitly based
on solid rules:
For example,
R1. base, limit, and increment are all integers --> op1
R2. ((limit-base)/increment) is an integer --> op2
(We can check if floor((limit-base)/increment)==((limit-base)/increment))
R3. Otherwise, --> op1
... and so on.

I think that R2 case is only our concern.

-- jo




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