CSR-DCF視頻目標跟蹤論文筆記（2）——關於濾波器Learning的推導（Augmented Lagrangian方法）

1. 論文基本信息

• 論文標題：Discriminative Correlation Filter with Channel and Spatial Reliability
• 做者：Alan Lukezic等
• 出處：CVPR，2017
• 文章連接：https://arxiv.org/abs/1611.08461
• 補充材料：https://www.semanticscholar.org/paper/Discriminative-Correlation-Filter-with-Channel-and-Luke%C5%BEi%C4%8D-Voj%C4%B1-%C5%99/7b485979c75b46d8c194868c0e70890f4a0f0ede
• 源碼連接：https://github.com/alanlukezic/csr-dcf

這篇筆記主要針對濾波器求解的推導過程進行分析（拉格朗日乘子法），主要參考內容是原文的補充材料，關於論文其餘部分創新點及其總體思路會在後續文章中進行分析。（筆記1的連接：http://blog.csdn.net/discoverer100/article/details/78182306）

2. 濾波器求解目標函數的構建

在多通道狀況下，目標函數爲

$$\begin{array}{c} \mathop {\arg \min }\limits_{\bf{h}} \sum\limits_{d = 1}^{{N_d}} {\left( {{{\left\| {{{\bf{f}}_d} \odot {{\bf{h}}_d} - {\bf{g}}} \right\|}^2} + \lambda {{\left\| {{{\bf{h}}_d}} \right\|}^2}} \right)} \\ {\rm{ = }}\mathop {\arg \min }\limits_{\bf{h}} \sum\limits_{d = 1}^{{N_d}} {\left( {{{\left\| {{\bf{\hat h}}_d^H{\rm{diag}}\left( {{{{\bf{\hat f}}}_d}} \right) - {{\hat g}_d}} \right\|}^2} + \lambda {{\left\| {{{{\bf{\hat h}}}_d}} \right\|}^2}} \right)} \end{array} \tag{1}$$
其中， $\bf{h}$表示濾波器， $d=1toN_d$表示 $N_d$個通道， $\bf{g}$表示指望的響應輸出， $\lambda$表示正則項用於防止過擬合（關於正則項爲何能夠防止過擬合能夠參考： http://www.cnblogs.com/alexanderkun/p/6922428.html）

根據上述(1)式，爲簡化推導過程，將多通道狀況改成單通道狀況模式，則目標函數爲

$$\begin{array}{c} \mathop {\arg \min }\limits_{\bf{h}} {\left\| {{\bf{f}} \odot {\bf{h}} - {\bf{g}}} \right\|^2} + \lambda {\left\| {\bf{h}} \right\|^2}\\ {\rm{ = }}\mathop {\arg \min }\limits_{\bf{h}} {\left\| {{{{\bf{\hat h}}}^H}{\rm{diag}}\left( {{\bf{\hat f}}} \right) - \hat g} \right\|^2} + \lambda {\left\| {{\bf{\hat h}}} \right\|^2} \end{array} \tag{2}$$
引入變量 $\bf{h}_c$並定義約束條件
$${{\bf{h}}_c} - {{\bf{h}}_m} = 0 \tag{3}$$
其中， ${{\bf{h}}_m} \equiv {\bf{m}} \odot {\bf{h}}$，而 $\bf{m}$表示論文中的空間置信圖（spatial reliability map），也能夠理解爲一個mask，具體概念能夠參考前面的一篇文章： http://blog.csdn.net/discoverer100/article/details/78182306，上述(3)式中引入的變量 $\bf{h}_c$能夠先不理會其物理意義，它的主要做用是讓算法可以收斂（論文原文表述：prohibits a closed-form solution），我的猜想：這裏的下標命名爲c，可能就是取constrained的第一個字母。

對(2)式引入上述約束條件，並進一步調整，獲得最終的目標函數

$$\begin{array}{l} \mathop {\arg \min }\limits_{{{\bf{h}}_c},{{\bf{h}}_m}} {\left\| {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - \hat g} \right\|^2} + \frac{\lambda }{2}{\left\| {{{{\bf{\hat h}}}_m}} \right\|^2}\\ s.t.\;{{\bf{h}}_c} - {{\bf{h}}_m} = 0 \end{array} \tag{4}$$
上述的正則項前面多出了一個係數 $1/2$，其主要意圖是求導數後係數能夠變爲 $1$，便於公式書寫。

這樣，公式(4)就是咱們推導的起始表達式。

3. 構建Lagrange表達式

根據上述目標函數，以及Augmented Lagrangian方法（參考Distributed optimization and statistical learning via the alternating direction method of multipliers），構建Lagrang表達式，以下

$${\cal L}\left( {{{{\bf{\hat h}}}_c},{\bf{h}},{\bf{\hat I}}\left| {\bf{m}} \right.} \right) = {\left\| {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - \hat g} \right\|^2} + \frac{\lambda }{2}{\left\| {{{\bf{h}}_m}} \right\|^2} + \left[ {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - {{{\bf{\hat h}}}_m}} \right) + \overline {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - {{{\bf{\hat h}}}_m}} \right)} } \right] + \mu {\left\| {{{{\bf{\hat h}}}_c} - {{{\bf{\hat h}}}_m}} \right\|^2} \tag{5}$$
其中，字母 $\bf{I}$表示Lagrange乘數，字母上面的橫槓表示 共軛矩陣，字母右上方的 $H$表示 共軛轉置矩陣，所以有規律： ${{\bf{\bar A}}^T} = {{\bf{A}}^H}$（後面的推導中可能同時存在兩種表示，須要留意）。將上述(5)式進行向量化表示，可得
$${\cal L}\left( {{{{\bf{\hat h}}}_c},{\bf{h}},{\bf{\hat I}}\left| {\bf{m}} \right.} \right) = {\left\| {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - \hat g} \right\|^2} + \frac{\lambda }{2}{\left\| {{{\bf{h}}_m}} \right\|^2} + \left[ {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right) + \overline {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} } \right] + \mu {\left\| {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right\|^2} \tag{6}$$
不難看出，上述(5)式到(6)式，主要變化就是將變量 ${{{{\bf{\hat h}}}_m}}$的表達式替換爲 ${\sqrt D {\bf{FMh}}}$，其中 $F$表示離散傅里葉變換矩陣，它至關於一個常量， $D$是 $F$的大小（ $F$是一個 $D \times D$的方陣）， $\bf{M}={{\rm{diag}}\left( {{\bf{m}}} \right)}$

將上述(6)式簡單表述爲四個項的和，爲

$${\cal L}\left( {{{{\bf{\hat h}}}_c},{\bf{h}},{\bf{\hat I}}} \right) = {{\cal L}_1} + {{\cal L}_2} + {{\cal L}_3} + {{\cal L}_4} \tag{7}$$
其中，
$$\left\{ \begin{array}{l} {{\cal L}_1} = {\left\| {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - \hat g} \right\|^2} = \left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right){\overline {\left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right)} ^T}\\ {{\cal L}_2} = \frac{\lambda }{2}{\left\| {{{\bf{h}}_m}} \right\|^2}\\ {{\cal L}_3} = {{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right) + \overline {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} \\ {{\cal L}_4} = \mu {\left\| {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right\|^2} = \mu \left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right){\overline {\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} ^T} \end{array} \right. \tag{8}$$

4. 開始優化，首先對h_c求偏導數

對上述公式(4)的優化能夠表述爲下面的迭代過程

$$\begin{array}{l} {\bf{\hat h}}_c^{{\rm{opt}}} = \mathop {\arg \min }\limits_{{{\bf{h}}_c}} L\left( {{{{\bf{\hat h}}}_c},{\bf{h}},{\bf{\hat I}}} \right)\\ {{\bf{h}}^{{\rm{opt}}}} = \mathop {\arg \min }\limits_{\bf{h}} L\left( {{\bf{\hat h}}_c^{{\rm{opt}}},{\bf{h}},{\bf{\hat I}}} \right) \end{array} \tag{9}$$
如今看關於變量 ${{\bf{\hat h}}_c}$的優化，須要令知足 ${\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{\cal L} \equiv 0$，也就是
$${\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_1} + {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_2} + {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_3} + {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_4} \equiv 0 \tag{10}$$
對各個份量求偏導數，有
$$\begin{aligned} {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_1} &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {\left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right){{\overline {\left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right)} }^T}} \right]\\ &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{{\left( {{\bf{\hat f}}} \right)}^H}{{{\bf{\hat h}}}_c} - {\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right){{{\bf{\hat g}}}^H} - {\bf{\hat g}}{\rm{diag}}{{\left( {{\bf{\hat f}}} \right)}^H}{{{\bf{\hat h}}}_c} + {\bf{\hat g}}{{{\bf{\hat g}}}^H}} \right]\\ &= {\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{\left( {{\bf{\hat f}}} \right)^H}{{{\bf{\hat h}}}_c} - {\rm{diag}}\left( {{\bf{\hat f}}} \right){{{\bf{\hat g}}}^H}-0+0\\ &= {\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{\left( {{\bf{\hat f}}} \right)^H}{{{\bf{\hat h}}}_c} - {\rm{diag}}\left( {{\bf{\hat f}}} \right){{{\bf{\hat g}}}^H} \end{aligned} \tag{11}$$

$$\begin{aligned} {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_{\rm{2}}} &= {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}\left[ {\frac{\lambda }{2}{{\left\| {{{\bf{h}}_m}} \right\|}^2}} \right]\\ &= {\rm{0}} \end{aligned} \tag{12}$$

$$\begin{aligned} {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_{\rm{3}}} &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right) + \overline {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} } \right]\\ &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {{{{\bf{\hat I}}}^H}{{{\bf{\hat h}}}_c} - {{{\bf{\hat I}}}^H}\sqrt D {\bf{FMh}} + {{{\bf{\hat I}}}^T}\overline {{{{\bf{\hat h}}}_c}} - {{{\bf{\hat I}}}^T}\overline {\sqrt D {\bf{FMh}}} } \right]\\ &= 0 - 0 + {{{\bf{\hat I}}}^T} - 0\\ &= {\bf{\hat I}} \end{aligned} \tag{13}$$

$$\begin{aligned} {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_{\rm{4}}} &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {\mu \left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right){{\overline {\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} }^T}} \right]\\ &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {\mu \left( {{{{\bf{\hat h}}}_c} \cdot {\bf{\hat h}}_c^H - {{{\bf{\hat h}}}_c} \cdot \sqrt D {{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H} - \sqrt D {\bf{FMh\hat h}}_c^H + \sqrt D {\bf{FMh}} \cdot \sqrt D {{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H}} \right)} \right]\\ &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {\mu \left( {{{{\bf{\hat h}}}_c} \cdot {\bf{\hat h}}_c^H - {{{\bf{\hat h}}}_c} \cdot \sqrt D {{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H} - \sqrt D {\bf{FMh\hat h}}_c^H + D{\bf{FMh}}{{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H}} \right)} \right]\\ &= \mu \left( {{{{\bf{\hat h}}}_c} - 0 - \sqrt D {\bf{FMh}} + 0} \right)\\ &= \mu {{{\bf{\hat h}}}_c} - \mu \sqrt D {\bf{FMh}} \end{aligned} \tag{14}$$

因而，上述公式(10)能夠寫成

$$\begin{array}{c} {\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{\left( {{\bf{\hat f}}} \right)^H}{{{\bf{\hat h}}}_c} - {\rm{diag}}\left( {{\bf{\hat f}}} \right){{{\bf{\hat g}}}^H} + 0 + {\bf{\hat I}} + \mu \left( {{{{\bf{\hat h}}}_c} - 0 - \sqrt D {\bf{FMh}} + 0} \right) \equiv 0\\ {\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{\left( {{\bf{\hat f}}} \right)^H}{{{\bf{\hat h}}}_c} - {\rm{diag}}\left( {{\bf{\hat f}}} \right){{{\bf{\hat g}}}^H} + {\bf{\hat I}} + \mu \left( {{{{\bf{\hat h}}}_c} - 0 - \sqrt D {\bf{FMh}} + 0} \right) \equiv 0 \end{array} \tag{15}$$
回顧公式(6)，咱們曾將變量 ${{{{\bf{\hat h}}}_m}}$的表達式替換爲 ${\sqrt D {\bf{FMh}}}$，如今咱們將它替換回來，得
$${\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{\left( {{\bf{\hat f}}} \right)^H}{{\bf{\hat h}}_c} - {\rm{diag}}\left( {{\bf{\hat f}}} \right){{\bf{\hat g}}^H} + {\bf{\hat I}} + \mu \left( {{{{\bf{\hat h}}}_c} - 0 - {{{\bf{\hat h}}}_m} + 0} \right) \equiv 0 \tag{16}$$
針對 ${{\bf{\hat h}}_c}$合併同類項，得

h^c⋅[diag(f^)diag(f^)H+μ]=μh^m+diag(f^)g^H−