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Dear Martin,Thank you so much. It works!Can we make it faster?It took 17 seconds for my problem butM1.LLL() took only 3 seconds. Of course I understand weare calculating extra matrix U.
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Dear Nils,Thank you so much for your comments.I consider Matrix E=[I,M1], where I is identity matrix.Then reduction of E took 100 seconds. Hence I am notgoing any advantage.
Regards,Santanu
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On Sunday, September 27, 2020 at 1:48:03 PM UTC-7, Santanu Sarkar wrote:Dear Nils,Thank you so much for your comments.I consider Matrix E=[I,M1], where I is identity matrix.Then reduction of E took 100 seconds. Hence I am notgoing any advantage.Try [10^b*M1,I] with b = 50 or so (it depends on your problem what a "large" number is). I'm not promising that it's faster, but it's worth a try. If you do not scale M1, then I would expect you're getting back different answers, unless all the vectors in the span of M1 already have very large norm.I now also see that you're looking at a very "overgenerated" lattice (span of 90 vectors in ZZ^6?) Such a lattice almost surely has a very small covolume. I'd expect that just a HNF is actually quite fast to compute. The transformation matrix in such a case is also highly non-unique. It's not so surprising that keeping track of it is very much more expensive in this case (although I'm not sure it explains the full factor that you see).I'd expect that something like 8 combinations of the 90 generating vectors already generate the same lattice. Now you're looking at LLL-reducing a 8x6 matrix rather than a 90x6 matrix!
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