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#17 algebra BOOTSTRAP fixed point  ← ↑ →  #171 Solve handles lists of expressions differently

#170 Axiom fails to solve "separable" system of equations
  
last edited 10 years ago by test1

Edit detail for #170 Axiom fails to solve "separable" system of equations revision 1 of 3
1 2 3
Editor:
Time: 2007/11/17 22:01:54 GMT-8
Note: 
changed:
-
Some examples:

\begin{axiom}
L := [ A = 2*X+Y, B = 2*Y+X, C = 2*U+V, D = 2*V+U];
solve(L, [X,Y])
\end{axiom}

However:

\begin{axiom}
solve(L,[X,Y,U,V])
\end{axiom}

Simpler:

\begin{axiom}
solve([a - b = 0, c - d = 0],[b])
linSolve([a - b, c - d],[b])
\end{axiom}

The operation 'solve' calls 'linSolve', which sets up the corresponding matrix
and vector and solves it using 'solve\$LinearSystemMatrixPackage'. This in turn
returns '"failed"', since the last columns of the matrix contain zeros, the
vector does not. In the example above, the matrix and vector are::

         +- 1+
   [mat= |   |,vec= [- a,d - c]]
         + 0 +


Note that 

\begin{axiom}
linSolve([a - b, 0],[b])
\end{axiom}

works.
The same happens, if the equation is not linear:

\begin{axiom}
L := [ A = 2*X^2+Y, B = 2*Y+X, C = 2*U+V, D = 2*V+U];
solve(L, [X, Y])
)set output algebra on
)set output tex off
solve(L, [X, Y, U, V])
)set output algebra off
)set output tex on
\end{axiom}

So, very probably, a fix would need to do two things:

- seperate the equations into those that do and those that don't contain the
  given variables.

- check whether those that don't contain the variables are contradicting.

- solve the others.

The second point is necessary, since

\begin{axiom}
L := [ A = P+Q, B = P-Q, C = 1, C = -1];
solve(L, [P,Q])
solve(L,[P,Q,C])
\end{axiom}

really has no solution. As far as I know, this would have to be done in the very last function defined in 'syssolp.spad', which is::

       -- general solver. Input in polynomial style  --
       solve(lr:L F,vl:L SE) ==
           empty? vl => empty()
           checkLinear(lr,vl) =>
                            -- linear system --
               soln := linSolve(lr, vl)
               soln case "failed" => []
               eqns: L EQ F := []
               for i in 1..#vl repeat
                   lhs := (vl.i::(P R))::F
                   rhs :=  rhs soln.i
                   eqns := append(eqns, [lhs = rhs])
               [eqns]

                         -- polynomial system --
           if R has GcdDomain then
             parRes:=triangularSystems(lr,vl)
             [[makeEq(map(makeR2F,f)\$PP2,vl) for f in pr]
                                                        for pr in parRes]
           else [[]]


The letter 'F' is a macro for 'FRAC POLY R' here. To check whether an equation contains a variable we have to check numerator and denominator of both sides of the equation with 'variables\$POLY R'. I do not know however, how to find out whether the equations independent of 'vl' are contradicting.

Submitted by : (unknown) at: 2007-11-17T22:01:54-08:00 (16 years ago)
Name :
Axiom Version :
Category : Severity : Status :
Optional subject :  
Optional comment : 

Some examples:


axiom
L := [ A = 2*X+Y, B = 2*Y+X, C = 2*U+V, D = 2*V+U];


Type: List(Equation(Polynomial(Integer)))


axiom
solve(L, [X,Y])

\begin{equation*}
\label{eq1}\left[ \right]?\end{equation*}

Type: List(List(Equation(Fraction(Polynomial(Integer)))))


However:


axiom
solve(L,[X,Y,U,V])

\begin{equation*}
\label{eq2}\begin{array}{@{}l}
\displaystyle
\left[ \left[{X ={{- B +{2 \  A}}\over 3}}, \:{Y ={{{2 \  B}- A}\over 3}}, \:{U ={{- D +{2 \  C}}\over 3}}, \: \right.
\
\
\displaystyle
\left.{V ={{{2 \  D}- C}\over 3}}\right] \right] 
\end{array}
\end{equation*}

Type: List(List(Equation(Fraction(Polynomial(Integer)))))


Simpler:


axiom
solve([a - b = 0, c - d = 0],[b])

\begin{equation*}
\label{eq3}\left[ \right]?\end{equation*}

Type: List(List(Equation(Fraction(Polynomial(Integer)))))


axiom
linSolve([a - b, c - d],[b])

\begin{equation*}
\label{eq4}\left[{particular = \mbox{\tt "failed"}}, \:{basis ={\left[ \right]?}}\right]\end{equation*}

Type: Record(particular: Union(Vector(Fraction(Polynomial(Integer))),"failed"),basis: List(Vector(Fraction(Polynomial(Integer)))))


The operation solve calls linSolve, which sets up the corresponding matrix
and vector and solves it using solve$LinearSystemMatrixPackage. This in turn
returns "failed", since the last columns of the matrix contain zeros, the
vector does not. In the example above, the matrix and vector are:

         +- 1+
   [mat= |   |,vec= [- a,d - c]]
         + 0 +





Note that 


axiom
linSolve([a - b, 0],[b])

\begin{equation*}
\label{eq5}\left[{particular ={\left[ a \right]?}}, \:{basis ={\left[ \right]?}}\right]\end{equation*}

Type: Record(particular: Union(Vector(Fraction(Polynomial(Integer))),"failed"),basis: List(Vector(Fraction(Polynomial(Integer)))))


works.
The same happens, if the equation is not linear:


axiom
L := [ A = 2*X^2+Y, B = 2*Y+X, C = 2*U+V, D = 2*V+U];


Type: List(Equation(Polynomial(Integer)))


axiom
solve(L, [X, Y])

\begin{equation*}
\label{eq6}\left[{\left[ \right]?}\right]\end{equation*}

Type: List(List(Equation(Fraction(Polynomial(Integer)))))


axiom
)set output algebra on
 


axiom
)set output tex off

solve(L, [X, Y, U, V])

                        2                   2           - D + 2C    2D - C
   (9)  [[X= - 2Y + B,8Y  + (- 8B + 1)Y + 2B  - A= 0,U= --------,V= ------]]
                                                            3          3


Type: List(List(Equation(Fraction(Polynomial(Integer)))))


axiom
)set output algebra off
 


axiom
)set output tex on


So, very probably, a fix would need to do two things:



    

  • seperate the equations into those that do and those that don't contain the
  given variables.
    • 

  • check whether those that don't contain the variables are contradicting.
    • 

  • solve the others.
    • 




The second point is necessary, since


axiom
L := [ A = P+Q, B = P-Q, C = 1, C = -1];


Type: List(Equation(Polynomial(Integer)))


axiom
solve(L, [P,Q])

\begin{equation*}
\label{eq7}\left[ \right]?\end{equation*}

Type: List(List(Equation(Fraction(Polynomial(Integer)))))


axiom
solve(L,[P,Q,C])

\begin{equation*}
\label{eq8}\left[ \right]?\end{equation*}

Type: List(List(Equation(Fraction(Polynomial(Integer)))))


really has no solution. As far as I know, this would have to be done in the very last function defined in syssolp.spad, which is:

       -- general solver. Input in polynomial style  --
       solve(lr:L F,vl:L SE) ==
           empty? vl => empty()
           checkLinear(lr,vl) =>
                            -- linear system --
               soln := linSolve(lr, vl)
               soln case "failed" => []
               eqns: L EQ F := []
               for i in 1..#vl repeat
                   lhs := (vl.i::(P R))::F
                   rhs :=  rhs soln.i
                   eqns := append(eqns, [lhs = rhs])
               [eqns]

                         -- polynomial system --
           if R has GcdDomain then
             parRes:=triangularSystems(lr,vl)
             [[makeEq(map(makeR2F,f)$PP2,vl) for f in pr]
                                                        for pr in parRes]
           else [[]]





The letter F is a macro for FRAC POLY R here. To check whether an equation contains a variable we have to check numerator and denominator of both sides of the equation with variables$POLY R. I do not know however, how to find out whether the equations independent of vl are contradicting.
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