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In the following sudoku the numbers on the borders indicate the first number less than 6 seen in that row/column. So for example, the 4 at the bottom means that the first number less than 6 seen from the bottom of column 8 is a 4. This is a link to the sudoku in Google Sheets. Enjoy!

grid to solve

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0
2
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645973218
712458396
893126754
156832479
237649581
984517632
329784165
461395827
578261943 enter image description here

The only places in R1 that can hold a 1, 2, or 3 are R1C8=1, R1C7=2, R1C6=3. (Putting those digits anywhere else in the row would contradict the numbers outside). Similarly, R7C1=3, R5C1=2, R9C1=5, R9C8=4, R9C9=3, R7C9=5, R5C9=1
6....3218
.........
8........
...8...7.
2..6.9..1
.....7...
3.9.....5
.........
5......43
enter image description here

The 1 in row 2 can only be in columns 1,2, or 3 (only 7 or 9 can be left of it); so the 1 in R1-3C4-6 must be in R3C3 (or it contradicts the outside clues in C5, C6).
The 1 on R4-6C4-6 can only be at R6C5 (if it's in R4, it's too high for the outside clues; there's already 1s in R5 and C4)
The 1 in R7 can't be in C2 (it's impossible to place a 1 in C3), C4 or C5 (already a 1 in those columns), C6 (it would require too many 6-9 in that minibox) or C8 (already a 1 in those columns). R7C7 = 1
R7-9C5 are 6,7,8, or 9 (for the outside column clue); at most one of R7-9C4 is 6-9. The 2 in C4 can't be in R1-7 (would require too many 6-9 in that column for the outside clue) or R8 (too far left for outside row clue). R9C4=2
The 2 in R7 can't be in C4-6 (already a 2 in that minibox), C7-9 (it would be impossible to place the 2 in R8); R7C2=2.
6....3218
.........
8..1.....
...8...7.
2..6.9..1
....17...
329...1.5
.........
5..2...43
enter image description here

The only place 4 can go in R7 is R7C6 (C4 is too low for outside column, C5 would contradict outside column, theres a 4 in R7-9C7-9)
The only place 4 can go in R4-6C4-6 in R5C5 (R4C5 is too far left; there's a 4 in C6; too low in R6C4).
R7C4 must be 6-9 (1-5 already in that row) R7-9C5 must be 6-9 (or it would contradict outside column clue); R9C6 is not 6-9, so be outside column clue, R9C6=1.
1 in C3 must be R8C3 (R1-4 is too high for outside column, other squares have 1 in row already)
4 in R7-9C1-3 is R8 (anywhere else would contradict outside column clue).
3 in R7-9C4-6 in R8C4 (anywhere else would contradict outside clue, or there's already a 3 in that row/column)
5 in R7-9C4-6 in R8C6 (anywhere else would contradict outside clue, or there's already a 5 in that row/column)
6....3218
.........
8..1.....
...8...7.
2..649..1
....17...
329..41.5
4.13.5...
5..2.1.43
enter image description here

The remaining cells in R4-6C4-6 must be R4C6=2, R6C4=5, R4C5=2
The only way to fill C6 in R2C6=8, R3C6=6
The only way to fill R7C4=7
The only way to fill R6C1=9, R4C1=1, R2C1=7
The only way to fill R1C3=5 (a 4 would contradict outside column clue)
R1C2=4 (outside row clue)
4 in R6 must be C3 (already a 4 in C2, anywhere else too far right for outside row clue)
Only way to fill C4 is R1C4=9, R2C4=4
Only way to fill R1C5=7. Outside column clue forces R2C5=5, R3C5=2
645973218
7..458...
8..126...
1..832.7.
2..649..1
9.4517...
3297.41.5
4.13.5...
5..2.1.43
enter image description here

Only place for 2 in R2 is R2C3 (R2C2 contra outside row clue, already a 2 in R1-3C7-9).
Outside row clue forces R2C2=1
R3C3=3, R3C2=9, R3C8=5.
Outside row clue R3C9=4
R3C7=7
4 in R4 must be C7 (already a 4 in C9, anywhere else too far left for outside row clue)
R4C3=6, R4C9=9 (outside row clue), R4C2=5
R2C9=6, R6C9=2, R8C9=7
enter image description here 2 in R8 must be R8C8 (already a 2 in C7, everywhere else is left of 5)
5 in C7 must be R5
9 in C8 must be R2. R2C7=3.
3 in R4-6C1-3 must be R5C2 (already in C3, R6 would contradict outside row clue). R6C2=8, R5C3=7, R9C3=8, R8C2=6, R9C2=7
Last 3 must be at R6C8
R5C8=8, R6C7=6, R7C8=6, R7C5=8, R8C5=9, R8C7=8, R9C5=6, R9C7=9
enter image description here

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    $\begingroup$ Could you provide an explanation of your solve path, or at least of the major deductions? Perhaps illustrated with pictures of intermediate steps? That is generally expected of answers to [grid-deduction] questions. $\endgroup$ – bobble Dec 5 '20 at 20:57
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    $\begingroup$ I wasn't keeping track of my deductions. I'll try to write up a solution tomorrow. $\endgroup$ – Mark Tilford Dec 6 '20 at 2:10
  • 1
    $\begingroup$ Okay, solution path is written up. $\endgroup$ – ralphmerridew Dec 6 '20 at 14:08

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