Ripple Effect (taken from Nikoli)
- The areas divided by bold lines are called "rooms". Fill in all empty cells with numbers under the following rules.
- Each room contains consecutive numbers starting from 1 (to $n$ where $n$ is the area of the room).
- If a number is duplicated in a row or a column, the space between the duplicated numbers must be equal to or larger than the value of the number.
3 Answers
Final solution
Step by step
First fill in the two single-cell rooms with 1. Consider the left-hand one of these, and the two-cell room near it. If 1 is on the right of that two-cell room, then 1 can't be anywhere in the tetromino between. So we fill that two-cell room with
1 2
. Then there's only one possible place for 2 in that tetromino between.
In the next zigzag tetromino to the left, we can see which cells must be 3,4 and which must be 1,2. In the triomino above, 2 can't be at the top. That tells us where the 1 and 2 are in that tetromino, and then we can fill in that whole triomino and the top left domino.
Now it's easy to fill in the rest of the three half-filled tetrominoes, and some stray 2s and 3s around the place:
And the rest goes smoothly to the end. (Let me know if I should add more details here, but I've shown now how the deductions work, and once half the grid is filled the rest becomes relatively easy.)
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$\begingroup$ You got the first :-) Nice first answer! $\endgroup$ Commented Sep 4, 2019 at 9:05
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$\begingroup$ Thanks but I didn't provide step-by-step pictures like you did $\endgroup$– user62289Commented Sep 4, 2019 at 9:11
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$\begingroup$ Could you put your solution in a spoiler block, please? $\endgroup$ Commented Sep 4, 2019 at 9:26
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$\begingroup$ I'm going to accept @Randal'Thor solution as it is more comprehensive, but +1 for fast solving! :D $\endgroup$– athinCommented Sep 4, 2019 at 10:16
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$\begingroup$ (sorry, "the" should be "there" in my first comment - too late to correct) $\endgroup$ Commented Sep 4, 2019 at 10:30
So I got a different answer than the accepted one, I assume that means I made a mistake somewhere. Can anyone help me spot it?
1 2 1
is possible but1 1
isn't? $\endgroup$