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Previous Puzzle:- Kyudoku :- Introduction

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 today's puzzle.

Note that this puzzle will be unsolvable. However, if you interchange 2 digits by their positions, the puzzle can uniquely be solved. Can you find how to solve this puzzle?

Note:- There is some brainwork you might have to do. You have to interchange 2 digits, such that the solution becomes unique. There may be a case when you interchange 2 digits and you found solutions, but the solution is not unique in that case.

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

Bonus:- There is an Easter Egg hidden in the puzzle. Can you find it? (Check the Title first) . Bonus:- There is a shaded box in the figure. Can you guess what number it can take?

Edit: @Smartest1here has provided a solution which contradicts mine, but his/her solution perfectly works. That, of course, does not make the puzzle unique, but I want a direct logical path from him/her to get to the solution. My solution, has a direct logical path. Also what I have in mind is 2 solutions, which also, is a problem =) .

Edit: I thought this puzzle would be interesting, so right now I am not concerned with uniqueness, as there are 3 solutions I found for now, 1 found by @Smartest1here. There are 2 more solutions I have in mind, the puzzle is still open for now.

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  • $\begingroup$ Can the shaded box be interchanged too? $\endgroup$ – Smartest1here Dec 25 '20 at 13:02
  • $\begingroup$ No, except the shaded box, you have to interchange 2 digits by their positions. $\endgroup$ – Anonymous Dec 25 '20 at 13:02
  • $\begingroup$ Is the number in the shaded box significant to the solution, if so, it should not be a bonus. $\endgroup$ – Smartest1here Dec 25 '20 at 13:06
  • $\begingroup$ No it is not, that it why it is given a bonus. Just consider it to be a barrier, it is of no use to the solution to the puzzle. $\endgroup$ – Anonymous Dec 25 '20 at 13:06
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    $\begingroup$ I'm not a downvoter, but I believe the downvotes are because this is a [grid-deduction] puzzle with multiple possible solutions - setters are expected to check their grid-deductions for uniqueness. $\endgroup$ – bobble Dec 26 '20 at 0:44
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Here is a solution:

enter image description here

Proof: for a moment let us assume the 3 and 5 is swapped:

If we then assume that C1R3 is circled we will get a contradiction in column 3: enter image description here

Then if we assume that C3R3 is circled we get a contradiction in that all 5s are eliminated: enter image description here

Then

C1R6 must be circled leading us to: enter image description here

Then a hypothetical leads us to a contradiction:

That there are no sixes that can be circled: enter image description here

So:

C6R2 must be circled leading us to: enter image description here

Then:

5,3 and 4 must be circled leading us to: enter image description here and then finally: enter image description here.

How I got the numbers to interchange:

I started of with these basic deductions:enter image description here

Then:

I basically repeated the same procedure as the steps above in the proof and I found the same contradictions until I reached: enter image description here

Lastly:

I saw that either 5 or 4 needed to be swapped with either a 2 or a 3 and with some experimentation I found the swap which lead me to a single solution so it seemed. Then I tried it with the proof procedure above to see whether it allowed for any new solutions with the former contradictions I had acheived which would disprove it but it did not affect the contradictions hence this must be the swap.

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  • $\begingroup$ Nevermind I will provide the explanations anyway. $\endgroup$ – Smartest1here Dec 25 '20 at 14:44
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    $\begingroup$ Not what I had in mind, but this perfectly works. The fact is, I was thinking in a different way, and you thought differently. Can you explain how you got that? because what I found as my answer was from direct logical deduction. First of all, what made you think you should interchange $3$ and $5$ only? $\endgroup$ – Anonymous Dec 25 '20 at 15:04
  • $\begingroup$ Don't wait for confirmation to provide explanations; you should be confident in your logical path to solve [grid-deduction] and post with that path presented. $\endgroup$ – bobble Dec 25 '20 at 15:18
  • $\begingroup$ Ok I analysed your approach, it is absolutely correct. But here's more to tell, there are more solutions. I understood why it's not pure logical deduction as approaching this differently can give you different solutions, especially you are allowed to interchange any digits. $\endgroup$ – Anonymous Dec 26 '20 at 5:06
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Not the answer to the grid but the bonus question:

The answer is:

0 as the number of the shaded grid as the numbers represent decimals of $\pi$

Also:

Is it related to the title?

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