More escaping to get around LaTeX equation parsing in Blackfriday

Fixes https://github.com/kaushalmodi/ox-hugo/issues/138
8 years ago · 549adc0f4f
parent 458d6549de
commit 549adc0f4f
4 changed files with 111 additions and 28 deletions
--- a/ox-blackfriday.el
+++ b/ox-blackfriday.el
@ -279,11 +279,33 @@ doesn't matter in equations.

  \"(c)\" -> \"( c)\"
  \"(r)\" -> \"( r)\"
-  \"(tm)\" -> \"( tm)\"."
+  \"(tm)\" -> \"( tm)\"
+
+https://gohugo.io/content-management/formats#solution
+https://github.com/kaushalmodi/ox-hugo/issues/138
+
+Need to escape the backslash in \"\\(\", \"\\)\", .. to make
+Blackfriday happy.  So:
+
+  \"\\(\" -> \"\\\\(\"
+  \"\\)\" -> \"\\\\)\"
+  \"\\\\=[\" -> \"\\\\\\=[\"
+  \"\\\\=]\" -> \"\\\\\\=]\"
+  \"\\|\" -> \"\\\\|\"
+
+and finally:
+
+  \"\\\\\" -> \"\\\\\\\\\\\\\"."
  (let* (;; _ -> \_, * -> \*
         (escaped-str (replace-regexp-in-string "[_*]" "\\\\\\&" str))
         ;; (c) -> ( c), (r) -> ( r), (tm) -> ( tm)
-         (escaped-str (replace-regexp-in-string "(\\(c\\|r\\|tm\\))" "( \\1)" escaped-str)))
+         (escaped-str (replace-regexp-in-string "(\\(c\\|r\\|tm\\))" "( \\1)" escaped-str))
+         ;; \( -> \\(, \) -> \\), \[ -> \\[, \] -> \\], \| -> \\|
+         (escaped-str (replace-regexp-in-string "\\(\\\\[]()[|]\\)" "\\\\\\1" escaped-str))
+         (escaped-str (replace-regexp-in-string
+                       "\\([^\\]\\)\\\\\\{2\\}[[:blank:]]*$" ;Replace "\\" at EOL with:
+                       "\\1\\\\\\\\\\\\\\\\\\\\\\\\"             ;"\\\\\\"
+                       escaped-str)))
    escaped-str))

 ;;;; Reset org-blackfriday--code-block-num-backticks
@ -557,11 +579,6 @@ INFO is a plist holding contextual information."
     ((memq processing-type '(t mathjax))
      (let* ((latex-frag (org-element-property :value latex-fragment))
             (frag (org-html-format-latex latex-frag 'mathjax info))
-             ;; https://gohugo.io/content-management/formats#solution
-             ;; Need to escape the backslash in "\(", "\)", .. to
-             ;; make Blackfriday happy.  So \( -> \\(, \) -> \\),
-             ;; \[ -> \\[ and \] -> \\].
-             (frag (replace-regexp-in-string "\\(\\\\[]()[]\\)" "\\\\\\1" frag))
             (frag (org-blackfriday-escape-chars-in-equation frag)))
        ;; (message "[ox-bf-latex-frag DBG] frag: %s" frag)
        frag))
--- a/test/site/content-org/all-posts.org
+++ b/test/site/content-org/all-posts.org
@ -2704,6 +2704,33 @@ one, but:
  exported Markdown file./
 - It has a different label (=\label{eq:2}= instead of =\label{eq:1}=);
  Mathjax *requires the equation labels to be unique*.
+** Blackfriday-specific escaping in equations :escaping:@upstream:blackfriday:
+:PROPERTIES:
+:EXPORT_FILE_NAME: equations-bf-escaping
+:END:
+{{{oxhugoissue(138)}}}
+*** =\|= → =\\|=
+$$
+C(w,b) = \frac{1}{2n} \sum_x{{\|y(x)-a\|}^2}
+$$
+*** =\\= at EOL → =\\\\\\=
+\begin{align}
+a^1  &= x \\
+a^2  &=  σ(W^2a^1 + b^2) \\
+a^3  &=  σ(W^3a^2 + b^3) \\
+⋯ \\
+a^L  &= σ(W^La^{L-1} + b^L) \\
+y  &= a^L \\
+\end{align}
+**** Same as above, but without space before the =\\= at EOL
+\begin{align}
+a^1  &= x\\
+a^2  &=  σ(W^2a^1 + b^2)\\
+a^3  &=  σ(W^3a^2 + b^3)\\
+⋯\\
+a^L  &= σ(W^La^{L-1} + b^L)\\
+y  &= a^L\\
+\end{align}
 * Lists                                                               :lists:
 ** List following a list
 :PROPERTIES:
--- a/test/site/content/posts/equations-bf-escaping.md
+++ b/test/site/content/posts/equations-bf-escaping.md
@ -0,0 +1,39 @@
+++
+title = "Blackfriday-specific escaping in equations"
+tags = ["equations", "escaping", "blackfriday"]
+categories = ["upstream"]
+draft = false
+++
+
+`ox-hugo` Issue #[138](https://github.com/kaushalmodi/ox-hugo/issues/138)
+
+
+## `\|` → `\\|` {#}
+
+\\[
+C(w,b) = \frac{1}{2n} \sum\_x{{\\|y(x)-a\\|}^2}
+\\]
+
+
+## `\\` at EOL → `\\\\\\` {#at-eol}
+
+\begin{align}
+a^1  &= x \\\\\\
+a^2  &=  σ(W^2a^1 + b^2) \\\\\\
+a^3  &=  σ(W^3a^2 + b^3) \\\\\\
+⋯ \\\\\\
+a^L  &= σ(W^La^{L-1} + b^L) \\\\\\
+y  &= a^L \\\\\\
+\end{align}
+
+
+### Same as above, but without space before the `\\` at EOL {#same-as-above-but-without-space-before-the-at-eol}
+
+\begin{align}
+a^1  &= x\\\\\\
+a^2  &=  σ(W^2a^1 + b^2)\\\\\\
+a^3  &=  σ(W^3a^2 + b^3)\\\\\\
+⋯\\\\\\
+a^L  &= σ(W^La^{L-1} + b^L)\\\\\\
+y  &= a^L\\\\\\
+\end{align}
--- a/test/site/content/real-examples/nn-intro.md
+++ b/test/site/content/real-examples/nn-intro.md
@ -49,7 +49,7 @@ Disclaimer
 a
 = \sigma(
    \left[ \begin{matrix}
-        w\_{1} & ⋯  & w\_{n} \\
+        w\_{1} & ⋯  & w\_{n} \\\\\\
    \end{matrix}\right]  ·
    \left[ \begin{array}{x} x\_1 \\ ⋮ \\ ⋮ \\ x\_n \end{array}\right] +
    b
@ -96,9 +96,9 @@ a
 \left[ \begin{array}{a} a\_1 \\ ⋮ \\ a\_s \end{array}\right]
 = \sigma(
    \left[ \begin{matrix}
-        w\_{1,1} & ⋯  & w\_{1,n} \\
-        ⋮ & ⋱  & ⋮  \\
-        w\_{s,1} & ⋯  & w\_{s,n} \\
+        w\_{1,1} & ⋯  & w\_{1,n} \\\\\\
+        ⋮ & ⋱  & ⋮  \\\\\\
+        w\_{s,1} & ⋯  & w\_{s,n} \\\\\\
    \end{matrix}\right]  ·
    \left[ \begin{array}{x} x\_1 \\ ⋮ \\ ⋮ \\ x\_n \end{array}\right] +
    \left[ \begin{array}{b} b\_1 \\ ⋮ \\ b\_s \end{array}\right]
@ -138,9 +138,9 @@ a
 \left[ \begin{array}{a} a^l\_1 \\ ⋮ \\ a^l\_{d\_l} \end{array}\right]
 = \sigma(
    \left[ \begin{matrix}
-        w^l\_{1,1} & ⋯  & w^l\_{1,d\_{l-1}} \\
-        ⋮ & ⋱  & ⋮  \\
-        w^l\_{d\_l,1} & ⋯  & w^l\_{d\_l,d\_{l-1}} \\
+        w^l\_{1,1} & ⋯  & w^l\_{1,d\_{l-1}} \\\\\\
+        ⋮ & ⋱  & ⋮  \\\\\\
+        w^l\_{d\_l,1} & ⋯  & w^l\_{d\_l,d\_{l-1}} \\\\\\
    \end{matrix}\right]  ·
    \left[ \begin{array}{x} a^{l-1}\_1 \\ ⋮ \\ ⋮ \\ a^{l-1}\_{d\_{l-1}} \end{array}\right] +
    \left[ \begin{array}{b} b^l\_1 \\ ⋮ \\ b^l\_{d\_l} \end{array}\right])
@ -177,12 +177,12 @@ a
 计算过程如下：

 \begin{align}
-a^1  &= x \\
-a^2  &=  σ(W^2a^1 + b^2) \\
-a^3  &=  σ(W^3a^2 + b^3) \\
-⋯ \\
-a^L  &= σ(W^La^{L-1} + b^L) \\
-y  &= a^L \\
+a^1  &= x \\\\\\
+a^2  &=  σ(W^2a^1 + b^2) \\\\\\
+a^3  &=  σ(W^3a^2 + b^3) \\\\\\
+⋯ \\\\\\
+a^L  &= σ(W^La^{L-1} + b^L) \\\\\\
+y  &= a^L \\\\\\
 \end{align}

 推断实际上就是一系列矩阵乘法与向量运算，一个训练好的神经网络可以高效地
@ -213,7 +213,7 @@ y  &= a^L \\
 (MeanSquareError)**

 \\[
-C(w,b) = \frac{1}{2n} \sum\_x{{\|y(x)-a\|}^2}
+C(w,b) = \frac{1}{2n} \sum\_x{{\\|y(x)-a\\|}^2}
 \\]

 前面的系数 \\(\frac 1 2\\) 是为了求导后简洁的形式而添加的， \\(n\\) 是使用样本
@ -286,7 +286,7 @@ MSE，而不是诸如"正确分类图像个数"的指标，是因为只有一个
 \\(w,b\\) 数目非常巨大

 \\[
-w →w' = w-η\frac{∂C}{∂w} \\
+w →w' = w-η\frac{∂C}{∂w} \\\\\\
 b → b' = b-η\frac{∂C}{∂b}
 \\]

@ -343,7 +343,7 @@ b → b' = b-η\frac{∂C}{∂b}
 | \\(∇C\_{W^l} = δ^l × {(a^{{l-1})}}T\\)    | 权值梯度计算公式 | BP3 |
 | \\(∇C\_b = δ^l\\)                         | 偏置梯度计算公式 | BP4 |

-当误差函数取MSE： \\(C = \frac 1 2 \|\vec{y} -\vec{a}\|^2= \frac 1 2
+当误差函数取MSE： \\(C = \frac 1 2 \\|\vec{y} -\vec{a}\\|^2= \frac 1 2
 [(y\_1 - a\_1)^2 + \cdots + (y\_{d\_L} - a\_{d\_L})^2]\\) ，激活函数取sigmoid时：

 | 计算方程                                       | 说明                                                           | 编号 |
@ -375,7 +375,7 @@ b → b' = b-η\frac{∂C}{∂b}
 \frac{∂C}{∂z^L} = \frac{∂C}{∂a^L} \frac{∂a^L}{∂z^L}  = ∇C\_a σ'(z^L)
 \\]

-而因为误差函数 \\(C = \frac 1 2 \|\vec{y} -\vec{a}\|^2= \frac 1 2
+而因为误差函数 \\(C = \frac 1 2 \\|\vec{y} -\vec{a}\\|^2= \frac 1 2
 [(y\_1 - a\_1)^2 + ⋯ + (y\_{d\_L} - a\_{d\_L})^2]\\) ，方程两侧对某个 \\(a\_j\\) 取偏导则
 有：

@ -428,10 +428,10 @@ z^{l+1}\_k = W^{l+1}\_{k,\*} ·a^l + b^{l+1}\_{k}  =   W^{l+1}\_{k,\*} · σ(z^{
 回代则有：

 \begin{align}
-δ^l\_j & = \sum\_{k=1}^{d\_{l+1}}  (δ^{l+1}\_k  \frac{∂z^{l+1}\_k}{∂z^{l}\_j})  \\
-& = σ'(z^l)   \sum\_{k=1}^{d\_{l+1}}     (δ^{l+1}\_k  w^{l+1}\_{kj}) \\
-& = σ'(z^l) ⊙  [(δ^{l+1}) · W^{l+1}\_{\*.j}] \\
-& = σ'(z^l) ⊙ [(W^{l+1})^T\_{j,\*} ·  (δ^{l+1}) ]\\
+δ^l\_j & = \sum\_{k=1}^{d\_{l+1}}  (δ^{l+1}\_k  \frac{∂z^{l+1}\_k}{∂z^{l}\_j})  \\\\\\
+& = σ'(z^l)   \sum\_{k=1}^{d\_{l+1}}     (δ^{l+1}\_k  w^{l+1}\_{kj}) \\\\\\
+& = σ'(z^l) ⊙  [(δ^{l+1}) · W^{l+1}\_{\*.j}] \\\\\\
+& = σ'(z^l) ⊙ [(W^{l+1})^T\_{j,\*} ·  (δ^{l+1}) ]\\\\\\
 \end{align}

 这里，对后一层所有神经元的误差权值之积求和，可以改写为两个向量的点积：