Last candidate 9 in D1 changed to solution

Last candidate 3 in D3 changed to solution

Last candidate 6 in D5 changed to solution

4 in D9 is last in row/box
5 in A3 is last in row
5 in C9 is last in column/box
5 in H5 is last in row/box
7 in E2 is last in column/box
8 in G1 is last in row
8 in J8 is last in column/box
9 in H3 is last in column/box

Naked Pair 2/7 on cells {D4|D6} means
- 2 can be removed from E6
- 7 can be removed from F4
- 7 can be removed from F6

Naked Pair 1/5 on cells {E1|F1} means
- 1 can be removed from B1
- 1 can be removed from C1
- 1 can be removed from H1

Naked Pair 1/4 on cells {J3|J5} means
- 1/4 can be removed from J6
- 1/4 can be removed from J7
- 1 can be removed from J9

Pair: between Box 7 and Row H:
- 2 can be removed from H7
- 2 can be removed from H8
- 2 can be removed from H9

If +4[B1] then -4[B5] forcing +4[J5] which removes all 4s from Box 7, so B1 cannot be 4

If +6[H4] then -6[A4] forcing +6[A2] which removes all 6s from Box 7, so H4 cannot be 6

Uniqueness Type 2 (Col) at EF16

The 3 in the roof cells (orange) stops a Deadly Pattern forming, so it must be placed in either E6 or F6. So..
The 3 in the roof cells (orange) stops a Deadly Pattern forming, so it must be placed in either E6 or F6. So..
The 3 in the roof cells (orange) stops a Deadly Pattern forming, so it must be placed in either E6 or F6. So..
The 3 in the roof cells (orange) stops a Deadly Pattern forming, so it must be placed in either E6 or F6. So..
The 3 in the roof cells (orange) stops a Deadly Pattern forming, so it must be placed in either E6 or F6. So..
The 3 in the roof cells (orange) stops a Deadly Pattern forming, so it must be placed in either E6 or F6. So..
- 3 can be removed from B6
- 3 can be removed from C6
- 3 can be removed from G6
- 3 can be removed from E5
- 3 can be removed from F4
- 3 can be removed from F5

Hidden Pair 3/5 in Box 5, on cells {E6|F6}
- removes 1 from E6
- removes 1 from F6

Pair: between Box 8 and Col 4:
- 3 can be removed from A4
- 3 can be removed from C4

3 in C8 is last in row

Naked Triple 2/6/7 on cells {H8|J7|J9} means
- 6 can be removed from G9
- 7 can be removed from H7
- 6/7 can be removed from H9

XYZ-Wing pattern. Hinge: G6 (1/4/6), wings G2 J5, therefore
- 1 can be removed from G4

Type 1: removing 1 at B2 because of AB25 and two strong links on 3

If +1[A5] then -1[J5] forcing +1[J3] which removes all 1s from Box 1, so A5 cannot be 1
Last candidate 3 in A5 changed to solution

3 in B2 is last in row/column/box

WXYZ Wing Type 1: The four cells {C4,C1,C6,F4} contain in total 1/2/4/8 with the hinge on C4.
Non-Restricted candidate 1 determines the 'wing' cells (in yellow) and one of those 1s must be true. Therefore...
- 1 can be removed from A4

2 cannot be set in the yellow cells {C7,C8,C9} and with the Remote Pair in A4 and B1 it means:
- 6 can be removed from A2
- 6 can be removed from B6

6 in A4 is last in row/box
6 in B1 is last in row/box

Last candidate 3 in G4 changed to solution

length=5, chain ends: G2 and J6
This proves 6 is the solution at one or possibly both ends of the chain
-6[G2]+1[G2]-1[J3]+4[J3]-4[J5]+1[J5]-1[H4]+7[H4]-7[J6]+6[J6]
- 6 can be removed from G6

6 in G2 is last in row
6 in J6 is last in column/box

6 in F9 is last in column
6 in H8 is last in row/box
7 in D6 is last in column
7 in H4 is last in box

Last candidate 2 in D4 changed to solution

W-Wing (single). Either there is a 9 in E5 or in G9. They are connected through 1 in J5/G6, therefore
- 9 can be removed from E9

(RULE 4): Uncoloured candidate 1 in G7 can see two different coloured candidates of 1 elsewhere.

length=5, chain ends: A7 and H9
This proves 1 is the solution at one or possibly both ends of the chain
-1[A7]+2[A7]-2[J7]+7[J7]-7[J9]+2[J9]-2[E9]+3[E9]-3[H9]+1[H9]
- 1 can be removed from B9
- 1 can be removed from H7

(RULE 4): Uncoloured candidate 1 in B5 can see two different coloured candidates of 1 elsewhere.

length=7, chain ends: A7 and E9
This proves 2 is the solution at one or possibly both ends of the chain
-2[A7]+1[A7]-1[A2]+2[A2]-2[C1]+4[C1]-4[H1]+2[H1]-2[H2]+1[H2]-1[H9]+3[H9]-3[E9]+2[E9]
- 2 can be removed from B9

length=6, chain ends: B9 and F8
This proves 7 is the solution at one or possibly both ends of the chain
-7[B9]+9[B9]-9[G9]+1[G9]-1[H9]+3[H9]-3[E9]+2[E9]-2[E8]+9[E8]-9[F8]+7[F8]
- 7 can be removed from B8

7 in F8 is last in column

Naked Triple 3/4/9 on cells {F7|G7|H7} means
- 9 can be removed from B7

length=6, chain ends: B8 and G9
This proves 9 is the solution at one or possibly both ends of the chain
-9[B8]+2[B8]-2[E8]+9[E8]-9[E5]+1[E5]-1[J5]+4[J5]-4[G6]+1[G6]-1[G9]+9[G9]
- 9 can be removed from B9
Last candidate 7 in B9 changed to solution

Last candidate 2 in J9 changed to solution

Last candidate 3 in E9 changed to solution
Last candidate 7 in J7 changed to solution

Last candidate 5 in E6 changed to solution
Last candidate 9 in F7 changed to solution
Last candidate 1 in H9 changed to solution

Last candidate 1 in E1 changed to solution
Last candidate 2 in E8 changed to solution
Last candidate 3 in F6 changed to solution
Last candidate 4 in G7 changed to solution
Last candidate 9 in G9 changed to solution
Last candidate 2 in H2 changed to solution

Last candidate 1 in A2 changed to solution
Last candidate 9 in B8 changed to solution
Last candidate 9 in E5 changed to solution
Last candidate 5 in F1 changed to solution
Last candidate 1 in G6 changed to solution
Last candidate 4 in H1 changed to solution
Last candidate 3 in H7 changed to solution

Last candidate 2 in A7 changed to solution
Last candidate 2 in C1 changed to solution
Last candidate 1 in J3 changed to solution
Last candidate 4 in J5 changed to solution

Last candidate 8 in B5 changed to solution
Last candidate 1 in B7 changed to solution
Last candidate 4 in C6 changed to solution

Last candidate 4 in B3 changed to solution
Last candidate 2 in B6 changed to solution
Last candidate 8 in C3 changed to solution
Last candidate 1 in C4 changed to solution
Last candidate 1 in F5 changed to solution

Last candidate 8 in F4 changed to solution