RW is right... You need only 2 locked candidates/box-line interactions to reduce the puzzle to
hidden singles only...
Pat is right too... If you considered a puzzle with
naked singles and
hidden singles left as being "solved" then just 1 locked candidate move is needed...
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7 in b7 locked in c1 => r124c1<>7
Afterwards, only
naked singles and
hidden singles are required to reach the solution...
But if like RW you don't like pencilmarks, you might not want to do any
naked single... So one more move to do:
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7 in b3 locked in r3 => r3c235<>7
Now you can reach the solution even if the only one move you ever know is called "
hidden single"...
And to start this puzzle without any pencilmarks:
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xx 6 5 | . . 4 | 9 1 .
xx . 4 | 9 . . | 3 5 .
3 xx xx | 5 xx 8 | ## ## .
----------+-----------+----------
xx . . | . 4 . | . . 3
5 . . | . 8 . | . . 7
2 . . | . 9 . | . . 1
----------+-----------+----------
. . . | 7 . 2 | 1 . 5
## 5 2 | 4 . . | 8 . 9
## 1 3 | 8 5 9 | . 6 .
Just notice the ## cells contain the only possible cells holding the 7 within the corresponding box... As a result, all the xx cells cannot hold a 7... Which leaves only 1 possible cell to hold 7 in box 1... Afterwards, you can fill in more cells in box 1 because there is only one possible placement for some digits within that box...