Puzzle #6 utilizes relationships between compartments to solve the puzzle. Puzzle #7 applies Setti's rule twice to combine inferences about numbers. Puzzle #9 uses large gaps and a hidden unique rectangle to make deductions early in solving. The document provides step-by-step solutions and strategies for three Diabolic Str8ts puzzles.
3. Start position
Diabolic #6 has very
nice properties in the
centre: several
compartments
influence each other.
Note: I’m only
showing the key
points of the solutions
here. For strategies
see link to my strategy
slides at the end.
4. The key insight
After the usual
removal work we
arrive at this position.
The relationships
between the green
compartments and
the compartments in
row E, as well as the
orange and blue fields
are key to solving the
puzzle.
5 in D4 sticks out.
5. Testing D4=789?
So let us assume that
D4≠5.
This removes 789
from D123/D789. In
turn this yields F9=9,
as all other 9s in the
green fields are
removed.
This would leave us
with…
6. Testing D4=789?
(continued)
… this position.
Because of the
relationship between
the red and orange
fields F3=36 too!
That leaves F2 and F8
with 78, which in turn
splits one of the
yellow compartments
in half.
D4≠789, i.e. D4=5
7. F3=789
Using the same
reasoning as before,
we can infer that F3
cannot be 3 or 6.
F3=36 would again
split one of the yellow
compartments in half.
With F3=789 many
cells can be solved.
8. Locked pairs
Next, we arrive at this
position.
Looking at B37 we find
that one of the cells
has to be 3 one of
the green cells has to
be 2 A8≠2.
Same for F78: one of
those cells has to be 6
one of E78 has to
be 5 E789=3456.
9. Solution
With that the puzzle is
solved.
The solution relied on
inferring ranges of
compartments
orthogonal to each
other. Sometimes this
is a useful technique.
11. Start position
Diabolic #7
The solution to Diabolic
#7 features some nice
combination of Setti’s
rule on different
numbers.
The first interesting
step is the application
of the large gap rule.
12. Mind the gap
F8=289 is a large gap
field that lets us
remove 2 from DE8. In
turn E8=15..9 lets us
remove 1 from D8.
We can also remove 4
and 5 from DE8, as 5 is
not reachable from F8
and with 5 gone, 4 in
E8 is unreachable too.
Next, with AB8=1234
DEF8 cannot be 123
DEF8=6789!
13. Where to start?
After the usual steps,
among them applying
Setti’s rule on 6 and
removing stranded 9s
from C34 (B4=9),
we get this position.
Where to start?
A4 seems like a good
idea: it has only two
candidates and
inhibits the formation
of an x-wing on 4 at
DF45.
14. Test A4=6?
So let’s test A4=6:
A4=6 A3=8, C3=6,
C7=4, C6=3c.
Because of the x-wing
on 4 at DF45 and C7=4
A9=4s C9=1s
C3=34.
This leads to a
contradiction.
Hence A4=4.
15. Setti’s rule on 5
and 9 combined
After removal and
applying Setti’s rule on
3 & 5 we arrive here:
Setti on 5 tells us
there must be a 5 in
every row, Setti on 9
tells us that one of
HJ6789 has a 9 one
of HJ1 must be 5
EF1≠5!
16. Test H1=5?
So let’s test H1=5:
Green: H1=5
H6=9, C6=6, C7=4
B7≠4 B8=4
Orange: H1=5
J1=2, E1=6
E7=2, A7=3 A8=2
AB8=24 which is
impossible.
Therefore H1=1 and
J1=5.
17. Test E1=6?
Using the very same
logic, we can test
E1=6:
Orange: E1=6 F1=4,
F5=3, F6=5, B6=3
B8=4
Green: E1=6 E7=2,
A7=3 A8=2
AB8=24 which is
impossible.
Therefore E2=2 and
the puzzle is solved.
18. Solution
Being able to combine
two applications of
Setti’s rule does not
happen that often, but
it can lead to powerful
conclusions like in this
puzzle.
20. Start position
The theme of Diabolic
#9 is large gaps and
their hidden
variations. It also
contains a tricky
hidden UR.
Due to the puzzle’s
geometry, we have to
do an elimination test
quite early.
21. Test J7=8?
After removing some
8s and 9s in J4..9
(large gap), let us test
J7=8:
J7=8 F7=7, B7=6
D7=5.
Also: F8=8 E8=6c
D8=5
As J7=8 leads to a
contradiction, we can
set J7=123.
22. Large Gap &
Hidden UR
GJ1 (orange) form a
large gap pair on 2: if
D1/E1=2, one of GJ1
would have to be 8,
which is out of range.
DE123 (green) forms a
hidden unique
rectangle: we can
remove 4 from E123.
If E123 were 234, we
could swap E123 with
D123 at any time and
have two solutions.
23. Hidden Gap
As A89 no longer
contains a 1, A1
together with A89
forms a large gap on 8:
if any of A2..6 were 8,
A89 would be 23 and
A1=1, a stranded digit.
Therefore we can
remove 8 from A2..6.
24. Setti on 6
X-Wing on 5
Applying Setti’s rule
on 6, we can remove 9
from A4.
Also, we can remove 5
from A2, because if
A2=5, CDEF2 cannot
contain a 6 either.
Afterwards, A6=5s,
J4=5s, and H3=5s give
us an x-wing on 5 at
DG78.
25. Solution
With this the puzzle
solves easily: another
large gap in E1, a
naked triple in row C,
some naked pairs,
locked compartments
etc.
26. 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
27. Diabolic Str8ts Puzzle #6, #7, and #9
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
