Some partial results to share
I am still concentrating on the second example
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214378695867295341935641872348962517591837264672514938123456789456789123789123456 morph of entry
2..........7.........64.8...4...2.......37...6.....9........7.........23.891..... known 17
No change so far for bands 1+2
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966164 valid sub puzzles 11 (1,083,938 in blue count)
5235 valid sub puzzles 10 clues
4 valid sub puzzles 9 clues
Let keep for later the count of valid 11 clues puzzles generated by the 5235+4 sub puzzles of lower size.
I introduced in the main road (valid sub
puzzles size 11) a filter based on the known valid band 3 possible sub puzzles
( 729 possibilities as of mladen's table in the first post).
The filter does not use directly that table, but a direct access to a three dimensions table of still valid cells after one, two or three cells have been assigned in band 3.
The main road run time falls to 10.5 seconds, always in parallel to 5 others batches (heavy processor consumption).
I had 371 039 puzzles to check (407 854 in the Gary McGuire' program, given by blue)
The next step will be to see how to handle the (5235 + 4) sub puzzles , with a possible bad surprise, but I show here after that we have room for optimism.
Other important steps will be
to optimize the start lot of UAs
to see what happens with a lower minimum of clues in band 3.
The expected distribution for solution grids having one or more 17 is for me the following
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6 93
5 7842
4 38127
3 238
46300
so the band3 with 6 clues is by far not the main case.
Regarding the processing of the (5235 + 4) sub puzzles, I am expecting things similar to that first case
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.1.3.......7..........4...234.9............6.........8 valid 10 clues sub grid for bands 1+2
still active UAs excluding pure band 3 UAs
(2 blank have been added to show the band 3
...............1.1.................................... ...............1.1......... i=0
........................11............................ ......11................... i=1
.....1.................................1.1............ ............11............. i=2
1............................................1.1...... .........1.1............... i=3
.........11..................................11....... ..................11....... i=4
......1.1.......................11.1.....11.1......... ........................1.1 i=5
........1..........11................................. ..1.....1.1......1..1....1. i=6
.............11....11................................. ..1.1...1.1...1..1......... i=7
......1.1......11..................................11. ........................111 i=8
..........1...1.....11..........................1.1... ...1.1....11............... i=9
............1.1.................1...............11.... ....11...............11.... i=10
A quick look to the band 3 in that table show a potential UA of degree 7
...............1.1.................................... ...............1.1......... i=0
........................11............................ ......11................... i=1
.....1.................................1.1............ ............11............. i=2
1............................................1.1...... .........1.1............... i=3
.........11..................................11....... ..................11....... i=4
......1.1.......................11.1.....11.1......... ........................1.1 i=5
............1.1.................1...............11.... ....11...............11.... i=10
So either we can solve band 3 with that small list of UAas,with no more cell in bands 1+2
or we take one clue in the left part of the table and finish as in the main road, 6 clues in band3
This is something that blue and I already used with success in the X+Y+27 post processing.