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Atcold
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Dec 6, 2022
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Thanks for this PR.
Please, go through all your edits and see if you removed any \_ combinations.
These are necessary sometimes with the KaTeX parser.
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| Where $\beta$ represents the inverse temperature parameter of the $\text{soft(arg)max}(\cdot)$. $\boldsymbol{X}^{\top}\in\mathbb{R}^{t \times n}$ is the transposed matrix representation of the set $\lbrace\boldsymbol{x}_i \rbrace\_{i=1}^t$, and $\boldsymbol{x}$ represents a generic $\boldsymbol{x}_i$ from the set. Note that the $j$-th row of $X^{\top}$ corresponds to an element $\boldsymbol{x}_j\in\mathbb{R}^n$, so the $j$-th row of $\boldsymbol{X}^{\top}\boldsymbol{x}$ is the scalar product of $\boldsymbol{x}_j$ with each $\boldsymbol{x}_i$ in $\lbrace \boldsymbol{x}_i \rbrace\_{i=1}^t$. | ||
| Where $\beta$ represents the inverse temperature parameter of the $\text{soft(arg)max}(\cdot)$. $\boldsymbol{X}^{\top}\in\mathbb{R}^{t \times n}$ is the transposed matrix representation of the set $\lbrace\boldsymbol{x}_i \rbrace\_{i=1}^t$, and $\boldsymbol{x}$ represents a generic $\boldsymbol{x}_i$ from the set. Note that the $j$-th row of $X^{\top}$ corresponds to an element $\boldsymbol{x}_j\in\mathbb{R}^n$, so the $j$-th row of $\boldsymbol{X}^{\top}\boldsymbol{x}$ is the scalar product of $\boldsymbol{x}_j$ with each $\boldsymbol{x}_i$ in $\lbrace \boldsymbol{x}_i \rbrace_{i=1}^t$. |
Owner
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| 其中$ \ beta $表示$ \ text {soft(arg)max}(\ cdot)$的逆温度参数。 $ \ boldsymbol {X} ^ {\ top} \ in \ mathbb {R} ^ {t \ times n} $是集合$ \ lbrace \ boldsymbol {x} _i \ rbrace \ _ {i = 1} ^ t $,而$ \ boldsymbol {x} $代表集合中的通用$ \ boldsymbol {x} _i $。 请注意,$ X ^ {\ top} $的第$ j $行对应于元素$ \ boldsymbol {x} _j \ in \ mathbb {R} ^ n $,因此$$的$ j $行 \ boldsymbol {X} ^ {\ top} \ boldsymbol {x} $是$ \ boldsymbol {x} _j $与$ \ lbrace中每个$ \ boldsymbol {x} _i $的标量乘积\ bracesymbol {x} _i \ rbrace \ _ {i = 1} ^ t $。 | ||
| 其中$\beta$表示$\text{soft(arg)max}(\cdot)$的逆温度参数。 $\boldsymbol{X}^{\top}\in\mathbb{R}^{t \times n}$是集合$\lbrace\boldsymbol{x}_i \rbrace\_{i=1}^t$的转置矩阵表示,而$\boldsymbol{x}$代表集合中的通用$\boldsymbol{x}_i$。 请注意,$X^{\top}$的第$j$行对应于元素$\boldsymbol{x}_j\in\mathbb{R}^n$,因此$\boldsymbol{X}^{\top}\boldsymbol{x}$的$j$行是$\boldsymbol{x}_j$与$\lbrace \boldsymbol{x}_i \rbrace_{i=1}^t$中每个$\boldsymbol{x}_i$的标量乘积。 |
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Here as well, we need \_.
Also, you removed \ bracesymbol {x} _i \ rbrace \ _ {i = 1} ^ t $ entirely.
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I found that the Chinese version was basically machine translated, which caused the latex syntax to be broken. Of course, there are a lot of unreasonable translation. This PR is mainly about fixing broken latex. I also did my best to fix some of the translations that were too much bullshit.