sos_decomposition1.m

 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.

実行結果

>> 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
    
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