This week's X puzzle by kst, is.gd/kst_661_X, requires practically all X techniques and has a purely analytic solution. It makes for a great textbook example. So, here goes the textbook!

Techniques:

* square rule, items 1,2,3

* skewed fish, items 2,3,4,7,8

* UR: when can and when can't they be used in X puzzles? 7,8,9

Solution:

1. X hi/lo 1/2 Just filling in numbers, so far. 23 in A12 due to gap or HP. We get our first two examples of

**square rule**:

-- J9 != 8 (otherwise 8 eliminated from both HJ8)

-- B2 != 3 (otherwise eliminated from A12)

2. X hi/lo 2/2 First example of a simple

**skewed fish**:

-- wing on 3, in diagonals /\. Eliminates 3 from the rest of columns 1 and 4 and makes 3 required in both cols (not decisive for this puzzle, so we're not filling it in).

We also see a beautiful example of

**advanced square rule:**

-- the NT 145 in col 3 makes 5 required on the two diagonal cells. This eliminates 5 from C7, G7 (and E5, if it weren't a black cell). The difference between this case of square rule and the above: both diagonals come into play here, where above we dealt with only one.

What really helps solve this puzzle, though, is the

**skewed 3-fish**on 2, H/\. It this weren't an X puzzle, we could eliminate 2 from the rest of columns 124. Not here, though: H and / intersect at H2, and 2 is a candidate in that cell (in 'regular' Str8ts, this can't happen). If H2=2, the fish logic collapses. We can think one notch deeper, though: if H2 != 2, we get a 3 fish of three non-intersecting compartments in 2 columns. That can't work. Hence H2=2. I call this (colloquially)

**"must-use-intersection fish".**

Let's fill this in and work out immediate consequences. The skewed wing on 3 collapses: 3. No 3 in C! Here we can use the

**square rule**again: C2!=45 (both required in \). Also, no 5 in the lower \ compartment (required in A1-D4). Plus, we get another nice

**skewed wing**(C/): A3!=4.

It's tempting to look into Settis 23. But let's first simplify the cells we already have candidates for.

4. Fishing on the right-hand side * Must-use-intersection fish: H8=8 (CH\)

* Wing on 9 (C\) --> H4=1, 9 in /

5. Settis 23

2 and 3 must be absent from col 5 but present in G --> 3 in G2-7 --> no 9 in G2-7 --> F6=9 6. Wing on 7

* Regular

**skewed wing**on 7, H\: eliminate 7 from rest of cols 79, no 1 in col 9 (7 required in D-J9). Solves B9, G9; single 7 in A.

* Fill in some more cells: several singles, NPs, ... Settis 68 col 1, Setti 5 E. 7. Uniqueness and X

*

**"Standard" URs cannot be used if one (or more) cells involved are on a diagonal.**It's tempting to eliminate 67 from F9 for uniqueness reasons FJ39. However, that's not valid. The diagonal can break the ambiguity. 7 is already eliminated from F9 due to X power (skewed wing above).

* If all involved cells are off-diagonal, standard URs can be used (see below).

* There are X-specific "skewed URs," though. See C34, G3F4, FG5. If FG5 were 45, the NPs on FG would make 45 ambiguous in row-space, column-space, and along the diagonal. We can't have that, FG5!=4 [I'm cheating, of course: we already know J5=4 due to the NP on J, so we don't need the skewed UR. It's valid, though.]

8. Generalized skewed wings

* The skewed wings so far were similar to regular fish in the sense that both sets of wingtips saw one another. That is not a requirement, though! Look at 6 in G (base), col 5 (antenna 1), \ (antenna 2). The two 'feelers' F5, J9 do not see each other. Still, we can eliminate 6 from all cells that see both feelers; in this case F9. Double-check: if F9=6, G5=6 via 5 but G7=6 via \. This is very useful to solve the puzzle. We've had another such skewed wing on the board for a while: F2 and B4 !=45 (base col 3, antennas \ and /)

* Regular skewed wing 5 (/5)

* UR EF89: F8!=34. This is valid because all involved cells are off-diagonal

* 3-fish 6 AH\: F7!=6. No intersections this time, so we can eliminate 6 from the rest of columns 679 (F7!=7 because of the wing on 7 above)

9. BUG We're now left with bi-value cells everywhere but H7. If H7!=6, we have two candidate cells for each unsolved digit in all rows, columns, and diagonals. That's a uniqueness problem called BUG, see www.sudokuwiki.org/BUG Hence H7=6, which solves.

If you're unsure/uncomfortable about BUG, it's not hard to show that H7!=5 (e.g., SI via col 7, G2, J2, J7=5). That, too, solves.

Thanks again, kst, for a great puzzle!