long integers in matlab
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Hi all,
how can we do (simple) operations in Matlab with integers that exceed intmax("uint64")?
I am aware that in Python this is a posibility and there's virtually no limit for int size. In Python 3:
>> a = 10**100;
>> b = 10**100 + 1;
>> b-a
1
Is this possible in MATLAB as well? I am just curious so I can understand Matlab a bit better. I have no use case in mind
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Accepted Answer
Stephen23
on 14 Jul 2022
Edited: Stephen23
on 14 Jul 2022
"what would be the recommended approach if I don't have access to Symbolic Math TB?"
You can download VPI by clicking on the big blue "DOWNLOAD" button in the top right corner:
unzipping it onto your MATLAB Search Path (e.g into the current directory), and then following its instructions.
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More Answers (2)
Dyuman Joshi
on 14 Jul 2022
You can use Java.
import java.math.*
y=BigInteger('10').pow(100)
z=BigInteger('10').pow(100).add(BigInteger('1'))
sub=z.subtract(y)
1 Comment
John D'Errico
on 14 Jul 2022
Edited: John D'Errico
on 14 Jul 2022
I always hated the java.math.BigInteger interface. Way too much of a kludge to just do simple arithmetic operations. So I wrote a version of VPI some time ago (called VPIJ, of course) that uses the Java tools.
The virtue of the Java tools is they are a bit faster than syms. Ok, a heck of a LOT faster. And VPIJ also screams compared to VPI, since I had to write all of VPI in MATLAB itself.
P = vpij(2)^521 - 1
P =
6864797660130609714981900799081393217269435300143305409394463459185543183397656052122559640661454554977296311391480858037121987999716643812574028291115057151
Q = sym(2)^521 - 1;
isprime(P)
ans =
logical
1
isprime(Q)
ans =
logical
1
timeit(@() isprime(P))
ans =
0.0028533
tic,isprime(Q),toc
ans =
logical
1
Elapsed time is 5.427998 seconds.
Note that I did not bother using timeit to check the time for the sym test, since timeit uses MULTIPLE tests. Same number, but often orders of magnitude difference in speed.
Sadly, Java seems to be slowly going away in MATLAB.
Steven Lord
on 14 Jul 2022
See Symbolic Math Toolbox.
ten = sym(10);
a = ten^100;
b = ten^100+1;
b-a
Note that performing the calculation then converting to a symbolic result won't work if the numbers get large enough. sym is perfectly happy to convert Inf to a symbolic value.
c1 = sym(10^500) % Already overflowed before sym was called
c2 = ten^500
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