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  • Really cute.
    October 1, 2015, 12:31 am
  • Very good!
    October 3, 2015, 4:49 pm
  • This is one of those I wish someone could give me tips on. How do you know how to start. There are no rows that take up the whole 15 or 16, which is what I usually try to start with.
    December 14, 2015, 10:17 pm
  • Hey Scooty, this one requires a lot of guesswork (at least for me, it does), but assuming it's symmetrical is the way to go.

    As for starting, in row 15, you know that the middle three squares are black. I don't know if I can explain it very well, but either counting or doing the math will give you that result. Doing the math, you have 7+1+1, (+1 for each space) =11. 11 is 4 less than the 15 columns, so 4 is your margin, for lack of a better term (you have wiggle room of four squares). 7 is greater than your four squares of wiggle room, by 3: the middle 3 squares.

    Or, to just count your way to that result, if you say, for the sake of figuring out what HAS to be black, say that the placement is crammed over to the left side as far as it can go, vs being crammed over to the right as far as it can go, you will end up with the three squares in the middle having to be black because they just don't have any other choice. No matter how that row is configured, you know with certainty that those are black.

    In this puzzle, you can do the same with row 7 and columns 2 and 14. Otherwise, I found myself just using symmetry and playing connect-the-dots.

    I hope that helps. :)
    December 15, 2015, 5:05 pm
  • Thanks. Help: symmetrical, R8/C1 C3 C7.
    February 10, 2018, 8:08 pm
  • >4 solutions.
    April 11, 2021, 5:10 am
  • It is not enough to count. You have to think to resolve.
    Thank you for the good riddle.
    June 9, 2021, 4:46 pm