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[ZH] Fixing the latex issue 13-3 #697

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[ZH] Fixing the latex issue 13-3 #697

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[ZH] Fixing the latex issue
JonathanSum Nov 25, 2020
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Update 13-3.md
JonathanSum Nov 25, 2020
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Update 13-3.md
JonathanSum Nov 25, 2020
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JonathanSum Nov 25, 2020
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10 changes: 8 additions & 2 deletions docs/zh/week13/13-3.md
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Expand Up @@ -21,7 +21,7 @@ translator: Jonathan Sum(😊🍩📙)
Unlike a sequence, it does not have an order.
- Hidden vector $\boldsymbol{h}$ is given by linear combination of the vectors in the set.
- We can express this as $\boldsymbol{X}\boldsymbol{a}$ using matrix vector multiplication, where $\boldsymbol{a}$ contains coefficients that scale the input vector $\boldsymbol{x}_{i}$. -->
在自我注意力机制中，我们有一个输入集$\lbrace\boldsymbol{x}\_{i}\rbrace^{t}_{i=1}$。不像序列那样，它没有顺序的。
在自我注意力机制中，我们有一个输入集$\lbrace\boldsymbol{x}\_{i}\rbrace^{t}\_{i=1}$。不像序列那样，它没有顺序的。
隐藏向量$\boldsymbol{h}$是由集之中的向量的线性组合得出来的。
我们可以用矩阵向量乘法来以$\boldsymbol{X}\boldsymbol{a}$去表达这个东西，这里$\boldsymbol{a}$包含一些会缩放向量$\boldsymbol{x}_{i}$的系数。

Expand Down Expand Up @@ -78,7 +78,13 @@ In the context of GCN, this means that if we have multiple incoming edges,*i.e.*
On the other hand, if we have only one incoming edge, this value gets smaller.
To remedy this issue of the value being proportionate to the number of incoming edges, we divide it by the number of incoming edges $d$.
We then apply rotation $\boldsymbol{V}$ to $\boldsymbol{X}\boldsymbol{a}d^{-1}$. -->
记得在自我注意力机制中，隐藏向量$\boldsymbol{h}$是以$\boldsymbol{X}\boldsymbol{a}$来计算出来，也就是说在$\boldsymbol{X}$中的行是以在$\boldsymbol{a}$中的因素缩放出来的。而图形卷积网络的内容中，这意思着这样，如果我们有多个由外传入来的边，比如在邻接向量$\boldsymbol{a}$中有多个1，{X}\boldsymbol{a}$就会变得很大。在另一面，如果我们只有一个传入来的边，这个值就会变得很细。去补救这个数值与由外传入来的边是成比例的问题，我们对它除以一个数，这个数就是由外传入来的边有多少的数量，这个数叫$d$。我们之后对$\boldsymbol{X}\boldsymbol{a}d^{-1}$使用一个旋转$\boldsymbol{V}$。
记得在自我注意力机制中，隐藏向量$\boldsymbol{h}$是以$\boldsymbol{X}\boldsymbol{a}$来计算出来，
也就是说在$\boldsymbol{X}$中的行是以在$\boldsymbol{a}$中的因素缩放出来的。而图形卷积网络的内容中，
这意思着这样，如果我们有多个由外传入来的边，比如在邻接向量$\boldsymbol{a}$中有多个1，$\_boldsymbol{X}\_boldsymbol{a}$就会变得很大。
在另一面，如果我们只有一个传入来的边，这个值就会变得很细。
去补救这个数值与由外传入来的边是成比例的问题，我们对它除以一个数，
这个数就是由外传入来的边有多少的数量，这个数叫$d$。我们之后对$\_boldsymbol{X}\_boldsymbol{a}d^{-1}$使用一个旋转$\_boldsymbol{V}$。


<!-- We can represent this hidden representation $\boldsymbol{h}$ for the entire set of inputs $\boldsymbol{x}$ using the following matrix notation: -->
我們可以以這個隱藏表示$\boldsymbol{h}$來以下方的矩陣符號來表達整個輸入集$\boldsymbol{x}$
Expand Down