Nested Lists and Maths Rendering

[aldenluthfi]

Description

I write my proofs in callouts on one of my blogs about math. The parser seems to miss some parts.

Input

Expected Output

> **PROOF**
> 
> Part 1: Proof that there are atleast *one* solution
> 
> 1. We have concluded that given $a$ and $b$, we can always find $m,n \in Z$ such that $\gcd(a,b) = am+bn$
> 2. Let $\gcd(a,b)=d$
> 3. $*d \mid c$* means $c = kd$ for some $k \in \Z$
> 4. We can rewrite $c=ax+by$ to $kd = ax+by$
> 5. We can always find $x,y \in \Z$ where $c=ax+by$ because from (1), we can always find $m, n \in Z$ such that $d = am+bn$ and
>     
>     $$
>     \begin{split}
>     c &= ax+by \\
>     kd &= k(am+bn)
>     \end{split}
>     
>     $$
>     
> 6. Therefore we can find atleast one pair of $x$ and $y$ such that $c = ax+by$ if $\gcd(a,b) \mid c$ in the form of $x=km$ and $y=kn$ where $k=\frac{c}{\gcd(a,b)}$ 
> 
> Part 2: Proof that there are *infinitely many* solutions
> 
> 7. From (6), suppose $x'$ and $y'$ is also a solution to $c = ax+by$
> 8. Therefore we can write $ax+by=ax'+by'$
> 9. From (8) we can rearrange to $a(x-x')=b(y'-y)$
> 10. Let $a'=\frac{a}{d}$ and $b'=\frac{b}{d}$, $a'$ and $b'$ are coprime because of the following proof of contradiction
>     1. Assume $a'=\frac{a}{d}$ and $b'=\frac{b}{d}$ are not coprime and $d = \gcd(a,b)$ 
>     2. there is $e> 1$ such that $e=\gcd(a',b')$
>     3. Therefore $\frac{a'}{e}=\frac{a}{ed}$ is an integer, same goes with $\frac{b'}{e}=\frac{b}{ed}$
>     4. $ed > d$ since $e > 1$
>     5. Because there is an integer larger than $d$ that divides $a$ and $b$, $d \neq \gcd(a,b)$
>     6. Therefore (a) is wrong, $a'$ and $b'$ are coprime $\square$
> 11. From (9) and (10), $a'(x-x')=b'(y'-y)$
> 12. From (11)
>     
>     $$
>     \begin{split}
>     y'-y& =\frac{a'(x-x')}{b'}\\
>     y' &=y + \frac{a'(x-x')}{b'}
>     \end{split}
>     $$
>     
> 13. From (11)
>     
>     $$
>     \begin{split}
>     x-x'& =\frac{b'(y'-y)}{a'}\\
>     x' &=x - \frac{b'(y'-y)}{a'}
>     \end{split}
>     $$
>     
> 14. from (11), $\frac{(x-x')}{b'} = \frac{(y'-y)}{a'}$
> 15. $x'$ is an integer if $\frac{(y'-y)}{a'}$ is an integer because $x,b'\in \Z$
> 16. $\frac{(y'-y)}{a'}$ is an integer if $x'$  is an integer because of the following proof
>     1. Assume $a' \mid (y'-y)$
>     2. Therefore $(y'-y) = a'r$ for some $r \in \Z$
>     3. From (b), $by'-by = a'r$
>     4. Because $by' = c - ax'$ and $by = c - ax$, thus $ax-ax'= a'r$
>     5. Dividing by $a$, $x-x'= \frac{r}{d}$
>     6. From (e) rearranging to $d(x-x')= r$, the statement is true if $x'$ is an integer because $d,x,r \in \Z$
>     7. Therefore $\frac{(y'-y)}{a'} \in \Z$ if $x'\in \Z$ $\square$
> 17. Let  $\frac{(x-x')}{b'} = \frac{(y'-y)}{a'} = t$
> 18. Rewrite $x'=x - \frac{tb}{d}$ and $y'= y + \frac{ta}{d}$
> 19. Therefore we can find infinitely many $x$ and $y$ such that $c = ax+by$ if $\gcd(a,b) \mid c$ in the form of $x=km - \frac{tb}{d}$ and $y=kn+ \frac{ta}{d}$ where $k=\frac{c}{\gcd(a,b)}$ for every integer $t \in \Z$ $\square$

Actual Output

> **PROOF**  
> Part 1: Proof that there are atleast _one_ solution  
>   
> 1. We have concluded that given $a$ and $b$, we can always find $m,n \in Z$ such that $\gcd(a,b) = am+bn$  
>   
> 2. Let $\gcd(a,b)=d$  
>   
> 3. $d \mid c$ means $c = kd$ for some $k \in \Z$  
>   
> 4. We can rewrite $c=ax+by$ to $kd = ax+by$  
>   
> 5. We can always find $x,y \in \Z$ where $c=ax+by$ because from (1), we can always find $m, n \in Z$ such that $d = am+bn$ and  
>   
> 6. Therefore we can find atleast one pair of $x$ and $y$ such that $c = ax+by$ if $\gcd(a,b) \mid c$ in the form of $x=km$ and $y=kn$ where $k=\frac{c}{\gcd(a,b)}$  
>   
> Part 2: Proof that there are _infinitely many_ solutions  
>   
> 1. From (6), suppose $x'$ and $y'$ is also a solution to $c = ax+by$  
>   
> 2. Therefore we can write $ax+by=ax'+by'$  
>   
> 3. From (8) we can rearrange to $a(x-x')=b(y'-y)$  
>   
> 4. Let $a'=\frac{a}{d}$ and $b'=\frac{b}{d}$, $a'$ and $b'$ are coprime because of the following proof of contradiction  
>   
> 5. From (9) and (10), $a'(x-x')=b'(y'-y)$  
>   
> 6. From (11)  
>   
> 7. From (11)  
>   
> 8. from (11), $\frac{(x-x')}{b'} = \frac{(y'-y)}{a'}$  
>   
> 9. $x'$ is an integer if $\frac{(y'-y)}{a'}$ is an integer because $x,b'\in \Z$  
>   
> 10. $\frac{(y'-y)}{a'}$ is an integer if $x' $ is an integer because of the following proof  
>   
> 11. Let $ \frac{(x-x')}{b'} = \frac{(y'-y)}{a'} = t$  
>   
> 12. Rewrite $x'=x - \frac{tb}{d}$ and $y'= y + \frac{ta}{d}$  
>   
> 13. Therefore we can find infinitely many $x$ and $y$ such that $c = ax+by$ if $\gcd(a,b) \mid c$ in the form of $x=km - \frac{tb}{d}$ and $y=kn+ \frac{ta}{d}$ where $k=\frac{c}{\gcd(a,b)}$ for every integer $t \in \Z$ $\square$

No pressure though, even the native notion export to md doesn't get it right

[creichelt-transporeon]

I have a similar problem, this is basically what i am calling

const blocks = await n2m.pageToMarkdown(pageId);
const markdown = n2m.toMarkdownString(blocks);

this is the notion page

this the output

> ## 📖 US-1 Create Capacity Plan  
>   
> As a warehouse manager, I need the ability to create Capacity Plan so that it could be used to generate slots for appointment scheduling.   
>   
> ### Acceptance Criteria  
>   
> 1. As a warehouse manager, when I submit a request to create a capacity plan, the system should perform the following:  
>   
>   
>   
> ## US-2 Update Capacity Plan  
>   
> As a warehouse manager, I need the ability to update an existing Capacity Plan to accommodate changes in scheduling requirements.  
>   
> ### Acceptance Criteria  
>   
> 1. As a warehouse manager, when I submit a request to update a capacity plan, the system should perform the following:  

the toMarkdownString is ignoring all children of a callout block and using directly it's 'parent' text
however the 'parent' text does not contain all deeper nested child items

for me the (native) export directly from notion works

[souvikinator]

Hey there @aldenluthfi @creichelt-transporeon,
v3 has a bunch of issues, and it doesn't handle the inconsistent representation of blocks from the API.

I would appreciate it if you could try v4 and let me know if it works as expected. If not, we would put this on priority in the upcoming release v4-alpha.6

[aldenluthfi]

Hey there @aldenluthfi @creichelt-transporeon, v3 has a bunch of issues, and it doesn't handle the inconsistent representation of blocks from the API.

I would appreciate it if you could try v4 and let me know if it works as expected. If not, we would put this on priority in the upcoming release v4-alpha.6

Sadly, no. the nested items are still gone :(

[souvikinator]

Got you @aldenluthfi , thanks for getting back. Can you share a copy of the notion page so we can dig down on the issue?

We'll ship the fixes in upcoming release

[aldenluthfi]

Sure, it is live on https://aldenluthfi.notion.site/Secret-Things-and-How-to-Count-Them-5080c01875f44596b1aed9b97ec2cb3b?pvs=143

[yashvikram30]

i have submitted a pr #155 for this issue. kindly review and merge

[souvikinator]

@yashvikram30 reviewed the PR it has a bunch of issues, and the PR doesn't cover all the rendering issues.
Let's schedule a call to review this around tomorrow night?