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Topic review - representation obtained by reduction? |
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greuel wrote: Hi, I guess 'division' is what you want. Thanks. Unfortunately, B is not necessarily a standard basis, and division appears to compute a standard basis of B before doing the division. For example, Code: > ring R=0,(x,y),dp; > ideal l = xy+1,y2+1; > poly f = x-y; > division(f,l); [1]: _[1,1]=-y _[2,1]=x [2]: _[1]=0 [3]: _[1,1]=1
The answer I want is Code: [1]: _[1,1]=0 _[2,1]=0 [2]: _[1]=x-y [3]: _[1,1]=1
If I read the documentation correctly, division computes a standard basis, then lifts to l; I want the result from the non-standard basis. Is there an option to turn that behavior off?
[quote="greuel"]Hi, I guess 'division' is what you want.[/quote] Thanks. Unfortunately, B is not necessarily a standard basis, and division appears to compute a standard basis of B before doing the division. For example,[code]> ring R=0,(x,y),dp; > ideal l = xy+1,y2+1; > poly f = x-y; > division(f,l); [1]: _[1,1]=-y _[2,1]=x [2]: _[1]=0 [3]: _[1,1]=1 [/code]The answer I [i]want[/i] is[code][1]: _[1,1]=0 _[2,1]=0 [2]: _[1]=x-y [3]: _[1,1]=1 [/code]If I read the documentation correctly, division computes a standard basis, then lifts to l; I want the result from the non-standard basis. Is there an option to turn that behavior off?
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Posted: Sat Nov 15, 2008 5:59 am |
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Hi, I guess 'division' is what you want.
Hi, I guess 'division' is what you want.
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Posted: Sat Nov 15, 2008 3:12 am |
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Lift yes, but... |
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bricken wrote: Thanks, that helps a great deal. I don't know that it's as efficient as I'd like, maybe you can say. To use lift to "recover" the H's, I understand that I need to type Code: r = reduce(f,B); lift(B,f-r);
When Singular performs the reduction, it computes H once. Then, when Singular computes the lift, it computes H a second time. I'd prefer to compute H only the first time, and recover it without having Singular go through it again.
[quote="bricken"]Hi John! Lift might do the job. http://www.singular.uni-kl.de/Manual/3-0-4/sing_235.htm Michael[/quote]Thanks, that helps a great deal. I don't know that it's as efficient as I'd like, maybe you can say. To use lift to "recover" the H's, I understand that I need to type[code]r = reduce(f,B); lift(B,f-r); [/code]When Singular performs the reduction, it computes H once.
Then, when Singular computes the lift, it computes H a second time.
I'd prefer to compute H only the first time, and recover it without having Singular go through it again.
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Posted: Thu Oct 30, 2008 11:13 pm |
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Hi John! Lift might do the job. http://www.singular.uni-kl.de/Manual/3-0-4/sing_235.htm Michael
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Posted: Thu Oct 30, 2008 9:52 am |
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representation obtained by reduction? |
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Suppose that B is an ideal, f is a polynomial, and r is obtained via r = reduce(f,B); Then there exists a list H of polynomials such that f == H1*B1 + ... + Hm*Bm + r
Does Singular have a function/option to determine H, after computing r?
Suppose that B is an ideal, f is a polynomial, and r is obtained via r = reduce(f,B); Then there exists a list H of polynomials such that f == H1*B1 + ... + Hm*Bm + r
Does Singular have a function/option to determine H, after computing r?
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Posted: Fri Oct 17, 2008 11:34 pm |
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