The value of skyscrapers and 2-string kites in solving Sudoku is that they can be used to remove a particular candidate from certain cells in a puzzle, and this may simplify or solve the puzzle. But the value of it to the Pen & Paper Solvers is the simplicity of spotting those eliminations that will help – if you know the constraints that allow you to quickly discard the situations that don’t apply or will not work.
I think that for best results it should be done in this order, after you have progressed as far as you can with the basics:
1) Identify the candidates that each box could be a kitebox for in your partly-completed puzzle.
2) Discard any candidates which do not figure in four boxes which form a rectangle.
3) For only the remaining candidates, look for possible skyscrapers. These are easier to spot than kites as there are fewer restrictions.
4) If you can’t find any skyscrapers, or what you did find does not solve the puzzle, look for 2-string kites, again only for the candidates and boxes that form a rectangle.
Let’s try this: 697.....2..1972.63..3..679.912...6.737426.95.8657.9.241486932757.9.24..6..68.7..9
- Kitebox1.png (12.24 KiB) Viewed 1755 times
Looking at where you are with this puzzle and drawing a scribble note in the form of a large Noughts & Crosses grid, it’s easy to map out the candidates that each box could possibly be a kitebox for, as shown. This immediately narrows the search to candidates 1, 5 & 8, the only ones which involve 4 boxes arranged in a rectangle: 5 in Boxes 1278, 8 in Boxes 2356 and there are three different options for candidate 1: 2356, 2389 & 5689.
Let’s try for a skyscraper first. We look for Columns with two cells (only) which have candidate 5, and the columns and cells are in separate boxes. We find this condition in c6 only – no skyscraper. Now we look for Rows with the same conditions in candidate 5. This applies only in r8 and again there is no skyscraper.
Now we look for Columns for candidate 8 – c9 only, and in Rows we find r2 & r5 with this condition, but they do not work together as a skyscraper.
Finally, in columns 6 & 9 we find the 1-candidate skyscraper {1c69} which eliminates candidate 1 from r1c78 => 1r3c9, which solves the puzzle in singles.
Now try this: ..46.2.....2.3154...14...263178692542495176..586324...4..1...6.1.....49...3.4.7..
Note: we have identified a Locked 3,5 Pair at r13c2 and candidates 7 & 8 are restricted to c1 in Box1, so r2c2 is limited to 6 or 9.
- Kitebox3.png (11.27 KiB) Viewed 1755 times
So what have we got here? Boxes 4 & 5 are complete but Box6 is also ruled out as a possible kitebox because all the unsolved candidates in it are confined to single rows. For the rest of the puzzle, candidates 3, 5, 6 & 7 are discarded from our search because they aren’t part of a rectangle, so we are left with only candidates 8 (in Boxes 2389) and 9 (in Boxes 1278.)
Following the same process as before we find no qualifying Columns or Rows to form a skyscraper for 9 and only r2 for candidate 8. Therefore there is no skyscraper.
So we turn to looking for possible kites in these two candidates.
Here we check each possible kitebox for a row string AND a column string that have only ONE instance of our candidate 8 outside of the kitebox. Within the kitebox our strings must come from cells in different rows AND columns. Our possible kitebox may have other instances of candidate 8 but the Intersecting Cell between our two strings within the kitebox must NOT have candidate 8. The intersecting Cell is either the cell which the two strings cross over, or the cell which is in line with or common to the starting cells of the two strings. If we can find such a situation coming out of one of our possible kiteboxes and there is an unsolved cell which can see BOTH the outside end cells of our two strings, then that cell cannot be an 8.
With these constraints in mind, what do we find? There are no Row strings possible for candidate 8 out of Box2, but in Box3 we find r2c19/r37c7 with the intersecting cell r2c7 already solved at 5. Almost! But r7c1 which can see both ends of the strings is already solved as 4, and it turns out that Boxes 8 & 9 can’t be strung in both directions.
That leaves us with candidate 9, which seems to have kiteboxes possible at Boxes 1, 2, 3, 7 & 8. We have already discarded Box3 as not part of a rectangle and we quickly find that 9 can’t be strung in Rows into Boxes 1 & 2. In Box7, with the Locked Pair we noted at r13c2 we find r27c2/r9c14, but our intersecting cell r9c2 could also be 9, so this is not a valid kite. Finally, in the last place we look (Box8) we find the 9Kite r7c25/r29c4. This time there are no complications with the intersecting cell 4r9c5, so r2c2=6 and this solves the puzzle.
My point is that this kind of search for an elimination can be a fairly simple, quick and rewarding process if its limitations are well-understood, and it seems to me a logical first try after the initial basics. The skill, if you like, is in seeing where it can or might work and recognising the situations that won’t work, and quickly moving on. And for me, whether it works is much more important than why it works, or how these things fit into the larger family of Turbot Fish. I’m sorry if I appear to be crashing through your doors with a different kind of attitude. My mission is to find the simplest way for ME to solve Sudoku with pen and paper, not to become a Sudoku Professor. My post is aimed at like-minded people. If my explanations appear verbose, it’s because I’m trying to describe the key ideas of these issues in plain English. I can see that some of you were or are running out of patience with me, and I know that working with me is a challenge. I’m 68 and hopefully I will go on being a challenge to you for a while yet!
Yogi
