This is the left side of the puzzle, with the rooms and gold numbers.
The gold numbers are the given clues.
**(1)**The 3 in 3×2 region can only fit one way; then the 1 and 2 regions also have only one way to fit.
(2) Then, the lower 3 region has only one way to be filled without breaking connectivity.
(3) And finally, the last two shaded cells can be placed with the "no two walls" rule.
This is, of course, the area at the bottom with the circles and triangles.
(1) Initial deductions based on the given unshaded cells and the no 2×2 rule.
(2) Noting that our path must pass through circles: on the left we must go through the top and bottom edges of the circle (or a 2x2 is created). We can also get to the goal on the right.
(3) Block loops from forming: near the goal, around the flat shaded 1×2 on the right, and around the tall shaded 2×1 on the left.
(4) No matter how we get to the goal, the single cell directly above the right circle must be shaded to stop a 2×2. This lets us finish off the puzzle.
Now we've got both of the minipuzzles on the left and right of the grid solved. Time to do a Nonogram!
First, to figure out whether we need to use shaded or unshaded cells. Calculating the totals, we get:
So we need to use the unshaded cells as the valid clues, and ignore the shaded cells.
(1) Some initial cells can be filled in using just the row clues.
(2) And then some more can be filled in with just the column clues, building off of that logic.
(3) Back to the row clues, we can narrow more placements down.
(4) And back to the column clues.
(5) One more trip to the row clues...
(6) And we're done!
Now it's time to use the Nonogram solution! As before, the shaded cells should be ignored. (This is clear from the clues that are 4 or more, which all happen to be shaded.)
Some initial deductions based on small 'constellations' of clues:
The loop edge on the left side of the top-left 3 cannot go left or it gets trapped. So it must go down, and this satisfies the 1 clue. (Also, some more logic building off of those constellations.)
Repeat the same logic on the right. On the top, the loop edge must escape by using the edge right of the 2. (We can also resolve the 2 just below the middle, and do things with it.)
Now, with a bit of 'don't-close-the-loop' logic, we can resolve the top half of the grid.
On the bottom right, the 3 must be ⊔-shaped rather than ⊐-shaped, so the right half can connect.
Meanwhile, in the middle-left, if we use the bottom edge of the 2, the loop there is forced to connect. This breaks either the 3 or the 2 below, since they will have nowhere to go.
In the previous step, I marked four loose ends. These must connect to each other, so this determines the "polarity" of the 3-3 in the bottom middle, and lets us complete the puzzle.
A new instruction!
"SET ALL LEMON AS BLACK AND RED AS WHITE, THEN SOLVE MASYU". So, I'll do that!
(1) Some initial deductions.
(2) With some 'don't-close-the-loop' logic, the left side can be completed.
(3) The top-middle black can't extend right, or it would make a tiny loop.
(4) Don't close the loop early again, and the puzzle is solved!
And another instruction:
The Masyu loop uses almost all of the squares. The unused ones, reading column-by-column, spell TEAL NURIKABE.
So, time to solve a Nurikabe from the teal numbers!
(1) The initial puzzle.
(2) Some basic deudctions just from 1 rooms, adjacent clues, and shaded cell connectivity.
(3) Some more deductions, extending the rooms. (Some cells have been marked for the next step.)
(4) Look at all the cells marked with "!!!". Each of these has only one room that can reach them -- and if it does, it cuts off some of the shaded cells.
(5) Next look at the 2x2 region of undetermined cells near the top. Either the top two there or the bottom two must be shaded; if the bottom two are, the 6 cannot extend left enough to prevent a 2x2 of shaded cells in the upper left.
(6) The top-right undetermined cell is impossible to reach, so it must be shaded. Then, the 3s must extend upwards to reach the two cells next to it, and the puzzle is solved.
Yet another instruction...
"HOLES TAUPE, TEES COLOR CYAN. MAKE A HERUGOLF SHAPE SUITE".
I'm not quite sure what "shape suite" means here just yet, but I can do the Herugolf portion. Let's do that:
A lot can be done with "this ball can only go this direction" logic -- in fact, the whole puzzle can be solved this way!
So now, what about the "shape suite"? The 20 pentominoes are specifically referred to as a 'suite of shapes', and they would fill the grid perfectly. This is probably a packing puzzle, then. But it's not immediately clear what constraint we have. I do note that we have 13 non-black squares in the shapes, and we also have 13 holes, 13 balls, and 13 unused cells in the Herugolf solution.
Actually, more specifically, there are 4 green squares and 9 white squares; there are also 4 balls that ended with 1 shot left, and 9 that used up all of their shots.
After a few other failed attempts to extract a message... let's try just packing them into the grid on the unused cells, and see where that gets us.
(1) The top-right L shape and the marked P shape both need an "empty-full-empty" pattern. There's only two places for that (though we don't know which is which yet). Then, the X can be placed, and then the W can be placed as well.
(2) There are now only two "two letters adjacent to each other" locations. This gives the placement of the F and the second L. (The second L must have its tip on the bottom rather than the top, or it would enclose an area of the wrong size in the corner.)
(3) There's only one way to satisfy the upper right corner. We also have to place the other L with its nub on the left, because it would enclose an area that isn't a multiple of 5 if we didn't.
(4) The T can't go on the
E, because then it would block off areas. So it has to go on the
N, and the Y has to go on the
E. Then, there's only one way to fit the remaining U pentominoes in.
The Next Puzzle
The four green-labelled squares say U, D, L, R. They're all different colors, so we can apply those letters to each of those colors...
...and solve the resulting Yajilin.
(1): Some initial deductions, not drawing loop segments yet.
(2): Drawing in some loop segments, and placing more dots where shaded cells can break the loop.
(3) There can't be a shaded cell on the very bottom of the 3↓ clue. This lets us draw a lot of the loop.
(4) Some "don't close the loop" logic on the left gives us most of the rest of the loop, and then a forced shaded cell in R3C9 finishes it off.
the shaded cells in the Yajilin spell PORTMANTEAU!
Somehow I don't think this puzzle genre will become popular enough to need a name besides "that one huge mashup". But, since I have the opportunity (and have been prompted to use it a few times, here's a fitting name:
HeyaNuriNonoSlitherMasyuNuriHeruYajilin, or "HNNSMNHY" for short (pronounced
/hn̦.sm̦.nʱi̤/ - good luck!)