Friday 29 March 2013

Friday Puzzles #202

I'm very very tired.  Tens sudoku.  Adjacent cells containing numbers summing to ten are both shaded. Adjacent cells which are not both shaded must not contain numbers summing to ten.  This is pretty hard.  Enjoy!
    #239 Tens Sudoku – rated hard
All puzzles © Tom Collyer 2009-13.


  1. I am not getting it here.Lets say there is an 8 in R2C4 which would lead to a 2 in both R2C3 and R2C5 since both are shaded and adjacent to R2C4.What then is the correct interpretation of the rules?

  2. No, that's not quite right. The rules definitely do not say if any two adjacent cells are both shaded, then the numbers placed must sum to 10. The two conditions given are NOT mutually converse.

    The correct interpretation for the example you give is less strong. Basically you conclude that at least (and indeed, because both are in the same row, at most) one of R2C3 and R2C5 contains a 2.

    1. The easiest way to think about it is that pairs summing to 10 must be arranged in dominoes (and possibly triminoes) in the shaded areas. You can do a quick thought experiment in this case to show that there are no triminoes. Working out what the layout of the dominoes actually is forms part of the puzzle.

  3. I have come a cropper in this puzzle.I am clueless on how to start after 20mins!.Since you have stated that 2 cells which are both shaded and adjacent need not neccessarily add upto a 10 the puzzle becomes even more increasingly complex and what is this domino ,trimino you refer to??!I have no idea what you are trying to convey me.Let me see if i can bifurcate or do something and complete this one.

  4. here is an example solution to one of the shaded rings (ignoring any external clues for now):


    Note that 9+2 is not 10. But 9 is adjacent to 1 and 2 is adjacent to 8, so all is well. If you draw a box round the pairs summing to 10, then you get some dominoes. In this case a horizontal [19] and [73] and a vertical [28] and [46].

    A trimino would look like:

    But as I say, these configurations are impossible in this puzzle.

    I'd also strongly discourage you from bifurcating. You can get a very long way on a wrong assumption before you reach a contradiction with this type.


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