BACF跟踪算法与代码一致的推导

论文地址

看BACF论文中的公式，有很多地方的推导感觉不对劲，推导出的结果也和代码不一致，自己又推导了一遍，得到了与代码一致的推导。

损失函数
$$E(\mathbf{h})=\frac{1}{2}\sum _{j=1}^T||\mathbf{y}(j)-\sum _{k=1}^K{\mathbf{h}}_k{\mathbf{P}}^T{\mathbf{x}}_k[\nabla {\tau }_j]|{|}_2^2+\frac{\lambda }{2}||{\mathbf{h}}_k|{|}_2^2$$
其中$T$为$\mathbf{x}$的像素个数。

转到频域，即论文的公式(4)，根据帕斯瓦尔定理，前面应该多一个$\frac{1}{T}$的系数。
$$\begin{gathered} E(\mathbf{h},\hat{\mathbf{g}})=\frac{1}{2T}||\hat{\mathbf{y}}-\hat{\mathbf{X}}\hat{\mathbf{g}}|{|}_2^2+\frac{\lambda }{2}||\mathbf{h}|{|}_2^2 \\ s.t\hat{\mathbf{g}}=\sqrt{T}({\mathbf{F}\mathbf{P}}^T\otimes {\mathbf{I}}_{\mathbf{K}})\mathbf{h} \end{gathered}$$
其中$\hat{}$表示傅里叶变换，$\hat{\mathbf{a}}=\sqrt{T}\mathbf{F}\mathbf{a}$，

$\hat{\mathbf{X}}=[diag({\mathbf{x}}_1{)}^T,...diag({\mathbf{x}}_K{)}^T]$，$\mathbf{h}=[{\mathbf{h}}_1^T,...,{\mathbf{h}}_K^T{]}^T$ , $\hat{\mathbf{g}}=[\widehat{{\mathbf{g}}_1^T},...,\widehat{{\mathbf{g}}_K^T}{]}^T$。

构造增广拉格朗日函数
$$\begin{array}{rl}\mathcal{L}(\hat{\mathbf{g}},\mathbf{h},\hat{\zeta })=& \frac{1}{2T}\Vert \hat{\mathbf{y}}-\hat{\mathbf{X}}\hat{\mathbf{g}}{\Vert }_2^2+\frac{\lambda }{2}\Vert \mathbf{h}{\Vert }_2^2\\ & +{\hat{\zeta }}^{\top }\left(\hat{\mathbf{g}}-\sqrt{T}\left(\mathbf{F}{\mathbf{P}}^{\top }\otimes {\mathbf{I}}_K\right)\mathbf{h}\right)\\ & +\frac{\mu }{2}{\Vert \hat{\mathbf{g}}-\sqrt{T}\left(\mathbf{F}{\mathbf{P}}^{\top }\otimes {\mathbf{I}}_K\right)\mathbf{h}\Vert }_2^2\end{array}$$
使用交替方向乘子法(ADMM)求解。
子问题$\mathbf{h}=\underset{\mathbf{h}}{\mathrm{argmin}}\mathcal{L}(\hat{\mathbf{g}},\mathbf{h},\hat{\zeta })$，对$\mathbf{h}$求偏导。
$$\begin{gathered} \frac{\partial \mathcal{L}}{\partial \mathbf{h}}=\lambda \mathbf{h}-\sqrt{T}({\mathbf{F}\mathbf{P}}^T\otimes {\mathbf{I}}_K{)}^T\hat{\zeta }-\mu \sqrt{T}({\mathbf{F}\mathbf{P}}^T\otimes {\mathbf{I}}_K{)}^T(\hat{\mathbf{g}}-\sqrt{T}({\mathbf{F}\mathbf{P}}^T\otimes {\mathbf{I}}_K)\mathbf{h}) \\ =\lambda \mathbf{h}-T\zeta -\mu T\mathbf{g}+\mu T\mathbf{h}=0 \end{gathered}$$
其中$\mathbf{g}=\frac{1}{\sqrt{T}}({\mathbf{P}\mathbf{F}}^T\otimes {\mathbf{I}}_K)\hat{\mathbf{g}}$, $\zeta =\frac{1}{\sqrt{T}}({\mathbf{P}\mathbf{F}}^T\otimes {\mathbf{I}}_K)\hat{\zeta }$，即傅里叶反变换。

解得
$$\mathbf{h}=\frac{\mu \mathbf{g}+\zeta }{\frac{\lambda }{T}+\mu }$$
与论文(6)式不一致。

子问题$\hat{\mathbf{g}}=\underset{\hat{\mathbf{g}}}{\mathrm{argmin}}\mathcal{L}(\hat{\mathbf{g}},\mathbf{h},\hat{\zeta })$， 由于每个像素的值独立，将问题分解为$T$个子问题，$t=1,2,...,T$。
$$\begin{array}{rl}\hat{\mathbf{g}}(t{)}^{\ast }=& \arg \underset{\hat{\mathbf{g}}(t)}{\min }\{\frac{1}{2T}{\Vert \hat{\mathbf{y}}(t)-\hat{\mathbf{x}}(t{)}^{\top }\hat{\mathbf{g}}(t)\Vert }_2^2\\ & +\hat{\zeta }(t{)}^{\top }(\hat{\mathbf{g}}(t)-\hat{\mathbf{h}}(t))\\ & +\frac{\mu }{2}\Vert \hat{\mathbf{g}}(t)-\hat{\mathbf{h}}(t){\Vert }_2^2\}\end{array}$$
求偏导，令为0
$$-\frac{1}{T}\hat{\mathbf{x}}(t)(\hat{\mathbf{y}}(t)-\hat{\mathbf{x}}(t{)}^T\hat{\mathbf{g}}(t))+\hat{\zeta }+\mu (\hat{\mathbf{g}}(t)-\hat{\mathbf{h}}(t))=0$$

$$(\hat{\mathbf{x}}(t)\hat{\mathbf{x}}(t{)}^T+\mu T\mathbf{I})\hat{\mathbf{g}}(t)=\hat{\mathbf{x}}(t)\hat{\mathbf{y}}(t)-T\hat{\zeta }+\mu T\hat{\mathbf{h}}(t)$$

这步得到的结果和论文公式(9)一致，也解答了$T$从哪推出来的疑惑。

得到
$$\begin{array}{rl}\hat{\mathbf{g}}(t{)}^{\ast }=& {\left(\hat{\mathbf{x}}(t)\hat{\mathbf{x}}(t{)}^{\top }+T\mu {\mathbf{I}}_K\right)}^{-1}\\ & (\hat{\mathbf{y}}(t)\hat{\mathbf{x}}(t)-T\hat{\zeta }(t)+T\mu \hat{\mathbf{h}}(t))\end{array}$$
根据Sherman-Morrison公式
$${\left(\mathbf{A}+{\mathbf{u}\mathbf{v}}^T\right)}^{-1}={\mathbf{A}}^{-1}-\frac{{\mathbf{A}}^{-1}\mathbf{u}{\mathbf{v}}^T{\mathbf{A}}^{-1}}{1+{\mathbf{v}}^T{\mathbf{A}}^{-1}\mathbf{u}}$$
令$\mathbf{q}=\hat{\mathbf{y}}(t)\hat{\mathbf{x}}(t)-T\hat{\zeta }(t)+T\mu \hat{\mathbf{h}}(t)$

得
$$\hat{\mathbf{g}}(t{)}^{\ast }=\frac{1}{\mu T}(\mathbf{I}-\frac{\hat{\mathbf{x}}(t)\hat{\mathbf{x}}(t{)}^T}{\mu T+\hat{\mathbf{x}}(t{)}^T\hat{\mathbf{x}}(t)})\mathbf{q}$$
整理一下应该是
$$\begin{array}{rl}\hat{\mathbf{g}}(t{)}^{\ast }=& \frac{1}{\mu T}(\hat{\mathbf{y}}(t)\hat{\mathbf{x}}(t)-T\hat{\mathbf{\zeta }}(t)+\mu T\hat{\mathbf{h}}(t))\\ & -\frac{\hat{\mathbf{x}}(t)}{\mu Tb}\left(\hat{\mathbf{y}}(t){\hat{s}}_{\mathbf{x}}(t)-T{\hat{s}}_{\zeta }(t)+\mu T{\hat{s}}_{\mathbf{h}}(t)\right)\end{array}$$
其中
$${\hat{s}}_{\mathbf{x}}(t)=\hat{\mathbf{x}}(t{)}^{\top }\hat{\mathbf{x}},{\hat{s}}_{\zeta }(t)=\hat{\mathbf{x}}(t{)}^{\top }\hat{\mathbf{\zeta }},{\hat{s}}_{\mathbf{h}}(t)=\hat{\mathbf{x}}(t{)}^{\top }\hat{\mathbf{h}},b={\hat{s}}_{\mathbf{x}}(t)+T\mu$$
再看看代码里的求解过程，是不是一致了？

 g_f = single(zeros(size(xf)));
    h_f = g_f;
    l_f = g_f;
    mu    = 1;
    betha = 10;
    mumax = 10000;
    i = 1;
    
    T = prod(use_sz);
    S_xx = sum(conj(model_xf) .* model_xf, 3);
    params.admm_iterations = 2;
    %   ADMM
    while (i <= params.admm_iterations)
        %   solve for G- please refer to the paper for more details
        B = S_xx + (T * mu);
        %B = S_xx + T*mu;
        S_lx = sum(conj(model_xf) .* l_f, 3);
        S_hx = sum(conj(model_xf) .* h_f, 3);
        g_f = (((1/(T*mu)) * bsxfun(@times, yf, model_xf)) - ((1/mu) * l_f) + h_f) - ...
            bsxfun(@rdivide,(((1/(T*mu)) * bsxfun(@times, model_xf, (S_xx .* yf))) - ((1/mu) * bsxfun(@times, model_xf, S_lx)) + (bsxfun(@times, model_xf, S_hx))), B);
        
        %g_f = (((1/(mu)) * bsxfun(@times, yf, model_xf)) - ((1/mu) * l_f) + h_f) - ...
        %    bsxfun(@rdivide,(((1/(mu)) * bsxfun(@times, model_xf, (S_xx .* yf))) - ((1/mu) * bsxfun(@times, model_xf, S_lx)) + (bsxfun(@times, model_xf, S_hx))), B);
        
        %   solve for H
        h = (T/((mu*T)+ params.admm_lambda))* ifft2((mu*g_f) + l_f);
        %h = (sqrt(T)/((mu*sqrt(T))+ params.admm_lambda))* ifft2((mu*g_f) + l_f);
        [sx,sy,h] = get_subwindow_no_window(h, floor(use_sz/2) , small_filter_sz);
        t = single(zeros(use_sz(1), use_sz(2), size(h,3)));
        t(sx,sy,:) = h;
        h_f = fft2(t);
        
        %   update L
        l_f = l_f + (mu * (g_f - h_f));
        
        %   update mu- betha = 10.
        mu = min(betha * mu, mumax);
        i = i+1;
    end