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Sudoku/Methods/Advanced/Simultaneous Bivalue Nishio/Sudoku Addicts Workbook/146

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Figure 1

The first "Fearsome Five" puzzle

Introduction[edit]

  • Raw puzzle.
  • SW Solver: Diabolical grade, overall score 622.
  • Stephens' estimated times: Improver: N/A, Expert: 280 mins, Genius: 93 mins.

Solving[edit]

This (Figure 1 on the right) shows the puzzle loaded into Hodoku. The seed cell, colored light orange, was chosen for this de novo analysis as r8c1={47}, 4 green and 7 red. As can be seen, the 4 choice, using basic simple strategies, produced a straightforward full solution, all cells have a green candidate. In my work, I do not consider this completely done, preferring to prove that the solution is unique. Normally, I can do that without an additional coloring, but this is not an ordinary sudoku. I saw no obvious proposed resolutions with red, indicating that this was actually a poor choice of seed. In ink, I'm usually much more careful, because I get one shot. Nevertheless, this worked. And then I ran SBN again.

Figure 2

Proving uniqueness[edit]

Figure 2 on the left: {38} in r3c1 was the seed I had chosen in ink in the book, and it was swift to resolve the seed. The red candidate leads quickly to a contradiction. The cycle that does that can be clearly seen, running from the source Box 1, to 4, to 9, to 7, and back to 1, where the pink cell has no possible red candidate. Hence r3c1=3. Resolving that and clearing the coloring, I return to the r8c1 seed, because I now have more support to complete the red coloring (presumably to a contradiction, if the solution already found is unique).

Figure 3

Figure 3 on the right shows the result. As expected from the first run, green completes. This time red can be extended, and it was quite stubborn, but eventually came to a contradiction. Because it's easy for me to make mistakes, I'm not 100% confident in my process and I get no solace from the automatic checking in Hoduku because the red chain is in error. When I remove a candidate in Hodoku that is the real deal, it's displayed in light red. As I was extending the red chain, I eliminated many candidates. Fairly quickly, I found confirmed resolutions. Those are shown as a green cell (some of these still show the eliminated candidates). The extra cell coloring is not part of what I routinely do, I do this so I can write about some of the process here.

The contradiction was in the cell shown in pink. There is no place for 9 in that box. The original red 9 in r6c8 was eliminated when that cell resolved to 7, as a result of the confirmed 9 in r6c2.

Comments[edit]

The sequence of operations is quite complex, but each step is simple. To understand this, it's probably necessary to do it. In a few places, I have given every step. That will be meaningless unless one is actually following along with the puzzle, marking the candidates and seeing, then, the interactions and how they eliminate or confirm candidates.

In relatively difficult sudoku, I've found, identifying hidden pairs or triples may be necessary. That is still considered basic strategy. I've started to mark pairs or triples, using a lighter color matching the chain involved. It's important to keep in mind that the two patterns are independent, elimination in one is not elimination in both, unless it actually is indentified as a common elimination.

As an example of common elimination, look at the red 9 in r6c8. Before r6c2 was confirmed as 9, that 9 was green. So the nines between them, in r6c56 are mutual eliminations. As well, once we have r6c8(red)=9, those can be eliminated because they are in the same cell as a green candidate). Wherever a proposed candidate in one chain sees its number in a cell in the other chain, the candidate can be eliminated there. I saw one or two examples in this puzzle of an unresolved but aligned pair of cells in one chain that pointed to a cell in the other chain, allowing a candidate elimination there. That was probably the most complicated step needed.

See also[edit]