# New puzzle for who has long boring days

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 ???

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 $$6$$.