Poll of the Day > What is this Sudoku voodoo?

(+) Yeah! (+)

I got stuck on a puzzle so I used a solver to learn what's up and the next step it gave me is this...

https://i.imgur.com/cGp9ZhY.jpg

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I love sudoku, but that just looks awful. I know what numbers go in what spots. But the lines dont really seem to be helping, tbh For me, at least
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Either the yellow group or the green group must exist. If the yellow group exists, then you have two 4s in the same row. Therefore, yellow group is wrong, and the correct answer is the green group.
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Amuseum posted...

Either the yellow group or the green group must exist. If the yellow group exists, then you have two 4s in the same row. Therefore, yellow group is wrong, and the correct answer is the green group.

I'm unfamiliar with this solving method

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It's pretty trivial, but can be hard to follow without visual aid (like colors or lines.) Basically find pairs that are mutually exclusive; sometimes if you keep looking, they form a complex path, like the example.

In this example, let's start with 4 in B1. There is exactly one other 4 in the same row. That means they are mutually exclusive; exactly one of them must be the answer. So assign them different groups or colors. Same thing if we look in the same column and same box.

Here the original 4 is in the green group, and all other 4's that are directly linked and mutually exclusive to it are in the yellow group. We continue this path with all mutually exclusive 4s that are linked to both of these groups. Every linked 4 will alternate between yellow or green.

Acknowledge that either the yellow or the green group is the correct answer. Any inconsistency for either group means that group is the wrong answer. Here the error is that two yellows will be in the same row. Therefore, yellow cannot be the correct answer. We choose the other group.
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if you didn't add the lines, that's probably just the scanning path

I think it's basically saying that you can split the 4s into the two groups, yellow and green. the 4s knock each other out, and if you split them apart based on collisions, you end up with two groups. somehow, if you do that, the yellow group still ends up with a collision but the green group does not, so the yellow group can be cancelled.

I have not used that approach before, but have definitely seen the pattern of "either this set of numbers wins, or this set of numbers wins, but there's no evidence to choose one over the other" like a million times.

I just don't know how they determine that those two 4s both belong in the same group despite colliding. and why wouldn't the colliding 4 just make a new third group by itself?

I think I get it; it inverts across each line. each line says: there is exactly 1 other 4 in this set, so that is a true dichotomy, and the other 4 can be colored opposite of me. if you do this, and yet there is still a collision somewhere between digits of the same color (polarity might be a better term), then that color/polarity is invalid and loses.

note that (I think) the line at the bottom is a separate attempted coloring that came up inconclusive, since no collisions were present after following the links.

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Sahuagin posted...

I just don't know how they determine that those two 4s both belong in the same group despite colliding. and why wouldn't the colliding 4 just make a new third group by itself?

This solution only works for two groups at a time. And the process to identify groups for this solution must see only two groups. The point is that selecting one number means the other number is excluded. It's binary XOR logic.

As I stated in previous post, find a number that appears exactly twice in the same region (row, column, or box). That means they are in two different groups. Follow one of these numbers but across a different region. Here you see that 4 in B1 has collisions in all three regions: same row B7; same column G1; same box C3. That means B1 is in one group, and its collisions will all be in the second group.

Now follow one of the second group. Like G1. It has another collision at G5. Thus G5 is opposite group to G1, means same group as B1. This is where you're confused? Do a simple check. If G1 is chosen, then B1 and G5 must be eliminated. The reverse is also true. If G1 is eliminated, B1 and G5 must be chosen. Thus you should come to understand that B1 and G5 are the same group, and G1 is the opposite group.

Now keep following G5. It has second collision at C5; hence C5 belongs to second group (along with B7, G1, and C3). As I stated in previous post, each sequential cell linked this way will alternate into yellow or green groups. Keep going until you find all valid members of both groups using the criterion that exactly two of that number in the same region.

Even though by now we have already identified the problem with the yellow group at row C. Two yellows at C3 and C5 means we can eliminate the yellow group. Actually another similar problem with yellow can be seen in the top-right box.
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Sahuagin posted...

note that (I think) the line at the bottom is a separate attempted coloring that came up inconclusive, since no collisions were present after following the links.

Those cells are inconclusive because more than two of the same number are found in the same region. The lines end at H5 and I3 because those numbers collide with too many of the same number in the same region. Three 4s at row H and three in column 3. Thus this solution would not work for this set of cells. It must be exactly two of the same number in any given region.
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If doing it in your head (how I like to do these puzzles), you can simplify all that by simple looking for the first either or situation. In this case either (think it was A4 or B1 HAVE to be 4 (they can both be 4 mind you, but one HAS to be 4). That is due to it being impossible for both to be the other options.

For head solving (if you even get to the point in a puzzle like this where you can't just find a number that has to be in a set place), you just choose one of the two options and follow it through, seeing how it would fill out the rest of the puzzle and force numbers into certain places.

Either spot you choose will automatically fill out a bunch of other numbers, and since BOTH end up actually being 4, either one would be correct and you wouldn't even have to backtrack etc.

I finished the rest of the puzzle quickly in my head after choosing 4 for B1, which finished that whole line, the line under it, and then it cascades from there solving the whole thing with no more either or moments.

Pretty easy puzzle over all, even doing it in your head.

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Sudoku? Isn't that what Japanese dudes do when they want to restore their honor or something?
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