zotmeister | Puzzle 26: Island Oasis

There's a delightful puzzle design out there with a long tradition of uninspiring names: the World Puzzle Championships have called it Corral; Nikoli calls it Bag. I was struggling to find a worthy title for it before trying to make my own. As it happened, glmathgrant beat me to it with The Inner Limits. I felt scooped. However, e suggested recasting it as an "island" puzzle, and in so doing unwittingly gave me an idea. As perhaps you've noticed, I like puzzles that present "exceptions to the rules", where the information provided at the onset is not necessarily trustworthy (as in Smullyanic Dynasty, where givens can lie) or where some element has a version that works differently (as in Seeking Syren, where one of the "potential nodes" is really Syren). I was able to find a way to squeeze that concept into this design, and it consequently gave me a title. I hereby present an original variation of this classic design. Echolocation fans should be particularly fond of this. The monochromatic grid on the left is an unsolved Island Oasis puzzle; the largely-familiarly-shaded version on the right is the unique solution to that puzzle. I could say I made the oasis a different shade of blue because it's fresh as opposed to salt, but I'd be lying: I made the oasis a different shade of blue because I could and I felt like it, and for no other reasons. Mangled together from the instructions to Echolocation and Islands in the Stream: Each cell of the grid is either "water" or "land"; the objective is to determine which for all cells. The land cells, which form the "island", must ALL be orthogonally contiguous; each water cell must be part of an orthogonally contiguous cluster of water cells incident on the outer border of the grid EXCEPT for exactly one - the "oasis" - which must be completely surrounded by land cells (including diagonal adjacencies). Cells containing numbers are land; each number is the quantity of land cells orthogonally radiant from it, including its own cell. Also mangled together: 1) Each cell of the puzzle grid is either a "land" cell or a "water" cell. Fill them all in to solve the puzzle. (Rather than using colored pencils, I recommend simply dotting the unnumbered land cells and shading the water cells as you go along.) 2) Each numbered cell is land. (No need to put dots in them.) 3) Think of the numbers as land surveys. Each number is how many land cells exist total between the one it's in and either the outer border or the first water cell it encounters in each of four directions (up, down, left, right). This total includes the cell the number is in, but not any water it's aligned with - the numbers are stricly land-only counts. 3) All the land cells must form a single "island". That's right - all of them; there's only one island in this island puzzle. All land cells in the grid must be orthogonally contiguous (a single polyomino). 4) Somewhere on the island is a freshwater oasis; in puzzle terms, there is a single water cell in the grid completely surrounded by land. All eight cells adjacent to this oasis - including diagonally - are land. It wouldn't be fresh if it were incident on the ocean, now, would it? 5) Apart from the oasis, no water may be enclosed by the island. There can't be any place you could draw a freehand loop around one or more non-oasis water cells where every point of the loop is on land, including sides and corners of land cells. Equivalently, if you were a fish in an arbitrary non-oasis water cell and could only swim from cell to cell up, down, left, and right one cell at a time (NEVER diagonally), you must be able to reach the outer border without crossing land. Remember the oasis is always a single cell. This fact means that much of the logic that applies to the standard version of this puzzle still functions here. One rule in particular that still holds is extremely valuable, and despite just how important it is to the puzzle, I explain it here anyway: (Start of sample puzzle solution) I start, just as I would with Echolocation, with the '2' on the bottom row. It must be land, since it's numbered; the cell to its right is also land, and that makes two land cells aligned with the '2', so the next cells out in each (orthogonal) direction must be water. I mark the cell left of the '2', the cell directly above the '2', and the cell directly right of the '6' as water. This, in turn, defines the '6': the only direction left for the '6' to find land in is up, and it needs every cell it can see to make '6', so the entire center column is land. My next deduction is based on that rule I advertised. Look diagonally up and left from that '2'. That cell cannot be land, and the reason why may shock you! Remember that in this puzzle, all land cells must form a single island. This means that this cell, if land, must be ultimately connected to the '2', some way, somehow. But if it is, then the water cell directly above the '2' would be landlocked! It can't be the oasis, either, since there's another water cell diagonally adjacent to it - the oasis must be completely surrounded by land. The cell directly above the lower-left corner must therefore be water. This can be applied as a general rule: nowhere in the puzzle can there be a checkered pattern of land and water in a 2×2 square. The two land squares would have to be connected, meaning one of those two water cells would be landlocked (we may not know which of the two would be surrounded, but it doesn't matter as either would be illegal) and could not be the oasis due to the other water cell. From there I examine the '5' in the leftmost column. Here I use a tactic I refer to as "limited domain", something that applies to plenty of puzzles. At this point I wouldn't know exactly how the '5' is to be satisfied, but I do know that the space it's in is limited enough that some part of it is certain. The '5' needs five land cells to be satisfied. Itself counts for one; it can get at most one above it and at most one below it. This leaves only right as a direction, and at least two more land cells to be seen. Therefore, there must be at least two land cells rightwards from the '5'. I now consider the oasis. The oasis must be a single water cell with a ring of land all the way around it. The oasis can't be along the border, otherwise it wouldn't be surrounded by land cells. There are exactly two cells left in the grid that could be it: diagonally in from the upper-right corner, and the one directly below that. But since they're next to each other, their potential land rings overlap - that is, I know of four cells that must be land either way. The center column is marked already, but I now know that both the cell below the upper-right corner and the cell below that one are land, as they are adjacent to both possible oasis cells. The '4' then settles the matter. The cell to the left of the '4' must be water, as otherwise the '4' would be aligned with five land cells. This eliminates one of the oasis options, leaving only one; filling in all the land around it defines the '3', the '4', and both '7's. Marking the three water cells needed to keep those from overloading, we're left with just the '5' again, whose domain is now perfectly limited - it needs both of the remaining cells to be land to meet its quota. (End of sample puzzle solution) Let me know what you think. I'm running low on prizes at the moment, but send in your solution anyway and maybe I'll scrounge up something. - ZM

Flat | Top-Level Comments Only

1) The 2 can see itself and the 6 (2 squares); the rest of the squares (EE, N, W of it) must be water. The square NW of it must be water, or there is a checkerboard.

2) The 5 can see at most 3 squares on the vertical, so it must see at least 2 more on the horizontal; the square E of it must be land.

3) By the top 7, there must be three land squares visible to it on the top row. Therefore, there are three land squares visible horizontally to the 3; the square S of it must be water.

4) The only place where the oasis can be is the square S of the 3; squares NE, E, SE, S of it are land.

5) The 3, 4, and lower 7 can see their fill; that forces W of 4, S of 4, WW of 3, W of lower 7 to be water.

6) The 5 can only see 3 squares horizontally; the squares N and S of it must be land for it to see the full 5.

DONE.

Geez, spoiler alert.

Sean

It was the solution to the 5x5 sample, that he didn't provide the solution like he normally does. Not the actual puzzle.

Didn't? Pah. Hadn't yet. I said it was coming soon... At any rate, it's there now, and reading your solution after the fact I see I took a different approach to the oasis than you did. Of course, yours is perfectly valid as well. The '3' was only there for symmetry - I don't use it in my solution, but it's educational to see how you've used it.

As it happens, I explain the whole "checkerboard" thing in my solution, so if anyone was confused by that part, check out my write-up. But do go back to ralphmerridew's after mine, as e demonstrates a useful technique that I didn't. - ZM

Congrats to Mark Tilford and GLmathgrant for their correct solutions. - ZM

Puzzle 26: Island Oasis

Solution to the sample

Re: Solution to the sample

Re: Solution to the sample

Re: Solution to the sample

no subject