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 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,
fricas eq1a:=((-Vr^3+Vr^2)*Vt+Vr^3-Vr^2)
Type: Polynomial(Integer)
fricas eq1:=((-Vr^3+Vr^2)*Vt+Vr^3-Vr^2)*R*p
Type: Polynomial(Integer)
fricas eq2:=-Vr*Vt^3+Vr*Vt^2-Vr*Vt+Vr*p
Type: Polynomial(Integer)
fricas eq3:=(R*Vr*Vt+R*Vr)*e+((R-1)*Vr+1)*Vt-R*Vr
Type: Polynomial(Integer)
fricas eq4:=-r^2+(Vt^2)+(Vr^2)
Type: Polynomial(Integer)
fricas Eqlista:=[eq1a,
Type: List(Polynomial(Integer))
fricas Eqlist:=[eq1,
Type: List(Polynomial(Integer))
fricas solve(Eqlista,
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 )
This looks the same as FriCAS answer. 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,
Type: List(Expression(Integer))
fricas s2 := zerosOf(Z^2 - Z + 1,
Type: List(Expression(Integer))
fricas s3a := zerosOf(-r^2 + Z + s2.1 - 1,
Type: List(Expression(Integer))
fricas s3b := zerosOf(-r^2 + Z + s2.2 - 1,
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?
( Note that you cannot simply convert a In the interpreter you can, since it knows about Thus, if your analysis is correct, you would have to write: lnp:=[dmpToHdmp(f)$PolToPol(lvar,POLY INT) for f in leq] Martin Maritn:Thanks for the help. I have made some progress; but I haven't made a firm determination yet. I think a practical, and perhaps design problem, is a lack a discrimination between variables and parameters in the simultanious polynomial solver. On a practical side it wastes time in an algorithm that is prone to wandering off for a long time. On the design side not having the equations ordered properly might confuse "groebner" and "groebSolve". I haven't found that part of the coding self evident. I take it that you think that dmpToHdmp orders by lv? That would be good, I haven't looked at it yet. Since I didn't assign lv initially in the code, I didn't find out about it till later. How do you access lv? I looked through the books but didn't find out how; I'm back to looking at cdr/car. Ray I'm a lamer on groebner bases...However: Line 78 of import PolToPol(lv,F) which has the effect that GroebnerSolve(lv,F,R) : C == T I don't really understand your question about accessing
groebSolve(leq: L DPoly,lvar:L OV) : L L DPoly ==
output(lv::OutputForm)$OutputPackage
lnp:=[dmpToHdmp(f) for f in leq]
and it will spit out the value of Questions:
Martin |
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last edited 8 years ago by test1 |
![\label{eq1}\left[ p , \: Vr , \: Vt , \: e \right]](images/2066658695132891724-16.0px.png "\label{eq1}\left[ p , \: Vr , \: Vt , \: e \right]")
![\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]")
![\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]")
![\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]")
![\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]")
![\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]")
![\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]")
![\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]")