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Sjudoku (sju = seven in Swedish).

Rules similar to Sudoku applies:

  • In each disk, the numbers 1,2,...7 should appear exactly once.

  • No line may contain duplicate digits (note that there are lines in 3 directions).

puzzle

Of course, there is a unique solution.

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    $\begingroup$ It may be of interest to non-Swedish-speakers to mention the pronunciation of “sju”. It’s notoriously difficult to describe, but very roughly, it’s somewhere between “hwoo”, “shwoo”, and “shoo”; for more detail and audio see en.wikipedia.org/wiki/Sj-sound . $\endgroup$ Nov 12 '20 at 10:22
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Completed grid:

enter image description here

Reasoning:

The puzzle was really straightforward; at no step did I need to do anything other than find a square which only had one possible fill-in. My break-in was the lower right square in the top circle, which is forced to be the 2 in that circle, since the middle row is blocked by the 2 at right, and the bottom left is blocked by the 2 at bottom left. All the rest of the deduction is in a similar vein.

That said, I did enjoy the puzzle, and like the layout. Would attempt again!

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    $\begingroup$ The puzzle is 6-way symmetric, so any deduction you place can immediately give you the 5 corresponding cells too. This is probably why it was so straightforward. $\endgroup$
    – Deusovi
    Nov 10 '20 at 22:26
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    $\begingroup$ @Deusovi Wish I had seen that :-) But there was never really a struggle point. You're right that the symmetry likely forced the lack of deductive schwerpunkts. $\endgroup$ Nov 10 '20 at 22:38
  • $\begingroup$ Does it automatically follow from the rules that a 5-line can be combined with the adjacent, parallel 2-line to make a group containing all 7 digits? Or is it by coincidence and/or design? $\endgroup$
    – Bass
    Nov 11 '20 at 9:24
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My answer:

answer

I can't explain the logic behind it.

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    $\begingroup$ Welcome to Puzzling! I have put your answer in a spoiler block. We also expect answers to puzzles like this to show at least some logic. $\endgroup$
    – bobble
    Nov 10 '20 at 22:36
  • $\begingroup$ Nice solve...welcome! $\endgroup$ Nov 10 '20 at 22:39
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The entire puzzle can be solved in four steps. At each step, there is only one option in the indicated box by considering where a certain number can go on within each circle. The colors in the image below correspond to the steps; the first step is green, the second is blue, and the third and fourth are orange and red, though at this point filling in the blanks is entirely trivial.

enter image description here

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    $\begingroup$ Nice - I will probably be back with a more challenging version. $\endgroup$ Nov 11 '20 at 19:46
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Each circle in the following order: center, then 1 o'clock, then clockwise. Circles in the same order: center, then 1 o'clock, then clockwise.

547126 324561 416753 762315 173642 251437

The inference rules were the very basic inference rules of Sudoku: draw the imaginary lines from the existing digits, see if this leaves just one place for a digit in a circle.

For example: from top 1 down left and from the bottom one up left only leaves one spot for a 1 in the 10 o'clock circle - the 3 o'clock position.

The second rule - if there are only two spots for a number in a circle and they are on a line, then you can still draw the imaginary line where this number can't be.

The last rule is simple exclusion - if there are only 2 or 3 free spots in a circle on in a line, see which numbers are missing in this circle, and check the lines that point a those spots against other circles. More often than not, you'll find that for one of the free sports, two of the remaining numbers are excluded.

Those rules are sufficient for the whole puzzle. They come straight for Sudoku and are topology agnostic. I have to say, a Sudoku puzzle than can be solved with those three inferences alone would be considered rather basic.


EDIT with a complete solve.

First, let's introduce some notation. I'll call circles (in the order above) X (center), A (1 o'clock), B, C, D, E, F and the positions within a circle - 0...6, where 0 is the center.

Think imaginary lines from a couple of filled cells in the indicated direction, and how they cover the circle where they intersect.

1@F6↙️ and 1@C4↖️ gives us 1@E2

4@A2↙️ and 4@D4➡️ => 4@C6

2@A1⬅️ and 2@D5↗️ => 2@F3

7@E5↘️ and 7@B3⬅️ => 7@D1

4@C6↖️↗️ and 4@D5↗️ => 4@X1

Exclusion in E => 4@E3

4@X1↘️ and 4@E3➡️ => 4@B0

3@C3↗️ and 3@F1↘️ => 3@B5

1@C4↗️ and 1@F6↘️ => 1@B1

2@D5➡️ => 2@B6

Exclusion in B => 5@B4

5@B4↖️ and 5@X0↖️ => 5@F2

1@B1↖️ and 1@E2➡️ => 1@A5

1@A5↙️ and 1@E2↙️ => 1@D0 5@X0↙️ => 5@D6

6@E6↘️ => 6@D3 Exclusion in D => 3@D2

Exclusion in the horizontal line at C1 =>6@C1

2@F3⬅️ and 2@B6⬅️ => 2@E0

5@D6↖️ => 5@E1

Exclusion in E => 3@E4

3@E4➡️ and 3@B5⬅️ => 3@X6

2@B6⬅️ and 2@A1↙️ => 2@X4

1@B1⬅️ => 1@X3

7@D1↗️ => 7@X2

Exclusion in X => 6@X5

7@X2↗️ and 7@E5➡️ => 7@A6

3@B5↗️ and 3@X6➡️ => 3@A0

6@B2↖️ => 6@A4

Exclusion in A => 5@A3

6@C1↖️ => 6@F0

4@E3↗️ => 4@F5

Exclusion in F => 7@F4

2@E0↘️ and 2@B6↗️ => 2@C2

5@A3↙️ => 5@C5

Exclusion in C => 7@C0

Solved. On the second go, I didn't even use the "two in line" rule.

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