chore: improve grammar.

src/rollim.org

@@ -323,15 +323,15 @@ Under this transformation, the recurrence relation becomes:
 
 \begin{align*}
   p_{i0} &= p_{0j} = m + n, \\
-  p_{ij} &= \min \begin{cases} p_{i-1,j-1} - (\mathbf{1}_{s_i \neq t_j} + 2) \\ p_{i-1,j} \\ p_{i,j-1} \end{cases}
+  p_{ij} &= \min \begin{cases} p_{i-1,j-1} + \mathbf{1}_{s_i \neq t_j} - 2 \\ p_{i-1,j} \\ p_{i,j-1} \end{cases}
 \end{align*}
 
 The above recurrence can be easily identified in the central function of the three train, which is
 folded over the table of the costs (table comparing the characters). For this one has to
 notice that we compare insertions and substitutions, and then we can do a min scan over the result
 to get the deletions, which yields a vectorized implementation.
 
-Now the only piece I cannot put together is the construction of the table of costs, which is done
+The only piece I cannot put together is the construction of the table of costs, which is done
 by reversing \(t\), but since the final result on \(p_{ij}\) is located in the bottom right corner,
 and we do a =foldr=, I would expect it to be \(s\) the one reversed. They both work, thought, as
 the following code shows:
@@ -345,7 +345,8 @@ the following code shows:
 #+RESULTS:
 : 1
 
-My hypothesis that this can be put together using this properties of the Levenshtein distance:
+My conjecture is that this can be understood using the following properties
+of the Levenshtein distance:
 
 - \(L(s,t) = L(t,s)\)
 - \(L(s,t) = L(\text{rev}(s),\text{rev}(t))\)