Convex Opt

[jikan1221]

Lagrange equation when tangent happend between constraint function with target function,the La'equation (construct by constraint + target function) will get extremum point, because the whole equation's Derivative equals 0 as the gradient will be offset among target and constraint.

the lambda is used to adapt to offset qualtity of gradients, so it extremum will be zero. $$\min f(x), x \in \mathbb{R}^n$$ s.t-> $$g_i(x)= (a_i)^T \times x + b_i \leq 0 ,i=1,2,3....z$$ $$L(x,\lambda) = f(x) + \sum \lambda_i g_i(x)$$ $$\Delta L(x,\lambda) = 0$$ $$-\Delta f(x) = \sum \lambda_i \Delta g_i(x)$$

because g_i are constraint set, the g_i*(x) is represent district area,the extremum point actually reference to only g_a,g_b, because only the two line contribute the offset direction gredient (the angel is converse to target function),and point always in these two lines's cross. $$g_a (x^x) = g_b (x^x)=0$$ $$\lambda_a \Delta g_a(x^x) + \lambda_b \Delta g_b(x^x)= -\Delta f(x),==> lamda ，a,b != 0$$ KKT condition！ complementary slackness $$L(x,\lambda,v)=f_o(x) + \sum \lambda_i f_i(x) + \sum v_i h_i(x)$$ primitive condition=>$$f_i(x)\leq 0,h_i(x)=0$$ duality condition=> $$\Delta_x L(x,\lambda,v)=0$$ $$complementary condition=>,\lambda_i f_i(x)=0$$ why the complementary condition always equals 0? consider that if lambda!=0,it means that the gredient in this line is contribute the offset to target function,and the point must be satisfy the constraint equation.if the point in the district area(f_i(x)!=0 and <0),the gredient of lines to this internal point will be useness,because the direction is not offset the target function,so the contribution is zero, lambda is 0 ...... $$primitive,minx max{\lambda,v} L(x,\lambda,v),\mathbb{s.t} \lambda \geq 0$$ $$duality, max{\lambda,v} g(\lambda,v)=$max{\lambda,v} min_x L(x,\lambda,v),\mathbb{s.t} \lambda \geq 0$$ $$L(x,\lambda,v)=f_o(x) + \sum \lambda_i f_i(x) + \sum v_i hi(x)$$ $$max{\lambda,v} L(x,\lambda,v)\geq L(x,\lambda,v) \geq minx L(x,\lambda,v)$$ $$A(x)= max{\lambda,v} L(x,\lambda,v)\geq L(x,\lambda,v) \geq min_x L(x,\lambda,v) =l(\lambda,v)$$ $$we can conclude->,A(x)\geq l(\lambda,v) $$ $$A(x)\geq mixx A(x) \geq max{\lambda,v}l(\lambda,v)\geq l(\lambda,v)$$ $$\mathbb{P}= mixx A(x)\geq max{\lambda,v}l(\lambda,v) =\mathbb{D}$$

[jikan1221]

conda and tensorflow env problem,using administer to install [Anaconda]

[jikan1221]

if dir missing file,delete the user/.condr

[jikan1221]

https://tensorflow.google.cn/install/source_windows#cpu

[jikan1221]

module 'pip' has no attribute 'pep425tags'

pip debug --verbose