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[Ajò Dimonios]

Hi everyone

[JC Van Hay]

15 singles; LC(6B1, 8B9)
The solutions of the 2 exclude {2r2c3, 2r3c7, 2r5c8, 2r8c2}

[Ajò Dimonios]

Hi everyone

[JC Van Hay]

or the following simpler 2-steps solution determined by the previous 1-step solution:

[JC Van Hay]

Hi Paolo,

your
step 6 is included in step 2;
step 8 is to be corrected, for example : {8C3, 6R2, (62)r4c1, 2C7} -> -8r6c7;
step 10 : r1c8=2.

JC

[Ajò Dimonios]

Hi JC

Sorry JC but I can not understand why Step 6 is already included in step 2 (Perhaps because , differed from you, I started from after basics). After inserting 7 in R4C4 and eliminating for locked candidates of 2 in R5C8, in cell R8C2 2 is still a possible candidate along with 4 and 8. In my opinion it is necessary to prove that 2 in R8C2 is false. Sorry but I also do not understand why step 8 needs to be corrected. The demonstration that 8 in R6C7 is false seems to me correct.

Ciao a Tutti
Paolo

[JC Van Hay]

Hi Paolo,

your step 2 : the solutions of your 5 constraints {(24)r28c5, 2C2B26} effectively exclude {2r3c7, 2r8c2}. To wit : 2r2c5 -> 2r3c2 , while 4r2c5 -> 2r8c2, EmptyRectangle(2C2, 2B6). Therefore 2r3c7 and 2r8c2 are excluded in both cases.

your step 8 contains in the second part the simultaneous implications +8r2c3 and +8r3c2[instead of +2r3c2 !]

JC

[Ajò Dimonios]

Hi JC

Sorry for the distraction in step 10 (clearly R1C8=2) and error of the second part of step 8 . Replace the whole step with
8 ) [(8)R6C7-(8)R6C3=(8)R5C2-(8)R3C2 and (8-2)R6C7 =(2)R4C7-(2)R4C13=(2)R5C2-(2)R3C2=>contradiction R3C2=Ø =>-(8)R6C7.=>contradiction R3C2=Ø =>-(8)R6C7.
As for Step 6 included in Step 2, I simply work step by step. Deleting 2 in R3C7 does not automatically involve eliminating 2 in R8C2, to eliminate candidate 2 in R8C2 after eliminating 2 in R3C7 you need the fin swordfish of the 2 in line 3,5 and 9 (fin R9C3 = 2). As you have shown with a more global view, R3C7=2 and R8C2=2 are biunivoca, when one is True the other is also true and viceversa ,[(2)R3C7-(2)R3C2; (2)R3C7-(2)R12C8=(2)R5C8-(2)R5C2=(2)R8C2 ; (2)R8C2-(2)R3C2 and (2)R8C2-(2)R8C5=(2)R2C5-(2)R3C46=(2)R3C7]. In practice when R3C7 = 2 => R8C2 = 2 and when R8C2 = 2 => R3C7 = 2. I consider it another step.
Ciao a Tutti
Paolo

[Ajò Dimonios]

Hi JC


JC Van Hay wrote:

[JC Van Hay]

Hi Paolo,

Using your constraints in step 2, one can write the following mixed block matrix :

[Ajò Dimonios]

Hi JC

Thank you for your excellent and precise explanation.

Anyway, I'm of the opinion that a single step is that of an advanced technique along with one or more basic techniques. If that was not the case, it would be absurd that always with a single step that includes many advanced techniques, one can get the solution.I think that when using more advanced techniques, each application represents a step. I have not found in this example any basics technique that proves that once proved to be false R3C7 = 2 attest to the falseness of R8C2 = 2 or vice versa.

Ciao a Tutti
Paolo

[Ajò Dimonios]

Hi JC

JC Van Hay Wrote: