# Tapa - Confluence

This is a Tapa puzzle.

Rules of Tapa:

• Shade some cells of the grid. Shaded cells should form a single [orthogonally] connected group; no 2×2 square should be fully shaded.
• Some cells have clues in them. These cells cannot be shaded.
• Clues give the runs of shaded cells in the eight touching cells, in no particular order. (This is like a Nonogram/Picross clue, but instead of a row or column, it 'measures' a square around the clue.)

Here's an example Tapa, with its solution:

For instance, around the central clue, there are two single black cells on the left two corners, and a "run" of three black cells on the right side. Similarly, around the bottom left clue there is a run of 3 (in the cells below, below-right, and right of the clue) and a run of 1 (the cell above the clue). In the top right, there is only a run of 5 (all five cells around the clue).

• I'm not familiar with the cited puzzles well enough to understand what's going on in this puzzle. Can @Deusovi state definitions of single connected group and run in terms of graph theory? I'm sorry, it's my lingo. – Galen Apr 29 '20 at 18:19
• @Galen Graph theory is overcomplicating it. The shaded cells must be connected (only considering orthogonal connections); this "connected" does have the same meaning as in graph theory. For a "run", take only the eight cells surrounding a clue, and look at the sizes of the connected groups of shaded cells within those eight. (I'll add an example.) – Deusovi Apr 29 '20 at 18:23
• If I understand the given example correctly, diagonals count for a black square to be connected to a clue but do not count for black squares to be connected to each other. – Galen Apr 29 '20 at 18:35
• The reference to nonogram/picross/etc is truly apt. This is in effect a collection of 8-long nonogram rows/columns woven together like chain mail/mesh. With the numbers so nicely encapsulated within the picture, as blank-cell clues at that. – humn May 2 '20 at 5:47
• Archaeologists have reconstructed the 81 × 81 pixel urpuzzle. It is recognized as an exquisite low-resolution jewel that sparkles with two handcrafted fonts at the quantum level. The smallest font delicately fits pairs of numerals within 9 × 9 settings. This timeless work of puzzle art neither wastes nor begs for a single pixel. – humn May 8 '20 at 16:11

## Solution

I started from the top right corner and essentially proceeded counterclockwise around the board, ending in the bottom right corner.

## Step by step deductions

Firstly consider the $$3,3$$ in the top right. Since there are only 8 cells around there, they must be two runs of 3 separated by two single blank cells.

The blank cells can't be above and below the middle, as that would contradict the $$3$$ below. They can't be top left and bottom right, as that would isolate the L-shaped run of 3 in the corner. Then we can fill in some unshaded cells around the $$3$$ below; after that the blank cells can't be top right and bottom left, as that would isolate some shaded cells on the right. So the blank cells are on the left and right, and now we've completed both the $$3,3$$ and the $$3$$ below it.

Now consider the $$2,4$$ to the left of that. We've already got

two shaded cells separated by a blank one, so we can definitely continue those shaded runs to at least length 2 going left. The top run can't be the length-2 one, as that would isolate the top shaded edge, so that's the length-4 one and the length-2 one along the bottom.

A few short deductions now:

The row of four shaded cells below $$2,4$$ and $$3,3$$ can only connect in one way: another shaded cell must be below.
The eight cells around the $$2,3$$ are divided by the $$6$$ in the bottom left and the blank cell in the top right, leaving two runs of 3. One of those must be completely shaded and the other two-thirds shaded; anyway we can shade the central cell of each one.
Looking at the $$1$$ at the top, we can't leave both cells to its right blank, as that would isolate the shaded top row. So the shaded cell there must be bottom right and all the others are blank.
Again the run of shaded cells from the top can only connect in one way.

Now consider the $$6$$.

There's a continuous run of 6 shaded cells around it, which must be everything except the cell with $$2,3$$ written in it and one of the ones next to that. So we can fill in 5 right away.
Since exactly one of those cells between both the $$6$$ and the $$2,3$$ is shaded, the other one to the right of $$2,3$$ must be shaded. Avoid a 2x2 square there. Now we've got the 1 for the $$1,2$$ while the 2 must be two of the three cells below it.

Meanwhile, for the $$5$$ in the bottom left,

we've already got the 3 above shaded, and at most one on the right can be unshaded (because of the $$7$$), so the bottom and bottom-left cells are unshaded and the right one is shaded.

For the $$7$$,

if the top-right and right cells are both shaded, then we're cut off from further advance to the right on all three rows there, so it can't be connected. Contradiction, so one of those two is the unshaded one, and we can shade the other four. That enables us to fill in everything to the left.
Finally, if the one just right of the $$7$$ is the unshaded one, then we can fill in above and to the right of that, but then there's no way of getting the $$1,2$$ and having everything connected.
So it must be the top-right one that's unshaded, and then everything else in the bottom right follows easily.

## Illustration

• Aw, but the bottom right corner is the interesting bit! – Deusovi Apr 29 '20 at 18:28
• That's correct, nicely done! (And nice illustration!) – Deusovi Apr 29 '20 at 19:31
• Nice rep :P I've looked at some old meta posts and wish to thank you for standing up for riddles. – Tom May 1 '20 at 6:16
• @Kalaivanan "Shaded cells should form a single [orthogonally] connected group" - so that cell must be shaded, to connect up the group. – Rand al'Thor May 4 '20 at 11:47
• @humn Thank you very much for the bounty! I used gifs to illustrate step-by-step grid-deductions once before, but this answer has put me on a trend and now I'm continuing to do it in some other recent answers. – Rand al'Thor May 8 '20 at 15:45