Come one, come all!
Come see the new type of logic puzzle, the Domidoku!

Here's a sample for all to try!


The process is simple!
First you solve the Sudoku and then you solve the Dominosa!

Clarification on Dominosa (the part everyone's confused on):
You must group all the numbers into pairs so that all the unique domino pieces show up once. (except the doubles and the zeros since it's also sudoku)
i.e. 1-2 through 8-9 will all show up once and only once. No more, no less.

Disclaimer: I am not responsible for any injuries, be they physical or mental, that occur while solving this puzzle.

  • 1
    $\begingroup$ what do the dark cells indicate? $\endgroup$ – Ivo Beckers Mar 1 '17 at 13:45
  • $\begingroup$ I also had to google what a Dominosa was. the link you provided doesn't really explain it $\endgroup$ – Ivo Beckers Mar 1 '17 at 13:46
  • $\begingroup$ At this current moment, I'll choose not to answer that. If it turns out that assistance is needed to solve this, then I'll elaborate. Yeah, I know I couldn't find a suitable link, feel free to replace it if you do. $\endgroup$ – dcfyj Mar 1 '17 at 13:46
  • $\begingroup$ I think I know what they mean now :) $\endgroup$ – Ivo Beckers Mar 1 '17 at 13:49
  • 3
    $\begingroup$ Dominosa link has been updated, courtesy of @techidiot. $\endgroup$ – dcfyj Mar 1 '17 at 13:58

I worked with @Techidiot's sudoku solution and solved the Dominosa:

enter image description here

The key to understand is that unlike regular Dominosa there are no tiles with same digits. and the grey cells indicate cells that shouldn't be included in the solution

  • $\begingroup$ Correct, will you be adding an explaination? (not a requirement, just curious) $\endgroup$ – dcfyj Mar 1 '17 at 14:24
  • $\begingroup$ Well. the first tile that immediately could be filled was the 35 in the bottom-right because of the dead end. Then you can cross out all connections between all other 35 pairs on the board.Then it was a matter of going through all number pairs and see which pairs only existed once and draw them. This results in a state where tiles could be placed immediately because of dead ends and these steps are repeated until solved $\endgroup$ – Ivo Beckers Mar 1 '17 at 14:27
  • $\begingroup$ Hi, I downvoted this by accident, and now the system won't let me change it. I'm sorry!!! $\endgroup$ – Woofmao Jul 25 '20 at 5:41

Final Solution(Though its already answered, I wanted to solve it on my own as a part of learning. Due to an ongoing problem with imgur, it took too long for uploading the solution)

enter image description here

SUDOKU Solution.

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Updated Better solution

Upon observing we get to know that, 5-8, 6-7 and 9-7 are unique. And we get ahead like this

enter image description here

Solution(Getting the idea from Ivo about the marked spots in the puzzle)

I am a novice to Grid puzzles and have specified all the steps I took while solving ...

1. We can simply fill 3-5 block just by looking.

enter image description here

2. I did a bit of a trial and errors, and I came up with the next cell as 3-9.

enter image description here

3. Now, that we have 3-5 and 3-9, we can cross of 3-4 at the bottom as there is no other way to go

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4. Since we have 9-3 pair, we can cross off 9-4 at the top

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5. Which gives 9-2 as well.

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6. Now, we have 9-3, 9-4, we get 9-8 as a new pair

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7. Also, we get 9-1.

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8. And then similarly, 9-6, 9-7, 9-5, 2-5, 4-7, 1-8
enter image description here

9. Now, as we already have 3-4, we get 4-2 as a new pair

enter image description here

10. Also, we already have 2-5, so we can knock off 5-1, 2-6,8-7(dead ends)

enter image description here

11. We have 8-7 and 7-4, so we get 7-5 as a new pair. Also, we can knock off 7-3, 8-6 at the top right.

enter image description here

12. We already have 7-4, 7-3, so we can knock off 7-6 which also gives 3-8, 5-4, 2-3 as dead ends

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13. Now, we have 8-7, 8-3, so we can knock off 8-4 which also gives 1-6, 3-6 as dead ends

enter image description here

14. We already have 8-4, so we can knock off 4-1 which also gives 8-2, 6-5 as dead ends

enter image description here

15. Now that we have 1-8 already, we can get 1-7 and then 1-3 at the top. Which gives 7-2, 6-4, 5-8, 1-2 as dead ends and completes the Domidoku.

enter image description here


  • $\begingroup$ Odd, your image isn't showing up in the answer, I have to click on it... $\endgroup$ – dcfyj Mar 1 '17 at 13:41
  • $\begingroup$ @dcfyj I can see it just fine? :-/ $\endgroup$ – Techidiot Mar 1 '17 at 13:43

Here is is. I hope I understood the rules of Dominosa.
I used my awesome Pinta skills to solve this.

Some of them are 2x2 squares instead of 1x2 rectangles because you can arrange 2 domino pieces in any position in those squares.

For sudoku I started from the top left 3x3 square. It is obvious where you can place an 8 and took it from there.
At one point, I admit, I took 2 leaps of faith and got it right for both of them.
For domino tiles, I tried to start from the bottom left corner. It's obvious that 3x5 go together. Then started incrementally 1x2, 1x3 and so on until something fit. Then some things became obvious.

Off topic a bit.

First I checked if this is not a trick question.
I tried first to see if the board can be covered by dominos like this.
Consider it as a chess board (9x9) where the top right corner is white.
This means 41 white squares and 40 black ones.
Then counted the blocked squares. I got 5 white and 4 black.
This means I have to place domino pieces on 72 squares where 36 are white and 36 are black. Since a domino piece covers a white and a black square, this is doable.

  • $\begingroup$ the 71 in the bottom left 2x2 square is impossible because there is a 71 already in 1st row 4th column so you know how to split the 2x2 square $\endgroup$ – Ivo Beckers Mar 1 '17 at 13:54
  • $\begingroup$ also I see two 53 pieces. i think you're not familiar with the rule in Dominosa that all tiles are unique $\endgroup$ – Ivo Beckers Mar 1 '17 at 13:55
  • $\begingroup$ S**t. This means I didn't read the rules properly. Thanks for the heads up. Back t the drawing board. $\endgroup$ – Marius Mar 1 '17 at 14:01
  • $\begingroup$ Updated the answer. $\endgroup$ – Marius Mar 1 '17 at 14:44
  • $\begingroup$ It's correct now, however, Ivo beat you to it. $\endgroup$ – dcfyj Mar 1 '17 at 15:58

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