Kyudoku: Introduction

Question

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	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/

Answer 1

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.

Final solution

Step by step

Answer 2

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.