This is by way of a reply to Leren’s post about kites in Keith’s thread about Finding the Possibility of Single-Candidate Eliminations, but is now more specifically about Kites. Leren posted some examples of Kite Puzzles which I have worked through. Generally it went well but there were some difficulties which I will go into later, but this is what I have distilled.

When you combine Keith’s rule with Leren’s suggestions it shakes down to a simple 3-step process:

1) For each candidate you want to look at, identify the boxes that could be a kitebox (multiple sites for the candidate, but not all in the same row or column.)

2) Check each possible kitebox for a row AND column that has only one instance of the candidate outside the kitebox.

(But is Keith right that the string ENDS must also be in this c-type box? I’m not so sure about that.)

3) Check to see if there is an unsolved cell with your candidate which can see BOTH the outside end cells of the two strings. If so then this cell CANNOT be that candidate.

The revelation for me was the idea of starting with the right kind of box and then looking for the two strings. I used to do it the other way around and got no-where. Another thing that has become clear that I had not realised before is that although the two strings of the kite can have only two instances of your candidate, the kitebox may have more than two. However, it does not seem to work if the two cells you use in your kite are in the same row or column in the kitebox. Apparently they have to be distinct in row AND column for the logic to work. Something else to watch for is that often there will be two options for the two strings for a given candidate in the same box. This came up in Puzzle3 below.

There is a parallel to this in the skyscraper: the walls of the skyscraper (that’s Columns if it is upright or upside-down) must have only two instances of your candidate, but the base may have more than two. Sorry, but RCB’s Solver exports definitions with zeros

Puzzle1: With basics this came through to here: 056070900040620080280000000000080000730000618005000020000000000070500049000048701

This one worked like a dream. I found a 3Kite in Box2: r38c5 / r2c16, so r8c1=6 stte

My jottings looked like this: 1-4578 2-4578 3-12578. I stopped when I found a kite that solved the puzzle with candidate 3.

With basics, Puzzle2 came to here: 361700295842395671050261483108526034625000018034100526400610852580002167216857349

Solved after 9Kite found in Box4: r6c16 / r47c2, r7c6 = 3 Worksheet was 3-8, 7-4, 9-4578

Puzzle3 came through to here: 320547906006213050045698230500472000007901025002805700214359678673184592050726143

My worksheet shows that I wasted a lot of time on this one: 1-1346 4-6 8-1346 9-4. This was because I got side-tracked by noticing that Box1 had two separate working kites for candidate 1, resulting in two different eliminations: r1c38 / r36c1 shows r6c8 is NOT 1, and r14c3 / r3c19 removes 1 from r4c9. What I did not realise at the time was that this means r4c9=9 and the puzzle can now be solved in singles.

I got Puzzle4 through to here: 081020600042060089056800240693142758428357916175689324510036892230008460860200000

For the four candidates I looked at, my Kite worksheet looked like this 1-2389 5-2389 7-2389 9-28 but I could not find a working kite anywhere.

The puzzle was solved when r2c4=7 after r2c4 set to 5 chained through to r2c7=5, which is not possible.

After trying for a skyscraper, I applied this Kite Hunt to another puzzle I got from Hodoku: 590036000006009000000800962807620490600090000059010206704003609960000100000960004

Because there are a lot of unsolved cells in this one it takes a lot of wading through and the only systematic way I could think of was to look at the possibilities for each cell, one-by-one. For candidate 1, Box1 looks like a possible kitebox, but is it? Which cells in it could be ‘strung?’ Here we see that none of the unsolved cells in rows 1 and 2 of Box1 could be strung as ROWS, because they have too many instances of candidate 1 in Boxes 2 & 3. Does this now rule Box1 out as a possible kitebox for candidate 1? No, because r2c1 can still be strung as a Column. Now we find the 1Kite r29c1 / r3c26 eliminating 1 from r9c6, which makes r7c4=1 and this solves the puzzle.

I’m putting this up as a thesis. Does my proposed 3-step process work for you?

I would welcome your input.

Yogi