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我有一個問題，其中有一個規範的解決方案，任何旋轉和鏡像（軸上）是另一種解決方案。 爲了避免由於旋轉引起的多重解決方案的問題，我已經在約束條件中設置了矢量應該儘可能與軸對齊。 「鏡像」解決方案現在是我的問題。基本上有些值可以有正解或負解。 這給我2^d解決方案的大小d的問題。 我嘗試使用assume並修正了一些始終爲正的值，這應該可以解決問題，但solve仍然是創建負面解決方案。在Maxima中避免一些解決方案
這是我到目前爲止的代碼：
/* A parameter w0 in (0,1] */
assume(w0>=0);
assume(w0<1);
d:2$
/* d+1 extra points, the is one at x=0, for a total fo d+2 points */
s:d+1$
/* The unknown are w (weights of the first component) and x_S vectors of dimension d */
v:append([w1],makelist(x[i-s*floor(i/s)+1,floor(i/s)+1],i,0,d*s-1));
/* The different constraints */
e:append(
/* The sum of weights is 1 */
[w0+s*w1-1=0],
/* Some of the components are 0 to be aligned with the axis */
flatten(makelist(makelist(x[i,j]=0,j,i+1,d),i,1,s-1)),
/* The mean is 0 */
makelist(sum(x[i,j],i,1,s)=0,j,1,d),
/* All vectors have length squared of x[1,1]^2, x[1,:] is skipped as only its first component is non-zero*/
makelist(sum(x[i,j]^2,j,1,d)-x[1,1]^2=0,i,2,s),
/* The diagonal of the covariance matrix is 1, w0*0 +w1*sum(x_i^2)=1*/
makelist(w1*sum(x[i,j]^2,i,1,s)-1=0,j,1,d),
/* The off-diagonal elements are 0. The dependancy on w1 can be eliminated since the equation is =0. */
flatten(makelist(makelist(sum(x[i,jj]*x[i,jj+j],i,1,s)=0,j,1,d-jj),jj,1,d-1))
);
/* THIS IS NOT WORKING AS I EXPECTED, I WANT SOLUTION ONLY WITH x[i,i]>0 */
assume(x[1,1]>0,x[2,2]>0);
solution:solve(e,v)$
number_solutions=length(solution);
有沒有辦法迫使solve探索只是問題的一些解決方案？
SOLUTION： 繼羅伯特的評論，我能得到「規範」的解決方案如下：
check_canonical(sol):=block([],
/* Extract the expression x[i,i]=... */
diag_expr0:makelist(sublist(sol,lambda([e],(if lhs(e)=x[i,i] then true else false))),i,1,d),
diag_expr1:flatten(diag_expr0),
/* Get the right hand side */
diag_expr2:makelist(rhs(diag_expr1[i]),i,1,d),
/* Check for the rhs to be positive */
pos_diag:sublist(diag_expr2,lambda([e],if e>0 then true else false)),
/* If all the elelment are positive, then this is a canonical solution */
if length(pos_diag)=d then true else false
)$
canonical_solution:flatten(sublist(solutions,check_canonical));
我不是千里馬的專家，但它的工作原理，但我認爲這將是避免探索不符合某些標準的解決方案更有趣。