Diabolic Str8ts Puzzle #2

Puzzle & solution bySlowThinker

Start positionWith the diabolic Str8ts series, I try to push the boundaries a bit. Normal strategies are not enough to solve these puzzles.

The 8s in this puzzle ensure that the numbers 1-7 appear in almost all rows and columns.

Solving…After applying the basic techniques, we arrive at the position on the right.C3 (red) contains a large gap pair, the green cells can be further reduced due to stranded digits, which in turn creates large gap pairs.In B5 (blue) 9 is a stranded digit.

Split compartmentA first look at BCD6 and BCD8 shows that they are split compartments, so we can examine each range individually, which removes the 3s from C68 and the 23s from B68.That in turn leads to B9=28 and a naked pair of BG9=28 (blue).

Locked compartmentsIn column 4 we have three locked compartments. The table shows their four possible arrangements:

As you can see, the red compartment cannot be 12, thus we can remove 2 from there. Also, we can remove 3 from the yellow compartment.

1 2 3 4 5 6 7 8 9

A x x

B x x

C x x

D x x

(optional solution step)

X-Wing on 4Next we have an X-Wing on 4 at HJ78 (marked green) that removes the 4s in the yellow cells.

Following this, we can

also remove 5 from HJ4,

due to the locked

compartment constraint

from the previous page.

(optional solution step)

Hidden pairNext we have the hidden pair 34 in C59 (marked blue).We can also remove 7 from E1..5 (green), because of E7 (red).

(optional solution step)

Setti’s rule on 3Applying Setti’s rule on 3, we find that the puzzle must contain a 3 in every column and thus in every row as well we can remove 9 from B1 and set G1=9S3

(optional solution step)

Hidden large gap pairNow here’s something you don’t see every day: a hidden large gap pair in B5 (marked blue).2 and 8 only occur in B5. And because of the size of the compartment, either 2 or 8 has to be part of it. Thus we can reduce B5 to 28.

(optional solution step)

Unique solution constraintWith 28 in B5, B9 and F9 (green) you may think that F5 cannot be 2 or 8. But this is wrong: due to the black cells at BF7, the cells are in different compartments and thus the uniqueness rule cannot be applied.

Unique solution constraintBut we can apply this rule to G3: if G3 were 1, then EF3 would be 23 and violate the unique rectangle rule together with EF2. G3 cannot be 1. In turn we can remove 2 from EF3, as G3 now has a large gap.

(optional solution step)

Further reductionsHere’s an interesting combination:Due to AB2 (green), AB4 (red) cannot be 45, as this violates the unique rectangle rule.Due to the locked compartments in column 4 (yellow), AB4 (red) cannot be 56. 5 can be removed in AB4! (optional solution step)

Again: Locked compartmentsWith the 5s gone in AB4, we can revisit the locked compartments in column 4:

What we are left with is that HJ4 (yellow) can only be 12.

1 2 3 4 5 6 7 8 9

A x x

B x x

(optional solution step)

A first testWe know that BCD6 and BCD8 are split compartments: either they are 123 or 567.If we assume C7=7 we get C3=2 which forces both compartments into the range of 567 which yields a direct contradiction, as both B6 and B8 would be 7

C7=6(optional solution step)

Setti’s rule on 66 appears in every column and therefore must appear in every row as well: row E contains a 6

and we can remove the 1s there.

row G contains a 6, but we cannot put this knowledge to good use right now.

(optional solution step)

Analysis of BCD68BCD68 (green) are split compartments and both cannot be 123 at the same time.Can they be 567 at the same time?The answer is no: BCD6=567 F6=123 andBCD8=567 F89=12 result in a naked triple 123 that removes all candidates from F2.

Analysis of BCD68We therefore know that BCD6 and BCD8 occupy different ranges One of BCD68 has to be 123. As the 1s, 2s, and 3s are in the same row in BCD68, we can remove them from the red cells (3s in row D, 2s in row C, 1s in row B).

X-Wing on 3 & 5 and then on 4That brings us to this position, where we have an X-Wing on 3 at EF23 (green) and an X-Wing on 5 at AB23 (blue). The 3s and 5s are removed in the yellow cells.This in turn leads to AB4=34 and an X-Wing on 4 at AB24.

Solving…From here on, the rest is easy:

F4=7pD9=4sA1=7sF8=1B8=7… and so on.

SolutionMost of the steps of this solution are not necessary to solve the puzzle.The key is the analysis of BCD68. Once you have established that at least one of those two compartments is 123, the puzzle is solved easily.

GlossaryLetters 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

Diabolic Str8ts Puzzle #2

Solution by SlowThinkerNote: 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