Cyclone -> "Obscured Clues" PAP (3/11/2013 3:15:09 AM)
I know I don't post very often here now, but I wanted to bring to your attention a variant of PAP that I find rather interesting. I start by showing you the puzzle, which I've redone in another program for the purposes of this demo. If you want to try it yourself, here is a direct PDF link. If you get lost, I've given a solve demo below the picture.
I recognize these images are large; only the starting position and the final solution will appear inline, and the rest will be linked for those who want to try to solve along with this demo.
The first thing you'll notice are the squares, which are called "obscured clues" (hereafter referred to mostly as squares or square clues). These clues are not known and are actually not needed to solve the puzzle; in essence, knowing those clues would make the puzzle way too easy. This is especially interesting in the bottom row, which has only a square as a clue.
Let's begin the way one would normally begin solving, by filling in known areas. For now, we're going to treat each black square - known to be a separate area of cells from any other number, per normal PAP rules - as a clue of "1" for the purposes of filling starting points (red don't come into play for this), and I'll explore more about squares later, not solving them for now. This is the result. (The black square in the third row from the bottom was used for the purpose of determining the lower-right cell being red; the cell above clearly cannot be.)
Now that we're out of obvious plays, let's start looking at the squares. Each square can be any number from 1-26 (left clues) or from 1-22 (top clues) based on the grid dimensions. I'm going to replace them with the actual clue determined via solving for simplicity, though you would normally just recognize that the square clue is solved. Because each square is a single set of cells, it can sometimes be very easy to solve for a square; in other cases, you must rely on a dot (cell that is not true) or a cell of another colour in the grid to tell you that a square clue can now be solved.
Let's solve for red. This is actually the easiest part of the puzzle, as all of the red cells are in the lower seven rows only. Also, six rows contain one red square each and every column of the puzzle has one red clue (number or square).
We'll start with the bottom row, which already has six red cells indicated. When solving for a square, any group of cells like this can automatically be connected as part of a single set of clues. This means that the entire row is filled, making the square in this row have a value of 26.
Using a similar technique, every single row can be connected in the middle two columns, resulting in clues of 4, 8, 12, 16, and 16 respectively in the first four rows. This also completes three squares in the columns, which from left to right are shown to be a 2, 7, and 7. This leaves us one final red square to solve for, in the second column from the right; by using the "2" clues nearby that are all completed it can be deduced that the group cannot extend to the next cell above, thus solving red in its entirety. We can also fill in one more black cell on the left side. You should be here at this point.
Sharp learners can see that I've solved one black square as a "4" already filled in on the left, in the fourth row from the bottom. Beyond this, we cannot solve for the left side yet, so we'll move to the right side of the grid.
Over here, look at row 13, with a square, 3, 3, and another square. On the right side, if you attempt to extend the first of the two filled cells to form a group of three, it connects to the other cell to make a group of four. Therefore, you know that all four cells are part of this obscured clue, and can fill them. Similar solving can be done for the 2 in the row below, but we're going to skip this and look at the third and fourth columns from the right. The top cells cannot stretch down to connect and form a 7; it would be too long. Therefore, we automatically know that the second and third sets must be connected as the square is the only remaining clue in each column, so go ahead and connect them. This also gives you two more cells, one as part of each of the "7" clues. DO NOT ASSUME THE SQUARE CLUES IN THE ROWS ARE SOLVED YET; we still have one unmarked column to the right.
The fifth and sixth rows from the bottom now have their "2" clues solved, and in the lower of the two rows, we can also solve the "1". This allows us to complete a few more cells with it, as well as solve more on the right side that gets narrowed in. Work on your own with basic clues (do not solve beyond these eight columns for any rows yet; there are some where you can continue, but I will pick up there) and you should eventually reach this point.
You may have noticed I've replaced several squares with "4" clues; these are solved just by working on the right side. If you are intrepid and have gone on past where I paused here, you may have further progress, but I will now resume solving for rows.
The "8" can now be started; three cells can be filled. Meanwhile, the row with 4,2,1,1,1,1,2,4 (the last "4" being a solved square) can have two cells filled in on the left thanks to progress on the right.
We still cannot proceed at all on the left, so we'll continue in the middle. Two columns can have a "1" clue bordered. Once this is done, it's possible to solve part of a "6" on the left side, allowing a group of false squares to be marked where a sole black "4" - similar to the right side - is given. This leaves the fifth row from the bottom and two related columns to be fully solved. More simple solving gets you this. Again, your progress may be slightly ahead; I avoid getting too far ahead for the sake of confusing you.
The ninth row from the top can now be started. On each side of the grid, there is a batch of 1,2,1 that can be filled in. This completes a lot of "4" clues in some columns, and THAT does something extremely useful; it puts a break between the square clues and the "7" clues on both the left and right sides of the grid. On each side, two obscured clues can be solved as "12" clues and connected, while the "7" clues can also be solved. More work results in both sides being completely solved (square clues all replaced with numbers).
There are only seven square clues left to be solved, and we can start by filling in more of the "8" clue smack dab in the middle. All six known cells are known to be a "1" clue, and can be closed. Following this, the second row from the bottom with NO RED in it (now marked as 4,2,1,1,2,4) can have one square of the "2" on each side filled. Moving to columns, we can now fill in seven of the required squares for each "8" clue. Back in the row we just visited, we can now solve each "2" clue, and we can start on the "3" clues in the row above. The known fills for the "3" clues will complete two columns and will lead us to a discovery that allows us to complete the puzzle.
The outer instances of obscured clues in the columns are now known to be "1" and the rest of those solved columns are completed appropriately. The "8" in the 4,8,4 is now able to be completed, and the remaining square clues in columns can be completed with the help of the row with the red group of four cells. After marking false squares in a few rows and affirming two squares as "1" clues (each row being 6,1,1,6), there is one final square to solve. All we have to do is connect and complete a "4" clue there. The rest is solved normally and you get what looks like a red carpet entrance to some home or festivity.
And that is how you solve an obscured clues puzzle. Please comment and discuss! If you want to download the PDF of the puzzle, the link is above at the start of the post.
Would love to see Conceptis try this someday to make those smaller puzzles just a tad more difficult.