This puzzle is part 5 of the Double feature series (first part here). The series will continue in "Double feature: Legal trouble".

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Rules of Nurikabe1

  • Shade some cells in the grid.
  • Numbered cells are unshaded.
  • Unshaded cells are divided into continuous regions, all of which contain exactly one number. The number indicates how many cells there are in that unshaded region.
  • Regions of unshaded cells cannot be adjacent to one another, but they may touch at a corner.
  • Shaded cells must all be orthogonally connected.
  • There are no groups of shaded cells that form a $2\times2$ square anywhere in the grid.

2. Fine threads of Nazis holding top-tier currency (7)
5. Turn over some journalism lab meat preserves (7)
7. Ancient Greek city moving to China (7)
9. Engulfed in boiling liquid (3)
11. Drive in a nail at the end of hotel's staircase support (8)
13. Taking in incomplete additional dose of sugar (8)

1. Source of blue dye from Manila (4)
2. Tuscan city welcoming new brownish colour (6)
3. Law exam returned by business intelligence teacher (5)
4. To start off, Sukarno is Indonesia's first president (4)
6. Foolish spectator's heart becoming extremely lonely (7)
8. Document written up from Latin manuscript (5)
10. Three lost in the middle of sandbar region in Germany (4)
12. Treatments for infectious disease (4)

1 Paraphrased from the original rules on Nikoli.

Solve both puzzles to answer the question: What is found in blood?

  • 3
    $\begingroup$ Ooh, finally, one of these is posted when I'm awake! $\endgroup$ – Deusovi Oct 17 '19 at 6:32
  • $\begingroup$ I really like this series, keep it up! $\endgroup$ – greenturtle3141 Oct 17 '19 at 13:42
  • $\begingroup$ @greenturtle3141 Glad to hear it! There are 7 more still to come :) $\endgroup$ – jafe Oct 17 '19 at 13:50

Solution, with cryptic clue explanations:

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  • $\begingroup$ (I spent about half an hour trying to find a clean logical path (without any deep case-work) through the Nurikabe, but couldn't get one, and it's currently Very Late o'clock. If someone else can find one, I'd love to see it!) $\endgroup$ – Deusovi Oct 17 '19 at 7:22
  • $\begingroup$ All correct, nice job! If you're interested in the intended path for the Nurikabe, I have it written down so I can post a self-answer in a few days if nobody else figures it out. $\endgroup$ – jafe Oct 17 '19 at 9:08

Self-answer to show the logical process.

Fill in the simple stuff in the beginning. Any cell that would merge two white areas is black, etc.
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The two cells in the top-right corner can't be reached. Only one available cell left in the top-right corner, so that has to be part of the 2.
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Some observations about the 9. First of all, it has to reach both the bottom left and the bottom right corner as no other number can reach that far. This takes 8 cells at a minimum.
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Neither of these squares on the right can be reached by the 9 without abandoning either of the corners. So it is the 4's responsibility to reach both of these. (We can already mark the cell directly to the right of the 4 as unshaded as that is the only way to reach the square marked red here.)
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From this follows that the 4 cannot extend to this square next to the 9. (This would require five cells at a minimum.)
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Long story short, consider these two squares on both sides of the 9. Only one of these can be reached by the 3, and no other numbers reach this far from the top part of the grid, and the 4 can't help because of the responsibilities shown above. So the 9 must reach at least one of these.
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We can now shade quite a few unreachable cells which cannot be reached by the 9 without abandoning all these responsibilities (i.e. reaching both corners, and reaching at least one of the squares next to the 9).
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The 9 cannot approach either corner from above, since that would leave an unreachable square at the bottom.
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So the cell horizontally towards the centre of both corners must be part of the 9. This means the cells directly above cannot be reached now.
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We now have two squares with 3/4 cells filled which can only be reached one way.
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This solves both the 5 in the top left and the 4 on the right.
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The 2 at the top is trivially solved now, and we know one additional cell of the 3 in the centre.
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The 9 cannot reach this square, so we know the 3 must extend this way.
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The 9 is now responsible for reaching both of these squares.
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Only one way to reach both of these squares at the bottom now.
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Final solution.
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  • $\begingroup$ Ah, that makes sense! I had gotten everything up to the part with the square just to the right of the 9. I think I considered that at one point, but not in tandem with the 4 needing to access that 2x2 near the top right. $\endgroup$ – Deusovi Oct 21 '19 at 7:34

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