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This puzzle is a hybrid of three different puzzles: Cross the Streams (h/t to athin for recalling this excellent puzzle type), Nurikabe, and Tapa. In this puzzle, as with all three of its progenitors, the goal is to shade cells in the grid such that the shaded cells form a single orthogonally connected region without any 2x2 square being wholly shaded. Three different types of clues are provided. Cross the Streams-style clues are given on the outside of the grid, and define the pattern of shaded squares in the given row/column. Nurikabe clues are given in the grid in black font, and indicate the number of orthogonally connected unshaded squares in same region as the clue; unshaded regions need not contain a Nurikabe-style clue, but will have at most one. Tapa clues are given in the grid in black font, and indicate the pattern of shaded squares around the clue. Wait, I used black font for both?!? Oh well, I guess you're going to have to figure out which are which! Cells with numbered clues are not shaded. I hope you enjoy!

Grid

Text Version

                * *  
              ? 3 3 *
        ?     ? * * 2   2
        2     ? 3 3 2 ? ?
      ? ? * * ? * * * ? *
      --------------------
2 ? 2| | | | | | | | | | |
     |--------------------
? ? 2| |5| | | | | | | | |
     |--------------------
2 ? 3| | | | | | | | | | |
     |--------------------
    *| | | | | | | | | | |
     |--------------------
    *| | |5| | | | | | | |
     |--------------------
    *| | | | | | | | | | |
     |--------------------
  ? ?| | | | | |4| | | | |
     |--------------------
? ? ?| | | | | | | | | | |
     |--------------------
* 2 *|3| | |4| | | | |3| |
     |--------------------
* 2 *| | | | | | | | | | |
     ---------------------
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9
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Step 1:

To start with, C6 does not have enough space to hold both 3s above the '4' in row 7. Therefore, the cells below the 4 must all be shaded. This, in turn, means that the 3 in C7 cannot go into the last 2 rows. Since there are also 2 3s in that column, the first 3 must occupy at least R2-3 and the second 3 in R6-7. This, in turn, fixes the first 3 of column 6 in R4-5 due to the 2 x 2 rule. In row 3, the last '3' must occupy R3C8 regardless of how that row is packed to either side. So, this gives us this:

CTS_1

Step 2:

Now, R3C10 cannot be shaded now since the ''3' cannot reach it. Following that, we can reason that the '2' in C10 cannot be in R1-2 or else, it will form an isolated black island or a 2 x 2 area. We can then further reason that R1C9 should be unshaded as well or else the '2' would get trapped.

CTS_2

From there, R2C6 must be shaded to satisfy the 2 and we can carry out a series of chain deductions by applying the "No 2 x 2 rule". From there, the '4' in R7C6 must be a Nurikabe clue. (now shaded in orange)

CTS_3

Step 3:

Now, we can shade some squares using Nonogram rules in the top left. R1C1-2 must both be shaded to give the '?' in R1 space to come out and join with the rest of the black cells.

CTS_4_1 Now, if R3C1 was unshaded, we run into a contradiction because that would force the '5' in R2C2 to be a Nurikabe clue and block the single black cell in R1C4, as seen below.

CTS_4_2

Therefore, that cell is shaded. After that, we need to ensure the connectivity of the black cells and we get to this:

CTS_4_3

Now, the '3' in R9C1 cannot be a Nurikabe clue. So, it must be a Tapa clue and the 3 cells next to it must be shaded. Similarly, the '5' in R5C3 must also be a Tapa clue.

CTS_5

Step 4:

Next, we focus on the right side of the grid, specifically C8. If R3C8 were to extend downwards, then R4C9 must be unshaded to prevent a 2 x 2 block. Now, we run into a problem. If R3-4C8 were a '2', then this forms an isolated black island. If it wasn't, then there isn't enough room for the remaining 2 '2's. Therefore, this R3C8 is unshaded and R3C9 is shaded. Additionally, this means that the first 2 in C8 must be in R5-6 and the second 2 must occupy at least R9 to prevent forming a 2 x2 group with the '3' in C7.

CTS_6_1

Now, the 2 in C10 cannot be in R4-5 since that would form an isolated black island. So that is unshaded and we get to:

CTS_6_2

Now, note that the '?' in C10 must occupy R8 or below. This means that the '3' in R9C9 cannot be a Tapa clue and must be a Nurikabe clue. Additionally, the '?' needs a way to connect to the rest of the black cells and this can only be through the cell in R8C6. This means R8C7 is shaded and we can finish up C8 and the '3' Nurikabe clue in one go.

CTS_6_3

Step 5:

Now, in R8, there are 3 '?'s, we need to satisfy that.

CTS_7_1

Next, we need to note the '4' in R9C4 must be a Nurikabe clue now. Also, the black cells in R8-10C2 need a way to connect to the rest of the black cells and this can only be through R7C4 (the bottom side is blocked):

CTS_7_2

Now, there is only one way to complete C5 and the '4' Nurikabe clue must be satisfied without blocking off a black cell, so we fill that in correctly and finish the grid:

CTS_7_3

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  • 5
    $\begingroup$ There is a tiny mistake in the last step...the shaded square under the 4 is isolated. $\endgroup$ – Jeremy Dover Dec 13 '20 at 16:32
  • 1
    $\begingroup$ @JeremyDover Argh sorry, I uploaded the wrong image. Will update in a while. Thanks! $\endgroup$ – Alaiko Dec 13 '20 at 20:17
  • $\begingroup$ Great job! I hope you enjoyed the puzzle. There was one logical leap which might have made the puzzle easier...the right side is easier to manage if, after step 2, you combine rot13(pbyhza 8'f pbafrphgvir 2'f, juvpu sbepr E7P8 gb or hafunqrq) with rot13(gur snpg gung fbzr pryy bs ebj 7 yrsg bs gur Ahevxnor 4 unf gb or funqrq) to show that rot13(nyy bs E7P8-P10 zhfg or hafunqrq). But there are many ways through, and you nailed it! $\endgroup$ – Jeremy Dover Dec 13 '20 at 20:36
  • $\begingroup$ @JeremyDover Ah ok, yeah I see it now. Great puzzle btw, combining 3 genres! $\endgroup$ – Alaiko Dec 13 '20 at 20:56

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