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# Sudoku/Methods/Advanced/Simultaneous Bivalue Nishio/Reddit/2019 08 03

## The puzzle[edit]

This is the discussion. The puzzle is on the right, as converted to Hodoku and with my SBN "seed." I started with an examination of box patterns. showing how I chose the seed.

link to SW solver loaded with the puzzle.

The OP has sensibly asked for a hint that will still allow practice at finding these patterns. I will go a little deeper, demonstrating a generic technique for preparing to find and use most advanced techniques. If I'm being careful, I will do this before using, say, AICs or what I call Simultaneous Bivalue Nishio (which is AIC, done a bit more thoroughly, with a heightened understanding not often explained). This readily identifies wings, skycrapers, and swordfish, besides giving me clues as to how to run a powerful and effective AIC.

This involves looking at unresolved candidates individually. In what houses do they live? Are they in chains or cycles? Strong links or weak links? I sometimes document this with a diagram. Here I will show diagrams, for fun. The interested reader should load the puzzle and look at this for themselves, and verify my notes. I also make mistakes, often. Extra credit for finding them!

Notice: missing >! box/line reduction!< for >!3 in Box 7!<. Fixed before proceeding.

The diagrams (code blocks) show the number of candidates in each box.

1: 4-box chain. 4587. No fish.

. . . 3 3 . 3 2 .

2 : All boxes but two. Many strong links. No fish.

. 4 2 . 3 . 2 5 2

3: Box cycle 5698, strong links outside, 1 inside. No fish.

. . . 2 5 4 2 3 2

4. 4-box cycle 4587 Notice same boxes as for {1}. No fish.

. . . 3 4 . 3 3 .

5. two 2-box pairs. No fish, obviously.

2 . . . 2 2 2 . .

6. Strong link chain boxes 123. All fish!!! A fish is not a cigar. I.e., any eliminations already done.

2 2 2 . . . . . .

7. 2-box pair. All fish, no cigar.

. . . 2 . 2 . . .

8. 4 box chain. Many strong links. No fish.

. 3 . . 2 . 2 4 .

9. 4 box cycle, plus two off to the side. No fish.

3 3 3 . 3 3 2 . .

Having done the above, I review it looking carefully for patterns of two or three blocks in two or three rows/columns. Just calling all such "fish" I mark my observations (or non-observation!)

Whatever pattern there is here is not likely to be a single-candidate pattern. [there is a pattern, all right, but it is not purely single-candidate, it is one that I might not see even if told it is there. Not immediately, anyway. that pattern involves a "hinge" cell that does not contain the candidate, but that links two cells that do.]

But from the above, I have a clear path forward. 1s and 4s share the same box pattern, and will likely reveal other patterns if explored with coloring. There are many pairs in Boxes 4 and 7 that lead to many proposed resolutions on either side. Which one to choose? When I see this many, I don't worry about "best." What I want is "good enough". So I pick the 3/4 pair!

Taking the puzzle into Hodoku, I color on r4c3, 4 green. I color the cell orange so I can find it! In tests like this, starting cell matters. It does not matter with a Medusa, which is an AIC with all strong links. But Medusa is actually rare. Because AICs will fully handle a Medusa, I don't look for them!

## Solution[edit]

As expected, this cracked the puzzle very quickly.

Unfortunately, those imprisoned in solvers that do not allow coloring may not be able to use this technique. In this case, interaction was not necessary for the primary solution, but to prove uniqueness, it's used.

I see now that I actually didn't complete the green chain. It doesn't matter. A much less extensive chain than I had can still be used for interactions. In fact, the "wrong" answer still generates interactions that can help complete the other chain.

In the limited solvers, where one can save the state of the sudoku, one might try alternate solutions, but comparison will be difficult, and it is in comparison that AICs really shine. I see that Sudoku Swami introduces AICs very early in his course. I don't yet see that he has found and expressed the full power of AICs, which is in *simultaneous* analysis. But I've just started to look at his work. He is extremely careful to be clear and to approach issues step-by-step.

Because I do not assume that any sudoku has a unique solution, even though ones that do not are extremely rare as to what is published (because computer checking is *almost always* done), I still need to show that the opposing choice leads to a contradiction.

## Proving uniqueness[edit]

So I run out the red candidates. As I do, I eliminate mutual eliminations and eventually find common resolutions.

Notice in row 8 there is a red 2 and a green 2, eliminating the 2 in r8c2, creating the common resolution, shown in light purple.

This creates another common resolution just above it, which, in turn, confirms the original green choice, proving uniqueness.

A way of saying this is "If A, then not B. If B, then not B also." As can be seen I have not completed either chain, but the A chain (green) merely has a couple of missed singles. If pursued, all B candidates (red) will end up, working with A, eliminating themselves.

This is like life. If we focus on what is clear, error eliminates itself, we don't need to attack it; in fact, simply accepting it as a proposition, and examining the consequences, often leads to reality. All roads lead to Rome.

This technique is often called "trial and error," or "guessing." I did not guess to find the answer, except in a limited way. I picked a pair, out of a number of possibilities. My operating hypothesis is that it is not necessary to pick the "best" pair. All roads lead to Rome. Really? So what happens if I pick a different pair?

## Choosing other pairs[edit]

I tried it. I used {13} in r5c2. This implies the 4 chain by a strong link, so I kept the same color, to make it easy to compare with the study above. Red led, all by its lonesome, to a contradiction, here is what it looked like. No place for 2 in box 9. Therefore green, the same as above.

I tried {39} in r9c3. This led quickly to a contradiction entirely within the left stack, that confirms the 4 we already know completes the puzzle by itself. This could, in fact be the "optimal choice," giving an unconditional resolution that we know will be essentially singles to the end.

I tried {23} in r9c2. Colored 2 green for compatibility (easy to see that). Same result, contradiction in left stack for red.

{12} in r8c2. Same as above.

{48} in r7c1 is even better! It immediately confirms the 4 in r4c3! Here is the logic:
If r7c1=4, **r4c3=4.** And if r7c1=8, r9c1=9, r9c3=3, **r4c3=4.** This one can be seen in the raw -- if we know where to look

I also tried every other pair in Boxes 4 and 7.

{89} in r9c1 is about as direct at the last one.
If r9c1=8, r7c1=4, **r4c3=4,** and if r9c1=9, r9c3=3, **r4c3=4.** Also easy to see.

{45} in r8c3. Resolves to **r4c3=4**. All pairs in those boxes lead to that conclusion, that cracks the puzzle. Glad I looked, I've been suspecting this. I don't know how general it is. With really difficult puzzles, I know that a poor choice of pair can lead to impasse.

So I may take some care before committing to a seed pair for SBN. However, here, I saw so many chains that I didn't bother looking for the ideal. Good Enough is Good Enough.

Looking back at the original pair choice, again, yes. I didn't notice it but the green choice was immediately confirmed entirely within the left stack, that would be singles to the end.

This is a technique for humans, that is tolerant of oversights, if we just keep on looking.