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. The 8s in this puzzle ensure that the numbers 1-7 appear in almost all rows and columns.
3. Solvingโฆ 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.
4. Split compartment A 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).
5. Locked compartments In 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. (optional solutionstep)
6. X-Wing on 4 Next 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 solutionstep)
7. Hidden pair Next we have the hidden pair 34 in C59 (marked blue). We can also remove 7 from E1..5 (green), because of E7 (red). (optional solutionstep)
8. Settiโs rule on 3 Applying Settiโs 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 solutionstep)
9. Hidden large gap pair Now hereโs something you donโt 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 solutionstep)
10. Unique solution constraint With 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.
11. Unique solution constraint But 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 solutionstep)
12. Further reductions Hereโs 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 solutionstep)
13. Again: Locked compartments With 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. (optional solutionstep)
14. A first test We 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 solutionstep)
15.
16. row G contains a 6, but we cannot put this knowledge to good use right now.(optional solutionstep)
17. Analysis of BCD68 BCD68 (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.
18. Analysis of BCD68 We 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).
19. X-Wing on 3 & 5 and then on 4 That 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.
20. Solvingโฆ From here on, the rest is easy: F4=7p D9=4s A1=7s F8=1 B8=7 โฆ and so on.
21. Solution Most 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.
22. Glossary Letters appended to steps indicate the last strategy used, just before filling in a field: No letter โฆ number was last candidate in field s โฆ single (last) candidate for that number in compartment c โฆ compartment range check d โฆ stranded (unreachable/impossible) digits removed h โฆ high/low range check across compartments p/t/q โฆ naked pair / naked triple / naked quadruple ph/th/qh โฆ hidden pair / hidden triple / hidden quadruple x โฆ X-wing (2 rows / 2 columns) w โฆ Swordfish (3 rows / 3 columns) j โฆ Jellyfish (4 rows / 4 columns) L โฆ large gap field Sx โฆ Settiโs rule (count the numbers rule) โ โxโ is the analysed number u โฆ unique rectangle y โฆ Y-Wing or XY-chains
23. Diabolic Str8ts Puzzle #2 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
