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last edited 9 years ago by test1 |
Edit detail for SandBox Shortcoming revision 1 of 2
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Time: 2007/11/18 18:35:33 GMT-8 |
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changed: - Please try the two solve statements at the bottom The first one runs in a short time but the second runs a long time. The only differenct is '*R*P' in eq1. Examining eq1a shows that in fact p=1 has a solution so there should be at least one solution to eq1. More generally the grobner should always converge; and this change seems minor. This a simplification from a more extended problem. The commented equations are a little more representative:: -- eq1:=((-2*R*Vr^3+R*Vr^2)*Vt+2*R*Vr^3-R*Vr^2)*aVr*p-Vt -- eq2:=-2*R*Vr*Vt^3+5*R*Vr*Vt^2-4*R*Vr*Vt+R*Vr*aVt*p -- eq3:=(R*Vr*Vt+R*Vr)*e+((R-1)*Vr+1)*Vt-R*Vr -- eq4:=-r^2+(Vt^2-Vt+1/4)*aVt+(Vr^2-Vr*R+1/4)*aVr Raymond E. Rogers \begin{axiom} list:=[p,Vr,Vt,e] eq1a:=((-Vr^3+Vr^2)*Vt+Vr^3-Vr^2) eq1:=((-Vr^3+Vr^2)*Vt+Vr^3-Vr^2)*R*p eq2:=-Vr*Vt^3+Vr*Vt^2-Vr*Vt+Vr*p eq3:=(R*Vr*Vt+R*Vr)*e+((R-1)*Vr+1)*Vt-R*Vr eq4:=-r^2+(Vt^2)+(Vr^2) Eqlista:=[eq1a,eq2,eq3,eq4] Eqlist:=[eq1,eq2,eq3,eq4] solve(Eqlista,list) \end{axiom} This command runs too long to display on this page:: solve(Eqlist,list) From billpage Wed Feb 1 22:54:20 -0600 2006 From: billpage Date: Wed, 01 Feb 2006 22:54:20 -0600 Subject: Maple result Message-ID: <20060201225420-0600@wiki.axiom-developer.org> It seems to me that Axiom does not really give a complete solution. Compare this to the result using Maple:: > solve(Eqlista,list); 2 -%1 - 1 + r [[p = 1, Vr = %1, Vt = 1, e = 1/2 ------------], 2 R (-1 + r ) 2 2 2 -2 %1 + r [p = r %1 + 1 - r , Vr = 1, Vt = %1, e = - ----------]] 2 r - 2 2 2 %1 := RootOf(-r + 1 + _Z ) Maple seems to agree that the 2nd problem is considerably more complex than the first, but finds the solution in only a few seconds:: > solve(Eqlist,list); [[p = 0, Vr = -r, Vt = 0, e = 1], [p = 0, Vr = r, Vt = 0, e = 1], [ p = 0, Vr = %3, Vt = %2, 2 %3 R %2 - %3 %2 - %3 R - %3 + %2 + 1 e = -1/3 --------------------------------------], %3 R 2 -%1 - 1 + r [p = 1, Vr = %1, Vt = 1, e = 1/2 ------------], 2 R (-1 + r ) 2 2 2 -2 %1 + r [p = r %1 + 1 - r , Vr = 1, Vt = %1, e = - ----------]] 2 r - 2 2 2 %1 := RootOf(-r + 1 + _Z ) 2 %2 := RootOf(_Z - _Z + 1) 2 2 %3 := RootOf(-r + _Z + %2 - 1) From rrogers Thu Feb 2 06:37:16 -0600 2006 From: rrogers Date: Thu, 02 Feb 2006 06:37:16 -0600 Subject: Re: Maple Message-ID: <20060202063716-0600@wiki.axiom-developer.org> Looks like your right, although I will have to back substitute to be sure. So there is a simpler, at least faster, resolution than the one axiom uses. What is _z ? BTW: I like axioms presentation better than Maple's. Those %1 just seem to obscure the answer. Does Maple have a certificate/trace facility? Does axiom's trace, which I can't do on the windows version, provide the same type of certificate? Since I am doing this for a circuit at work, it might be nice to include this in the Doc's; not necessary. My company is on a ISO9000 kick and throwing some chum in the water, extra paper, might satisfy them. They know I have a critical view of some things, like explaining circuits to HR people. From BillPage Thu Feb 2 08:59:38 -0600 2006 From: Bill Page Date: Thu, 02 Feb 2006 08:59:38 -0600 Subject: Re:Maple Message-ID: <20060202085938-0600@wiki.axiom-developer.org> In Maple's solution '_Z' is a free variable. 'RootOf' is like 'zerosOf' in Axiom. The '%1', '%2' symbols denote common subexpressions that Maple decided would simplify printing the output. So for example: \begin{axiom} s1 := zerosOf(-r^2 + 1 + Z^2, Z) s2 := zerosOf(Z^2 - Z + 1, Z) s3a := zerosOf(-r^2 + Z + s2.1 - 1, Z) s3b := zerosOf(-r^2 + Z + s2.2 - 1, Z) \end{axiom} As far as I can understand your question ... No neither Maple nor Axiom have any kind of "certificate/trace" facility. But back substituting the answers back into the equations to check the solutions is a good idea in any computer algebra system! Maybe you could describe in more detail what you mean by a "certificate/trace" facility? What functionality would you expect? From unknown Thu Feb 2 09:20:53 -0600 2006 From: unknown Date: Thu, 02 Feb 2006 09:20:53 -0600 Subject: Re:Maple Message-ID: <20060202092053-0600@wiki.axiom-developer.org> In some Maple programs you can ask for a "proof" (which I think is sometime called a certificate) and it will print out enough intermediate steps to satisfy (some) people. I have been thinking that traceing the subroutine that resolves the S-expressions would give the succesion of simpler, in the grobner well ordering sense, polynomial equations. That should be sufficient. Back substitution vefies the answers, but doesn't prove that you have all of the answers; which in worst-case analysis is needed (you might have overlooked a worst case). Ray From rrogers Sat Feb 11 12:47:35 -0600 2006 From: rrogers Date: Sat, 11 Feb 2006 12:47:35 -0600 Subject: groebsol.spad compile ? Message-ID: <20060211124735-0600@wiki.axiom-developer.org> I want to modify groebsol.spad like so:: -- general solve, gives an error if the system not 0-dimensional groebSolve(leq: L DPoly,lvar:L OV) : L L DPoly == -- lnp:=[dmpToHdmp(f) for f in leq] -- replacement lnp:=[(f :: HDMP(lvar,POLY INT)) for f in leq] Which I suspect is more correct. While the statement seems to work from the console on: L DistributedMultivariatePolynomial The compiler objects? Some hints/assistance would be appreciated. Ray From rrogers Sat Feb 11 12:57:30 -0600 2006 From: rrogers Date: Sat, 11 Feb 2006 12:57:30 -0600 Subject: groebsol.spad compile ? Message-ID: <20060211125730-0600@wiki.axiom-developer.org> In-Reply-To: <20060211124735-0600@wiki.axiom-developer.org> Well I will try again:: I am trying to modify groabsol.spad like so groebSolve(leq: L DPoly,lvar:L OV) : L L DPoly == -- lnp:=[dmpToHdmp(f) for f in leq] lnp:=[(f :: HDMP(lvar,POLY INT)) for f in leq] Which I believe is more correct HDMP works on L DMP from the console but the compiler objects Help would be appreciated. From kratt6 Mon Feb 13 03:21:32 -0600 2006 From: kratt6 Date: Mon, 13 Feb 2006 03:21:32 -0600 Subject: Message-ID: <20060213032132-0600@wiki.axiom-developer.org> So you are saying that 'f' should be converted to a 'HDMP' using the variables given by 'lvar' rather than 'lv'? ('PolToPol(lv, F)' is importet in 'GROEBSOL') Note that you cannot simply convert a 'DMP' to a 'HDMP' using the '::' notation for 'coerce', since there is no operation 'coerce' with signature 'DMP -> HDMP'. This is the reason why there is a package 'polToPol', although I find this a great nuisance. I think that the real reason behind this is that you cannot "extend" domains in Axiom, as you can in Aldor. In the interpreter you can, since it knows about 'PolToPol'. Thus, if your analysis is correct, you would have to write:: lnp:=[dmpToHdmp(f)$PolToPol(lvar,POLY INT) for f in leq] Martin From rrogers Mon Feb 13 07:06:20 -0600 2006 From: rrogers Date: Mon, 13 Feb 2006 07:06:20 -0600 Subject: groebsol.spad compile Message-ID: <20060213070620-0600@wiki.axiom-developer.org> In-Reply-To: <20060213032132-0600@wiki.axiom-developer.org> Maritn:: Thanks for the help. I have made some progress; but I haven't made a firm determination yet. I think a practical, and perhaps design problem, is a lack a discrimination between variables and parameters in the simultanious polynomial solver. On a practical side it wastes time in an algorithm that is prone to wandering off for a long time. On the design side not having the equations ordered properly might confuse "groebner" and "groebSolve". I haven't found that part of the coding self evident. I take it that you think that dmpToHdmp orders by lv? That would be good, I haven't looked at it yet. Since I didn't assign lv initially in the code, I didn't find out about it till later. How do you access lv? I looked through the books but didn't find out how; I'm back to looking at cdr/car. Ray From kratt6 Tue Feb 14 09:52:32 -0600 2006 From: kratt6 Date: Tue, 14 Feb 2006 09:52:32 -0600 Subject: Message-ID: <20060214095232-0600@wiki.axiom-developer.org> I'm a lamer on groebner bases... However: Line 78 of 'groebsol.spad' says:: import PolToPol(lv,F) which has the effect that 'dmpToHdmp(f)' is the same as 'dmpToHdmp(f)$PolToPol(lv,F)'. Here, 'lv' is a parameter to the package 'GROEBSOL' as you can see in line 37 of 'groebsol.spad', that reads:: GroebnerSolve(lv,F,R) : C == T I don't really understand your question about accessing 'lv'. If you use 'groebSolve' from the interpreter, you would have to supply the value to 'lv' as parameter to 'GroebnerSolve', since you have to package call 'groebSolve'... To access 'lv', for example, you could modify 'groebSolve' as follows:: groebSolve(leq: L DPoly,lvar:L OV) : L L DPoly == output(lv::OutputForm)$OutputPackage lnp:=[dmpToHdmp(f) for f in leq] and it will spit out the value of 'lv' every time 'groebSolve' is called... Questions: * did you find a bug (i.e., wrong result), or did you experience slow response, or what are you trying to accomplish * could you PLEASE post on axiom-developer. It is difficult for me to follow here. And the discussions here are not archived! Martin
Please try the two solve statements at the bottom
The first one runs in a short time but
the second runs a long time. The only
differenct is *R*P in eq1. Examining eq1a shows that
in fact p=1 has a solution so there should be at least
one solution to eq1. More generally the grobner
should always converge; and this change seems minor.
This a simplification from a more extended problem.
The commented equations are a little more representative:
-- eq1:=((-2*R*Vr^3+R*Vr^2)*Vt+2*R*Vr^3-R*Vr^2)*aVr*p-Vt -- eq2:=-2*R*Vr*Vt^3+5*R*Vr*Vt^2-4*R*Vr*Vt+R*Vr*aVt*p -- eq3:=(R*Vr*Vt+R*Vr)*e+((R-1)*Vr+1)*Vt-R*Vr -- eq4:=-r^2+(Vt^2-Vt+1/4)*aVt+(Vr^2-Vr*R+1/4)*aVr
Raymond E. Rogers
fricas
list:=[p,Vr, Vt, e]
![\label{eq1}\left[ p , \: Vr , \: Vt , \: e \right]](images/2066658695132891724-16.0px.png "\label{eq1}\left[ p , \: Vr , \: Vt , \: e \right]") | (1) |
fricas
eq1a:=((-Vr^3+Vr^2)*Vt+Vr^3-Vr^2)
 | (2) |
Type: Polynomial(Integer)
fricas
eq1:=((-Vr^3+Vr^2)*Vt+Vr^3-Vr^2)*R*p
 | (3) |
Type: Polynomial(Integer)
fricas
eq2:=-Vr*Vt^3+Vr*Vt^2-Vr*Vt+Vr*p
 | (4) |
Type: Polynomial(Integer)
fricas
eq3:=(R*Vr*Vt+R*Vr)*e+((R-1)*Vr+1)*Vt-R*Vr
 | (5) |
Type: Polynomial(Integer)
fricas
eq4:=-r^2+(Vt^2)+(Vr^2)
 | (6) |
Type: Polynomial(Integer)
fricas
Eqlista:=[eq1a,eq2, eq3, eq4]
![\label{eq7}\begin{array}{@{}l}
\displaystyle
\left[{{{\left(-{{Vr}^{3}}+{{Vr}^{2}}\right)}\ Vt}+{{Vr}^{3}}-{{Vr}^{2}}}, \: \right.
\
\
\displaystyle
\left.{{Vr \ p}-{Vr \ {{Vt}^{3}}}+{Vr \ {{Vt}^{2}}}-{Vr \ Vt}}, \: \right.
\
\
\displaystyle
\left.{{{\left({R \ Vr \ Vt}+{R \ Vr}\right)}\ e}+{{\left({{\left(R - 1 \right)}\ Vr}+ 1 \right)}\ Vt}-{R \ Vr}}, \: \right.
\
\
\displaystyle
\left.{-{{r}^{2}}+{{Vt}^{2}}+{{Vr}^{2}}}\right]](images/5176661650590976721-16.0px.png "\label{eq7}\begin{array}{@{}l}
\displaystyle
\left[{{{\left(-{{Vr}^{3}}+{{Vr}^{2}}\right)}\ Vt}+{{Vr}^{3}}-{{Vr}^{2}}}, \: \right.
\
\
\displaystyle
\left.{{Vr \ p}-{Vr \ {{Vt}^{3}}}+{Vr \ {{Vt}^{2}}}-{Vr \ Vt}}, \: \right.
\
\
\displaystyle
\left.{{{\left({R \ Vr \ Vt}+{R \ Vr}\right)}\ e}+{{\left({{\left(R - 1 \right)}\ Vr}+ 1 \right)}\ Vt}-{R \ Vr}}, \: \right.
\
\
\displaystyle
\left.{-{{r}^{2}}+{{Vt}^{2}}+{{Vr}^{2}}}\right]") | (7) |
Type: List(Polynomial(Integer))
fricas
Eqlist:=[eq1,eq2, eq3, eq4]
![\label{eq8}\begin{array}{@{}l}
\displaystyle
\left[{{\left({{\left(-{R \ {{Vr}^{3}}}+{R \ {{Vr}^{2}}}\right)}\ Vt}+{R \ {{Vr}^{3}}}-{R \ {{Vr}^{2}}}\right)}\ p}, \: \right.
\
\
\displaystyle
\left.{{Vr \ p}-{Vr \ {{Vt}^{3}}}+{Vr \ {{Vt}^{2}}}-{Vr \ Vt}}, \: \right.
\
\
\displaystyle
\left.{{{\left({R \ Vr \ Vt}+{R \ Vr}\right)}\ e}+{{\left({{\left(R - 1 \right)}\ Vr}+ 1 \right)}\ Vt}-{R \ Vr}}, \: \right.
\
\
\displaystyle
\left.{-{{r}^{2}}+{{Vt}^{2}}+{{Vr}^{2}}}\right]](images/2513204169870541425-16.0px.png "\label{eq8}\begin{array}{@{}l}
\displaystyle
\left[{{\left({{\left(-{R \ {{Vr}^{3}}}+{R \ {{Vr}^{2}}}\right)}\ Vt}+{R \ {{Vr}^{3}}}-{R \ {{Vr}^{2}}}\right)}\ p}, \: \right.
\
\
\displaystyle
\left.{{Vr \ p}-{Vr \ {{Vt}^{3}}}+{Vr \ {{Vt}^{2}}}-{Vr \ Vt}}, \: \right.
\
\
\displaystyle
\left.{{{\left({R \ Vr \ Vt}+{R \ Vr}\right)}\ e}+{{\left({{\left(R - 1 \right)}\ Vr}+ 1 \right)}\ Vt}-{R \ Vr}}, \: \right.
\
\
\displaystyle
\left.{-{{r}^{2}}+{{Vt}^{2}}+{{Vr}^{2}}}\right]") | (8) |
Type: List(Polynomial(Integer))
fricas
solve(Eqlista,list)
![\label{eq9}\begin{array}{@{}l}
\displaystyle
\left[{
\begin{array}{@{}l}
\displaystyle
\left[{p ={{{{\left(e + 1 \right)}\ {{r}^{4}}}+{{\left(-{2 \ e}- 2 \right)}\ {{r}^{2}}}+ 2}\over 2}}, \:{Vr = 1}, \: \right.
\
\
\displaystyle
\left.{Vt ={{{{\left(e + 1 \right)}\ {{r}^{2}}}-{2 \ e}}\over 2}}, \:{{{{\left({{e}^{2}}+{2 \ e}+ 1 \right)}\ {{r}^{2}}}-{2 \ {{e}^{2}}}- 2}= 0}\right]](images/1025163051637173290-16.0px.png "\label{eq9}\begin{array}{@{}l}
\displaystyle
\left[{
\begin{array}{@{}l}
\displaystyle
\left[{p ={{{{\left(e + 1 \right)}\ {{r}^{4}}}+{{\left(-{2 \ e}- 2 \right)}\ {{r}^{2}}}+ 2}\over 2}}, \:{Vr = 1}, \: \right.
\
\
\displaystyle
\left.{Vt ={{{{\left(e + 1 \right)}\ {{r}^{2}}}-{2 \ e}}\over 2}}, \:{{{{\left({{e}^{2}}+{2 \ e}+ 1 \right)}\ {{r}^{2}}}-{2 \ {{e}^{2}}}- 2}= 0}\right]") | (9) |
Type: List(List(Equation(Fraction(Polynomial(Integer)))))
This command runs too long to display on this page:
solve(Eqlist,list)
Maple result --billpage, Wed, 01 Feb 2006 22:54:20 -0600 reply
It seems to me that Axiom does not really give a complete
solution. Compare this to the result using Maple:
> solve(Eqlista,list);
2
-%1 - 1 + r
[[p = 1, Vr = %1, Vt = 1, e = 1/2 ------------],
2
R (-1 + r )
2
2 2 -2 %1 + r
[p = r %1 + 1 - r , Vr = 1, Vt = %1, e = - ----------]]
2
r - 2
2 2
%1 := RootOf(-r + 1 + _Z )
Maple seems to agree that the 2nd problem is considerably more complex than the first, but finds the solution in only a few seconds:
> solve(Eqlist,list);
[[p = 0, Vr = -r, Vt = 0, e = 1], [p = 0, Vr = r, Vt = 0, e = 1], [
p = 0, Vr = %3, Vt = %2,
2 %3 R %2 - %3 %2 - %3 R - %3 + %2 + 1
e = -1/3 --------------------------------------],
%3 R
2
-%1 - 1 + r
[p = 1, Vr = %1, Vt = 1, e = 1/2 ------------],
2
R (-1 + r )
2
2 2 -2 %1 + r
[p = r %1 + 1 - r , Vr = 1, Vt = %1, e = - ----------]]
2
r - 2
2 2
%1 := RootOf(-r + 1 + _Z )
2
%2 := RootOf(_Z - _Z + 1)
2 2
%3 := RootOf(-r + _Z + %2 - 1)
Looks like your right, although I will have to back substitute to be sure.
So there is a simpler, at least faster, resolution than the one axiom uses.
What is _z ?
BTW: I like axioms presentation better than Maple's. Those %1 just seem to obscure the answer.
Does Maple have a certificate/trace facility? Does axiom's trace, which I can't do on the windows version, provide the same type of certificate?
Since I am doing this for a circuit at work, it might be nice to include this in the Doc's; not necessary.
My company is on a ISO9000 kick and throwing some chum in the water, extra paper, might satisfy them.
They know I have a critical view of some things, like explaining circuits to HR people.
In Maple's solution _Z is a free variable. RootOf is like
zerosOf in Axiom. The %1, %2 symbols denote common
subexpressions that Maple decided would simplify printing the
output.
So for example:
fricas
s1 := zerosOf(-r^2 + 1 + Z^2,Z)
![\label{eq10}\left[{{\sqrt{{4 \ {{r}^{2}}}- 4}}\over 2}, \: -{{\sqrt{{4 \ {{r}^{2}}}- 4}}\over 2}\right]](images/7950421336390128995-16.0px.png "\label{eq10}\left[{{\sqrt{{4 \ {{r}^{2}}}- 4}}\over 2}, \: -{{\sqrt{{4 \ {{r}^{2}}}- 4}}\over 2}\right]") | (10) |
Type: List(Expression(Integer))
fricas
s2 := zerosOf(Z^2 - Z + 1,Z)
![\label{eq11}\left[{{{\sqrt{- 3}}+ 1}\over 2}, \:{{-{\sqrt{- 3}}+ 1}\over 2}\right]](images/7352284782113384751-16.0px.png "\label{eq11}\left[{{{\sqrt{- 3}}+ 1}\over 2}, \:{{-{\sqrt{- 3}}+ 1}\over 2}\right]") | (11) |
Type: List(Expression(Integer))
fricas
s3a := zerosOf(-r^2 + Z + s2.1 - 1,Z)
![\label{eq12}\left[{{-{\sqrt{- 3}}+{2 \ {{r}^{2}}}+ 1}\over 2}\right]](images/7755415474326685577-16.0px.png "\label{eq12}\left[{{-{\sqrt{- 3}}+{2 \ {{r}^{2}}}+ 1}\over 2}\right]") | (12) |
Type: List(Expression(Integer))
fricas
s3b := zerosOf(-r^2 + Z + s2.2 - 1,Z)
![\label{eq13}\left[{{{\sqrt{- 3}}+{2 \ {{r}^{2}}}+ 1}\over 2}\right]](images/6349050927215050588-16.0px.png "\label{eq13}\left[{{{\sqrt{- 3}}+{2 \ {{r}^{2}}}+ 1}\over 2}\right]") | (13) |
Type: List(Expression(Integer))
As far as I can understand your question ... No neither Maple nor Axiom have any kind of "certificate/trace" facility. But back substituting the answers back into the equations to check the solutions is a good idea in any computer algebra system!
Maybe you could describe in more detail what you mean by a "certificate/trace" facility? What functionality would you expect?
In some Maple programs you can ask for a "proof" (which I think is sometime called a certificate) and it will print out enough intermediate steps to satisfy (some) people. I have been thinking that traceing the subroutine that resolves the S-expressions would give the succesion of simpler, in the grobner well ordering sense, polynomial equations. That should be sufficient. Back substitution vefies the answers, but doesn't prove that you have all of the answers; which in worst-case analysis is needed (you might have overlooked a worst case).Ray
I want to modify groebsol.spad like so:-- general solve, gives an error if the system not 0-dimensional groebSolve(leq: L DPoly,lvar:L OV) : L L DPoly == -- lnp:=[dmpToHdmp(f) for f in leq] -- replacement lnp:=[(f :: HDMP(lvar,POLY INT)) for f in leq] Which I suspect is more correct. While the statement seems to work from the console on: L DistributedMultivariatePolynomial The compiler objects? Some hints/assistance would be appreciated. RayWell I will try again:
I am trying to modify groabsol.spad like so
groebSolve(leq: L DPoly,lvar:L OV) : L L DPoly ==
-- lnp:=[dmpToHdmp(f) for f in leq]
lnp:=[(f :: HDMP(lvar,POLY INT)) for f in leq]
Which I believe is more correct
HDMP works on L DMP from the console but the compiler objects
Help would be appreciated.
So you are saying thatf should be converted to a HDMP using the variables given by lvar rather than lv?
(PolToPol(lv, F) is importet in GROEBSOL)
Note that you cannot simply convert a DMP to a HDMP using the :: notation for coerce, since there is no operation coerce with signature DMP -> HDMP. This is the reason why there is a package polToPol, although I find this a great nuisance. I think that the real reason behind this is that you cannot "extend" domains in Axiom, as you can in Aldor.
In the interpreter you can, since it knows about PolToPol.
Thus, if your analysis is correct, you would have to write:
lnp:=[dmpToHdmp(f)$PolToPol(lvar,POLY INT) for f in leq]
Martin
Maritn:Thanks for the help. I have made some progress; but I haven't made a firm determination yet. I think a practical, and perhaps design problem, is a lack a discrimination between variables and parameters in the simultanious polynomial solver. On a practical side it wastes time in an algorithm that is prone to wandering off for a long time. On the design side not having the equations ordered properly might confuse "groebner" and "groebSolve". I haven't found that part of the coding self evident. I take it that you think that dmpToHdmp orders by lv? That would be good, I haven't looked at it yet. Since I didn't assign lv initially in the code, I didn't find out about it till later. How do you access lv? I looked through the books but didn't find out how; I'm back to looking at cdr/car.
Ray
I'm a lamer on groebner bases...However: Line 78 of groebsol.spad says:
import PolToPol(lv,F)
which has the effect that dmpToHdmp(f) is the same as dmpToHdmp(f)$PolToPol(lv,F). Here, lv is a parameter to the package GROEBSOL as you can see in line 37 of groebsol.spad, that reads:
GroebnerSolve(lv,F,R) : C == T
I don't really understand your question about accessing lv. If you use groebSolve from the interpreter, you would have to supply the value to lv as parameter to GroebnerSolve, since you have to package call groebSolve... To access lv, for example, you could modify groebSolve as follows:
groebSolve(leq: L DPoly,lvar:L OV) : L L DPoly ==
output(lv::OutputForm)$OutputPackage
lnp:=[dmpToHdmp(f) for f in leq]
and it will spit out the value of lv every time groebSolve is called...
Questions:
- did you find a bug (i.e., wrong result), or did you experience slow response, or what are you trying to accomplish * could you PLEASE post on axiom-developer. It is difficult for me to follow here. And the discussions here are not archived!
Martin