These are three-dimensional yajilin puzzles. In each puzzle, the four squares depict the layers of a $4\times4\times4$ cube.
- Shade some cells on each layer. The numbered cells show how many shaded cells (not includong numbered cells) are in the direction of the arrow.
- Some numbers have been replaced with question marks to make the puzzle more difficult. Cells with question marks behave the same as other numbered cells.
- Diagonal arrows point to squares on other layers. An up-left arrow points to smaller-numbered layers, and a down-right arrow points at higher-numbered layers. For example, a down-right arrow on layer 2 points to cells in the same row and same column on layers 3 and 4.
- Shaded cells cannot be adjacent to another shaded cell (even those on different levels). The shaded cells are allowed to touch the numbered cells, however.
- Unshaded unnumbered cells on each layer are all adjacent to one another in 2D (i.e. each layer is treated separately).
- Make a single loop in 3D space which goes through every unshaded, unnumbered cell.
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$\begingroup$ What is a '?' ? $\endgroup$– JMPCommented Aug 9, 2019 at 6:52
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$\begingroup$ Oops, that was missing from the description. Added now, sorry about that. $\endgroup$– JafeCommented Aug 9, 2019 at 6:54
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$\begingroup$ "Unshaded cells on each layer are all connected in 2D (i.e. each layer is treated separately)." does that include numbered cells? $\endgroup$– Omega KryptonCommented Aug 9, 2019 at 7:22
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$\begingroup$ "The numbered cells show how many shaded cells are in the direction of the arrow." - does that include numbered cells, or only user-shaded cells? $\endgroup$– BirjolaxewCommented Aug 9, 2019 at 7:40
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1$\begingroup$ @Birjolaxew The unshaded cells are all connected to each other by placement, they do not have to be connected by a line. The line can leave and enter a layer as often as it wants. $\endgroup$– w lCommented Aug 9, 2019 at 8:23
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2 Answers
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This is a pencil-and-paper-games full solution... yay
answered Aug 9, 2019 at 12:12
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1$\begingroup$ Pen and paper is a good strategy for this +1. $\endgroup$– hexominoCommented Aug 9, 2019 at 12:16
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Here is the answer to the first one at least (apologies for the crude drawing)
