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At least for this research, Sage may have been better. This code follows the random matrix example to show that Sage gives at least consistent results in one o
by jasongrout 12y ago
At least for this research, Sage may have been better. This code follows the random matrix example to show that Sage gives at least consistent results in one of the cases they pointed out.
Here's the public SageMathCloud worksheet: https://cloud.sagemath.com/projects/49a2531d-9d02-42c9-9db6-f9551fbfa59e/files/2014-10-24-212837.sagews https://cloud.sagemath.com/projects/49a2531d-9d02-42c9-9db6-...
I wish they would have supplied those matrices they printed out in machine-readable format so we could easily put those in Sage too.
- williamstein 12y agoHere's another worksheet with some symbolic integrals from the article: https://cloud.sagemath.com/projects/4a5f0542-5873-4eed-a85c-a18c706e8bcd/files/support/ams-article-mathematica-bugs.sagews https://cloud.sagemath.com/projects/4a5f0542-5873-4eed-a85c-... I wrote the det code in Sage that you're calling there (though in Sage-6.4 it'll likely be replaced by a call to FLINT). It computes det(A) in a very interesting way, which is asymptotically massively faster than Mathematica. To compute det(A), choose a random vector v and solve Ax = v using a p-adic lifting algorithm (the one in https://cs.uwaterloo.ca/~astorjoh/iml.html https://cs.uwaterloo.ca/~astorjoh/iml.html). One can prove --using Cramer's rule--that the lcm of the denominators of the entries of x will then be a divisor d of det(A), and with high probability one expects that det(A)/d is a tiny integer. One can then provably (using the Hadamard bound) find det(A) by working modulo a few additional primes and using the Chinese Remainder theorem.