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enter image description here

Tilepaint (adapted from Nikoli)

  • The areas enclosed by bold lines are called "Tiles" and color the tiles with the following rules.
  • The cells of a tile must all be colored with black or left uncolored.
  • A number tells the number of black cells there has to be to the right or downward in the line/column.
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  • $\begingroup$ These were both a lot of fun! (I do feel like I missed a global break-in on the second one though...) $\endgroup$ – Deusovi Aug 31 at 4:11
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For the first one:

Note that exactly one of the two groups of 4 in column 6 must be shaded, because otherwise you can't get 5 cells in that column.

Now, use the two adjacent "1" clues for rows 4 and 5. All of the columns but the ninth have them as part of the same box - so none of those can be shaded.

enter image description here

Now, there's only one way to make 5 in many of the columns:

enter image description here

Next, check the rows

with the 7 and 8 clues. If the single box near the right of the 7-row is shaded, then both the top boxes of row 6 and 9 are shaded -- but that can't happen, because it breaks the 5 clue on top. So that box is unshaded (and therefore so is the other size-1 box in that column).

Then, look at the 5 and 8 rows. Both need exactly one more shaded cell. At most one of the top left two boxes can be shaded, and at most one of the bottom left two boxes can be shaded. So those account for the shaded cells in those rows, and all the others in those rows can be shaded; the rest resolves through fairly trivial logic.

enter image description here


For the second puzzle:

First, look at the second Γ from the bottom. Can it be shaded? If so, the regions just above it and below it in the 4 column must be shaded -- but then the bottom right Γ is shaded (from the rightmost 3 column) and unshaded (from the 3 row). So that Γ cannot be shaded.

Using the third row from the bottom gives us two more unshaded cells. enter image description here

With a bit more trivial logic from there, we can determine a few more regions:

enter image description here

If

the top of column 5 is shaded, then column 4 cannot be satisfied. If the top left region is unshaded, then row 2 cannot be satisfied (because the rest of row 1 must be shaded). If the 3 in row 6 is shaded, then the 4 under it cannot be satisfied (because there are only size-2 and size-3 regions in the 4 row).
enter image description here

Now, check the column

with the 7: it has 8 shadeable cells total, so the unshaded cell must be one of the size-1 regions. Some basic deductions give this:
enter image description here

And finally

look at the columns with the 4 and 3 on the right; because of how they interact with row 4, either both of their top regions or both of their middle regions must be shaded. The top breaks row 2, so it must be the middle. The rest of the puzzle resolves with fairly basic deductions.
enter image description here

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  • $\begingroup$ oops, sorry for sniping you for 11 mins! $\endgroup$ – Omega Krypton Aug 31 at 4:05
  • $\begingroup$ Great job, as always! :D Glad you're enjoying them -- unfortunately, actually the second one doesn't have any "global break-in", the trick is to find the starting point and how to progress as you already exactly did $\endgroup$ – athin Aug 31 at 7:25
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Obviously there's some problem with the ISO of the images, so here dark should be bright, and bright should be shaded XD

Here is #1

enter image description here

Short explanation for #1:

Testing briefly for the two 1s along the Y-axis shows that the 1's must contribute to two different columns, or else an impossibility would be reached, as shown in the red part below.
enter image description here
Others are done through pure logic and a bit guessing with the red and dark grey parts in the solution

Kind of a fluke for me, but I made #2:

enter image description here

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  • $\begingroup$ Unfortunately, I guess your solution for #1 is incorrect -- take a look at row 7 :) +1 tho for fast solving! $\endgroup$ – athin Aug 31 at 7:21

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