Kurodoko (taken from Nikoli)
- Colour cells black according to the following rules.
- The numbers on the board show the number of white cells from the number to the nearest black cell or to the frame of the board (horizontal + vertical).
- The cells with numbers cannot be black, they are and remain white.
- Black cells cannot touch horizontally or vertically, and all white cells must be connected.
Solved the big one!
For my screenshots, a small dot is "i know there is no wall here".
one can see that adding a dot (so not a wall) in between, forces an invalid solution, so it has to be a wall.
If we assume the fours have a wall between them, the rightmost four has to have a free square below it no matter what (red). Then we need a wall (blue), or the three will see too much. We need to have a free space next to the wall (purple). However, the top three now will need a free square to the left no matter what (yellow). This means that the lower four now has too many visible squares, so there cannot be a wall between the fours.
Then it follows that there cannot be a wall between the fives, as this makes the leftmost four have too many visible squares.
Using the same logic, none of the squares on the edge of the two fours can have any walls, as that would give them too many visible squares, so adding the red dots as known free spaces.
The lower four cannot look left or right, as that would give it all possible squares, and then it has to have a wall on top that will give one of the other fours too many squares. So the lower four should have walls on both sides.
After adding the walls and marking the squares around as known free, one can see that the lower four cannot extend two down as it then will get one extra square. So the lower four is now given. This also forces the upper three to have a free spot to the right.
The rightmost three cannot extend further up, or it will see four squares, so add a wall there. This means that it has to extend down, so add a free square and then a wall below. This wall cannot have wall neighbours, so then we add free squares around, and the lower three is now also completed. Can then add walls around that one (not shown).
As for the fours, it cannot be a wall where I put the red dot, as that would separate the dot below from the rest. So it will have to be free, and then add a wall to the right of the fours as they are now done.
From this it then becomes ok straight forward. Make sure to not create any separated dots, and just count the given dots/free spaces when reaching the higher numbers.
I think the answer to the first one is
First notice that the two 3s in the second column must include each other's squares. If the squares either above or below are white then those to the left and right must be black, which is disallowed. Hence, the squares above and below these 3s must be coloured black.
In the second row the square to the left of the first 3 must be left white. If the square to the right of this 3 is white, then we have four in a row, which is disallowed. Hence, the square to the right of this 3 is black.
Now assume that the square below the second 3 in the second row is white. Then the square to the left of this is also white and we can start to fill in the other black squares (I've indicated with light blue below).
However, we quickly run into a contradiction. The second 3 in the second row must be in a column of 3s but the square just above cannot now be coloured black and the square two below can also not be coloured black. Hence, we have a contradiction making our initial assumption wrong, so we must have the following
If we analyse the 3 in the 6th row, we find that the square that is three above it must be coloured white but it turns out that the square two above it must also be coloured white. This follows from trying to develop the board around the two 3s together in the 2nd column and finding that, no matter what path we choose, this square will necessarily be white. It follows that the square just above this 3 must be black.
Now, the squares which are mutually adjacent to the 3s in the 6th and 7th rows cannot both be coloured white (leads to adjacent black square colouring) and cannot both be coloured black (3 in the 7th row would be cut off). Hence, one is coloured black and one is coloured white. This means that the square to the left of the 3 in the 6th row must be white and the square to the left of that must be black as follows
From there, the logic is not too difficult. We can see the squares which must be white next to the first 3 in the 7th row and we must ensure that the white section in the bottom left hand corner is not cut off from the rest. Using the second part of rule 4 that "all white cells must be connected" allows us to fill in the rest of the grid uniquely.