Solution to Diabolic Str8ts #8 puzzle (http://is.gd/slowthinker_diabolic_str8ts_8c) and its tough variation Diabolic #8b (http://is.gd/slowthinker_diabolic_str8ts_8d)

1. Diabolic Str8ts Puzzle #8
Puzzle & solution by
SlowThinker

2. Start position
With the diabolic
Str8ts series, I try to
push the boundaries a
bit. Normal strategies
are not enough to
solve these puzzles.
Diabolic #8 is easier
than its variation #8b.
So let’s look at
Diabolic #8 first. The
start position is on the
right.

3. Unique solution
constraint
In EF12 we find a
unique rectangle on
23 which removes 2
and 3 from E1.
Once E1≠23 FJ1 are a
hidden pair on 23.
Hence J1=23 and
E1=4s.
(We will see in Diabolic
#8b that E1=4 is crucial for
the easier difficulty of the
puzzle.)

4. Setti on 4
Next we can set C6=4
due to 4s appearing in
all rows.

5. X-Wing on 5
Setti on 8
HJ35 forms an x-wing
on 5 that removes the
5s from the yellow
fields. This leads to
G6=5s.
Next with Setti on 8
(border marks) we can
remove the 5s from
CE8, as 8 has to
appear in column 8.

6. X-Wing on 7
HJ34 forms an x-wing
on 7 that removes the
7s from the yellow
fields. This leads to
G1=7s and D1=6s.
This solves quite many
fields.
Note that we can also
remove 1 from EH3,
because the x-wing on
7 makes 7 a sure
candidate in D..J3.

7. X-Wing on 5
Another x-wing on 5,
this time at CE47 that
results in E9=6x. That
in turn solves another
5 cells.
Next we have a naked
pair BE3=28 that
results in G3=6p.

8. Chain test:
E5=1?
At the end we need
one chain. Let’s have a
look at E5=1:
E5=1  H1=1s 
C1=8, C2=7 
Orange: B2=8, B3=2
Blue: C7=5, E7=8
Which leaves no
candidates in E3.
Therefore E5 cannot
be 1.

9. Solution
Once we know E5=8,
the puzzle solves
easily.
So this puzzle was not
too difficult. Let’s
examine its much
tougher variation.

10. Diabolic #8b
In this variation the
puzzle is exactly the
same, but for one
change: instead of
F3=4 we now have
F2=2.
As we will see this
small change greatly
increases the difficulty
of the puzzle.

11. Hidden triple
The main difference is
that F7 now contains
45 as additional
candidates and
therefore 78 remains
in HJ7. Also, E1 no
longer can be set to 4.
EFJ1 now contain the
hidden triple 234.

12. Chain test:
HJ7=78
Let’s test H7=8.
H7=8  H1=1 
D1=6c  D9=7, D8=8,
D7=9.
But: with H7=8 D7
cannot be 9 as it
would be a stranded
digit.
 HJ7 ≠ 78.

13. Range test:
D..J3=1-6?
Let’s assume that D..J3
has the range 1-6.
 J3≠7  J4=7s 
H4=8
With H4=8 we get
D9=7 (just like in the
test before)  There
is no 7 left in row H!
Hence D..J3’s range
cannot be 1-6 and we
can remove 1 from the
blue fields.

14. Setti on 8
Again we can remove
the 5s in CE8, as Setti
on 8 tells us that there
must be a 8 in every
column.

15. Test H9=7?
H9=7 inhibits an x-
wing from forming, so
let’s test it:
H9=7  H4=8  H1=1
H9=7  D9=6  D1=9
But this cannot be.
The numbers are out
of range. Thus H9≠7.
Next we get an x-wing
on 7 at HJ34.

16. Test C6=1?
Next let’s test C6=1
using Setti on 4 and 7:
C6=1  C12=78S7,
E6=4S4  G6=5s 
G1=7s  C1=8, C2=7,
B2=8, B3=2  B4=6
C6=1  C8=6c, E6=4S4
 C4=35, E1=2 
E4=1 (b/c of B4=6)
But these two
numbers are out of
range  C6=4!

17. Test A7=3?
As there are only two
3s in row A, let’s test
A7=3:
A7=3  A8=4s
A7=3  J7=2  H7=1
 HJ8=34
So, A7=3 results in a
contradiction.
Therefore we can set
A4=3.

18. Test AB3=12?
B3=2  B4=6
(a)  J4=7, H4=8 
H1=1  C1=78c
(b) Because of H1=1
 D1=6  D7=9 and
because B4=6  C4=5
we get C7=67
Thus we have a
contradiction in
C12=78 and C78=67.
Hence B3=8 and A3=9.
Next: X-wing on 8
removes 8 from GH5.

19. Test C7=3?
There are only two 3s
in row C, so let’s test
C7=3:
C7=3  D7=678 
D1=19  C1=7
C7=3  J7=2, H7=1 
HJ8=43  A8=5 
E7=5s  G6=5s 
G1=7s
This leads to a
contradiction  C7≠3
which solves a few
more cells.

20. Test D789=678?
D789=678  D1=19
D1=19  C1=7 
C8=6, C7=5, C4=1 
E4=5  E6=79
D1=19  G1=6s 
G3=5  G6=79
No 5 left in EFG6 
C789 ≠ 678  C7=9
Plus: we can remove
G6=6 as this again
yields D1=19. Thus we
know that D1=6s!

21. Test E5=1?
There are only two 1s
in column 5  let’s
test E5=1. The green
cells solve quickly:
E5=1  E6=9s, E7=8s,
F5=8s, F6=7, F7=6
From there:
F7=6  B7=4  B5=6
 J5=5
F7=6  F9=5  H9=4
 H5=5
Contradiction, thus:
H5=1s & E4=1s!

22. Solution
Finally this solves the
puzzle. The main
obstacle was that E1
was undetermined
and not equal to 4, as
in the easy variation.
There are of course
other ways to solve
this puzzle. I tried to
show tests & chains
that are not too long.

23. Glossary
Letters appended to steps indicate the last strategy used, just before filling in a field:
• No letter … number was last candidate in field
• s … single (last) candidate for that number in compartment
• c … compartment range check
• d … stranded (unreachable/impossible) digits removed
• h … high/low range check across compartments
• p/t/q … naked pair / naked triple / naked quadruple
• ph/th/qh … hidden pair / hidden triple / hidden quadruple
• x … X-wing (2 rows / 2 columns)
• w … Swordfish (3 rows / 3 columns)
• j … Jellyfish (4 rows / 4 columns)
• L … large gap field
• Sx … Setti’s rule (count the numbers rule) – ‘x’ is the analysed number
• u … unique rectangle
• y … Y-Wing or XY-chains

24. Diabolic Str8ts Puzzle #8
Solution by SlowThinker
Note: there are other (maybe easier) ways to solve this puzzle.
View & download my strategy slides from:
http://slideshare.net/SlowThinker/str8ts-basic-and-advanced-strategies
or from Google Docs:
http://is.gd/slowthinker_str8ts_strategy