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Diabolic Str8ts #1

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Solution to Diabolic Str8ts #1

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Diabolic Str8ts #1

  1. 1. Diabolic Str8ts Puzzle #1<br />Solution & puzzle bySlowThinker<br />
  2. 2. Start position<br />With the diabolic str8ts series, I try to push the boundaries a bit. Ordinary strategies are not enough to solve these puzzles.<br />
  3. 3. Solving…<br />After applying the basic techniques, we arrive at the position on the right.<br />In D7 and E7 9 is a stranded digit, as D7 contains the only 7 and E7 the only 8 in ABCDE7 <br />E9=9d<br />D5=9s<br />
  4. 4. Split unique UR<br />HJ7 is a split compartment: either it is 234 or it is 789. But in the upper range it would form a unique rectangle with HJ6 <br />to avoid the UR J7 must not be 78 <br />J7=2349<br />H7=348<br />
  5. 5. Setti on 7<br />There’s a 7 in every row, therefore there must be a 7 in every column too:<br />D7=7S7<br />In column 2 we can only deduce that one of B2 and J2 must contain a 7.<br />
  6. 6. Setti on 8<br />Row C does not contain an 8  at least one column must not have an 8 either  column 4 is the only possible column <br />FGHJ4=4567S8<br />D4=2h<br />A4=9h<br />
  7. 7. X-Wing & Setti<br />Again omitting the basic elimination steps, we arrive at this position:<br />There’s an X-wing (marked green) on 3 that removes the 3 from E3 (marked red).<br />Applying Setti’s rule on 4, we find that every row must have a 4  F1 != 9<br />
  8. 8. Setti’s rule on 6& an easy test<br />B1 is the only 6 in row B. If there’s no 6 there, there’s also no 6 in F2 and J2. What if B1=7?<br />B1=7  F2=4S6<br />but also:<br />B1=7  C1=6  F1=4<br />Hence: B1!=7  B2=7<br />Note that we can remove 6 from J2 with very similar logic.<br />
  9. 9. Analysis<br />Due to Setti on 6 we have a link between B1 and F2. There’s also a link between G1 and J7, because of Setti on 2 (marked border).<br />We also have the green cells, which are mostly about 234 and the yellow area mostly about 689.<br />At the intersection: J3<br />
  10. 10. Testing J3=2<br />So let us test J3=2, because it has an impact on J7:<br />J3=2  J7=9  G1=9S2<br />But also:<br />J3=2  E3=4, E2=3  F2=4c  B1=9S6<br />Which is impossible. Hence:<br />J3=6<br />
  11. 11. Solving…<br />With the naked triple in J468 (yellow) we can remove 47 from J1.<br />This gives us anX-wing on 4 at EF12, which removes 4 from F4.<br />
  12. 12. Final test: J7=9<br />Keeping in mind the links we already discovered, we find that J7 cannot be 9:<br />J7=9  G2=9, B1=6  F2=6 and F4=7<br />But also:<br />J7=9  H7=8  H6=7, J6=8, J8=4 and J4=7<br />Thus: J7=2<br />The rest is easy.<br />
  13. 13. Solution<br />There are of course other (maybe easier) ways to solve this puzzle. This solution emphasized links between cells and different areas of the board to test and eliminate candidates.<br />
  14. 14. Glossary<br />Letters appended to steps indicate the last strategy used, just before filling in a field:<br />No letter … number was last candidate in field<br />s … single (last) candidate for that number in compartment<br />c … compartment range check<br />d … stranded (unreachable/impossible) digits removed<br />h … high/low range check across compartments<br />p/t/q … naked pair / naked triple / naked quadruple<br />ph/th/qh … hidden pair / hidden triple / hidden quadruple<br />x … X-wing (2 rows / 2 columns)<br />w … Swordfish (3 rows / 3 columns)<br />j … Jellyfish (4 rows / 4 columns)<br />L … large gap field<br />Sx … Setti’s rule (count the numbers rule) – ‘x’ is the analysed number<br />u … unique rectangle<br />y … Y-Wing or XY-chains<br />
  15. 15. Diabolic Str8ts Puzzle #1<br />Solution by SlowThinker<br />Note: there are other (maybe easier) ways to solve this puzzle.<br />View & download my strategy slides from:<br />http://slideshare.net/SlowThinker/str8ts-basic-and-advanced-strategies<br />or from Google Docs:<br />http://is.gd/slowthinker_str8ts_strategy<br />