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% Examples of use of Groebner code. % In the Examples 1 - 3 the polynomial ring for the ideal operations % (variable sequence, term order mode) is defined globally in advance. % Example 1, Linz 85. torder ({q1,q2,q3,q4,q5,q6},lex)$ groebner {q1, q2**2 + q3**2 + q4**2, q4*q3*q2, q3**2*q2**2 + q4**2*q2**2 + q4**2*q3**2, q6**2 + 1/3*q5**2, q6**3 - q5**2*q6, 2*q2**2*q6 - q3**2*q6 - q4**2*q6 + q3**2*q5 - q4**2*q5, 2*q2**2*q6**2 - q3**2*q6**2 - q4**2*q6**2 - 2*q3**2*q5*q6 + 2*q4**2*q5*q6 - 2/3*q2**2*q5**2 + 1/3*q3**2*q5**2 + 1/3*q4**2*q5**2, - q3**2*q2**2*q6 - q4**2*q2**2*q6 + 2*q4**2*q3**2*q6 - q3**2*q2**2*q5 + q4**2*q2**2*q5, - q3**2*q2**2*q6**2 - q4**2*q2**2*q6**2 + 2*q4**2*q3**2*q6**2 + 2*q3**2*q2**2*q5*q6 - 2*q4**2*q2**2*q5*q6 + 1/3*q3**2*q2**2 *q5**2 + 1/3*q4**2*q2**2*q5**2 - 2/3*q4**2*q3**2*q5**2, - 3*q3**2*q2**4*q5*q6**2 + 3*q4**2*q2**4*q5*q6**2 + 3*q3**4*q2**2*q5*q6**2 - 3*q4**4*q2**2*q5*q6**2 - 3*q4**2*q3**4*q5*q6**2 + 3*q4**4*q3**2*q5*q6**2 + 1/3*q3**2*q2**4*q5**3 - 1/3*q4**2*q2**4*q5**3 - 1/3*q3**4*q2**2*q5**3 + 1/3*q4**4*q2**2*q5**3 + 1/3*q4**2 *q3**4*q5**3 - 1/3*q4**4*q3**2*q5**3}; % Example 2. (Little) Trinks problem with 7 polynomials in 6 variables. trinkspolys:={45*p + 35*s - 165*b - 36, 35*p + 40*z + 25*t - 27*s, 15*w + 25*p*s + 30*z - 18*t - 165*b**2, - 9*w + 15*p*t + 20*z*s, w*p + 2*z*t - 11*b**3, 99*w - 11*s*b + 3*b**2, b**2 + 33/50*b + 2673/10000}$ trinksvars := {w,p,z,t,s,b}$ torder(trinksvars,lex)$ switch varopt; off varopt; groebner trinkspolys; groesolve ws; % Example 3. Hairer, Runge-Kutta 1, 6 polynomials 8 variables. torder({c2,c3,b3,b2,b1,a21,a32,a31},lex); groebnerf{c2 - a21, c3 - a31 - a32, b1 + b2 + b3 - 1, b2*c2 + b3*c3 - 1/2, b2*c2**2 + b3*c3**2 - 1/3, b3*a32*c2 - 1/6}; % The examples 4 and 5 use automatic variable extraction. % Example 4. torder gradlex$ g4:= groebner{b + e + f - 1, c + d + 2*e - 3, b + d + 2*f - 1, a - b - c - d - e - f, d*e*a**2 - 1569/31250*b*c**3, c*f - 587/15625*b*d}; hilbertpolynomial g4; glexconvert(g4,gvarslast,newvars={e},maxdeg=8); % Example 5. off varopt; torder({u0,u2,u3,u1},lex)$ groesolve({u0**2 - u0 + 2*u1**2 + 2*u2**2 + 2*u3**2, 2*u0*u1 + 2*u1*u2 + 2*u2*u3 - u1, 2*u0*u2 + u1**2 + 2*u1*u3 - u2, u0 + 2*u1 + 2*u2 + 2*u3 - 1}, {u0,u2,u3,u1}); % Example 6. (Big) Trinks problem with 6 polynomials in 6 variables. torder(trinksvars,lex)$ btbas:= groebner{45*p + 35*s - 165*b - 36, 35*p + 40*z + 25*t - 27*s, 15*w + 25*p*s + 30*z - 18*t - 165*b**2, -9*w + 15*p*t + 20*z*s, w*p + 2*z*t - 11*b**3, 99*w - 11*b*s + 3*b**2}; % The above system has dimension zero. Therefore its Hilbert polynomial % is a constant which is the number of zero points (including complex % zeros and multipliticities); hilbertpolynomial ws; % Example of Groebner with numerical postprocessing. on rounded;off varopt; groesolve(trinkspolys,trinksvars); off rounded; % Additional groebner operators. % Reduce one polynomial wrt the basis of big Trinks. The result 0 % is a proof for the ideal membership of the polynomial. torder(trinksvars,lex)$ preduce(45*p + 35*s - 165*b - 36,btbas); % The following examples show how to work with the distributive % form of polynomials. torder({u0,u1,u2,u3},gradlex)$ gsplit(2*u0*u2 + u1**2 + 2*u1*u3 - u2,{u0,u1,u2,u3}); torder(trinksvars,lex)$ gsort trinkspolys; gspoly(first trinkspolys,second trinkspolys); gvars trinkspolys; % Tagged basis and reduction trace. A tagged basis is a basis where % each polynomial is equated to a linear combination of the input % set. A tagged reduction shows how the result is computed by using % the basis polynomials. % First example for tagged polynomials: show how a polynomial is % represented as linear combination of the basis polynomials. % First I set up an environment for the computation. torder(trinksvars,lex)$ % Then I compute an ordinary Groebner basis. bas:=groebner trinkspolys$ % Next I assign a tag to each basis polynomial. taggedbas:=for i:=1:length bas collect mkid(p,i)=part(bas,i); % And finally I reduce a (tagged) polynomial wrt the tagged basis. preducet(new=w*p + 2*z*t - 11*b**3,taggedbas); % Second example for tagged polynomials: representing a Groebner basis % as a combination of the input polynomials, here in a simple geometric % problem. torder({x,y},lex)$ groebnert {circle=x**2 + y**2 - r**2,line=a*x + b*y}; % In the third example I enter two polynomials that have no common zero. % Consequently the basis is {1}. The tagged computation gives me a proof % for the inconsistency of the system which is independent of the % Groebner formalism. groebnert {circle1=x**2 + y**2 - 10,circle2=x**2 + y**2 - 2}; % Solve a special elimination task by using a blockwise elimination % order defined by a matrix. The equation set goes back to A.M.H. % Levelt (Nijmegen). The question is whether there is a member in the % ideal which depends only on two variables. Here we select x4 and y1. % The existence of such a polynomial proves that the system has exactly % one degree of freedom. % The first two rows of the term order matrix define the groupwise % elimination. The remaining lines define a secondary local % lexicographical behavior which is needed to construct an admissible % ordering. f1:=y1^2 + z1^2 -1; f2:=x2^2 + y2^2 + z2^2 -1; f3:=x3^2 + y3^2 + z3^2 -1; f4:=x4^2 + z4^2 -1; f5:=y1*y2 + z1*z2; f6:=x2*x3 + y2*y3 + z2*z3; f7:=x3*x4 + z3*z4; f8:=x2 + x3 + x4 + 1; f9:=y1 + y2 + y3 - 1; f10:=z1 + z2 + z3 + z4; eqns:={f1,f2,f3,f4,f5,f6,f7,f8,f9,f10}$ vars:={x2,x3,y2,y3,z1,z2,z3,z4,x4,y1}$ torder(vars,matrix, mat((1,1,1,1,1,1,1,1,0,0), (0,0,0,0,0,0,0,0,1,1), (1,0,0,0,0,0,0,0,0,0), (0,1,0,0,0,0,0,0,0,0), (0,0,1,0,0,0,0,0,0,0), (0,0,0,1,0,0,0,0,0,0), (0,0,0,0,1,0,0,0,0,0), (0,0,0,0,0,1,0,0,0,0), (0,0,0,0,0,0,1,0,0,0), (0,0,0,0,0,0,0,0,1,0))); first reverse groebner(eqns,vars); % For a faster execution we convert the matrix into a % proper machine code routine. on comp; torder_compile(levelt,mat( (1,1,1,1,1,1,1,1,0,0), (0,0,0,0,0,0,0,0,1,1), (1,0,0,0,0,0,0,0,0,0), (0,1,0,0,0,0,0,0,0,0), (0,0,1,0,0,0,0,0,0,0), (0,0,0,1,0,0,0,0,0,0), (0,0,0,0,1,0,0,0,0,0), (0,0,0,0,0,1,0,0,0,0), (0,0,0,0,0,0,1,0,0,0), (0,0,0,0,0,0,0,0,1,0))); torder(vars,levelt)$ first reverse groebner(eqns,vars); % For a homogeneous polynomial set we compute a graded Groebner % basis with grade limits. We use the graded term order with lex % as following order. As the grade vector has no zeros, this ordering % is functionally equivalent to a weighted ordering. torder({x,y,z},graded,{1,1,2},lex); dd_groebner(0,10,{x^10*y + y*z^5, x*y^12 + y*z^6}); dd_groebner(0,50,{x^10*y + y*z^5, x*y^12 + y*z^6}); dd_groebner(0,infinity,{x^10*y + y*z^5, x*y^12 + y*z^6}); % Test groebner_walk trinkspolys := {45*p + 35*s - 165*b - 36, 35*p + 40*z + 25*t - 27*s, 15*w + 25*p*s + 30*z - 18*t - 165*b**2, - 9*w + 15*p*t + 20*z*s, w*p + 2*z*t - 11*b**3, 99*w - 11*s*b + 3*b**2, b**2 + 33/50*b + 2673/10000}$ trinksvars := {w,p,z,t,s,b}$ torder(trinksvars,gradlex)$ gg:=groebner trinkspolys$ g:=groebner_walk gg$ on div$ g; on varopt; g1:=solve({first g},{b}); g0:=sub({first g1},g); solve({ second g0},{w}); solve({third g0},{p}); solve({part(g0,4)},{z}); solve({part(g0,5)},{t}); solve({part(g0,6)},{s}); g0:=sub({second g1},g); solve({second g0},{w}); solve({third g0},{p}); solve({part(g0,4)},{z}); solve({part(g0,5)},{t}); solve({part(g0,6)},{s}); % Example after the book "David Cox, John Little, Donal O'Shea: % "Ideals, Varieties and Algorithms", chapter 2, paragraph 8, example 3. % This example was given by Shigetoshi Katsura (Japan). off groebopt;torder({x,y,z,l},lex); g:=groebner{3*x^2+2*y*z-2*x*l,2*x*z-2*y*l,2*x*y-2*z-2*z*l,x^2+y^2+z^2-1}$ gdimension g; gindependent_sets g; clear g, gg, trinkspolys, trinksvars$ end;