## 致病和治病同时传播的传染病

### 两种传染病抵消机制的简化描述

\begin{align}
q^{\left(1\right)}=p\left(1_{j},t\right) = \frac{\eta^{\left(1\right)}_{j} \left(t\right) – \eta^{\left(-1\right)}_{j} \left(t\right)}{\eta^{\left(1\right)}_{j} \left(t\right) + \eta^{\left(-1\right)}_{j} \left(t\right)}\theta\left(\eta^{\left(1\right)}_{j} \left(t\right) – \eta^{\left(-1\right)}_{j} \left(t\right)\right) \notag \\
q^{\left(-1\right)}=p\left(-1_{j},t\right) = \frac{\eta^{\left(-1\right)}_{j} \left(t\right) – \eta^{\left(1\right)}_{j} \left(t\right)}{\eta^{\left(1\right)}_{j} \left(t\right) + \eta^{\left(-1\right)}_{j} \left(t\right)}\theta\left(\eta^{\left(-1\right)}_{j} \left(t\right) – \eta^{\left(1\right)}_{j} \left(t\right)\right) \notag \\
p\left(0_{j},t\right) = 1-\frac{\left|\eta^{\left(1\right)}_{j} \left(t\right) – \eta^{\left(-1\right)}_{j} \left(t\right)\right|}{\eta^{\left(1\right)}_{j} \left(t\right) + \eta^{\left(-1\right)}_{j} \left(t\right)} = 1- q^{\left(1\right)} -q^{\left(-1\right)}
\end{align}

1. 如果$$k_{j}=0$$，则没有影响
2. 如果$$k_{j}=1$$（这个时候可以约定$$\eta^{\left(1\right)}_{j}=1,\eta^{\left(-1\right)}_{j}=0$$），则采用乘性传播
3. \begin{align}
\eta^{\left(1\right)}_{i}\left(t\right) = \eta^{\left(1\right)}_{i}\left(t-1\right) + \omega^{\left(1\right),j}_{i}\eta^{\left(1\right)}_{j}\left(t-1\right)
\end{align}
或者加性传播
\begin{align}
\eta^{\left(1\right)}_{i}\left(t\right) = \eta^{\left(1\right)}_{i}\left(t-1\right) + \omega^{\left(1\right),j}_{i}
\end{align}

4. 如果$$k_{j}=-1$$（这个时候可以约定$$\eta^{\left(1\right)}_{j}=0,\eta^{\left(-1\right)}_{j}=1$$），则采用乘性传播
5. \begin{align}
\eta^{\left(-1\right)}_{i}\left(t\right) = \eta^{\left(-1\right)}_{i}\left(t-1\right) + \omega^{\left(-1\right),j}_{i}\eta^{\left(-1\right)}_{j}\left(t-1\right)
\end{align}
或者加性传播
\begin{align}
\eta^{\left(1\right)}_{i}\left(t\right) = \eta^{\left(1\right)}_{i}\left(t-1\right) + \omega^{\left(-1\right),j}_{i}
\end{align}

\begin{align}
\eta^{\left(k_{j}\right)}_{i}\left(t\right) = \eta^{\left(k_{j}\right)}_{i}\left(t-1\right) + \omega^{\left(k_{j}\right),j}_{i}\eta^{\left(k_{j}\right)}_{j}\left(t-1\right)\left(k^{j}\right)^{2}
\end{align}

\begin{align}
\eta^{\left(k_{j}\right)}_{i}\left(t\right) = \eta^{\left(k_{j}\right)}_{i}\left(t-1\right) + \omega^{\left(k_{j}\right),j}_{i}\left(k^{j}\right)^{2}
\end{align}

## 概念网络上的高效学习方式

1. 给定$$\vec{S}$$之后在第$$m$$步的累计成本为
2. \begin{align}
C\left(\vec{S}, m\right) = \sum_{j=1}^{m} C_{s_{j}}
\end{align}

3. 给定$$\vec{S}$$之后在第$$m$$步的累计价值为
4. \begin{align}
V\left(\vec{S}, m\right) = \sum_{j=1}^{m} V_{s_{j}}
\end{align}

1. 认识下层字$$j$$，以一定的概率降低上层字$$i$$的学习成本，$$\omega^{j, \left(\uparrow\right)}_{i}$$，和结构矩阵的元素$$a^{j}_{i}$$有关。原则上可以不遵循结构矩阵，来自于其他实证关系。做为一个简化模型，我们可以假设$$\omega^{j, \left(\uparrow\right)}_{i} = 0 \mbox{或者} a^{j}_{i}$$。更复杂的需要考虑理据性。
2. 认识上层字$$j$$，以一定的概率降低下层字$$i$$的学习成本$$\omega^{j, \left(\downarrow\right)}_{i}$$，和结构矩阵的元素$$a^{i}_{j}$$有关。原则上可以不遵循结构矩阵，来自于其他实证关系。做为一个简化模型，我们可以假设$$\omega^{j, \left(\downarrow\right)}_{i} = 0$$。更复杂的需要考虑理据性。
3. 有了这个改变学习成本的概率，我们再来看学习成本改变的值。采用递归定义：
4. \begin{align}
c_{i}\left(t\right) = \left(1-k_{i}\right) \cdot \left[\Pi_{u}\left(1-k_{u}\omega^{u,\left(\downarrow\right)}_{i}\right)a^{i}_{u}\right]\cdot \left\{1+\sum_{d}\left(1-k_{d}\omega^{d\left(\uparrow\right)}_{i}\right) \cdot \left[1+\left(1-k_{d}\right)c_{d}\left(t\right)\right]a^{d}_{i}\right\}.
\end{align}

\begin{align}
c_{i}\left(t\right) = \left(1-k_{i}\right) \cdot \left\{1+\sum_{d}\left[1+\left(1-k_{d}\right)c_{d}\left(t\right)\right]a^{d}_{i}\right\}.
\end{align}