多項式
が与えられたとき,
となる
を求める問題を解く.ただし,
である.
>> sos_decomposition1
SeDuMi 1.3 by AdvOL, 2005-2008 and Jos F. Sturm, 1998-2003.
Alg = 2: xz-corrector, theta = 0.250, beta = 0.500
Put 5 free variables in a quadratic cone
eqs m = 6, order n = 6, dim = 16, blocks = 3
nnz(A) = 12 + 0, nnz(ADA) = 36, nnz(L) = 21
it : b*y gap delta rate t/tP* t/tD* feas cg cg prec
0 : 3.22E+000 0.000
1 : 0.00E+000 5.94E-001 0.000 0.1844 0.9000 0.9000 1.27 1 1 1.6E+000
2 : 0.00E+000 3.17E-002 0.000 0.0534 0.9900 0.9900 1.26 1 1 7.7E-001
3 : 0.00E+000 1.60E-006 0.000 0.0001 1.0000 1.0000 1.00 1 1 3.9E-004
4 : 0.00E+000 7.26E-013 0.000 0.0000 1.0000 1.0000 1.00 1 3 1.4E-010
iter seconds digits c*x b*y
4 0.5 13.4 1.9109846621e-013 0.0000000000e+000
|Ax-b| = 3.0e-013, [Ay-c]_+ = 9.1E-013, |x|= 1.1e+000, |y|= 5.8e+000
Detailed timing (sec)
Pre IPM Post
2.130E-001 4.500E-001 7.000E-002
Max-norms: ||b||=0, ||c|| = 4,
Cholesky |add|=0, |skip| = 1, ||L.L|| = 1.
sol =
yalmiptime: 0.7690
solvertime: 0.8010
info: 'No problems detected (SeDuMi-1.3)'
problem: 0
dimacs: [3.0116e-013 0 0 1.9185e-013 1.9110e-013 1.8475e-012]
>> double(Q)
ans =
2.0000 0.0000 -0.2636
0.0000 3.5272 1.0000
-0.2636 1.0000 4.0000
>> min(eig(double(Q))) %%% Q の最小固有値
ans =
1.9509
>> sdisplay(z)
ans =
'x1^2'
'x1*x2'
'x2^2'
>> sdisplay(z'*double(Q)*z)
2.0000*x1^4+3.0000*x1^2*x2^2+2.0000*x1*x2^3+4.0000*x2^4