Accurate Scale Estimation With IoU and Distance Between Centroids for Object Tracking
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摘要: 通过分析基于交并比(Intersection over union, IoU)预测的尺度估计模型的梯度更新过程, 发现其在训练和推理过程仅将IoU作为度量, 缺乏对预测框和真实目标框中心点距离的约束, 导致外观模型更新过程中模板受到污染, 前景和背景分类时定位出现偏差. 基于此发现, 构建了一种结合IoU和中心点距离的新度量NDIoU (Normalization distance IoU), 在此基础上提出一种新的尺度估计方法, 并将其嵌入判别式跟踪框架. 即在训练阶段以NDIoU为标签, 设计了具有中心点距离约束的损失函数监督网络的学习, 在线推理期间通过最大化NDIoU微调目标尺度, 以帮助外观模型更新时获得更加准确的样本. 在七个数据集上与相关主流方法进行对比, 所提方法的综合性能优于所有对比算法. 特别是在GOT-10k数据集上, 所提方法的AO、$S{R}_{0.50}$和$ S{R}_{0.75} $三个指标达到了65.4%、78.7%和53.4%, 分别超过基线模型4.3%、7.0%和4.2%.Abstract: This paper first analyzes the gradient update process of the scale estimation model of intersection over union (IoU) prediction in detail, and finds that when the IoU is used as a metric in the training and inference process, the target scale estimation in the tracking process is inaccurate due to the absence of the constraint on the distance between the two centroids. As a result, the template is polluted in the updating process of the object appearance model, which cannot discriminate the target and environment. With this insight, we propose a new metric NDIoU (normalization distance IoU) that combines the IoU and distance between two centroids to estimate the target scale and proposes a new scale estimation method, which is embedded into the discriminative tracking framework. Using NDIoU as the label to supervise the distance between centroids, it is incorporated into the loss function to facilitate the learning of the network. During online inference, NDIoU is maximized to fine-tune the target scale. Finally, the proposed method is embedded into the discriminative tracking framework and compared with related state-of-the-art methods on seven data sets. The extensive experiments demonstrate that our method outperforms all the state-of-the-art algorithms. Especially, on the GOT-10k dataset, our method achieves 65.4%, 78.7% and 53.4% on the three metrics of AO, $S{R}_{0.50}$ and $ S{R}_{0.75} $, which are better than the baseline by 4.3%, 7.0% and 4.2%, respectively.
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近年来, 张量方法在控制理论及其应用的各个领域得到了广泛的应用, 如基于张量数据表示的深度学习模型[1]、基于非负张量分析的时序链路预测方法[2]、局部图像描述中基于结构张量的HDO (Histograms of dominant orientations)算法[3]、二维解析张量投票算法[4]等.
张量动力系统已经广泛应用于Volterra系统识别[5]、张量乘积TP (Tensor product)模型变换[6]、人体动作识别[7-8]和塑性模型[9]等各个领域.因此, 由于在张量常微分方程解中的关键作用, 张量指数函数的计算已经成为一个重要的研究领域.
考虑如下常微分方程[9]
$$ \begin{align}\label{equation1-1} \left\{ \begin{array}{ll} \dot{\mathcal{Y}}(t)=\mathcal{A}\mathcal{Y}(t) \\ {\mathcal{Y}}(t_0)=\mathcal{Y}_{0} \end{array} \right. \end{align} $$ (1) 这里$\mathcal{A}$和$\mathcal{Y}_{0}$是给定张量, 一般是非对称的, 那么关于系统(1)的张量指数函数${\rm exp}(\cdot)$有如下唯一解:
$$ \begin{align}\label{equation1-2} {\mathcal{Y}}(t)={\rm exp}[(t-t_0)\mathcal{A}]{\mathcal{Y}}_{0} \end{align} $$ (2) 对于一般的张量$\mathcal{A}$, 它的指数函数可以表示为它的级数形式:
$$ \begin{align*}{\rm exp}(\mathcal{A})=\sum\limits_{n=0}^\infty \frac{1}{n!}\mathcal{A}^n\end{align*} $$ 目前计算张量指数函数(2)通常使用的方法是截断法[10], 其实质就是将张量指数函数展开成无穷级数后, 取前$n_{\rm{max}}$项, 得到近似解, 即:
$$ \begin{align*}{\rm exp}(\mathcal{A}) \approx \sum\limits_{n=0}^{n_{\rm{max}}} \frac{1}{n!}\mathcal{A}^n\end{align*} $$ 张量指数函数选取的项数$n_{\rm{max}}$受限于所需的精度:
$$ \begin{align*}\frac{1}{n_{\rm{max}}!}\|\mathcal{A}^{n_{\rm{max}}}\|\le \epsilon_{\rm{tol}}\end{align*} $$ 显然有截断法的精度与张量指数函数选取的项数$n_{\rm{max}}$有关, 保留的项数越多, 精度越高, 但是需要进行的张量乘积次数也就越多; 保留的项数越少, 计算量越少, 但是精度也就越低.因此, 计算张量指数函数的截断法有待改进.
为研究上述问题, 本文首次定义了张量的一种广义逆, 并以此为基础构造了张量广义逆Padé逼近GITPA (Generalized inverse tensor Padé approximation)的一种$\varepsilon$-算法. GITPA方法的优势在于:在计算过程中, 不必用到张量的乘积, 也不要计算张量的逆, 另外, 该方法对奇异张量也是适用的.目前在国内外, 关于计算张量的逆还没有找到一种比较可行的计算方法.作为GITPA方法的一个重要应用, 本文在后面给出计算张量指数函数的数值实验, 来说明$\varepsilon$-算法的有效性.
本文组织如下:第1节简单介绍本文用到的张量基础知识, 并定义张量的一种广义逆; 第2节, 首先给出广义逆张量Padé逼近的定义, 并以此为基础给出张量$\varepsilon$-算法; 第3节将$\varepsilon$-算法用来计算张量指数函数值, 并与通常使用的的级数截断法相比较; 最后给出简单的小结.
1. 一种张量广义逆
1.1 张量基础知识
本节介绍本文将要用到的张量基础知识.张量是一种多维数组, 其中向量是一阶张量, 矩阵是二阶张量, 特别地, 一个$ p $阶张量拥有$ p $个下标, 是$ p $个拥有独立坐标系的向量空间的外积(张量积).一个$ p $阶$ n_1\times n_2\times \cdots \times n_p $维张量可以表示为:
$$ \mathcal{A} = (a_{i_{1}i_{2}\cdots i_{p}})\in {\bf C}^{n_{1}\times n_{2}\times\cdots \times n_{p}} $$ 文献[11]提出了张量的切片方法.对一个三阶张量, 可以固定其中任意一个下标, 从而得到该张量的一种表示形式.设$ \mathcal{A} = (a_{i_{1}i_{2}i_{3}})\in {\bf C}^{2\times 2\times3} $, 固定其第三个下标, 则张量可以表示为
$$ \begin{align*} \mathcal{A} = \, &\left[ \begin{array}{c|c|c} \mathcal{A}_{1} &\mathcal{A}_{2}& \mathcal{A}_{3} \end{array}\right] = \nonumber\\ &\left[ \begin{array}{cc|cc|cc} a_{111} & a_{121} & a_{112} & a_{122} & a_{113} & a_{123}\\ a_{211} & a_{221} & a_{212} & a_{222} & a_{213} & a_{223}\\ \end{array} \right] \end{align*} $$ 下面定义两个三阶张量的$ t $-积, 可以通过递归自然地推广到高阶张量的$ t $-积.
定义1[12]. (块循环矩阵)令$ \mathcal{A} \in {\bf R}^{l\times p\times n} $, 则$ \mathcal{A} $的块循环矩阵被定义为
$$ \begin{equation*} bcirc(\mathcal{A}) = {\left[ \begin{array}{ccccc} \mathcal{A}_{1} & \mathcal{A}_{n} & \mathcal{A}_{n-1} & \cdots & \mathcal{A}_{2} \\ \mathcal{A}_{2} & \mathcal{A}_{1} & \mathcal{A}_{n} & \cdots & \mathcal{A}_{3} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \mathcal{A}_{n} & \mathcal{A}_{n-1} & \cdots & \mathcal{A}_{2} & \mathcal{A}_{1} \\ \end{array} \right]}_{ln\times pn} \end{equation*} $$ 定义一个展开算子: $ unfold(\cdot)$[12], 用以下方式将一个${p\times m\times n}$的张量展开成一个${pn\times m}$的矩阵:
$$ \begin{equation*} unfold(\mathcal{B}) = \left[ \begin{array}{c} \mathcal{B}_{1}^{\rm{T}} \mathcal{B}_{2}^{\rm{T}} \cdots \mathcal{B}_{n}^{\rm{T}}\\ \end{array} \right]^{\rm{T}} \end{equation*} $$ 这里$fold(\cdot)$[12]是它的逆算子, 它会将一个$ {pn\times m} $的矩阵转化成$ {p\times m\times n} $的张量.因此,
$$ \begin{equation*} fold(unfold(\mathcal{B})) = \mathcal{B} \end{equation*} $$ 定义2[12-13]. (张量$ t $-积)令$ \mathcal{A} $是一个$ l\times p\times n $的张量, $ \mathcal{B} $是一个$ p\times m\times n $的张量, 则张量$ t $-积$ \mathcal{A}*\mathcal{B} $将得到一个$ l\times m\times n $的张量, 定义如下:
$$ \begin{equation*} \mathcal{A}\ast \mathcal{B} = fold(bcirc(\mathcal{A}) \cdot unfold(B)) \end{equation*} $$ 例1. 设$ \mathcal{A}, \mathcal{B}\in {\bf R}^{2\times 2\times3} $, 现固定其第三个下标, 分别产生
$$ \begin{align*} \begin{array}{rl} \mathcal{A} = &\left[ \begin{array}{cc|cc|cc} 1 & 2 & 5 & 6 & 9 & 10 \\ 3 & 4 & 7 & 8 & 11 & 12 \\ \end{array} \right]\\ \mathcal{B} = &\left[ \begin{array}{cc|cc|cc} 1 & 2 & 4 & 3 & 1 & 0 \\ 3 & 4 & 2 & 1 & 0 & 1 \\ \end{array} \right]\end{array} \end{align*} $$ 则由定义2得到
$$ \begin{align*} \begin{array}{rl} \mathcal{A}\ast\! \mathcal{B} = \\&fold\left[ \begin{array}{c} \left[ \begin{array}{ccc} \mathcal{A}_{1} & \mathcal{A}_{3} & \mathcal{A}_{2} \\ \mathcal{A}_{2} & \mathcal{A}_{1} & \mathcal{A}_{3} \\ \mathcal{A}_{3} & \mathcal{A}_{2} & \mathcal{A}_{1} \\ \end{array} \right]\cdot\left[ \begin{array}{c} \mathcal{B}_{1} \\ \mathcal{B}_{2} \\ \mathcal{B}_{3} \\ \end{array} \right] \\ \end{array} \right] = \\ &fold\left[ \begin{array}{c} \mathcal{A}_{1}\mathcal{B}_{1}+\mathcal{A}_{3}\mathcal{B}_{2}+\mathcal{A}_{2}\mathcal{B}_{3} \\ \mathcal{A}_{2}\mathcal{B}_{1}+\mathcal{A}_{1}\mathcal{B}_{2}+\mathcal{A}_{3}\mathcal{B}_{3} \\ \mathcal{A}_{3}\mathcal{B}_{1}+\mathcal{A}_{2}\mathcal{B}_{2}+\mathcal{A}_{1}\mathcal{B}_{3} \\ \end{array} \right] = \\ &\left[ \begin{array}{cc|cc|cc} 68 & 53 & 40 & 49 & 72 & 81 \\ 90 & 75 & 62 & 71 & 94 & 103 \\ \end{array} \right] \end{array} \end{align*} $$ 下面定义张量的范数.令$ \mathcal{A}\in {\bf C}^{n_{1}\times n_{2}\times\cdots \times n_{p}} $, 张量的范数就等于它所有元素平方和的平方根[11], 即
$$ \begin{align} \|\mathcal{A}\| = \sqrt{\sum^{n_{1}}_{i_{1} = 1}\sum^{n_{2}}_{i_{2} = 1}\cdots\sum^{n_{p}}_{i_{p} = 1}a^2_{i_{1}i_{2}\cdots i_{p}}} \end{align} $$ (3) 这和矩阵的$ Frobenius $-模是类似的.两个大小相同的张量$ \mathcal{A} $, $ \mathcal{B}\in {\bf C}^{n_{1}\times n_{2}\times\cdots \times n_{p}} $的内积等于它们对应元素的乘积的和[11], 即
$$ \begin{align} (\mathcal{A}, \mathcal{B}) = \sum^{n_{1}}_{i_{1} = 1}\sum^{n_{2}}_{i_{2} = 1}\cdots\sum^{n_{p}}_{i_{p} = 1}a_{i_{1}i_{2}\cdots i_{p}}b_{i_{1}i_{2}\cdots i_{p}} \end{align} $$ (4) 于是显然成立$ (\mathcal{A}, \mathcal{A}^\ast) = \|\mathcal{A}\|^2 $, 其中记号"$ \ast $''表示取复共轭.
1.2 张量广义逆
参考复数、向量和矩阵的倒数, 容易得到以下结论:
1) 若$ b $是一个复数, $ b\in {\bf C} $, 则$ bb^\ast = |b|^2 $, $ {1}/{b} = b^{-1} = {b^\ast}/{|b|^2} $;
2) 若$ {\pmb v} $是一个向量, $ {\pmb v}\in {\bf C}^n $, 则$ {\pmb v}\cdot {\pmb v}^\ast = |{\pmb v}|^2 $, $ {1}/{{\pmb v}} = {\pmb v}^{-1} = {{\pmb v}^\ast}/{|{\pmb v}|^2} $ (见Graves-Morris [14]);
3) 若$ A = a_{ij} $, $ B = b_{ij}\in {\bf C}^{s\times t} $, 则$ A\cdot B = \sum^s_{i = 1}\sum^t_{j = 1}(a_{ij}b_{ij})\in {\bf C} $, $ {1}/{A} = A_r^{-1} = {A^\ast}{\|A\|^2} $ (见顾传青[15-17]).因此, 下面张量的广义逆可以看作是向量和矩阵的推广:
定义3. 令$ \mathcal{A}, \mathcal{B}\in {\bf C}^{n_1\times n_2\times \cdots\times n_p} $, 其内积为
$$ \begin{equation*} (\mathcal{A}, \mathcal{B}) = \sum\limits^{n_{1}}_{i_{1} = 1}\sum\limits^{n_{2}}_{i_{2} = 1}\cdots\sum\limits^{n_{p}}_{i_{p} = 1}a_{i_{1}i_{2}\cdots i_{p}b_{i_{1}i_{2}\cdots i_{p}}} \end{equation*} $$ $ \mathcal{A}\cdot\mathcal{A}^\ast = \parallel\mathcal{A}\parallel^2 $, 则张量$ \mathcal{A} $的广义逆被定义为
$$ \begin{align} \begin{aligned} \mathcal{A}_r^{-1}& = \frac{1}{\mathcal{A}} = \frac{\mathcal{A}^\ast}{\parallel\mathcal{A}\parallel^2}, \mathcal{A}\neq 0, \mathcal{A}\in {\bf C}^{n_1\times n_2\times \cdots\times n_p} \end{aligned} \end{align} $$ (5) 其中张量的范数由式(3)给出.
引理1. 令$ \mathcal{A}, \mathcal{B}\in {\bf C}^{n_1\times n_2\times \cdots\times n_p}, \mathcal{A}, \mathcal{B}\neq0 $且$ b\in {\bf R}, b\neq0 $, 则下列关系成立:
1) $ \frac{b}{\mathcal{A}} = \frac{1}{\mathcal{B}}\Longleftrightarrow\mathcal{A} = b\mathcal{B} $;
2) $ (\mathcal{A}^{-1}_r)^{-1}_r = \mathcal{A}, (b\mathcal{A})^{-1}_r = \frac{1}{b}\mathcal{A}^{-1}_r $.
证明. 根据定义3, 结论2)是显然的.现在证明结论1).因为$ \mathcal{A}, \mathcal{B}\neq0 $, 从$ \mathcal{A} = b\mathcal{B} $, 容易推得$ \frac{b}{\mathcal{A}} = \frac{1}{\mathcal{B}} $.利用定义3, 得到引理1的左端, $ \frac{b\mathcal{A}^\ast}{\|\mathcal{A}\|^2} = \frac{\mathcal{B}^\ast}{\|\mathcal{B}\|^2} $, 即
$$ \begin{align*} \mathcal{A}^\ast = \frac{\parallel\mathcal{A}\parallel^2}{b\parallel\mathcal{B}\parallel^2}\mathcal{B}^\ast, \mathcal{A} = \frac{\parallel\mathcal{A}\parallel^2}{b\parallel\mathcal{B}\parallel^2}\mathcal{B} \end{align*} $$ 因此
$$ \begin{align*} \parallel\mathcal{A}\parallel^2 = \mathcal{A}\cdot\mathcal{A}^\ast = \frac{\parallel\mathcal{A}\parallel^4\parallel\mathcal{B}\parallel^2}{b^2\parallel\mathcal{B}\parallel^4} \end{align*} $$ 即$ b^2 = \frac{\parallel\mathcal{A}\parallel^2}{\parallel\mathcal{B}\parallel^2} $, 所以$ \mathcal{A} = b\mathcal{B} $.
2. 张量广义逆$ \pmb{\varepsilon} $-算法
2.1 广义逆张量Padé逼近的定义
设$ f(t) $是给定的张量多项式, 其系数为张量, 即
$$ \begin{equation} f(t) = \mathcal{C}_0+\mathcal{C}_1t+\mathcal{C}_2t^2+\cdots+\mathcal{C}_nt^n+\cdots \end{equation} $$ (6) $$ \begin{equation*} \mathcal{C}_i = (c_i^{(i_1i_2\cdots i_p)})\in {\bf C}^{n_{1}\times n_{2}\times\cdots \times n_{p}}, t\in {\bf C} \end{equation*} $$ 定义4. 令$ C^{n_1\times n_2\times \cdots\times n_p}[t] $是一个$ p $阶张量的多项式集合, 维数分别为$ n_1\times n_2\times \cdots\times n_p $.一个张量多项式$ \mathcal{A}(t) = (a_{i_1i_2 \cdots i_p}(t))\in {\bf C}^{n_1\times n_2\times \cdots\times n_p}[t] $是$ m $阶的, 表示为$ \partial\mathcal{A}(t) = m $, 若$ \partial(a_{i_1i_2 \cdots i_p}(t))\leq m $对于所有$ i_i = 1, 2, \cdots, n_i, 1\leq i\leq p $都成立, 且$ \partial(a_{(i_1i_2 \cdots i_p)}(t)) = m $对于某些$ i_i = 1, 2, \cdots, n_i, 1\leq i\leq p $成立.
定义5. 定义$ [\frac{n}{2k}] $型广义逆张量Padé逼近(GITPA)是一个张量有理函数
$$ \begin{equation} \mathcal{R}(t) = \frac{\mathcal{P}(t)}{q(t)} \end{equation} $$ (7) 其中, $ \mathcal{P}(t) $是一个张量多项式, $ q(t) $是一个数量多项式, 满足以下条件:
1) $ \partial{ \mathcal{P}(t)}\leq n, \partial{q(t)} = 2k; $
2) $ q(t)\mid \parallel\mathcal{P}(t)\parallel^2; $
3) $ q(t)f(t)-\mathcal{P}(t) = O(t^{n+1}); $
其中, $ \mathcal{P}(t) = (p_{(i_1i_2 \cdots i_p)}(t))\in {\bf C}^{n_1\times n_2\times \cdots \times n_p} $, 其范数$ \|\mathcal{P}(t)\|^2 $由式(3)给出, 而整除性条件2)表明分母数量多项式$ q(t) $能够整除分子张量多项式$ \mathcal{P}(t) $范数的平方.
2.2 张量$ \pmb\varepsilon $-算法
给定张量多项式(6), 根据张量广义逆(5), 定义张量$ \varepsilon $-算法如下:
$$ \begin{align*} &\varepsilon_{-1}^{(j)} = 0 , j = 0, 1, 2, \cdots \end{align*} $$ (8) $$ \begin{align*} &\varepsilon_0^{(j)} = \sum\limits_{i = 0}^j C_i z^i , j = 0, 1, 2, \cdots \end{align*} $$ (9) $$ \begin{align*} &\varepsilon_{k+1}^{(j)} = \varepsilon_{k-1}^{(j+1)}+(\varepsilon_k^{(j+1)}-\varepsilon_k^{(j)})^{-1} , j, k\ge0 \end{align*} $$ (10) 类似于矩阵情况, 根据式(8)$ \, \sim\, $(10), 上述张量$ \varepsilon $-算法组成了一个称为$ \varepsilon $-表的二维数组:
其中, 根据$ \varepsilon $-算法产生的元素$ \varepsilon_{k+1}^{(j)} $, 它的上标$ j $表示斜行数, 下标$ k+1 $表示列数.每个元素的产生与一个菱形有关, 例如, 由$ \varepsilon_0^{(1)}, \varepsilon_1^{(0)}, \varepsilon_1^{(1)}, \varepsilon_2^{(0)} $所组成的菱形, 现在要计算$ \varepsilon_2^{(0)} $, 由式(10), 将$ \varepsilon_1^{(1)} $减去$ \varepsilon_1^{(0)} $, 再取广义逆(3), 然后将结果加上$ \varepsilon_0^{(1)} $, 从而得到$ \varepsilon_2^{(0)} $.
定理1. (恒等定理)利用张量广义逆(5), 根据$ \varepsilon $-算法(8)$ \, \sim\, $(10)构造$ \varepsilon_{2k}^{(j)} $, 如果在计算过程中没有出现分母为零的情形, 则广义逆Padé逼近$ [\frac{j+2k}{2k}]_f $存在, 且成立恒等式:
$$ \begin{equation} \varepsilon_{2k}^{(j)} = \left[\frac{j+2k}{2k}\right]_f, j, k\ge 0 \end{equation} $$ (11) 证明. 该证明与矩阵情况类似, 可以参考文献[15].
设张量指数函数为
$$ \begin{equation} {\rm exp}(\mathcal{A}t) = \mathcal{I}+\mathcal{A}t+\frac{1}{2!}\mathcal{A}^2t^2+\frac{1}{3!}\mathcal{A}^3t^3+\cdots \end{equation} $$ (12) 给出计算张量指数函数(9)的张量$ \varepsilon $-算法如下:
算法1. (计算张量指数函数的$ \varepsilon $-算法):
输入:张量$ \mathcal{A} $、自变量$ t $和需要逼近的阶数$ j_{\rm{max}} $和$ k_{\rm{max}} $ (偶数)的值.
1) 计算$ \varepsilon $-表的第一列,
$ \varepsilon_{-1}^{(j)} = 0, j = 0, 1, 2, \cdots, j_{\rm{max}}+1 $.
2) 计算$ \varepsilon $-表的第二列,
$ \varepsilon_0^{(j)} = \sum_{i = 0}^j \frac{1}{i!}\mathcal{A}^it^i, j = 0, 1, 2, \cdots, j_{\rm{max}} $.
3) 逐列计算$ \varepsilon $-表的第三列至第$ k_{\rm{max}}+2 $列,
$ \varepsilon_{k+1}^{(j)} = \varepsilon_{k-1}^{(j+1)}+(\varepsilon_k^{(j+1)}-\varepsilon_k^{(j)})^{-1}, j, k\ge0 $.
输出:计算结果$ \varepsilon_{2k}^{(j)} = [\frac{j+2k}{2k}]_e $, 即为所求张量指数的$ [\frac{j+2k}{2k}] $型广义逆Padé逼近, 其中在第2)步计算$ \mathcal{A}^i $时用到定义1给出的张量$ t $-积.
2.3 张量$ \pmb\varepsilon $-算法的常用格式
给定张量指数函数(12), 下面给出张量$ \varepsilon $-算法的几个常用格式:
格式Ⅰ: $ \varepsilon_2^{-1} = [\frac{1}{2}]_e = \varepsilon_{0}^{(0)}+(\varepsilon_1^{(0)}-\varepsilon_1^{(-1)})^{-1} $, 其中
$$ \begin{align*} &\varepsilon_{0}^{(0)} = \mathcal{I}, \varepsilon_{0}^{(1)} = \mathcal{I}+\mathcal{A}t\\ &\varepsilon_{1}^{(0)} = (\varepsilon_0^{(1)}-\varepsilon_0^{(0)})^{-1} = \frac{1}{\mathcal{A}t}\\ &\varepsilon_{0}^{(-1)} = 0, \varepsilon_{1}^{(-1)} = (\varepsilon_0^{(0)}-\varepsilon_0^{(-1)})^{-1} = \frac{1}{\mathcal{I}} \end{align*} $$ 格式Ⅱ: $ \varepsilon_2^{0} = [\frac{2}{2}]_e = \varepsilon_{0}^{(1)}+(\varepsilon_1^{(1)}-\varepsilon_1^{(0)})^{-1} $, 其中
$$ \begin{align*} &\varepsilon_{1}^{(1)} = (\varepsilon_0^{(2)}-\varepsilon_0^{(1)})^{-1} = \frac{2}{\mathcal{A}^2t^2}\\ &\varepsilon_{0}^{(2)} = \mathcal{I}+\mathcal{A}t+\frac{\mathcal{A}^2t^2}{2} \end{align*} $$ 格式Ⅲ: $ \varepsilon_2^{1} = [\frac{3}{2}]_e = \varepsilon_{0}^{(2)}+(\varepsilon_1^{(2)}-\varepsilon_1^{(1)})^{-1} $, 其中
$$ \begin{align*} &\varepsilon_{1}^{(2)} = (\varepsilon_0^{(3)}-\varepsilon_0^{(2)})^{-1} = \frac{6}{\mathcal{A}^3t^3}\\ &\varepsilon_{0}^{(3)} = \mathcal{I}+\mathcal{A}t+\frac{\mathcal{A}^2t^2}{2}+\frac{\mathcal{A}^3t^3}{6} \end{align*} $$ 格式Ⅳ: $ \varepsilon_4^{-1} = [\frac{3}{4}]_e = \varepsilon_{2}^{(0)}+(\varepsilon_3^{(0)}-\varepsilon_3^{(-1)})^{-1} $, 其中
$$ \begin{align*} &\varepsilon_{3}^{(0)} = \varepsilon_{1}^{(1)}+(\varepsilon_2^{(1)}-\varepsilon_2^{(0)})^{-1}\\ &\varepsilon_{3}^{(-1)} = \varepsilon_{1}^{(0)}+(\varepsilon_2^{(0)}-\varepsilon_2^{(-1)})^{-1} \end{align*} $$ 格式Ⅴ: $ \varepsilon_4^{0} = [\frac{4}{4}]_e = \varepsilon_{2}^{(1)}+(\varepsilon_3^{(1)}-\varepsilon_3^{(0)})^{-1} $, 其中
$$ \begin{align*} &\varepsilon_{3}^{(1)} = \varepsilon_1^{(2)}+(\varepsilon_2^{(2)}-\varepsilon_2^{(1)})^{-1} \end{align*} $$ 格式Ⅵ: $ \varepsilon_4^{1} = [\frac{5}{4}]_e = \varepsilon_{2}^{(2)}+(\varepsilon_3^{(2)}-\varepsilon_3^{(1)})^{-1} $, 其中
$$ \begin{align*} &\varepsilon_{3}^{(2)} = \varepsilon_1^{(3)}+(\varepsilon_2^{(3)}-\varepsilon_2^{(2)})^{-1} \end{align*} $$ 例2. 设张量$ \mathcal{A}\in {\bf R}^{2\times 2\times 2} $, 其张量指数函数为
$$ \begin{equation*} \label{equation3-8} \begin{aligned} {\rm exp}(\mathcal{A}t) = \, &\left[ \begin{array}{cc|cc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right] +\\ &\left[ \begin{array}{cc|cc} 0 & 1 & 0 & 2 \\ 0 & -2 & 0 & -1 \\ \end{array} \right]t+\\ &\left[ \begin{array}{cc|cc} 0 & -2 & 0 & -\dfrac{5}{2} \\ 0 & \dfrac{5}{2} & 0 & 2 \\ \end{array} \right]t^2 +\end{aligned} \end{equation*} $$ $$ \begin{align} &\left[ \begin{array}{cc|cc} 0 & \dfrac{13}{6} & 0 & \dfrac{7}{3} \\ 0 & -\dfrac{7}{3} & 0 & -\dfrac{13}{6} \\ \end{array} \right]t^3+\\ &\left[ \begin{array}{cc|cc} 0 & -\dfrac{5}{3} & 0 & -\dfrac{41}{24} \nonumber\\ 0 & \dfrac{41}{24} & 0 & \dfrac{5}{3} \nonumber\\ \end{array} \right]t^4+\cdots = \\ &\mathcal{I}+\mathcal{A}t+\frac{1}{2!}\mathcal{A}^2t^2+ \frac{1}{3!}\mathcal{A}^3t^3+\\ &\frac{1}{4!}\mathcal{A}^4t^4+\cdots \end{align} $$ (13) 计算$ [\frac{2}{2}] $型GITPA.
由常用格式Ⅱ得
$$ \begin{equation*} \begin{aligned} &\varepsilon_2^{(0)} = \left[\frac{2}{2}\right]_e = \varepsilon_{0}^{(0)}+(\varepsilon_1^{(0)}-\varepsilon_1^{(-1)})^{-1} = \\ &\mathcal{I}+\mathcal{A}t+\frac{1}{\frac{2}{\mathcal{A}^2t^2}-\frac{1}{\mathcal{A}t}} = \\ &\frac{\left[ \begin{array}{cc|cc} a_1 & a_2 & 0 & a_3\\ 0 & a_1-a_3 & 0 & -a_2\\ \end{array} \right]}{a_1} = \frac{\mathcal{P}_2(t)}{q_2(t)} \end{aligned} \end{equation*} $$ 其中
$$ \begin{array}{rl} a_1 = &41+140t+125t^2\\ a_2 = &40t^2+41t\\ a_3 = &155t^2+82t \end{array} $$ 由定义5, $ \varepsilon_2^{(0)} = \frac{\mathcal{P}_2(t)}{q_2(t)} $是式(13)的$ [\frac{2}{2}] $型GITPA, 满足以下条件:
1) $ \partial{\mathcal{P}_2(t)} = 2, \partial{q_2(t)} = 2; $
2) $ q_2(t)\mid \parallel\mathcal{P}_2(t)\parallel^2; $这里
$$ \begin{align*} \|\mathcal{P}_2(t)\|^2 = \, &2(a_1^2+a_2^2+a_3^2)-2a_1a_3 = \\ &2(a_1^2+205t^2a_1-a_1a_3) = \\ &2a_1(a_1+205t^2-a_3) \end{align*} $$ 3) $ q_2(t){\rm exp}(\mathcal{A}t)-\mathcal{P}_2(t) = O(t^{3}) $.
3. 数值实验
本节计算张量$\varepsilon$-算法的近似值和精确值的误差, 并将本文的方法与计算张量指数函数的截断法进行比较.首先给出目前通常使用的无穷序列截断法:
算法2. (无穷序列截断法[10])
输入:张量$\mathcal{A}$、自变量$t$和误差限$\epsilon_{\rm{tol}}$的值.
1) 初始化$n=0$和${\rm exp}(\mathcal{A}t):=\mathcal{I}$.
2) $n:=n+1$.
3) 计算$\frac{t^n}{n!}$和$\mathcal{A}^n$.
4) 将该项计算结果加到
$$ {\rm exp}(\mathcal{A}t):={\rm exp}(\mathcal{A}t)+\frac{t^n}{n!}\mathcal{A}^n $$ 5) 判断是否停止, 若
$$ \begin{equation*} \frac{\|\mathcal{A}^n\|t^n}{n!} <\epsilon_{\rm{tol}} \end{equation*} $$ 则停止, 否则回到第2)步.
输出: ${\rm exp}(\mathcal{A}t)$.
例3. 设张量$\mathcal{A}\in {\bf R}^{2\times 2\times 2}$, $\mathcal{A}$中的元素为
$$ a_{121}=\frac{1}{2}, a_{221}=-\frac{2}{3}, a_{122}=-\frac{1}{2}, a_{222}=\frac{2}{3} $$ 其余均为0, 它的指数函数展开式为
$$ \begin{equation}\label{equation4-1} \begin{aligned} {\rm exp}(\mathcal{A}t)=\, &\left[ \begin{array}{cc|cc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right] +\\ &\left[ \begin{array}{cc|cc} 0 & \dfrac{1}{2} & 0 & \dfrac{2}{3} \\[2mm] 0 & -\dfrac{2}{3} & 0 & -\dfrac{1}{2} \\ \end{array} \right]t+\\ &\left[ \begin{array}{cc|cc} 0 & -\dfrac{1}{3} & 0 & -\dfrac{15}{72} \\[2mm] 0 & \dfrac{15}{72} & 0 & \dfrac{1}{3} \\ \end{array} \right]t^2+\\ &\left[ \begin{array}{cc|cc} 0 & \dfrac{19}{144} & 0 & \dfrac{43}{324} \\[2mm] 0 & -\dfrac{43}{324} & 0 & -\dfrac{19}{144} \\ \end{array} \right]t^3+\\ &\left[ \begin{array}{cc|cc} 0&-\dfrac{25}{648}&0&-\dfrac{128}{3315}\\[2mm] 0&\dfrac{128}{3315}&0&\dfrac{25}{648}\\ \end{array} \right]t^4+\cdots=\\ &\mathcal{I}+\mathcal{A}t+\frac{1}{2!}\mathcal{A}^2t^2+ \frac{1}{3!}\mathcal{A}^3t^3+\\ &\frac{1}{4!}\mathcal{A}^4t^4+\cdots \end{aligned} \end{equation} $$ (14) 用$\varepsilon$-算法和截断法计算该指数函数.
3.1 $\pmb{[\frac{4}{4}]}$型GITPA的数值实验
由常用格式V计算$[\frac{4}{4}]$型GITPA, 数值实验结果见表 1.
表 1 $[\frac{4}{4}]$型GITPA-算法数值实验Table 1 The numerical experiment of $[\frac{4}{4}]$ type GITPA-algorithm$x$ $(1, 2, 1)$ $(2, 2, 1)$ $(1, 2, 2)$ $(2, 2, 2)$ $ RES $ 0.2 $E$值 0.08766299 0.87955329 0.12044671 -0.08766299 $5.69\times10^{-13}$ $A$值 0.08766327 0.87955283 0.12044717 -0.08766327 0.4 $E$值 0.15420167 0.78130960 0.21869040 -0.15420167 $3.74\times10^{-10}$ $A$值 0.15420895 0.78129804 0.21870196 -0.15420895 0.6 $E$值 0.20408121 0.70078192 0.29921808 -0.20408121 $1.40\times10^{-8}$ $A$值 0.20412606 0.70071136 0.29928864 -0.20412606 0.8 $E$值 0.24081224 0.63444735 0.36555265 -0.24081224 $1.63\times10^{-7}$ $A$值 0.24096630 0.63420702 0.36579298 -0.24096630 1 $E$值 0.26715410 0.57953894 0.42046106 -0.26715410 $1.01\times10^{-6}$ $A$值 0.26753925 0.57894247 0.42105753 -0.26753925 表 1中由常用格式V计算的近似值记作$A$值, 根据截断法即算法2取指数函数展开式(14)前15项计算得到的结果为精确值, 记作$E$值.表 1中分别列出张量下标为(1, 2, 1), (2, 2, 1), (1, 2, 2), (2, 2, 2)在点0.2, 0.4, 0.6, 0.8, 1相应的近似值与精确值.表 1中最后一列表示近似值与精确值残差的范数的平方, 记作$RES$, 其计算公式如下:
$$ \begin{equation*} RES(t)=\|{\rm exp}(\mathcal{A}t)-\left[\frac{4}{4}\right]_{e^{\mathcal{A}t}}(t)\|^2 \end{equation*} $$ 其中, 模范数$\|\cdot\|$由式(3)定义.
表 1显示, 张量$\varepsilon$-算法得到的近似值能够达到较高的计算精度.可以发现, 自变量的值越接近0, 张量$\varepsilon$-算法的逼近效果越好, 计算结果基本符合理论预期.
3.2 两个算法数值实验的比较
首先, 在张量指数函数(14)中令$t=2$, 取前15项计算得到的精确值为
$$ \begin{equation*} \left[ \begin{array}{cc|cc} 1 & ~~0.3098~~&~~0 & ~~0.5932 \\ 0 & ~~0.4068~~&~~0 & ~~-0.3098 \\ \end{array} \right] \end{equation*} $$ 根据算法1分别计算$[\frac{2}{2}], [\frac{4}{4}], [\frac{6}{6}]$型GITPA, 并与截断算法2进行比较, 得到的数值实验结果见表 2.
表 2 算法1和算法2的数值实验比较Table 2 The numerical experiment comparison of Algorithm 1 and Algorithm 2$[\frac{j+2k}{2k}]$ 张量$\varepsilon$-算法 $n_{\rm{max}}$ $\sum^{n_{\rm{max}}}_{n=0}\frac{1}{n!}\mathcal{A}^nt^n $ $a_{121}$ $a_{221}$ $a_{122}$ $a_{222}$ $a_{121}$ $a_{221}$ $a_{122}$ $a_{222}$ $[\frac{2}{2}]$ 0.4235 0.3513 0.6487 -0.4235 1 1.0000 -0.3333 1.3333 -1.0000 $[\frac{4}{4}]$ 0.3049 0.4141 0.5859 -0.3049 2 -0.3333 1.0556 -0.0556 0.3333 $[\frac{6}{6}]$ 0.3098 0.4068 0.5932 -0.3098 3 0.7222 -0.0062 1.0062 -0.7222 - - - - 4 0.1049 0.6116 0.3884 -0.1049 - - - - 5 0.3931 0.3234 0.6766 -0.3931 - - - - 6 0.2810 0.4355 0.5645 -0.2810 - - - - 7 0.3184 0.3981 0.6019 -0.3184 - - - - 8 0.3075 0.4090 0.5910 -0.3075 - - - - 9 0.3103 0.4062 0.5938 -0.3103 - - - - 10 0.3097 0.4069 0.5931 -0.3097 - - - - 11 0.3098 0.4067 0.5933 -0.3098 - - - - 12 0.3098 0.4068 0.5932 -0.3098 表 2显示, $[\frac{6}{6}]$型GITPA计算的数值结果与截断法取前13项相加的数值结果的精确度相近, 由此说明本文给出的张量$\varepsilon$-算法是有效的.
3.3 两个算法计算复杂度的比较
下面从计算复杂度的角度来对两种算法进行比较.由于需要自乘, 前两维的维数必须相等.对于一个$l\times l\times n$的三阶张量$\mathcal{A}$, 1次张量$t$-积的计算复杂度为$ {\rm O}(l^3n^2)$; 而1次张量的广义逆等价于2次张量的数乘计算, 其计算复杂度为$ {\rm O}(l^2n)$.因此, 随着张量维数的增大, 计算张量$t$-积显然要比计算张量的广义逆付出更大的成本.
根据张量$\varepsilon$-算法计算$[\frac{6}{6}]$型GITPA所使用的是张量指数函数(14)的前7项, 需要进行5次张量$t$-积和21次广义逆计算, 而达到相同精度, 截断法需要取前13项, 需要进行11次张量$t$-积和12次数乘运算.两种算法的计算复杂度比较见表 3.从表 3可以看出:在计算过程中, 张量$\varepsilon$-算法计算复杂度为$5l^3n^2+42l^2n$, 截断法计算复杂度为$11l^3n^2+12l^2n$.
表 3 算法1和算法2的计算复杂度分析Table 3 The analysis of computational complexity of Algorithm 1 and Algorithm 2计算一次 $\varepsilon$-算法: $[\frac{6}{6}]$型 截断法:取13项 复杂度 计算 计算 $t$-积运算 $l^3n^2$ 5 11 数乘运算 $l^2n$ 21 12 范数运算 $l^2n$ 21 0 总和 $5l^3n^2+42l^2n$ $11l^3n^2+12l^2n$ 例4. 为了比较两种算法所需的时间, 在Matlab中选取$'seed~'=1:100$, 随机生成张量$\mathcal{A}$如下: $3\times 3\times 3, 10\times 10\times 10, 20\times 20\times 20, 30\times 30\times 30, 40\times 40\times 40$各100个张量, 分别使用算法1和算法2计算其张量指数函数(12), 其中, 本文给出的算法1计算的是$[\frac{6}{6}]$型GITPA, 算法2计算的是指数函数展开的前13项之和.两个算法计算所需时间见表 4, 表中数据均是由内存8 GB, 主频2.50 GHz, 操作系统Windows 10, 处理器Inter(R) Core(TM) i5-7300HQ的笔记本电脑的Matlab (R2015b)得到.
表 4 不同维数下两种算法的运行时间表(s)Table 4 The consuming time of two algorithms in different dimensions (s)张量维数 张量$\varepsilon$-算法运行时间 截断法运行时间 $ 3~\times ~3\times ~3$ 1.755262 1.103832 $10\times 10\times 10$ 2.272663 2.164860 $20\times 20\times 20$ 3.831094 4.805505 $30\times 30\times 30$ 6.419004 10.545785 $40\times 40\times 40$ 15.814063 30.012744 图 1显示, 算法1计算$[\frac{6}{6}]$型GITPA与算法2即截断法计算前13项之和相比, 在维数较低的情况下(维数低于$10\times 10\times 10$), GITPA的运行时间略高于截断法; 当维数较高时, 本文的算法1逐渐显示出优越性, 可以很好地降低计算时间.
图 1 不同维数下两种算法运行时间直方图(s)Fig. 1 The time-consuming comparison histogram of two algorithms in different dimensions (s)4. 结论
目前, 国内外还没有看到计算张量的逆和广义逆的有效算法, 本文提出的一种张量广义逆即定义3是一个实用的计算方法, 它是同一类型的向量广义逆(Graves-Morris [14])和矩阵广义逆(顾传青[15-16, 18])在张量上的推广.在该张量广义逆的基础上, 本文得到了计算张量指数函数(12)的张量$\varepsilon$-算法.从计算张量指数函数(14)的数值实验来看, 与目前通用的无穷序列截断法相比较, 在计算精度和计算复杂度上具有一定的优势, 在张量维数比较大时, 这个优势会更加明显.下面的研究工作从两个方面进行, 一是用本文提出的广义逆张量Padé逼近(GITPA)方法来考虑控制论中的模型简化问题, 二是对张量$\varepsilon$-算法的稳定性进行探讨.
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图 1 IoU相同但中心点距离不同的情况(红色代表候选的边界框, 绿色代表真实边界框)
Fig. 1 Same IoU while different distances between centroids (Red represents the candidate bounding box, and green represents the ground-truth bounding box)
图 3 IoU和中心点距离对应视频帧数的统计
Fig. 3 The number statistics of video frame corresponding to IoU and distances between centroids
图 4 在视频序列Dinosaur上跟踪的结果可视化
Fig. 4 Visualization of tracking results on the video sequence Dinosaur
图 5 本文方法(ASEID)在OTB-100数据集上与相关方法的比较
Fig. 5 Comparison of the proposed method (ASEID) with related algorithms on OTB-100 dataset
图 6 OTB-100 数据集不同挑战性因素影响下的成功率图
Fig. 6 Success plots on sequences with different challenging attributes on OTB-100 dataset
图 7 OTB-100 数据集不同挑战性因素影响下的精度图
Fig. 7 Precision plots on sequences with different challenging attributes on OTB-100 dataset
图 8 本文方法与相关方法的可视化比较
Fig. 8 Visualization comparison of the proposed method and related trackers
图 9 OTB-100数据集中的失败案例(绿色框代表真实框, 红色框代表本文算法的预测框)
Fig. 9 Failure cases in OTB-100 dataset (The green box represents ground truth box, and the red box represents the prediction box)
图 10 GOT-10k数据集中的失败案例(在GOT-10k的测试集中, 由于只能拿到测试视频序列的第一帧的真实框, 因此第一帧的标记代表被跟踪目标)
Fig. 10 Failure cases in GOT-10k dataset (In GOT-10k test set, only the ground truth in the first frame of the test dataset can be obtained. Therefore, the bounding box of the first frame represents the tracked target)
表 1 OTB-100数据集上的消融实验
Table 1 Ablation study on OTB-100 dataset
方法 AUC (%) Precision (%) Norm.Pre (%) 帧速率(帧/s) 多尺度搜索 68.4 88.8 83.8 21 IoU 68.4 89.4 84.2 35 NDIoU 69.8 91.3 87.3 35 
下载: 导出CSV
表 4 在GOT-10k数据集上与SOTA方法的比较(%)
Table 4 Compare with SOTA trackers on GOT-10k dataset (%)
DCFST[32] PrDiMP50[15] KYS[17] SiamFC++[13] D3S[43] Ocean[12] ROAM[44] ATOM[9] DiMP50 (基线)[14] ASEID (本文) $ \mathit{S}{\mathit{R}}_{0.50}$ 68.3 73.8 75.1 69.5 67.6 72.1 46.6 63.4 71.7 78.7 $ \mathit{S}{\mathit{R}}_{0.75} $ 44.8 54.3 51.5 47.9 46.2 — 16.4 40.2 49.2 53.4 $ \mathit{A}\mathit{O}$ 59.2 63.4 63.6 59.5 59.7 61.1 43.6 55.6 61.1 65.4
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