New puzzle for who has long boring days

Question

This is a Sujiken puzzle

4  
9  x  
2  x  x  
8  x  x  5  
3  x  x  7  x  
7  x  x  3  4  x  
x  x  x  x  x  x  x  
x  x  x  x  x  x  5  x  
x  9  8  x  7  x  x  1  2  

Who is the fastest to solve it ???

Answer 1

I put it into a neater format:

• Left column:

  missing digits are $1, 5, 6$, and $5$ can only be in A1.

• Bottom row:

  missing digits are $3, 4, 6$, and $3$ can only be in F1.

• Bottom-right square:

  $7$ can only be in B2; we know A2, A3 are $1,6$, so B3, C2, C3 are $2, 3, 4$, where 3 can't be C3 and 4 can't be C2.

• Long diagonal:

  $7$ can only be in C7. Now we can't have any more $7$s anywhere in the whole thing:
  
  

• Middle bottom box:

  $5$ can only be in E3, $9$ must be in row 2, $8$ must be in column F.

Now putting $4$ in B3 leads to

contradiction (red numbers added under this assumption, and now there's no possible place for $4$ in the bottom row):

So $4$ is in C3. Now putting $3$ in B3 leads to

contradiction (red numbers added under this assumption, second-longest diagonal now can't contain either $8$ or $3$ even though it's eight long):

So $2$ is in B3 and $3$ is in C2. Now:

left-middle square, $2$ can only be in C6. If we put $8$ in F2, again we get a contradiction (row 3 now can't contain $7,8,9$ even though it's seven long):

So $8$ is in F3. If $4$ is in G1, then

left-middle square, $4$ can only be in B5, but then $4$ can't be anywhere in the middle-bottom square. Contradiction, so $6$ is in G1 (yay, we placed the first $6$) and $4$ is in D1.

Consider the left middle square.

• If $4$ is in B5, then we get this contradiction (nothing can be in F2):
  
  
  
  So $4$ is in B6.

• If $5$ is in C4, then we get this contradiction:
  
  
  
  So $5$ is in B5.

• Whatever is in A3 ($1$ or $6$) must also be in C4, so that's not $9$, which means C5 is $9$.

Now

F2 can only be $1$, so A2 and B4 and D3 are $6$ while A3 and C4 are $1$, so B7 is $3$.

Long diagonal:

$3$ can only be in G3, $9$ must be F4 or H2 so D2 can't be $9$, so E2 is $9$ and D2 is $2$. Then $9$ can only be in F4, H2 must be $8$, B8 must be $1$, and finally E5 is $6$.

Final solution