Interesting Sudoku Puzzle

Sorry for the mess. Many of my pages were mangled when I converted CLASSIC Site pages into NEW Site pages. My <pre> to </pre> constructs were stripped, and indented paragraphs were used instead.

This is an interesting puzzle because it is solved mainly by using phantoms and exclusions.

Notice where several boxes already have a row of three numbers filled in, leaving just the other two rows in those boxes.

B1, B2, B8, B9 are that way, and others will quickly join them, with either rows or columns within the box filled by three numbers.

I'll start the solution, and leave the rest to you.

The 6 in R3,C7 means 6 can only occur in R2 of B1.

That means the only other place for a 6 in the top boxes is in

R1 of B2. R1,C4 is occupied, and there's a 6 in C6, so we get:

(R1,C5) = 6 by phantom isolation.

Likewise, the 3 if R7,C3 means 3 can only occur in R8 of B9.

That means the only other place for a 3 in the bottom boxes is in

R9 of B8. R9,C6 is occupied, and there's a 3 in C4, so we get:

(R9,C5) = 3 by phantom isolation.

Both 4 and 9 must be phantoms in R2 of B2 because of R1 in B1.

There's a 4 in C9, so we quickly get these three numbers:

(R3,C8) = 4 by exclusion

(R3,C9) = 9 by what's left (of 4,9)

(R5,C7) = 4 by isolation (gives a box with 3-number column)

Likewise, 5 and 9 must be phantoms in R8 of B8 because of R9 in B9.

There's a 9 in C1, so we quickly get these three numbers:

(R7,C2) = 9 by exclusion

(R7,C1) = 5 by what's left (of 5,9)

(R5,C3) = 9 by isolation (gives a box with 3-number column)

Notice that now R3 and R7 both have just 3-numbers left, and all in individual boxes, B1 and B9.

So we know that B1 will have 1,2,7 in R3 of B1, leaving 5,6,8 in R2 of B1.

Likewise, we know that B9 will have 2,6,8 in R7, leaving 1,3,7 in R8.

These phantoms lead to many more finds, such as (R9,C5) = 3.

As you can see, this puzzle is solved mainly by box content,

lots of phantoms, lots of exclusions, and a few isolations.