I try to perform this simple calculation, but the programm always gives me incorrect results. If I use only integers, then I get 96
var=$(((100/24)*24))
even if I try to use bc:
var=$(bc <<< "scale=100;100/24*24")
I get something like this:
99.99999999999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999984;
How to let make the program count the correct way? So that I could get the same values as from the calculator?
This is the result of rounding, to integers (100/24 -> 4 in bash) and floating point (bc lacks a "round" function). See this SO question for related information and links.
It's very hard to provide a general solution here: Most solutions depend on your exact use case, which is probably not what you posted in your question (otherwise you could just put var=100 and move on).
One technique is to reorder the problem (var=$((24*100/24))), which would keep all operations integers.
Another is to multiply by a large offset (var_millionths=$((1000000*(100/24)*24))) and divide out as needed, which would keep enough precision that Bash could round the result well.
Finally, you could (ab)use printf to perform the rounding step for you, because you want the number to be round in base 10 (i.e. when printed):
printf "%.0f\n" $(BC_LINE_LENGTH=0 bc <<< "scale=100;100/24*24")
Related
Let us say I have a program called foo. It takes 1 argument (e.g. foo 42) and it spits out a single line containing a numerical value (e.g. 2034).
I want to find the optimal value for the input argument, so the output is minimized. Currently I do that by hand by (semi-)binary search, but that is slow - especially if foo is slow.
Is there an automated tool to find the optimal value for the input?
We assume the values are continuous, and it is OK to find a local optimum, so something that uses gradient descent would be ideal (especially if it can take n inputs).
As an example you can use this for foo:
perl -e 'print (1+(shift() - 10203040506)**2)'
In the example I want the tool to return 10203040506 since it gives the smallest value.
First version is now available: https://gitlab.com/ole.tange/tangetools/-/tree/master/find-optimal
It uses Nelder-Mead and it only works well on floats. It does very bad on integer values.
If you know of a "Nelder-Mead" for integers, let me know - especially if there is sample Python code for it.
What is the typical approach in LUA (before the introduction of integers in 5.3) for dealing with calculated range values in for loops? Mathematical calculations on the start and end values in a numerical for loop put the code at risk of bugs, possibly nasty latent ones as this will only occur on certain values and/or with changes to calculation ordering. Here's a concocted example of a loop not producing the desire output:
a={"a","b","c","d","e"}
maybethree = 3
maybethree = maybethree / 94
maybethree = maybethree * 94
for i = 1,maybethree do print(a[i]) end
This produces the unforuntate output of two items rather than the desired three (tested on 5.1.4 on 64bit x86):
a
b
Programmers unfamiliar with this territory might be further confused by print() output as that prints 3!
The application of a rounding function to the nearest whole number could work here. I understand the approximatation with FP and why this fails, I'm interested in what the typical style/solution is for this in LUA.
Related questions:
Lua for loop does not do all iterations
Lua: converting from float to int
The solution is to avoid this reliance on floating-point math where floating-point precision may become an issue. Or, more realistically, just be aware of when you are using FP and be mindul of the precision issue. This isn’t a Lua problem that requires a Lua-specific solution.
maybethree is a misnomer: it is never three. Your code above is deterministic. It will always print just a and b. Since the maybethree variable is less than three, of course the for loop would not execute 3 times.
The print function is also behaving as defined/expected. Use string.format to show thr FP number in all its glory:
print(string.format("%1.16f", maybethree)) -- 2.9999999999999996
Still need to use calculated values to control your for loop? Then you already mentioned the answer: implement a rounding function.
I am trying to compare two vectors.
v1 = {0.520974 , 0.438171 , 0.559061}
v2 = [0.520974 , 0.438171 , 0.559061}
I write v1 to a file, read and that's v2. For some reason when I compare the two vectors, I am getting false!
When I do: v1[0]-v2[0] I get 4.3123e-8
Thanks,
Double values, unlike integers, are fragile against write and read. That means, the information that represents them in a string is not necessarily complete.
The leading reason of that is rounding: it's like if you had 1/7 and wanted to write it on a paper in the same format as in your question, you'd get:
0.142857
That's exact to 6 decimal places, but no more than that, and the difference shows up. The only difference in the computer is that it counts in binary (and rounds in binary, too), and is further complicated by the fact that at output (or input) you coerce that into decimal (or back, respectively) and round again at each step. All of those are sources of little errors.
If you want to be able to save and reload your doubles exactly (on the same machine), do it in their native binary representation using a write and read. If you want them to be human-readable, you need to sacrifice the exact reconstruction. You'd then need to compare them up to a little allowed deviation.
In Mma, for example, I want to calculate
1.0492843824838929890231*0.2323432432432432^3
But it does not show the full precision. I tried N or various other functions but none seemed to work. How to achieve this? Many thanks.
When you specify numbers using decimal point, it takes them to have MachinePrecision, roughly 16 digits, hence the results typically have less than 16 meaningful digits. You can do infinite precision by using rational/algebraic numbers. If you want finite precision that's better than default, specify your numbers like this
123.23`100
This makes Mathematica interpret the number as having 100 digits of precision. So you can do
ans=1.0492843824838929890231`100*0.2323432432432432`100^3
Check precision of the final answer using Precision
Precision[ans]
Check tutorial/ArbitraryPrecisionNumbers for more details
You may do:
r[x_]:=Rationalize[x,0];
n = r#1.0492843824838929890231 (r#0.2323432432432432)^3
Out:
228598965838025665886943284771018147212124/17369643723462006556253010609136949809542531
And now, for example
N[n,100]
0.01316083216659453615093767083090600540780118249299143245357391544869\
928014026433963352910151464006549
Sometimes you just want to see more of the machine precision result. These are a few methods.
(1) Put the cursor at the end of the output line, and press Enter (not on the numeric keypad) to copy the output to a new input line, showing all digits.
(2) Use InputForm as in InputForm[1.0/7]
(3) Change the setting of PrintPrecision using the Options Inspector.
I am interested in use or created an script to get error rounding reports in algorithms.
I hope the script or something similar is already done...
I think this would be usefull for digital electronic system design because sometimes it´s neccesary to study how would be the accuracy error depending of the number of decimal places that are considered in the design.
This script would work with 3 elements, the algorithm code, the input, and the output.
This script would show the error line by line of the algorithm code.
It would modify the algorith code with some command like roundn and compare the error of the output.
I would define the error as
Errorrounding = Output(without rounding) - Output round
For instance I have the next algorithm
calculation1 = input*constan1 + constan2 %line 1 of the algorithm
output = exp(calculation1) %line 2 of the algorithm
Where 'input' is the input of n elements vector and 'output' is the output and 'constan1' and 'constan2' are constants.
n is the number of elements of the input vector
So, I would put my algorithm in the script and it generated in a automatic way the next algorithm:
input_round = roundn(input,-1*mdec)
calculation1 = input*constant1+constant2*ones(1,n)
calculation1_round = roundn(calculation1,-1*mdec)
output=exp(calculation1_round)
output_round= roundn(output,-1*mdec)
where mdec is the number of decimal places to consider.
Finally the script give the next message
The rounding error at line 1 is #Errorrounding_calculation1
Where '#Errorrounding' would be the result of the next operation Errorrounding_calculation1 = calculation1 - calculation1_round
The rounding error at line 2 is #Errorrounding_output
Where 'Errorrounding_output' would be the result of the next operation Errorrounding_output = output - output_round
Does anyone know if there is something similar already done, or Matlab provides a solution to deal with some issues related?
Thank you.
First point: I suggest reading What Every Computer Scientist Should Know About Floating-Point Arithmetic by David Goldberg. It should illuminate a lot of issues regarding floating-point computations that will help you understand more of the intricacies of the problem you are considering.
Second point: I think the problem you are considering is a lot more complicated than you realize. You are interested in the error introduced into a calculation due to the reduced precision from rounding. What you don't realize is that these errors will propagate through your computations. Consider your example:
output = input*C1 + C2
If each of the three operands is a double-precision floating-point number, they will each have some round-off error in their precision. A bound on this round-off error can be found using the function EPS, which tells you the distance from one double-precision number to the next largest one. For example, a bound on the relative error of the representation of input will be 0.5*eps(input), or halfway between it and the next largest double-precision number. We can therefore estimate some errors bounds on the three operands as follows:
err_input = 0.5.*eps(input); %# Maximum round-off error for input
err_C1 = 0.5.*eps(C1); %# Maximum round-off error for C1
err_C2 = 0.5.*eps(C2); %# Maximum round-off error for C2
Note that these errors could be positive or negative, since the true number may have been rounded up or down to represent it as a double-precision value. Now, notice what happens when we estimate the true value of the operands before they were rounded-off by adding these errors to them, then perform the calculation for output:
output = (input+err_input)*(C1+err_C1) + C2+err_C2
%# ...and after reordering terms
output = input*C1 + C2 + err_input*C1 + err_C1*input + err_input*err_C1 + err_C2
%# ^-----------^ ^-----------------------------------------------------^
%# | |
%# rounded computation difference
You can see from this that the precision round-off of the three operands before performing the calculation could change the output we get by as much as difference. In addition, there will be another source of round-off error when the value output is rounded off to represent it as a double-precision value.
So, you can see how it's quite a bit more complicated than you thought to adequately estimate the errors introduced by precision round-off.
This is more of an extended comment than an answer:
I'm voting to close this on the grounds that it isn't a well-formed question. It sort of expresses a hope or wish that there exists some type of program which would be interesting or useful to you. I suggest that you revise the question to, well, to be a question.
You propose to write a Matlab program to analyse the numerical errors in other Matlab programs. I would not use Matlab for this. I'd probably use Mathematica, which offers more sophisticated structural operations on strings (such as program source text), symbolic computation, and arbitrary precision arithmetic. One of the limitations of Matlab for what you propose is that Matlab, like all other computer implementations of real arithmetic, suffers rounding errors. There are other languages which you might choose too.
What you propose is quite difficult, and would probably require a longer answer than most SOers, including this one, would be happy to contemplate writing. Happily for you, other people have written books on the subject, I suggest you start with this one by NJ Higham. You might also want to investigate matters such as interval arithmetic.
Good luck.