Crosses and Circles











up vote
12
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favorite












Place two crosses on two cells of each row and column of this 9×9 board, and circles elsewhere, so that the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses.



enter image description here










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  • This is almost like a nonogram!
    – JGibbers
    Nov 30 at 15:31










  • Nitpick: the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses. Both sentences have the same content ("indicates the number of circles between its two crosses") so it's confusing: it implies the row and column numbers mean different things but they actually have the same meaning.
    – Voile
    2 days ago

















up vote
12
down vote

favorite












Place two crosses on two cells of each row and column of this 9×9 board, and circles elsewhere, so that the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses.



enter image description here










share|improve this question






















  • This is almost like a nonogram!
    – JGibbers
    Nov 30 at 15:31










  • Nitpick: the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses. Both sentences have the same content ("indicates the number of circles between its two crosses") so it's confusing: it implies the row and column numbers mean different things but they actually have the same meaning.
    – Voile
    2 days ago















up vote
12
down vote

favorite









up vote
12
down vote

favorite











Place two crosses on two cells of each row and column of this 9×9 board, and circles elsewhere, so that the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses.



enter image description here










share|improve this question













Place two crosses on two cells of each row and column of this 9×9 board, and circles elsewhere, so that the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses.



enter image description here







combinatorics






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asked Nov 30 at 12:24









Freddy Barrera

33318




33318












  • This is almost like a nonogram!
    – JGibbers
    Nov 30 at 15:31










  • Nitpick: the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses. Both sentences have the same content ("indicates the number of circles between its two crosses") so it's confusing: it implies the row and column numbers mean different things but they actually have the same meaning.
    – Voile
    2 days ago




















  • This is almost like a nonogram!
    – JGibbers
    Nov 30 at 15:31










  • Nitpick: the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses. Both sentences have the same content ("indicates the number of circles between its two crosses") so it's confusing: it implies the row and column numbers mean different things but they actually have the same meaning.
    – Voile
    2 days ago


















This is almost like a nonogram!
– JGibbers
Nov 30 at 15:31




This is almost like a nonogram!
– JGibbers
Nov 30 at 15:31












Nitpick: the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses. Both sentences have the same content ("indicates the number of circles between its two crosses") so it's confusing: it implies the row and column numbers mean different things but they actually have the same meaning.
– Voile
2 days ago






Nitpick: the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses. Both sentences have the same content ("indicates the number of circles between its two crosses") so it's confusing: it implies the row and column numbers mean different things but they actually have the same meaning.
– Voile
2 days ago












4 Answers
4






active

oldest

votes

















up vote
7
down vote



accepted










Answer is probably this (confessing that this is trial and error):




enter image description here




Approach:




I have started where Row:Col differences are 0:0 (which is either 6,6/9,9 or 6,9/9,6)

Using the walls as an aid, I have used it to identify the adjacent X

i.e. using 6,9 to identify 6,8 as gap for Row 6 is 0.

Similarly using 9,6 to identify 8,6

Next confirmed Xs in-order are 8,1 ; 4,1 ; 4,8


From from 6,9 the next guesses are 5,9 or 7,9 which has a lengthy iteration process ...... (to-be-completed when the time allows)







share|improve this answer






























    up vote
    5
    down vote













    Starting from top row




    (3rd, 7th) column,
    (2, 4),
    (3, 5),
    (1, 8),
    (7, 9),
    (8, 9),
    (2, 4),
    (1, 6),
    (5, 6)


    | | |X| | | |X| | | 3
    | |X| |X| | | | | | 1
    | | |X| |X| | | | | 1
    |X| | | | | | |X| | 6
    | | | | | | |X| |X| 1
    | | | | | | | |X|X| 0
    | |X| |X| | | | | | 1
    |X| | | | |X| | | | 4
    | | | | |X|X| | | | 0
    3 4 1 4 5 0 3 1 0






    share|improve this answer










    New contributor




    aivirai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.

























      up vote
      4
      down vote













      At the risk of running over my lunch break at work:



      Solution: (verified by other answers)




      SolutionPic




      This is a pure logic solution, similar to Sudoku and others. I forget the name of the exact type of puzzle, but the trick is to fill the entire grid with crosses (+) and only fill in circles (O) where they MUST be.



      For example:




      In row 4, there must be 6 O between +. Therefore, there will be 2 possible sets that will satisfy the condition.
      LogicExplanation
      As you can see, the O in yellow MUST be as such.




      Using this trick, we can fill out rows 4 and 8 as well as columns 2, 4, and 5.




      Step1




      A-ha! We have our first set of + in column 5. But what next?




      Looking back at column 4 and applying the trick (since we already have 2 O), the bottom entry in column 4 must be O...
      Then following the logic condition, row 9 and column 6 are immediately solved, as follows.
      Step2




      The rest is an exercise of repetition. Apologies for the bad Excel snaps and some poor formatting, and hope there is no confusion (or worse, a mistake!).



      Cheers,
      KF






      share|improve this answer








      New contributor




      KarmaFodder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.


















      • Could you explain this a bit more: "A-ha! We have our first set of + in column 5." I follow your logic that in column 5 we have O in rows 4, 5, 6, 8. I can see how logically you'd have a O in row 2 as well, but how do you eliminate rows 1 & 7 as possible +?
        – Dean
        Dec 1 at 5:41




















      up vote
      3
      down vote













      Solution




      o o x o o o x o o
      o x o x o o o o o
      o o x o x o o o o
      x o o o o o o x o
      o o o o o o x o x
      o o o o o o o x x
      o x o x o o o o o
      x o o o o x o o o
      o o o o x x o o o




      Approach:




      Starting with the 4th row, I placed an X on the left, then 6 o's, then another x, then a final o. In the two columns with x's in, I then placed o's in the cells which couldn't have x's in. From this, I was then able to place o's in other cells which couldn't have x's in, and x's in cells which had to have x's in. Eventually the grid was filled and I checked to ensure each row and column met their criteria. The question did ask, after all, to put o's where there weren't x's.







      share|improve this answer










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      RBZ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        7
        down vote



        accepted










        Answer is probably this (confessing that this is trial and error):




        enter image description here




        Approach:




        I have started where Row:Col differences are 0:0 (which is either 6,6/9,9 or 6,9/9,6)

        Using the walls as an aid, I have used it to identify the adjacent X

        i.e. using 6,9 to identify 6,8 as gap for Row 6 is 0.

        Similarly using 9,6 to identify 8,6

        Next confirmed Xs in-order are 8,1 ; 4,1 ; 4,8


        From from 6,9 the next guesses are 5,9 or 7,9 which has a lengthy iteration process ...... (to-be-completed when the time allows)







        share|improve this answer



























          up vote
          7
          down vote



          accepted










          Answer is probably this (confessing that this is trial and error):




          enter image description here




          Approach:




          I have started where Row:Col differences are 0:0 (which is either 6,6/9,9 or 6,9/9,6)

          Using the walls as an aid, I have used it to identify the adjacent X

          i.e. using 6,9 to identify 6,8 as gap for Row 6 is 0.

          Similarly using 9,6 to identify 8,6

          Next confirmed Xs in-order are 8,1 ; 4,1 ; 4,8


          From from 6,9 the next guesses are 5,9 or 7,9 which has a lengthy iteration process ...... (to-be-completed when the time allows)







          share|improve this answer

























            up vote
            7
            down vote



            accepted







            up vote
            7
            down vote



            accepted






            Answer is probably this (confessing that this is trial and error):




            enter image description here




            Approach:




            I have started where Row:Col differences are 0:0 (which is either 6,6/9,9 or 6,9/9,6)

            Using the walls as an aid, I have used it to identify the adjacent X

            i.e. using 6,9 to identify 6,8 as gap for Row 6 is 0.

            Similarly using 9,6 to identify 8,6

            Next confirmed Xs in-order are 8,1 ; 4,1 ; 4,8


            From from 6,9 the next guesses are 5,9 or 7,9 which has a lengthy iteration process ...... (to-be-completed when the time allows)







            share|improve this answer














            Answer is probably this (confessing that this is trial and error):




            enter image description here




            Approach:




            I have started where Row:Col differences are 0:0 (which is either 6,6/9,9 or 6,9/9,6)

            Using the walls as an aid, I have used it to identify the adjacent X

            i.e. using 6,9 to identify 6,8 as gap for Row 6 is 0.

            Similarly using 9,6 to identify 8,6

            Next confirmed Xs in-order are 8,1 ; 4,1 ; 4,8


            From from 6,9 the next guesses are 5,9 or 7,9 which has a lengthy iteration process ...... (to-be-completed when the time allows)








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 30 at 13:26

























            answered Nov 30 at 13:17









            Kryesec

            66410




            66410






















                up vote
                5
                down vote













                Starting from top row




                (3rd, 7th) column,
                (2, 4),
                (3, 5),
                (1, 8),
                (7, 9),
                (8, 9),
                (2, 4),
                (1, 6),
                (5, 6)


                | | |X| | | |X| | | 3
                | |X| |X| | | | | | 1
                | | |X| |X| | | | | 1
                |X| | | | | | |X| | 6
                | | | | | | |X| |X| 1
                | | | | | | | |X|X| 0
                | |X| |X| | | | | | 1
                |X| | | | |X| | | | 4
                | | | | |X|X| | | | 0
                3 4 1 4 5 0 3 1 0






                share|improve this answer










                New contributor




                aivirai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






















                  up vote
                  5
                  down vote













                  Starting from top row




                  (3rd, 7th) column,
                  (2, 4),
                  (3, 5),
                  (1, 8),
                  (7, 9),
                  (8, 9),
                  (2, 4),
                  (1, 6),
                  (5, 6)


                  | | |X| | | |X| | | 3
                  | |X| |X| | | | | | 1
                  | | |X| |X| | | | | 1
                  |X| | | | | | |X| | 6
                  | | | | | | |X| |X| 1
                  | | | | | | | |X|X| 0
                  | |X| |X| | | | | | 1
                  |X| | | | |X| | | | 4
                  | | | | |X|X| | | | 0
                  3 4 1 4 5 0 3 1 0






                  share|improve this answer










                  New contributor




                  aivirai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.




















                    up vote
                    5
                    down vote










                    up vote
                    5
                    down vote









                    Starting from top row




                    (3rd, 7th) column,
                    (2, 4),
                    (3, 5),
                    (1, 8),
                    (7, 9),
                    (8, 9),
                    (2, 4),
                    (1, 6),
                    (5, 6)


                    | | |X| | | |X| | | 3
                    | |X| |X| | | | | | 1
                    | | |X| |X| | | | | 1
                    |X| | | | | | |X| | 6
                    | | | | | | |X| |X| 1
                    | | | | | | | |X|X| 0
                    | |X| |X| | | | | | 1
                    |X| | | | |X| | | | 4
                    | | | | |X|X| | | | 0
                    3 4 1 4 5 0 3 1 0






                    share|improve this answer










                    New contributor




                    aivirai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    Starting from top row




                    (3rd, 7th) column,
                    (2, 4),
                    (3, 5),
                    (1, 8),
                    (7, 9),
                    (8, 9),
                    (2, 4),
                    (1, 6),
                    (5, 6)


                    | | |X| | | |X| | | 3
                    | |X| |X| | | | | | 1
                    | | |X| |X| | | | | 1
                    |X| | | | | | |X| | 6
                    | | | | | | |X| |X| 1
                    | | | | | | | |X|X| 0
                    | |X| |X| | | | | | 1
                    |X| | | | |X| | | | 4
                    | | | | |X|X| | | | 0
                    3 4 1 4 5 0 3 1 0







                    share|improve this answer










                    New contributor




                    aivirai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|improve this answer



                    share|improve this answer








                    edited Nov 30 at 13:29









                    gabbo1092

                    4,684738




                    4,684738






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                    answered Nov 30 at 13:11









                    aivirai

                    1007




                    1007




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                    New contributor





                    aivirai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    aivirai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                        up vote
                        4
                        down vote













                        At the risk of running over my lunch break at work:



                        Solution: (verified by other answers)




                        SolutionPic




                        This is a pure logic solution, similar to Sudoku and others. I forget the name of the exact type of puzzle, but the trick is to fill the entire grid with crosses (+) and only fill in circles (O) where they MUST be.



                        For example:




                        In row 4, there must be 6 O between +. Therefore, there will be 2 possible sets that will satisfy the condition.
                        LogicExplanation
                        As you can see, the O in yellow MUST be as such.




                        Using this trick, we can fill out rows 4 and 8 as well as columns 2, 4, and 5.




                        Step1




                        A-ha! We have our first set of + in column 5. But what next?




                        Looking back at column 4 and applying the trick (since we already have 2 O), the bottom entry in column 4 must be O...
                        Then following the logic condition, row 9 and column 6 are immediately solved, as follows.
                        Step2




                        The rest is an exercise of repetition. Apologies for the bad Excel snaps and some poor formatting, and hope there is no confusion (or worse, a mistake!).



                        Cheers,
                        KF






                        share|improve this answer








                        New contributor




                        KarmaFodder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.


















                        • Could you explain this a bit more: "A-ha! We have our first set of + in column 5." I follow your logic that in column 5 we have O in rows 4, 5, 6, 8. I can see how logically you'd have a O in row 2 as well, but how do you eliminate rows 1 & 7 as possible +?
                          – Dean
                          Dec 1 at 5:41

















                        up vote
                        4
                        down vote













                        At the risk of running over my lunch break at work:



                        Solution: (verified by other answers)




                        SolutionPic




                        This is a pure logic solution, similar to Sudoku and others. I forget the name of the exact type of puzzle, but the trick is to fill the entire grid with crosses (+) and only fill in circles (O) where they MUST be.



                        For example:




                        In row 4, there must be 6 O between +. Therefore, there will be 2 possible sets that will satisfy the condition.
                        LogicExplanation
                        As you can see, the O in yellow MUST be as such.




                        Using this trick, we can fill out rows 4 and 8 as well as columns 2, 4, and 5.




                        Step1




                        A-ha! We have our first set of + in column 5. But what next?




                        Looking back at column 4 and applying the trick (since we already have 2 O), the bottom entry in column 4 must be O...
                        Then following the logic condition, row 9 and column 6 are immediately solved, as follows.
                        Step2




                        The rest is an exercise of repetition. Apologies for the bad Excel snaps and some poor formatting, and hope there is no confusion (or worse, a mistake!).



                        Cheers,
                        KF






                        share|improve this answer








                        New contributor




                        KarmaFodder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.


















                        • Could you explain this a bit more: "A-ha! We have our first set of + in column 5." I follow your logic that in column 5 we have O in rows 4, 5, 6, 8. I can see how logically you'd have a O in row 2 as well, but how do you eliminate rows 1 & 7 as possible +?
                          – Dean
                          Dec 1 at 5:41















                        up vote
                        4
                        down vote










                        up vote
                        4
                        down vote









                        At the risk of running over my lunch break at work:



                        Solution: (verified by other answers)




                        SolutionPic




                        This is a pure logic solution, similar to Sudoku and others. I forget the name of the exact type of puzzle, but the trick is to fill the entire grid with crosses (+) and only fill in circles (O) where they MUST be.



                        For example:




                        In row 4, there must be 6 O between +. Therefore, there will be 2 possible sets that will satisfy the condition.
                        LogicExplanation
                        As you can see, the O in yellow MUST be as such.




                        Using this trick, we can fill out rows 4 and 8 as well as columns 2, 4, and 5.




                        Step1




                        A-ha! We have our first set of + in column 5. But what next?




                        Looking back at column 4 and applying the trick (since we already have 2 O), the bottom entry in column 4 must be O...
                        Then following the logic condition, row 9 and column 6 are immediately solved, as follows.
                        Step2




                        The rest is an exercise of repetition. Apologies for the bad Excel snaps and some poor formatting, and hope there is no confusion (or worse, a mistake!).



                        Cheers,
                        KF






                        share|improve this answer








                        New contributor




                        KarmaFodder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.









                        At the risk of running over my lunch break at work:



                        Solution: (verified by other answers)




                        SolutionPic




                        This is a pure logic solution, similar to Sudoku and others. I forget the name of the exact type of puzzle, but the trick is to fill the entire grid with crosses (+) and only fill in circles (O) where they MUST be.



                        For example:




                        In row 4, there must be 6 O between +. Therefore, there will be 2 possible sets that will satisfy the condition.
                        LogicExplanation
                        As you can see, the O in yellow MUST be as such.




                        Using this trick, we can fill out rows 4 and 8 as well as columns 2, 4, and 5.




                        Step1




                        A-ha! We have our first set of + in column 5. But what next?




                        Looking back at column 4 and applying the trick (since we already have 2 O), the bottom entry in column 4 must be O...
                        Then following the logic condition, row 9 and column 6 are immediately solved, as follows.
                        Step2




                        The rest is an exercise of repetition. Apologies for the bad Excel snaps and some poor formatting, and hope there is no confusion (or worse, a mistake!).



                        Cheers,
                        KF







                        share|improve this answer








                        New contributor




                        KarmaFodder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.









                        share|improve this answer



                        share|improve this answer






                        New contributor




                        KarmaFodder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                        answered Nov 30 at 21:13









                        KarmaFodder

                        411




                        411




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                        New contributor





                        KarmaFodder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                        KarmaFodder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.












                        • Could you explain this a bit more: "A-ha! We have our first set of + in column 5." I follow your logic that in column 5 we have O in rows 4, 5, 6, 8. I can see how logically you'd have a O in row 2 as well, but how do you eliminate rows 1 & 7 as possible +?
                          – Dean
                          Dec 1 at 5:41




















                        • Could you explain this a bit more: "A-ha! We have our first set of + in column 5." I follow your logic that in column 5 we have O in rows 4, 5, 6, 8. I can see how logically you'd have a O in row 2 as well, but how do you eliminate rows 1 & 7 as possible +?
                          – Dean
                          Dec 1 at 5:41


















                        Could you explain this a bit more: "A-ha! We have our first set of + in column 5." I follow your logic that in column 5 we have O in rows 4, 5, 6, 8. I can see how logically you'd have a O in row 2 as well, but how do you eliminate rows 1 & 7 as possible +?
                        – Dean
                        Dec 1 at 5:41






                        Could you explain this a bit more: "A-ha! We have our first set of + in column 5." I follow your logic that in column 5 we have O in rows 4, 5, 6, 8. I can see how logically you'd have a O in row 2 as well, but how do you eliminate rows 1 & 7 as possible +?
                        – Dean
                        Dec 1 at 5:41












                        up vote
                        3
                        down vote













                        Solution




                        o o x o o o x o o
                        o x o x o o o o o
                        o o x o x o o o o
                        x o o o o o o x o
                        o o o o o o x o x
                        o o o o o o o x x
                        o x o x o o o o o
                        x o o o o x o o o
                        o o o o x x o o o




                        Approach:




                        Starting with the 4th row, I placed an X on the left, then 6 o's, then another x, then a final o. In the two columns with x's in, I then placed o's in the cells which couldn't have x's in. From this, I was then able to place o's in other cells which couldn't have x's in, and x's in cells which had to have x's in. Eventually the grid was filled and I checked to ensure each row and column met their criteria. The question did ask, after all, to put o's where there weren't x's.







                        share|improve this answer










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                          up vote
                          3
                          down vote













                          Solution




                          o o x o o o x o o
                          o x o x o o o o o
                          o o x o x o o o o
                          x o o o o o o x o
                          o o o o o o x o x
                          o o o o o o o x x
                          o x o x o o o o o
                          x o o o o x o o o
                          o o o o x x o o o




                          Approach:




                          Starting with the 4th row, I placed an X on the left, then 6 o's, then another x, then a final o. In the two columns with x's in, I then placed o's in the cells which couldn't have x's in. From this, I was then able to place o's in other cells which couldn't have x's in, and x's in cells which had to have x's in. Eventually the grid was filled and I checked to ensure each row and column met their criteria. The question did ask, after all, to put o's where there weren't x's.







                          share|improve this answer










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                            up vote
                            3
                            down vote










                            up vote
                            3
                            down vote









                            Solution




                            o o x o o o x o o
                            o x o x o o o o o
                            o o x o x o o o o
                            x o o o o o o x o
                            o o o o o o x o x
                            o o o o o o o x x
                            o x o x o o o o o
                            x o o o o x o o o
                            o o o o x x o o o




                            Approach:




                            Starting with the 4th row, I placed an X on the left, then 6 o's, then another x, then a final o. In the two columns with x's in, I then placed o's in the cells which couldn't have x's in. From this, I was then able to place o's in other cells which couldn't have x's in, and x's in cells which had to have x's in. Eventually the grid was filled and I checked to ensure each row and column met their criteria. The question did ask, after all, to put o's where there weren't x's.







                            share|improve this answer










                            New contributor




                            RBZ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                            Solution




                            o o x o o o x o o
                            o x o x o o o o o
                            o o x o x o o o o
                            x o o o o o o x o
                            o o o o o o x o x
                            o o o o o o o x x
                            o x o x o o o o o
                            x o o o o x o o o
                            o o o o x x o o o




                            Approach:




                            Starting with the 4th row, I placed an X on the left, then 6 o's, then another x, then a final o. In the two columns with x's in, I then placed o's in the cells which couldn't have x's in. From this, I was then able to place o's in other cells which couldn't have x's in, and x's in cells which had to have x's in. Eventually the grid was filled and I checked to ensure each row and column met their criteria. The question did ask, after all, to put o's where there weren't x's.








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                            New contributor




                            RBZ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                            share|improve this answer



                            share|improve this answer








                            edited Nov 30 at 17:54









                            gabbo1092

                            4,684738




                            4,684738






                            New contributor




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                            answered Nov 30 at 17:40









                            RBZ

                            311




                            311




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                            New contributor





                            RBZ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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