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grandy52
1 year ago

r5c3 can be only 9. There’s a 3/4 couple in r4c2/3. Then you can make a lot in r4 (e.g., you get to ne6 in r4c1).

Hamburger02
1 year ago
Reply to  grandy52

r5c3 can be only 9.

Why?

grandy52
1 year ago
Reply to  Hamburger02

Naked Single. 1/2/6 form a triple in the box/column 1. Stay 3/4/9 and r5c3 sees 3 and 4 in the line.

grandy52
1 year ago
Reply to  Laminat03

Yes row/column=line/column

Hamburger02
1 year ago

How to continue?

This often happens where logically it does not go on, but two possibilities could be correct. Then you just have to play both versions, whether they work or not. Then you have the right version.

Here:

at the bottom right, one of the two circles must have a 4. Which one doesn’t know. I just put it in the top and all the other red figures are logically out of it without producing a contradiction. That was probably the right thing to do and you can do it.

grandy52
1 year ago
Reply to  Hamburger02

Well, each Soduku solver of the halfway what gives up would not be satisfied with so much bifurcation. Since the fact that one has not objected to this point does not necessarily mean that one does not meet it later. There is a much simpler, more logical variant that I have shown in my answer.

W0nd3rful
1 year ago

On the right side of the 5 instead of 9 a 4 would have to

W0nd3rful
1 year ago
Reply to  W0nd3rful

What do I write completely wrong, forget it