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Diabolic Str8ts #6, #7, and #9


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Solutions to Diabolic Str8ts puzzles #6 (, Diabolic #7 (, and Diabolic #9 (

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Diabolic Str8ts #6, #7, and #9

  1. 1. Diabolic Str8ts Puzzles #6, #7, and #9 Puzzles & solutions by SlowThinker
  2. 2. Diabolic Str8ts Puzzle #6
  3. 3. Start positionDiabolic #6 has verynice properties in thecentre: severalcompartmentsinfluence each other.Note: I’m onlyshowing the keypoints of the solutionshere. For strategiessee link to my strategyslides at the end.
  4. 4. The key insightAfter the usualremoval work wearrive at this position.The relationshipsbetween the greencompartments andthe compartments inrow E, as well as theorange and blue fieldsare key to solving thepuzzle.5 in D4 sticks out.
  5. 5. Testing D4=789?So let us assume thatD4≠5.This removes 789from D123/D789. Inturn this yields F9=9,as all other 9s in thegreen fields areremoved.This would leave uswith…
  6. 6. Testing D4=789?(continued)… this position.Because of therelationship betweenthe red and orangefields  F3=36 too!That leaves F2 and F8with 78, which in turnsplits one of theyellow compartmentsin half. D4≠789, i.e. D4=5
  7. 7. F3=789Using the samereasoning as before,we can infer that F3cannot be 3 or 6.F3=36 would againsplit one of the yellowcompartments in half.With F3=789 manycells can be solved.
  8. 8. Locked pairsNext, we arrive at thisposition.Looking at B37 we findthat one of the cellshas to be 3  one ofthe green cells has tobe 2  A8≠2.Same for F78: one ofthose cells has to be 6 one of E78 has tobe 5  E789=3456.
  9. 9. SolutionWith that the puzzle issolved.The solution relied oninferring ranges ofcompartmentsorthogonal to eachother. Sometimes thisis a useful technique.
  10. 10. Diabolic Str8ts Puzzle #7
  11. 11. Start positionDiabolic #7The solution to Diabolic#7 features some nicecombination of Setti’srule on differentnumbers.The first interestingstep is the applicationof the large gap rule.
  12. 12. Mind the gapF8=289 is a large gapfield that lets usremove 2 from DE8. Inturn E8=15..9 lets usremove 1 from D8.We can also remove 4and 5 from DE8, as 5 isnot reachable from F8and with 5 gone, 4 inE8 is unreachable too.Next, with AB8=1234DEF8 cannot be 123 DEF8=6789!
  13. 13. Where to start?After the usual steps,among them applyingSetti’s rule on 6 andremoving stranded 9sfrom C34 (B4=9),we get this position.Where to start?A4 seems like a goodidea: it has only twocandidates andinhibits the formationof an x-wing on 4 atDF45.
  14. 14. Test A4=6?So let’s test A4=6:A4=6  A3=8, C3=6,C7=4, C6=3c.Because of the x-wingon 4 at DF45 and C7=4 A9=4s  C9=1s C3=34.This leads to acontradiction.Hence A4=4.
  15. 15. Setti’s rule on 5and 9 combinedAfter removal andapplying Setti’s rule on3 & 5 we arrive here:Setti on 5 tells usthere must be a 5 inevery row, Setti on 9tells us that one ofHJ6789 has a 9  oneof HJ1 must be 5 EF1≠5!
  16. 16. Test H1=5?So let’s test H1=5:Green: H1=5 H6=9, C6=6, C7=4 B7≠4  B8=4Orange: H1=5 J1=2, E1=6 E7=2, A7=3  A8=2 AB8=24 which isimpossible.Therefore H1=1 andJ1=5.
  17. 17. Test E1=6?Using the very samelogic, we can testE1=6:Orange: E1=6  F1=4,F5=3, F6=5, B6=3 B8=4Green: E1=6  E7=2,A7=3  A8=2 AB8=24 which isimpossible.Therefore E2=2 andthe puzzle is solved.
  18. 18. SolutionBeing able to combinetwo applications ofSetti’s rule does nothappen that often, butit can lead to powerfulconclusions like in thispuzzle.
  19. 19. Diabolic Str8ts Puzzle #9
  20. 20. Start positionThe theme of Diabolic#9 is large gaps andtheir hiddenvariations. It alsocontains a trickyhidden UR.Due to the puzzle’sgeometry, we have todo an elimination testquite early.
  21. 21. Test J7=8?After removing some8s and 9s in J4..9(large gap), let us testJ7=8:J7=8  F7=7, B7=6 D7=5.Also:  F8=8  E8=6c D8=5As J7=8 leads to acontradiction, we canset J7=123.
  22. 22. Large Gap &Hidden URGJ1 (orange) form alarge gap pair on 2: ifD1/E1=2, one of GJ1would have to be 8,which is out of range.DE123 (green) forms ahidden uniquerectangle: we canremove 4 from E123.If E123 were 234, wecould swap E123 withD123 at any time andhave two solutions.
  23. 23. Hidden GapAs A89 no longercontains a 1, A1together with A89forms a large gap on 8:if any of A2..6 were 8,A89 would be 23 andA1=1, a stranded digit.Therefore we canremove 8 from A2..6.
  24. 24. Setti on 6X-Wing on 5Applying Setti’s ruleon 6, we can remove 9from A4.Also, we can remove 5from A2, because ifA2=5, CDEF2 cannotcontain a 6 either.Afterwards, A6=5s,J4=5s, and H3=5s giveus an x-wing on 5 atDG78.
  25. 25. SolutionWith this the puzzlesolves easily: anotherlarge gap in E1, anaked triple in row C,some naked pairs,locked compartmentsetc.
  26. 26. 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
  27. 27. Diabolic Str8ts Puzzle #6, #7, and #9Solution by SlowThinkerNote: there are other (maybe easier) ways to solve this puzzle.View & download my strategy slides from: from Google Docs: