More on Array Logic

Now, I could keep going step by step with my wingdings, and in fact I did over here in the googledoc of my rough draft.  But assuming that anyone reading this blog already does a lot of sudokus, I've already pointed out most of what I wanted to.

Here's what the puzzle looked like after I'd gone through it once.

M	7	k	f	T	O	8	3	J
s	s	s	x	s	s	s	s	s
	M		7 8	f	T		O	J
T J	7	T	8			f	O	M
f	JJ		O		7	T		M
M	f	7		3	k	J	8	T
M 8		8	T	k	O	J	f
J	k		f	7	8	3	M	T
7 /k	O	J	7 /M	T	f	M 8	3	8 3
7 /k	8	M	3 /J	O	3 /J		T	f
3	T	f	k	8	7 /M	M O	J	O

The X over the anchors is just an indicator that that array is entirely solved.

Yellow spaces are Antis, blue are knowns, light green are blockades, orange are Antis after placing the red smiley in ML and pink the Anti's after placing the blue smiley in TL.





















If we keep going, we'll get to a place where the Top zone looks something like this:

8	M		7 8	f	T		O	J
T 8	7	T	8			f	O	M
f	J		O		7	T		M

Do you how the parallel pairs of computer mice in TLl and TCl will give us a pair of mice in TRb(c/r)?  And of course, looking at the whole puzzle, we see another mouse in MRtc, which means that the mouse in TR has to fall into TRbr.

8	M		7 8	f	T		O	J
T 8	7	T	8			f	O	M
f	J		O		7	T	8	M 8

That's our X-wing, Parallel Pair, whatever you may call it, in action.  And while we might have figured out the same thing by looking at row Tb and seeing that the mouse couldn't go in each of the other spaces because of information in the zones below, marking the pairs in the squares made the deduction more noticeable.

More Array Logic

Note of importance! If you're not using Chrome, you may not see the wingdings.  You'll see other symbols instead.  Sorry about that.

I've worked out this whole puzzle at the googledoc I linked earlier, but I am only going to highlight some of the steps here.

M	7	k
s	s	s
	M		7	f	T			J
	7					f		M
			O		7	T		M
M	f	7		3	k		8
M			T	k	O
	k			7		3	M
7 /k		J			f	M
7 /k		M					T
3			k	8		M	J

At this point, I'd done two symbols and started the third.  Two things happen here.  In MC, we get a pair that is an ANGLE.  Angles don't affect anything else in the array right away, the way that arrows do, but they make great places to bifurcate.

Another thing that happens at this step is that we found two pairs in BL that share the same two spaces.  Most sudoku books call this a "twin".  (A few of the early ones call pairs twins and vice versa.) I think of it as a "subset" -- or a Bobbsey, but that is because I am old and it amuses me.

Whatever you call it, the formation which shows up in BLtl and BLml, combined with the Known in BLbl, creates what I call a "Blockade".  Blockades are only blockades when you're considering an array that does not include any of their members.  In other words, BLl is a blockade for everything EXCEPT the zigzags, the ampersands, and the trashcan.

There's one more thing to notice.  The two ampersands in BLtl and BLml form an ARROW, but unlike the arrow we saw previously, we can't make any kind of headway from it.  There isn't enough information in the Top Zone to limit the ampersands in TL to any two spaces, much less to a single space.  I call that an ARROW IN THE AIR.  When you're doing a harder puzzle, and start to get stuck, it's well worth your time to look for arrows in the air, because it's easy to miss their effects when you're switching from one kind of logic to another.  They can even be used to resolve twoshares elsewhere in the same square when you've been writing faster than you've been thinking.