Original Puzzle

F6 8 (C5,D2,E9)

G8 8 (B7,E9,H1)

I4 8 (C5,F6,G8)

A3 8 (B7,C5,H1)

E2 4 (D9,E4)

D1 1 (E6,F5,H1)

C2 1 (B9,H3)

D3 6 (A2)

F2 5 (what's left in box)

F5 6 (D3,E7)

D5 3 (C4,E3)

I9 6 (C8,E7,H4)

G1 6 (D6,H4,I9)

G3 2 (Phan 2's H5/6, I8)

I3 5 (C1,F2)

I1 4 (E2)

B2 2 (F1,G3) Hard part, Phantom 5,9 in D8,E8 because of 5,9 in row F. Thus, both 5,9 are blocked elsewhere in column 8. The 5 at C1 blocks all of row C, so phantom 5’s must exist at either A7 or A9, blocking row A elsewhere.

H6 5 (5's blocked by A?,C1,I3) Where can 5 go in column 6 ?? Only at H6. A? means the phantom 5 in either A7 or A9.

G7 5 (H6,I3)

A9 5 (A7 blocked by G7) Either A7 or A9 must be a 5, so G7 forces 5 at A9.

G4 1 (what's left in row G)

A5 1 (B9,C2,G4,E6)

H5 2 (I8) 2 in I8 blocks bottom row, forcing 2 at H5.

C9 2 (D7,I8)

I7 1 (B9,F8)

I6 3 (D5) 3 in D5 forces 3 at I6.

I5 9 (what's left in box)

I2 7 (what's left in row I)

H2 9 (what's left in box)

E5 5 (phan 5/9 pair in D4,E5) 9 in I5 eliminates the possibility of E5 being a 9, so it must be 5.

D4 9 (what's left in box)

D8 5 (what's left in row D)

E8 9 (what's left in row E)

B4 5 (A9,C1,E5,H6)

A4 7 (what's left in col 4)

B5 4 (what's left in col 5)

A6 2 (B2,C9)

C6 9 (what's left in box)

B3 7 (phan 4/7 pair in B3,C3) 4 in B5 blocks 4 in B3, so B3 must be 7.

C3 4 (what's left in col 3)

A7 9 (C6,E8,G9)

B1 9 (A7)

A1 3 (what's left in box)

A8 4 (what's left in row A)

B8 3 (what's left in row B)

C7 7 (what's left in box)

F7 3 (phan 3/7 in row F) C7 is 7, so F7 can NOT be 7 of the 3/7 pair. So, F7 must be 3.

F9 7 (what's left in box)

(row H completed by columns) A8,D9 give H7=4; C7,F9 give H8=7; B8,F7 give H9=3. Puzzles Solved.