# Kyudoku: Introduction

This is a Kyudoku Puzzle. I wrote the rules myself, though I did not invent the puzzle.

There will be numbers in a grid. You just have to circle nine unique numbers (1 to 9) such that each row and column has sum of 9 or less. In some puzzles, one or more circled numbers may already be given.

Here is the real puzzle.

9 1 2 7 6 7
9 9 5 5 4 1
5 7 9 4 3 8
4 6 8 5 1 9
9 5 8 7 2 8
1 8 6 2 8

Note: This is not a puzzle of my own. There are puzzles similar to this you will find in the internet, but it has not been introduced that well in PSE before.

Original puzzle from: https://www.brainzilla.com/logic/kyudoku/

• Hello, I want to clarify one thing, as you stated that you "made the rules yourself", does it mean you are the original creator of the puzzle type? Also, I have seen this type before too in other sites (though it's actually quite rare.) In other note, this type of puzzle is already "mentioned" before here in PSE: puzzling.stackexchange.com/q/89477/28719 though it's a puzzle-identification question. Commented Dec 24, 2020 at 14:25
• No, I am not the creator. I just wrote up the rule myself. (You won't find the rules in Nikoli). Commented Dec 24, 2020 at 14:47

## 2 and 3

The 2 is already circled, so we can grey out all the other 2s in the grid.

There's only one 3 in the grid, so circle it and grey out everything else that follows from the 2 and 3 in their rows/columns.

## 8 and 9

There are only three ungreyed 8s in the grid now (on the fourth and fifth rows), so one of them must be circled.

There's also two 9s in those rows; assuming either of those 9s circled, we would know exactly which 8 is circled, and we can quickly obtain a contradiction. So both of those 9s can be greyed out.

There are only three ungreyed 9s in the grid now (top left corner), so one of them must be circled.

## 6 and 7

There are only three ungreyed 6s and three ungreyed 7s in the grid now.

If the 6 in the top row is circled, then the 7 in the fifth row must be circled, and then the 8 in the fourth row must be circled, and we end up with no options for circled 4, contradiction.

If the 6 in the fourth row is circled, then after some greying out we find that the circled 7 must be in the top row and the circled 5 must be in the second row, so there's no options for circled 9, contradiction.

Now we know that

the circled 6 must be in the bottom row, then the circled 8 must be on the right side, then the circled 7 must be in the fourth column, then the circled 4 and 5 must be on the left side ... the deductions fall like dominoes.

## Step by step

Here is the answer you are looking for:

First of all some obvious deductions, the crosses are the excluded numbers:

Then I tried:

The 4 in column 1 as a hypothetical number ringed and using some deductions this lead me to this:

Then with this hypothetical:

I saw that if 5 was circled it would lead to this contradiction in column 3 so 5 is crossed out:

Then I figured out

That the 5 in column 1 must be circled and that there needs to be 3 numbers that are shared with another number in the columns and 3 in the rows so the 6 in row 6 must be circled then some basic deductions to finish us off leads us to:

Lastly:

Since there is only one answer for this puzzle and they hypothetical leads to the solution 4 must be the right answer and this must be the solution.

• Note that this is solvable without Trial and Error, notice that grid-deduction tag. Commented Dec 24, 2020 at 7:39
• @Smartest1here, could you crop all the extra stuff out of your images? The top and bottom empty areas just distract from the actual grid Commented Dec 24, 2020 at 16:24
• I don't really think it is worth the effort considering my question is no longer checkmarked- no one will look at it. Commented Dec 24, 2020 at 16:40
• People may want to look at another solve path. Also, could you please not upload pictures like this in the first place? When you submit an answer it should have nice, useable pictures. Commented Dec 24, 2020 at 18:00