Brian Challenger’s Super Fiendish 7


The review of Challenger’s Super Fiendish collection opens with a long post, revisiting some advanced methods not seen here recently.

Somewhat fiendish is the stonewalling in basic. The bypass gets nowhere, and the hidden triple alone removes candidates.

Here is the hidden triple before removals on the line marked grid.

 

 

 

 

 

 

 

An XY wing brings a hidden single in c4.

 

 

 

 

 

 

 

Next is this finned 4-wing removing 4r7c3 that sees fin 4r8c2 in fin box SW.

Alternatively on the X-panel, the same candidates form a 4-chain confirming ANL generating the same effects.

 

 

 

 

The removal enables a second XY ANL, a 598-wing, that generates an r6s17 naked pair.

 

 

 

 

 

The pair has dramatic effects, including a long time not seen favorite of mine, the Sue de Coq below.

 

 

A description of the contents of intersect SWc1, based on the SW remainder ALS 589 and 24, is

SWc1 =

1(2+4)(8+9).

In the c1 remainder, 8 or 9 must come from cell r5c1, and the other 8 must be discarded. The ALS 247 provides a complementing 2 or 4. Besides the c3 naked triple 124,

a third XY wing  provides the trigger for the remaining collapse.

 

Brian Challenger does not provide answers in Sudoku Super Fiendish. That’s a break with tradition, and an inconvenience for sysudokies who trace their solutions. A trace shows you exactly where you departed from the solution. Solutions are provided here for the review puzzles. Here is the SF 7 solution. Hopefully, you could ignore it.

Next post reports on 17 and 27. One falls in the bypass. Find out which before next Tuesday. No previews on Challenger’s Super Fiendish, remember?

 

 

 

 

About Sudent

My real name is John Welch. I'm a happily married, retired professor, father of 3 wonderful daughters and granddad to 7 fabulous grandchildren. Sudoku analysis and illustration is a great hobby and a healthy mental challenge.
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