基于拟共形理论的分数阶多尺度微分同胚图像配准

图1 单元中心分割: $\Omega_{i,j}$

设 $\overrightarrow D = \overrightarrow D \big( {PX} \big) \in \mathbb{R}^{{n^2} \times 1}$ 为离散化的目标图像, $\overrightarrow T = \overrightarrow T \big( {PX + PU} \big) \in \mathbb{R}^{{n^2} \times 1}$ 为离散化的目标图像. 其中 $P \in \mathbb{R}^{2{n^2} \times 2{{(n+1)}^2}}$ 是从 $U$ 的节点网络转移到单元中心位置的平均矩阵. 因此, 对于数据项的离散, 可以得到如下式子

$\begin{equation} S(\boldsymbol{u}) \approx {h^2}{\big( {\overrightarrow T \big( {PX + PU} \big) - \overrightarrow D \big( {PX} \big)} \big)^T}\big( {\overrightarrow T \big( {PX + PU} \big) - \overrightarrow D \big( {PX} \big)} \big). \end{equation}$

(3.3)

带分数阶正则项 $R_0(\boldsymbol{u})$ 的离散化

接下来, 我们对带分数阶导数的正则项 ${R_0}(\boldsymbol{u}) = \int_\Omega {{{\left| {{\nabla ^\alpha }\boldsymbol{u}} \right|}^2}{\rm d}\boldsymbol{x}}$ 进行离散化处理.

$\begin{equation} {R_0}(\boldsymbol{u}) = \int_\Omega {{{\left| {{\nabla ^\alpha }\boldsymbol{u}} \right|}^2}{\rm d}\boldsymbol{x}} \\ = \int_\Omega {{{\big( {\frac{{{\partial ^\alpha }{u_1}}}{{\partial {x_1}^\alpha }}} \big)}^2} + } {\big( {\frac{{{\partial ^\alpha }{u_1}}}{{\partial {x_2}^\alpha }}} \big)^2} + {\big( {\frac{{{\partial ^\alpha }{u_2}}}{{\partial {x_1}^\alpha }}} \big)^2} + {\big( {\frac{{{\partial ^\alpha }{u_2}}}{{\partial {x_2}^\alpha }}} \big)^2}{\rm d}\boldsymbol{x}, \\ \end{equation}$

(3.4)

根据图1(b) 及数值积分中的中点法则, 可以得到

$\begin{equation*} \left\{\begin{aligned} &\int_{\Omega_{i,j}^{x_1}}{{\big( {\frac{{{\partial ^\alpha }{u_l}}}{{\partial {x_1}^\alpha }}} \big)}^2}{\rm d}\boldsymbol{x} \approx h^2{{\big( {\partial _{{x_1},\alpha }^{i + \frac{1}{2},j}{u_l}} \big)}^2} \quad 1\leqslant j\leqslant n-1, & l=1,2,\\ &\int_{\Omega_{i,j}^{x_1}}{{\big( {\frac{{{\partial ^\alpha }{u_l}}}{{\partial {x_1}^\alpha }}} \big)}^2}{\rm d}\boldsymbol{x}\approx \frac{h^2}{2}{{\big( {\partial _{{x_1},\alpha }^{i + \frac{1}{2},j}{u_l}} \big)}^2} \quad j=0,n, & l=1,2.\\ \end{aligned}\right. \end{equation*}$

同理, 对于 $\int_{\Omega_{i,j}^{x_2}}{{\big( {\frac{{{\partial ^\alpha }{u_l}}}{{\partial {x_2}^\alpha }}} \big)}^2}{\rm d}\boldsymbol{x}$ , $l=1,2$ , 可以得到类似的结果.

另外, 对于 ${\partial _{{x_1},\alpha }{u_l}}$ 和 ${\partial _{{x_2},\alpha }{u_l}}$ , $l=1,2$ , 用

$\begin{equation*} \partial _{x{}_1,\alpha }^{i + \frac{1}{2},j}{u_l} \approx \frac{{\partial _{{x_{{}_1}},\alpha }^{i,j}{u_l} + \partial _{{x_{{}_1}},\alpha }^{i + 1,j}{u_l}}}{2}\quad \partial _{x{}_2,\alpha }^{i,j + \frac{1}{2}}{u_l} \approx \frac{{\partial _{{x_{{}_2}},\alpha }^{i,j}{u_l} + \partial _{{x_{{}_2}},\alpha }^{i,j + 1}{u_l}}}{2} \end{equation*}$

来估计.

因此, 可以得到

$\begin{equation*} \begin{aligned} &{R_0}(\boldsymbol{u}) = \int_\Omega {{{\left| {{\nabla ^\alpha }\boldsymbol{u}} \right|}^2}{\rm d}\boldsymbol{x}} \\ &\approx\sum\limits_{l = 1}^2 {\sum\limits_{i = 0}^{n - 1}{{\sum\limits_{j = 1}^{n - 1}}}{{h^2}{{\big( {\partial _{{x_1},\alpha }^{i + \frac{1}{2},j}{u_l}} \big)}^2}} +\sum\limits_{l = 1}^2{\sum\limits_{i = 0}^{n - 1} {\frac{{{h^2}}}{2}{{\big( {\partial _{{x_{{}_1}},\alpha }^{i + \frac{1}{2},0}{u_l}} \big)}^2}}}+ \sum\limits_{l = 1}^2{\frac{{{h^2}}}{2}{{\big( {\partial _{{x_{{}_1}},\alpha }^{i + \frac{1}{2},n}{u_l}} \big)}^2}}}\\ &+\sum\limits_{l = 1}^2 {\sum\limits_{j = 0}^{n - 1} {\sum\limits_{i = 1}^{n - 1} {{h^2}{{\big( {\partial _{{x_{{}_2}},\alpha }^{i,j + \frac{1}{2}}{u_l}} \big)}^2}} + \sum\limits_{l = 1}^2{\sum\limits_{j = 0}^{n - 1}}{\frac{{{h^2}}}{2}{{\big( {\partial _{{x_{{}_2}},\alpha }^{0,j + \frac{1}{2}}{u_l}} \big)}^2}} +\sum\limits_{l = 1}^2{\frac{{{h^2}}}{2}{{\big( {\partial _{{x_{{}_2}},\alpha }^{n,j + \frac{1}{2}}{u_l}} \big)}^2}}}}\\ &\approx \frac{1}{4}{h^2}\left[ {\sum\limits_{i = 0}^{n - 1} {\sum\limits_{j = 1}^{n - 1} {{{\big( {\partial _{{x_1},\alpha }^{i,j}{u_1}\!+\! \partial _{{x_1},\alpha }^{i + 1,j}{u_1}} \big)}^2}\! +\! \sum\limits_{i = 0}^{n - 1} {\frac{1}{2}{{\big( {\partial _{{x_{{}_1}},\alpha }^{i,0}{u_1}\! +\! \partial _{{x_{{}_1}},\alpha }^{i + 1,0}{u_1}} \big)}^2}\! +\! \frac{1}{2}{{\big( {\partial _{{x_{{}_1}},\alpha }^{i,n}{u_1}\! +\! \partial _{{x_{{}_1}},\alpha }^{i + 1,n}{u_1}} \big)}^2}} } } } \right] \\ &+ \frac{1}{4}{h^2}\left[ {\sum\limits_{i = 0}^{n - 1} {\sum\limits_{j = 1}^{n - 1} {{{\big( {\partial _{{x_1},\alpha }^{i,j}{u_2}\! +\! \partial _{{x_1},\alpha }^{i + 1,j}{u_2}} \big)}^2}\! +\! \sum\limits_{i = 0}^{n - 1} {\frac{1}{2}{{\big( {\partial _{{x_{{}_1}},\alpha }^{i,0}{u_2}\! +\! \partial _{{x_{{}_1}},\alpha }^{i + 1,0}{u_2}} \big)}^2}\! +\! \frac{1}{2}{{\big( {\partial _{{x_{{}_1}},\alpha }^{i,n}{u_2}\! +\! \partial _{{x_{{}_1}},\alpha }^{i + 1,n}{u_2}} \big)}^2}} } } } \right] \\ &+\frac{1}{4}{h^2}{\left[ {\sum\limits_{j = 0}^{n - 1} {\sum\limits_{i = 1}^{n - 1} {{{\big( {\partial _{{x_{{}_2}},\alpha }^{i,j}{u_{\text{1}}}\! +\! \partial _{{x_{{}_2}},\alpha }^{i,j + 1}{u_{\text{1}}}} \big)}^2}\! +\! \sum\limits_{j = 0}^{n - 1} {\frac{1}{2}{{\big( {\partial _{{x_{{}_2}},\alpha }^{0,j}{u_{\text{1}}}\! +\! \partial _{{x_{{}_2}},\alpha }^{0,j + 1}{u_{\text{1}}}} \big)}^2}\! +\! \frac{1}{2}{{\big( {\partial _{{x_2},\alpha }^{n,j}{u_{\text{1}}}\! +\! \partial _{{x_2},\alpha }^{n,j + 1}{u_{\text{1}}}} \big)}^2}} } } } \right]} \\ &+\frac{1}{4}{h^2}\left[ {\sum\limits_{j = 0}^{n - 1} {\sum\limits_{i = 1}^{n - 1} {{{\big( {\partial _{{x_{{}_2}},\alpha }^{i,j}{u_{\text{2}}}\! +\! \partial _{{x_{{}_2}},\alpha }^{i,j + 1}{u_{\text{2}}}} \big)}^2}\! +\! \sum\limits_{j = 0}^{n - 1} {\frac{1}{2}{{\big( {\partial _{{x_{{}_2}},\alpha }^{0,j}{u_{\text{2}}}\! +\! \partial _{{x_{{}_2}},\alpha }^{0,j + 1}{u_{\text{2}}}} \big)}^2}\! +\! \frac{1}{2}{{\big( {\partial _{{x_{{}_2}},\alpha }^{n,j}{u_{\text{2}}}\! +\! \partial _{{x_{{}_2}},\alpha }^{n,j + 1}{u_{\text{2}}}} \big)}^2}} } } } \right] \\ &=\frac{{{h^2}}}{4}{{\big( {{\partial _\alpha }U} \big)^T}{A^T}GA\big( {{\partial _\alpha }U} \big)}, \\ \end{aligned} \end{equation*}$

其中有关矩阵 $A$ , $G$ 的详细信息, 请参见附录 A.

根据文献 [30,31] 中的 Grunwald 近似法, 对于 $\frac{{{\partial ^\alpha }{u_l}(x)}}{{\partial {x_p}^\alpha }}(p = 1,2;l = 1,2)$ 可以按以下方式离散化

$\begin{equation} \frac{{{\partial ^\alpha }{u_l}\big( {{x_{i,j}}} \big)}}{{\partial {x_p}^\alpha }} = \delta _{p - }^\alpha {u_l}\big( {{x_{i,j}}} \big) + O(h), \end{equation}$

(3.5)

其中 $\delta _{1 - }^\alpha {u_l}\big( {{x^{_{i,j}}}} \big) = \frac{1}{{{h^\alpha }}}\sum\limits_{q = 1}^i {\rho _q^{(\alpha)}{u_l}\big( {{x^{_{i - q + 1,j}}}} \big)}$ , $\delta _{2 - }^\alpha {u_l}\big( {{x^{_{i,j}}}} \big) = \frac{1}{{{h^\alpha }}}\sum\limits_{q = 1}^{j + 1} {\rho _q^{(\alpha)}{u_l}\big( {{x^{_{i,j - q + 1}}}} \big)}$ , $\rho _0^{(\alpha)} = 1$ , $\rho _q^{(\alpha)} = \big( {1 - \frac{{q + \alpha }}{q}} \big)\rho _{q - 1}^{(\alpha)}$ .

记 ${U_j} = {(u_{_1}^{0,j},u_1^{1,j},\cdots,u_{_1}^{n,j})^T} \in {\mathbb{R}^{({n + 1}) \times 1}},u_{_l}^{0,j} \triangleq {u_l}\big( {{x^{0,j}}} \big),j = 0,1, \cdots, n$ , 根据 (3.5) 式则有

$\begin{equation} \frac{{{\partial ^\alpha }{U_j}}}{{\partial {x_1}^\alpha }} = {\Big( {\frac{{{\partial ^\alpha }u_1^{0,j}}}{{{{\partial {x_1}^\alpha }} }},\frac{{{\partial ^\alpha }u_1^{1,j}}}{{{{\partial {x_1}^\alpha }} }},\cdots,\frac{{{\partial ^\alpha }u_{_1}^{n,j}}}{{{{\partial {x_1}^\alpha }} }}} \Big)^T} \approx {B_{n + 1,\alpha }^h} {U_j}, \end{equation}$

(3.6)

其中

$\begin{equation} {B_{n + 1,\alpha }^h} =\frac{1}{{{h^\alpha }}}{B_{n + 1,\alpha }}=\frac{1}{{{h^\alpha }}}\left( {\begin{array}{*{20}{c}} \begin{gathered} \rho _1^{(\alpha)} \hfill \\ \rho _2^{(\alpha)} \hfill \\ \end{gathered} &\begin{gathered} \rho _0^{(\alpha)} \hfill \\ \rho _1^{(\alpha)} \hfill \\ \end{gathered} &\begin{gathered} 0 \hfill \\ \rho _0^{(\alpha)} \hfill \\ \end{gathered} \\ \vdots & \vdots & \ddots \\ \begin{gathered} \rho _n^{(\alpha)} \hfill \\ \rho _{n + 1}^{(\alpha)} \hfill \\ \end{gathered} &\begin{gathered} \rho _{n - 1}^{(\alpha)} \hfill \\ \rho _n^{(\alpha)} \hfill \\ \end{gathered} &\begin{gathered} \cdots \hfill \\ \cdots \hfill \\ \end{gathered} \end{array}\begin{array}{*{20}{c}} \begin{gathered} \cdots \hfill \\ \cdots \hfill \\ \ddots \hfill \\ \rho _1^{(\alpha)} \hfill \\ \end{gathered} &\begin{gathered} 0 \hfill \\ 0 \hfill \\ \vdots \hfill \\ \rho _0^{(\alpha)} \hfill \\ \end{gathered} \\ {\rho _2^{(\alpha)}}&{\rho _1^{(\alpha)}} \end{array}} \right) \in {\mathbb{R}^{(n + 1) \times (n + 1)}}. \end{equation}$

(3.7)

由以上分数阶离散理论可知 ${\partial _\alpha }U = \frac{1}{h^{\alpha}}CU$ . 有关矩阵 $C$ 的详细信息,请参见附录 A.

因此, 对带分数阶导数的正则项 ${R_0}\big( u \big)$ 的离散, 我们可以得出以下结论

$\begin{equation} {R_0}\big( u \big) \approx \frac{{{h^{2-2\alpha}}}}{4}{U^T}{C^T}{A^T}GACU. \end{equation}$

(3.8)

正则项 $R_1(\boldsymbol{u})$ 的离散化

最后, 是对正则项 ${R_1}\big( \boldsymbol{u} \big)$ 的离散. 根据文献 [11] 可以得到

$\begin{equation} {R_1}\big( \boldsymbol{u} \big) \approx \frac{{{h^2}}}{4}\phi \big( {\vec r\big( U \big)} \big){{\rm e}^T}, \end{equation}$

(3.9)

其中 $\phi \big( {\vec r\big( U \big)} \big) = \big( {\phi \big( {\vec r{{\big( U \big)}_1}} \big), \cdots, \phi \big( {\vec r{{\big( U \big)}_{4{n^2}}}} \big)} \big)$ , $e = \big( {1,1, \cdots 1} \big) \in {\mathbb{R}}^{4{n^2}}$ . $\vec r\big( U \big)$ 为离散化的形变场 $\boldsymbol{h}$ 的 Beltrami 系数的模的平方, 详见文献 [11] 中的附录 B.

最终, 结合 (3.7), (3.8) 和 (3.9) 式, 我们可以得到模型 (3.1) 的离散化形式如下

$\begin{equation} \begin{aligned} {U^n} &\triangleq \arg \mathop {\inf }\limits_{\boldsymbol{u} \in {{\left[ {H_0^\alpha ( \Omega )} \right]}^2}} J\big( U \big) \\ &= \arg \mathop {\inf }\limits_{\boldsymbol{u} \in {{\left[ {H_0^\alpha ( \Omega )} \right]}^2}} {\lambda _n}{h^2}{\big( {\overrightarrow T \big( {PX + PU} \big) - \overrightarrow D \big( {PX} \big)} \big)^T}\big( {\overrightarrow T \big( {PX + PU} \big) - \overrightarrow D \big( {PX} \big)} \big) \\ & + {\gamma _n}\frac{{{h^{2-2\alpha}}}}{4}{U^T}{C^T}{A^T}GACU + {\Theta _n}\frac{{{h^2}}}{4}\phi \big( {\vec r\big( U \big)} \big){{\rm e}^T}. \end{aligned} \end{equation}$

(3.10)

3.2.2 离散化问题 (3.10) 的最优化方法

在数值计算中, 我们选择线性搜索法求解无约束优化问题 (3.10). 为了保证搜索方向是一个下降方向, 我们采用 Gauss-Newton 方向. 另外, 使用 Gauss-Newton 法时, 我们不需要计算二阶项, 可以加速 CPU 时间消耗.

$J( U ):{\mathbb{R}}^{2{{( {n + 1} )}^2}} \to \mathbb{R}$ 是两次连续可微分的, ${U^i} \in {{\mathbb{R}}^{2{{( {n + 1} )}^2}}}$ , 且近似 Hessian 矩阵 $H( {{U^i}} )$ 是正定的. 用二次逼近的方法将 $J( U)$ 在 ${U^i}$ 处展开

$\begin{equation} J\big( {{U^i} + s} \big) \approx {q^i}(s) = J\big( {{U^i}} \big) + {d_J}{\big( {{U^i}} \big)^T}s + \frac{1}{2}{s^T}H{\big( {{U^i}} \big)^T}s, \end{equation}$

(3.11)

其中 $s = U - {U^i}$ , ${d_J}\big( {{U^i}} \big) = \nabla J\big( {{U^i}} \big)$ . 同时, 由 (3.11) 式可以得到

$\begin{equation} {U^{i + 1}} = {U^i} - {\left[ {H\big( {{U^i}} \big)} \right]^{ - 1}}{d_J}\big( {{U^i}} \big). \end{equation}$

(3.12)

为了保证 Gauss-Newton 法的全局收敛性, 我们采用了线性搜索法. 其迭代过程如下

$\begin{equation} {U^{i + 1}} = {U^i} - {\theta _i}{\left[ {H\big( {{U^i}} \big)} \right]^{ - 1}}{d_J}\big( {{U^i}} \big), \end{equation}$

(3.13)

其中 ${\theta _i}$ 表示步长.

接下来, 我们将研究近似的 Hessian 矩阵 $H\big( {{U^i}} \big)$ 、搜索方向, 步长及算法终止标准.

Hessian 矩阵 $H$ 估计

我们分别考虑 (3.10) 式中 $J (U)$ 的三个项.

首先, 是离散化的 SSD 项

$\begin{equation} {h^2}{\big( {\overrightarrow T \big( {PX + PU} \big) - \overrightarrow D \big( {PX} \big)} \big)^T}\big( {\overrightarrow T \big( {PX + PU} \big) - \overrightarrow D \big( {PX} \big)} \big), \end{equation}$

(3.14)

其梯度和 Hessian 矩阵分别为

$\begin{equation} \left\{ \begin{aligned} &{d_1} = 2{h^2}{P^T}\overrightarrow T _{\widetilde U}^T{\big( {\overrightarrow T \big( {\widetilde U} \big) - \overrightarrow D } \big)^T} \in {\mathbb{R}^{2{{(n+1)}^2} \times 1}}, \\ &{H_1} = 2{h^2}{P^T}\big( {\overrightarrow T _{\widetilde U}^T{{\overrightarrow T }_{\widetilde U}} + \sum\limits_{l = 1}^{{n^2}} {{{\big( {\overrightarrow T \big( {\widetilde U} \big) - \overrightarrow D } \big)}_l}{\nabla ^2}{{\big( {\overrightarrow T \big( {\widetilde U} \big) - \overrightarrow D } \big)}_l}} } \big)P, \end{aligned} \right. \end{equation}$

(3.15)

其中 $\widetilde {U} = PX + PU$ , $\overrightarrow T _{\widetilde {U}} = \frac{{\partial \overrightarrow T ( \widetilde {U} )}}{{\partial \widetilde {U}}}$ .

对于 ${H_1}$ , 我们无法确保它是半正定的. 如果它不是正定的, 我们可能得不到下降方向. 因此, 我们考虑省略 ${H_1}$ 的二阶项, 得到 (3.14) 式的近似 Hessian 矩阵

$\begin{equation} \widehat {{H_1}} = 2{h^2}{P^T} \big( {\overrightarrow T _{\widetilde {U}}^T}{{\overrightarrow T }_{\widetilde {U}}} \big)P. \end{equation}$

(3.16)

另外, 对于离散带分数阶导数的正则项 $\frac{{{h^{2-2\alpha}}}}{4}{U^T}{C^T}{A^T}GACU$ , 其梯度和 Hessian 矩阵分别为

$\begin{equation} \left\{ \begin{aligned} &{d_2} = \frac{h^{2-2\alpha}}{2}{C^T}{A^T}GACU \in \mathbb{R}^{2{{(n+1)}^2} \times 1}, \\ &{H_2} =\frac{h^{2-2\alpha}}{2}{C^T}{A^T}GAC \in {\mathbb{R}}^{2{{(n+1)}^2} \times 2{{(n+1)}^2}}. \end{aligned} \right. \end{equation}$

(3.17)

最后, 对于 $J\big( U \big)$ 的第三项 $\frac{{{h^2}}}{4}\phi \big( {\vec r\big( U \big)} \big){{\rm e}^T}$ , 其梯度和 Hessian 矩阵 $H$ 分别为

$\begin{equation} \left\{ \begin{aligned} &{d_3} = \frac{h^2}{4}{\rm d}{{\vec r}^T}{\rm d}\phi \big( {\vec r} \big) \in {\mathbb{R}}^{2{{(n+1)}^2} \times 1}, \\ &{H_3} =\frac{h^2}{4}\big( {{\rm d}{{\vec r}^T}{d^2}\phi \big( {\vec r} \big){\rm d}\vec r + \sum\limits_{l = 1}^{4{n^2}} {{{\left[ {{\rm d}\phi \big( {\vec r} \big)} \right]}_l}{\nabla ^2}{{\vec r}_l}} } \big) \in {\mathbb{R}^{2{{(n+1)}^2} \times 2{{(n+1)}^2}}}. \end{aligned} \right. \end{equation}$

(3.18)

为了从 (3.18) 式中提取半正定部分, 我们省略其二阶项, 得到近似的 Hessian 矩阵为

$\begin{equation} \widehat {{H_3}} = \frac{h^2}{4}{h^2}{\rm d}{\vec r^T}{d^2}\phi \big( {\vec r} \big){\rm d}\vec r. \end{equation}$

(3.19)

因此, 对于 (3.10) 式中的函数 $J\big( U \big)$ , 我们可以得到它的梯度矩阵为

$\begin{equation} {d_J}^n = \lambda^n{d_1} + \gamma^n{d_2} + \Theta^n{d_3}, \end{equation}$

(3.20)

以及近似的 Hessian 矩阵

$\begin{equation} H^n = \lambda^n \widehat {{H_1}} + \gamma^{n}{H_2} + \Theta^n\widehat {{H_3}}. \end{equation}$

(3.21)

搜索方向

每次迭代时, 利用式 (3.20) 和 (3.21) 式, 我们需要求解高斯-牛顿系统, 找到 (3.10) 式的搜索方向

$\begin{equation} H\delta U = - {d_J}, \end{equation}$

(3.22)

其中 $\delta U$ 是搜索方向. 在我们的实验过程中, 我们使用带有对角线预处理的极小化残差算法 (MINRES)^[32,33]来求解这个线性系统.

步长和终止标准

我们使用标准化的 Armijo 线搜索策略, 通过反向追踪来找到合适的步长 $\theta$ . 在算法过程中, 我们还需要确保 $\vec r\big( U \big)$ 小于 1. 我们分别选择 $T0=10^{-8}$ 和 $Tol=10^{-12}$ 作为步长 $\theta$ 和 $\theta\left\| \delta U \right\|$ 的下限^{[34⇓⇓-37]}. 因此给出以下算法 1

算法 1

输入: 初始化 ID=0, $\theta=1$ , $T0=10^{-8}$ , $Tol=10^{-12}$ , $\eta=10^{-4}$ , $U$ , $\delta U$ ;

a) 计算 $J(U)$ 和 $d_J$ ;

$\text { b) while } \theta\|\delta U\| \geqslant T o l \text {, 则 } U^{n e w}=U+\theta\|\delta U\| \text {; }$

$if \vec{r}\left(U^{n e w}\right) \leqslant 1, 则$

$if \theta \geqslant T 0, 退出 while 循环, 执行 c );$

$else \ if \theta＜T 0, 执行 d );$

$结束 if 判断;$

更新 $\theta=\frac{\theta}{2}$ ;

结束 while 循环;

c) while $\theta\left\| \delta U \right\|\geqslant Tol$ ,则计算 $J(U^{new})$ ;

if $J(U^{new})\leqslant J(U)+\theta\eta {d_J}^T\delta U$ , 则

if $\theta\geqslant T0$ , 结束算法, 输出 $U=U^{new}$ ;

else if $\theta < T0$ , 执行 d);

结束 if 判断;

更新 $\theta=\frac{\theta}{2}$ , $U^{new}=U+\theta\delta U$ ;

结束 while 循环;

d) ID=1, $U=U^{new}$ ;

输出: ID, $U$ .

根据文献 [38], 给定

$\begin{equation*} \begin{aligned} &{\rm (i.a)} \left\| J(U^{i+1})-J(U^{i}) \right\|\leqslant\tau_J(1+\left\| J(U^0) \right\|), \\ &{\rm (i.b)} \left\| {\boldsymbol{h}}^{i+1}-{\boldsymbol{h}}^{i}\right\|\leqslant\tau_W\left\| (1+{\boldsymbol{h}}^0) \right\|,\\ &{\rm (i.c)} \left\| d_J \right\|\leqslant\tau_G(1+\left\| J(U^0) \right\|), \\ &{\rm (ii)} d_J\leqslant {\rm eps},\\ &{\rm (iii)} i\geqslant {\rm MaxIter}. \\ \end{aligned} \end{equation*}$

其中 eps 是机器精度, MaxIter 是外部迭代的最大次数. 我们设置 $\tau_J=10^{-3}$ , $\tau_W=10^{-2}$ , $\tau_G=10^{-2}$ , MaxIter=500. 如果 (i)(ii) 或 (iii) 中的任何一项得到满足, 迭代就会终止. 因此利用 Armijo 搜索得到的 Gauss-Newton 算法可由下述算法 2 给出

算法 2

输入: 目标图像 $D(\cdot)$ , 待配准图像 $T(\cdot)$ , $U^0$ , $\Theta$ , $\gamma$ , $\alpha$ ;

a) $i=0$ , $U^i=U^0$ ;

b) 根据 (3.10) 式计算 $J(U^i)$ , 根据 (3.20) 和 (3.21) 式计算 ${d_J}^i$ 和 $H^i$ ;

c) while (i)--(iii) 均不满足, 则根据 $H^i\delta U^i=-{d_J}^i$ 得到 $U^i$ 和 $\delta U^i$ , 执行算法 1, 得到 $U^{i+1}$ 和 ID;

if ID=1, 则结束该算法, 输出 $U=U^{i+1}$ 和 ID;

else $i=i+1$ , 计算 $J(U^i)$ , ${d_J}^i$ 和 $H^i$ ;

结束 if 循环;

结束 while 循环;

输出: ID, $U$ .

3.2.3 多尺度图像配准算法

在本章的最后, 给出对应的二维多尺度微分同胚算法 3

算法 3

输入: 最大尺度数 $K_M$ , $\lambda_{0}$ , $\gamma_0$ , $\Theta_0$ , $\alpha$ ;

for $scale = 1:K_M$

a) 令 $\boldsymbol{u}=\boldsymbol{0}$ , 更新 $\lambda_{scale}= \lambda_0\times a^{scale-1}$ , $\gamma=\frac{\gamma_0}{\lambda_{scale}}$ , $\Theta=\frac{\Theta_0}{\lambda_{scale}}$ ;

b) 由算法 2 得到 $\boldsymbol{u}^{scale}$ ;

c) 更新 $\boldsymbol{h}_{scale}=\boldsymbol{x}+\boldsymbol{u}^{scale}$ , $\overline{{\boldsymbol{h}}_{scale}}=\overline{{\boldsymbol{h}}_{scale-1}}\circ \boldsymbol{h}_{scale}$ , 计算 $T_{scale}(\cdot)=T_{scale-1}\circ\overline{{\boldsymbol{h}}_{scale}}(\cdot)$ ;

end

输出: 配准结果 $T\circ\overline{{\boldsymbol{h}}_{K_M}}(\cdot)$ .

4 数值实验

4.1 实验数据及对比算法

为验证算法的有效性, 本文主要采用以下几类图像数据及对比算法

实验数据

(i) 合成图像数据: A-A, A-R;

(ii) 自然图像数据: Watermelon, Pineapple-Pepper;

(iii) 医学图像数据: Liver MRI, Brain MRI, Hand CT.

对比算法: TV model^[39], Demons, the Multimodality Nonrigid Demon (MND)^[40,41].

4.2 评价指标

在本文中, 主要选取相对残差平方和 ( $Re-SSD$ ) 和网格重叠个数 (MFN) 两个指标来判断配准结果的好坏, 其表达式分别为

$\begin{equation} Re-SSD=\frac{SSD(T(\boldsymbol{x}+\boldsymbol{u}(\boldsymbol{x})),D)}{SSD(T,D)}, \end{equation}$

(4.1)

其中 $SSD(T,D)=\frac{1}{2}\sum_{i,j}(T_{i,j}-D_{i,j})^2$ ;

$\begin{equation} MFN(\boldsymbol{u})=\#(\det J(\boldsymbol{u})\leqslant0), \end{equation}$

(4.2)

其中 $\det(J(\boldsymbol{u}))=(1 + \frac{{\partial {u_1}}}{{\partial {x_1}}})(1+\frac{\partial{u_2}}{\partial{x_2}})-{\frac{\partial{u_1}}{\partial{x_2}}}{\frac{\partial{u_2}}{\partial{x_1}}}$ , $\#(A)$ 表示集合 $A$ 中元素个数.

4.3 数值实验

4.3.1 合成图像配准

为了验证所提算法在图像配准中的有效性, 在此部分以 A-A 和 A-R 两对典型合成数据为例, 以 $Re-SSD$ 和 $MFN$ 为指标来判断所提算法在合成图像中配准结果的好坏, 并给出了对应的配准结果示意图, 形变场 $\boldsymbol{h}$ 及对应的 $Re-SSD$ 随尺度的变化趋势, 结果如图2-5 所示. 从图中可以看出, 随着尺度的增加, 误差 $Re-SSD$ 在逐渐减小, 最终误差分别为 3.86 $\%$ 和 6.86 $\%$ , 这说明配准结果 $T\circ{\overline{\boldsymbol{h}_K}}{(\cdot)}$ 与目标图像 $D(\cdot)$ 较为接近. 另外, 如表1 所示, 通过与其它三种算法进行对比, 结果表明, 所提出的算法误差较小并且能够有效地避免网格重叠.

图2

图2 A-A配准结果

图3

图3 A-A 在不同尺度下的配准结果

图4

图4 A-R 在不同尺度下的配准结果

图5

图5 A-R在不同尺度下的配准结果

表1 合成图像配准中四种不同配准算法的配准结果定量比较

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4.3.2 自然图像配准

在本节中, 我们对两个不同的自然图像对“西瓜、菠萝-辣椒”进行了配准.

使用所提出的算法对其进行配准, 定量评估指标和最终配准结果见图6-9. 为了体现所提算法的竞争力, 还采用了另外三种算法同时对上述自然图像进行配准, 并将配准结果与所提算法的最终结果进行比较; 比较结果如表2 所示. 表2 显示, 与其他三种算法相比, 所提出的算法具有很强的竞争力.

图6

图6 Watermelon 配准结果

图7

图7 Watermelon 在不同尺度下的配准结果

图8

图8 Pineapple 配准结果

图9

图9 Pineapple 在不同尺度下的配准结果

表2 自然图像配准中四种不同配准算法的配准结果定量比较

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4.3.3 医学图像配准

在这个部分, 我们对三个不同的医学图像对“肝脏、大脑和手部”进行了配准.

使用所提算法对上述医学图像对进行配准, 定量评价指标及最终配准结果如图10-15 所示. 另外, 为体现所提算法的竞争性, 同时用其他三种算法对上述医学图像进行配准, 将得到的配准结果与所提算法的配准结果进行对比, 比较结果如表3 所示. 由表3 可知, 本文所提算法与其他 3 种算法相比, 具有较强的竞争力.

图10

图10 Liver 配准结果

图11

图11 Liver 在不同尺度下的配准结果

图12

图12 Brain 配准结果

图13

图13 Brain 在不同尺度下的配准结果

图14

图14 Hand 配准结果

图15

图15 Hand 在不同尺度下的配准结果

表3 医学图像配准中四种不同配准算法的配准结果定量比较

新窗口打开| 下载CSV

注 4.1 TV 模型是算法 3 中 $\alpha=1$ 的特例. 从表1-3 的对比可以看出, $\alpha$ 取整数虽然能加速算法的 CPU 时间消耗, 但计算精度要低很多. 本文更关心于精度方法的提升, 这也是为什么引入分数阶正则项 (非局部算子) 的重要原因.

5 结论

本文提出了一种基于拟共形理论的分数阶微分同胚多尺度方法, 该方法有效地避免了网格重叠, 同时在没有先验正则项的情况下找到了微分同胚图像配准的最优解, 并且通过数值实验验证了所提算法的有效性.

附录 A 3.2.1 节中矩阵 $A, G, C$ 的计算

令矩阵 ${A_1}={I_2} \otimes {I_{n + 1}} \otimes \partial _n^1 \in {R^{2n(n+1) \times 2{{(n+1)}^2}}}$ , ${A_2} = {I_2} \otimes \partial _n^1 \otimes {I_{n + 1}} \in {R^{2n(n+1) \times 2{{(n+1)}^2}}}$ ,

其中 $\partial _n^1 = \left[ {\begin{array}{*{20}{c}} 1&1&{}&{}&{} \\ {}&1&1&{}&{} \\ {}& \cdots & \cdots & \cdots &{} \\ {}&{}&1&1&{} \\ {}&{}&{}&1&1 \end{array}} \right] \in {R^{n,n + 1}}$ , $\otimes$ 表示克罗内克积,则

$A = \left[ {\begin{array}{*{20}{c}} {A_1}&{{0_{2n(n+1) \times 2{{(n+1)}^2}}}} \\ {{0_{2n(n+1) \times 2{{(n+1)}^2}}}}&{A_2} \end{array}} \right] \in {R^{4n(n+1) \times 4{{(n+1)}^2}}}.$

设矩阵 $G_1$ 满足 ${G_{{1_{i + 1 + jn,i + 1 + jn}}}} = \left\{ \begin{gathered} 1,{\text{ }}0 \leqslant i \leqslant n - 1,{\text{ }}1 \leqslant j \leqslant n - 1 \hfill \\ \frac{1}{2},{\text{ }}0 \leqslant i \leqslant n - 1,{\text{ }}j = 0,n \hfill \\ \end{gathered} \right.,{\text{ }}{G_{{1_{i,j}}}} = 0,{\text{ }}i \ne j$ , 矩阵 $G_2$ 满足 ${G_{{2_{i + 1 + j(n+1),i + 1 + j(n+1)}}}} = \left\{ \begin{gathered} 1,{\text{ }}1 \leqslant i \leqslant n - 1,{\text{ }}0 \leqslant j \leqslant n - 1 \hfill \\ \frac{1}{2},{\text{ }}i = 0,n,{\text{ }}0 \leqslant j \leqslant n - 1 \hfill \\ \end{gathered} \right.,{\text{ }}{G_{{2_{i,j}}}} = 0,{\text{ }}i \ne j$ , 则

$G = \left[ {\begin{array}{*{20}{c}} {{G_1}}&0&0&0 \\ 0&{{G_1}}&0&0 \\ 0&0&{{G_2}}&0 \\ 0&0&0&{{G_2}} \end{array}} \right] \in {R^{4n(n+1) \times 4n(n+1)}}.$

令矩阵 $C_1={{I_2} \otimes {I_{n + 1}} \otimes {B_{n + 1,\alpha }}}\in {R^{ 2(n+1)^2\times 2(n+1)^2}}$ , ${C_2} = \left[ {{I_2} \otimes {B_{n + 1,\alpha }} \otimes {I_{n + 1}}} \right](list,:)$ $\in {R^{ 2(n+1)^2\times 2(n+1)^2}}$ ,其中 $list = \big( 1 + 0(n+1), \cdots, 1 + n(n+1), \cdots, n + 1 + 0(n+1), \cdots, n + 1 + n(n+1) \big)^T\in{R^{(n+1)^2\times 1}}$ ,则

$C=\Bigg[ {\begin{array}{*{20}{c}} {{C_1}} \\ {{C_2}} \end{array}} \Bigg]\in {R^{4(n+1)^2\times 2(n+1)^2}}.$

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