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To celebrate the Pi-Day (3/14) adequately, a challenging math puzzle must not be missing.

enter image description here

Rules:

  • Fill in the numbers 1-9 exactly once in every row, column, and region.
  • On top of that, you need to use $\pi$ exactly three times in every row, column and region to fill in the remaining gaps.

What's the solution for this $\pi$doku?

Additional challenge: Try to find a "creative way" to write Pi each time you're using it! For example: $4\cdot\left(\left(\frac{1}{2}\right)!\right)^2$

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  • 3
    $\begingroup$ Welcome to PSE! Do take a look at the tour page to acquaint yourself with the site, and earn a badge. $\endgroup$
    – Phylyp
    Mar 14, 2018 at 17:27
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    $\begingroup$ Also, very nice and smart puzzle! $\endgroup$
    – Phylyp
    Mar 14, 2018 at 17:27
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    $\begingroup$ I got to be the 3rd upvote for the below answer, and the 14th for the (awesome) puzzle - seems quite fitting for today :) $\endgroup$
    – puzzledPig
    Mar 14, 2018 at 20:40
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    $\begingroup$ (Clever additional challenge, but moving the goalposts makes answerers redo work they've already done once, for little gain.) $\endgroup$
    – Rubio
    Mar 15, 2018 at 4:42
  • 1
    $\begingroup$ @Rubio I'm sorry. I will keep that in mind for future posts! $\endgroup$
    – NAMELESS
    Mar 15, 2018 at 11:48

2 Answers 2

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For the sake of completeness, there are actually 3 possible solutions. Using process of elimination and deduction can get you to this point:

enter image description here

One solution is given by Sid already:

enter image description here

But two more are:

enter image description here
(Notice the threes stay put, but the nines, pis, and all but 1 five move)

and:

enter image description here
(Notice the threes and fives in the middle are flipped)

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    $\begingroup$ If we impose the condition that there can only be one solution. R9C7 has 3 options - 3,5,9; R8C9, R6C9 and R6C9 only has 2 options - 3,5. Thus to avoid having a rectangle containing 3 and 5s(which results in 2 solutions), R9C7 has to be 9 $\endgroup$
    – Ariana
    Mar 15, 2018 at 5:55
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    $\begingroup$ @Ariana If we impose the condition that there can only be one solution, take any of the above solutions and you're done. $\endgroup$
    – boboquack
    Mar 15, 2018 at 6:35
  • $\begingroup$ @boboquack But how would you eliminate the other 2 solutions? $\endgroup$
    – Ariana
    Mar 15, 2018 at 6:36
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    $\begingroup$ @Ariana the other solutions are eliminated because you have assumed there is one solution. $\endgroup$
    – boboquack
    Mar 15, 2018 at 6:37
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Ariana
    Mar 15, 2018 at 7:34
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Here's the solution: I will provide the explanations within the next 24 hours because now I have to go to bed.

enter image description here

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