Solution to Diabolic Str8ts #3
(http://is.gd/slowthinker_diabolic_str8ts_3)
and solution to Weekly Extreme Str8ts #63.
Watch me solve it on YouTube: http://youtu.be/uE25Kdop9wM
2. Start positionExtreme #63 Let’s have a quick look at the key points in Extreme #63 first. I was able to solve it rather quickly. The discussion assumes you are familiar with the strategies discussed. See my strategy slides for more information (link on last slide.)
3. Lockedcompartments Many puzzles have columns with two 4 cell compartments. These are locked compartments, which can be arranged in 6 ways (3 shown + reverse green and orange.) The last line (red) shows the sure candidates. It pays off to remember this, as this pattern occurs often.
4. X-Wing on 5 After the basic candidate removal, we encounter an X-Wing on 5 (marked blue) that removes the 5s from the yellow cells. A direct consequence is B9=5s.
5. More lockedcompartments Whenever you see compartments occupying the same range, you should investigate closer. Here E23 cannot be 23, because it would wipe out E7. Thus E3=14 and E2=235.
6. Turbo Setti1/4 This puzzle has an interesting property: except for row E, all other rows have a single compartment with 7 or more cells. So in 8 rows 3..7 are surecandidates. And as A8 and J6 are 8, we can actually say that in 8 rows 3..8 are sure candidates.
7. Turbo Setti2/4 What about the columns? Columns 2, 3, 5, 7, 8 have at least 8 cells ( 2..8 are sure candidates) and in columns 4 & 6 234678 are sure candidates. Intersect with the sure candidates in the rows and we have the candidates 34678.
8. Turbo Setti3/4 We now know that the candidates 34678 appear in at least 8 rows and at least 7 columns. Thus: If one of these numbers does not appear in one of column 1 or 9, it must appear in the other one. If it appears in row E, it must appear in columns 1 and 9.
9. Turbo Setti4/4 Let’s look at our candidates 34678: Thus we immediately know that 4 and 8 must appear in column 1 and that 7 must not appear in column 1.
10. Solving… Here’s the result: GHJ1=2345 and CD1=89. Btw, by the same logic we now know that 3 must appear in row E. Following up: J3=9p C8=9s
11. Setti on 2 After an X-wing on 3 at GJ12, which sets G9=4, and some more basic removal, we find ourselves at this position. Row E is filled out now, so we can easily conclude that 2 appears in 8 rows 2 must be in GHJ1 H1=24 H2=5s
12. Solution That’s it. The rest is the menial job of basic candidate removal. You can watch me solve the puzzle (in a slightly different way) on YouTube here: http://youtu.be/uE25Kdop9wM
14. Start positionDiabolic #3 With the diabolic Str8ts series, I try to push the boundaries a bit. Normal strategies are not enough to solve these puzzles. This one has a beautiful hidden unique rectangle.
15. Mind the gap C2 is almost a large gap field, but not quite. But wait: D2 has 3 as its highest candidate. If F1 or G1 would be 7, then we would have a 2 in C1 or D1, which is impossible No 7 in FG1 Following up this leads to G56=89.
16. Hidden unique rectangle Let us analyse HJ34. HJ4 is a split compartment. Hence we can analyse both ranges separately.The upper range (78) is of no interest. But the lower range is a unique rectangle. Let’s have a closer look.
17. Hidden unique rectangle If HJ34 would look like this, we immediately know that one of J34 must be 1. But which one? J3 or J4? The compartment that can influence this decision is A..F4: in this position it must contain a 3, otherwise we could always exchange HJ3 and HJ4.
18. Hidden unique rectangle To sum up, either HJ4=78 or A..F4 contains a 3. But what if A..F4 contains a 9, that is, it has a range of 4 to 9? Then HJ4 would not be 78 and we would not have a 3 in AF..4!! AF..4 must not contain a 9.
19. Setti on 3 Moving on, we arrive at this position. Because of Setti on 3 there’s a 3 in every row. C12 and C45 have mutually exclusive ranges, thus we can remove 1 from both compartments. The same is true for G12 and G89.
20. Setti on 5 The same is true for Setti on 5. Again there has to be a 5 in every row and G12 and G89 have mutually exclusive ranges. Thus we can remove 7 from G1 and G9.
21. Applying some logic Seeing so many 2s and 3s next to each other in columns 1&2 should make one curious. Let’s see: G2 cannot be 3, as G2=3 G1=2, C2=2, D2=1 D1=2 G2 cannot be 2 either, as G2=2 G1=3, C2=3, D2=1 C1=2 D1=1
22. X-Wing on 2 Which brings us to this position and an X-Wing on 2 at CE45 that removes the 2 from H4. And with that the puzzle is solved. Not that hard, no?
23. Solution The unique solution constraint is a very powerful technique for solving Str8ts puzzles. You just have to develop an eye for it. I hope this example has whet your appetite.
24. 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
25. Weekly Extreme Str8ts Puzzle #63 Diabolic Str8ts Puzzle #3 Solution by SlowThinker Note: there are other (maybe easier) ways to solve these puzzles. 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
