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Thursday 19-Sep-2019

... by: numpl_npm

| _ 6 8 | _ _ 5 | _ 4 _ | A
| _ 9 7 | 8 _ 4 | 5 _ _ | B
| 5 4 _ | _ _ _ | _ _ 2 | C

| 4 8 9 | 6 _ _ | 7 _ _ | D
| 6 5 _ | _ 7 _ | _ 1 _ | E
| 7 _ _ | _ _ 3 | _ _ _ | F

| _ _ 5 | _ 4 6 | _ _ _ | G
| 8 _ 4 | 9 _ _ | 6 _ _ | H
| 9 7 6 | 5 _ _ | 4 _ _ | J

+7B3 and +8D2 => +3E3 +1C3 +1F2 +1G1 +1A7 +1B5 ... contadiction
So either -7B3 or -8D2.

Thursday 19-Sep-2019

... by: numpl_npm

BobW's 1st exocet leads to ...

| _ 6 8 | _ _ 5 | _ 4 _ | A
| _ 9 _ | 8 _ 4 | 5 _ _ | B
| 5 4 _ | _ _ _ | _ _ 2 | C

| 4 _ 9 | 6 _ _ | 7 _ _ | D
| 6 5 _ | _ 7 _ | _ 1 _ | E
| 7 _ _ | _ _ 3 | _ _ _ | F

| _ _ 5 | _ 4 6 | _ _ _ | G
| 8 _ 4 | 9 _ _ | 6 _ _ | H
| 9 7 6 | 5 _ _ | 4 _ _ | J

Almost MSLS found.

u 123 / n 456789 --- ( u 9 + n 7 ) - ( o 15 ) = 1
. _ _ _ . _ 3 _ . _ _ _ .
. _ _ _ . _ 2 2 . _ 3 3 .
. _ _ _ . _ 1 1 . _ 2 1 .
. _ _ _ . _ u u . _ u u .

| _ 6 8 | _ _ 5 | _ 4 _ | A
| _ 9 _ | 8 o 4 | 5 o o | B n 6 7?
| 5 4 _ | _ _ _ | _ _ 2 | C

| 4 _ 9 | 6 o o | 7 o o | D n 5 8?
| 6 5 _ | _ 7 _ | _ 1 _ | E
| 7 _ _ | _ _ 3 | _ _ _ | F

| _ _ 5 | _ 4 6 | _ _ _ | G
| 8 _ 4 | 9 o o | 6 o o | H n 5 7
| 9 7 6 | 5 o o | 4 o o | J n 8

It's not MSLS. But either (7 not in rowB) or (8 not in rowD).
In both case, it's MSLS and -123A5 -13C5.

Then +9A5 +6C5, and to the solution with basics.

Is this idea valid?

Tuesday 17-Sep-2019

... by: numpl_npm

Sorry
When +7H6 and +4D4 ...
-->
When +7H6 and +4E4 ...

Tuesday 17-Sep-2019

... by: numpl_npm

BobW's 1st exocet leads to ...

| _ 6 8 | _ _ 5 | _ 4 _ | A
| _ 9 _ | 8 _ 4 | 5 _ _ | B
| 5 4 _ | _ _ _ | _ _ 2 | C

| 4 _ 9 | 6 _ _ | 7 _ _ | D
| 6 5 _ | _ 7 _ | _ 1 _ | E
| 7 _ _ | _ _ 3 | _ _ _ | F

| _ _ 5 | _ 4 6 | _ _ _ | G
| 8 _ 4 | 9 _ _ | 6 _ _ | H
| 9 7 6 | 5 _ _ | 4 _ _ | J

Then (+7G4|+7H6 and +(2|4)E4) => In any case, +9E6 +6C5
And to the solution with basics.

When +7G4 => +7C6 +9E6
> +7A9 +7H8 +5H9 +8(E7|E9) +3D9 +3E3 +1C3 +1G1 +1J9 +1A7 +89C78 +6C5

When +7H6 and +2E4 => +3E3 +2J6 +8J5 +1J9 +3J8 +3D9 +13AC7 +8C8 +6C5
> +789G789 +2H8 +5D8 +5F5 +9E6

When +7H6 and +4D4 => +4F9 +6F8 +5F5 +9E6 +6C5

Monday 16-Sep-2019

... by: numpl_npm

BobW's 1st exocet ...

| _ _ 8 | a _ 5 | c 4 _ | A
| z 9 _ | 8 _ _ | 5 _ _ | B
| 5 _ _ | a _ _ | c _ 2 | C

| _ _ 9 | 6 b b | 7 d d | D
| 6 5 x | _ 7 _ | e 1 _ | E
| 7 _ x | _ _ 3 | e _ _ | F

| o o _ | _ 4 _ | _ _ _ | G
| 8 y _ | 9 _ _ | 6 _ _ | H
| 9 7 _ | 5 _ _ | 4 _ _ | J

1 in xx and 1 not in y --> 1 in oo, 1 in bb, 1 in aa, 1 in cc --> 1 in z
2 in xx and 2 not in y --> 2 in oo, 2 in ee, 2 in bb, 2 in aa --> 2 in z
3 in xx and 3 not in y --> 3 in oo, 3 in aa, 3 in dd, 3 in cc --> 3 in z
4 in xx --> y in 4

Monday 16-Sep-2019

... by: rybinpv

Hello!
3 6 8 2 9 5 1 4 7
2 9 7 8 1 4 5 3 6
5 4 1 3 6 7 8 9 2
4 1 9 6 8 2 7 5 3
6 5 3 4 7 9 2 1 8
7 8 2 1 5 3 9 6 4
1 2 5 7 4 6 3 8 9
8 3 4 9 2 1 6 7 5
9 7 6 5 3 8 4 2 1
time: 0:11:47.17
python: 3.7

Sunday 15-Sep-2019

... by: BobW

Examining this puzzle further, after the eliminations resulting from the first exocet 1234 EF3, the grid is as follows:


1 2 3 4 5 6 7 8 9
+---------------+---------------+-------------------+
A| 123 6 8 | 1237 1239 5 | 139 4 1379 |
B| 123 9 1237 | 8 1236 4 | 5 367 1367 |
C| 5 4 137 | 137 1369 179 | 1389 36789 2 |
+---------------+---------------+-------------------+
D| 4 1238 9 | 6 1258 128 | 7 2358 358 |
E| 6 5 23 | 24 7 289 | 2389 1 3489 |
F| 7 128 12 | 124 1589 3 | 289 25689 45689 |
+---------------+---------------+-------------------+
G| 123 123 5 | 1237 4 6 | 12389 23789 13789 |
H| 8 123 4 | 9 123 127 | 6 2357 1357 |
J| 9 7 6 | 5 1238 128 | 4 23 13 |
+---------------+---------------+-------------------+


At this point my solver identifies 14 prospective exocets.
These three each lead to a basics solution:
128 F23, D5 E7
123 J89, G1 H5
123 J89, G1 H6

These four are invalid, proven by contradiction:
123 AB1, E3 H2
1379 A79, B1 C6
123 GH2, A1 E3
1238 HJ5, A4 D6

The remaining seven do not lead to contradiction, but are insufficient to yield a solution using basics only:
123 AB1, E3 G2
1379 C46, B3 A7
1379 A79, B2 C4
123 EF3, A1 H2
123 EF3, B1 H2
2489 E46, D1 F7
123 GH2, A1 F3

Sunday 15-Sep-2019

... by: BobW

I've now found an alternative second exocet to the one found by SuDokuFan that also leads to a basics solution.
1st exocet as before 1234EF3, H2, B1.
2nd exocet: 123J89, G1, H6
Only one elimination: -7H6, but this is enough to reduce the puzzle to basics.

Sunday 15-Sep-2019

... by: SuDokuFan

Using two Exocets to achieve a basics only solution:
Use BobW’s Exocet as described below: 1234EF3, H2, B1.
Second Exocet: 1236 A12, C5, B8.
Eliminations: -2A12, -9C5, -7B89.
Solver places a 6 in cell A2- therefore one of (C5, B8) must contain 6- -6C8, -6B56, therefore +6C5, +6B9 (by Exocet definition), leads to +3B8, +3C4, +3A1, stte

Saturday 14-Sep-2019

... by: BobW

Further experimentation with my solver, and non-junior exocets, leads to a solution by assuming the exocet is valid until a contradiction proves it false.

Exocet (unproven):
Base: E3 F3 1234; Targets: B1: 1234 H2: 1234
Eliminations:
-4E3 -4F3 -4B1 -4H2 -6G2

This then leads to a solution involving 11 inference chains/nets.

Saturday 14-Sep-2019

... by: Frans Goosens

To David Filmer

Here an example of 2 solutions.
B8 and F3 in 12475 logical steps
F4 and F7 in 5219 logical steps



#365--------------------------Solution
008 005 040-------------368 295 147
090 800 500-------------297 814 536
500 000 002-------------541 367 892

009 600 700-------------419 682 753
650 070 010-------------653 479 218
700 003 000-------------782 153 964

000 040 000-------------125 746 389
800 900 600-------------834 921 675
970 500 400-------------976 538 421

Total solving time is : 120 sec.
Number of logical steps is : 12475




With trial and error

Combination F4=124 and F7=289

************************************************************
F4=1 F7=2 Wrong, Undo calculation

All reset to initial position

************************************************************
F4=1 F7=8 Wrong, Undo calculation

All reset to initial position

************************************************************
F4=1 F7=9 No solution, Fixed

Combination 2-digits cells

A7=1 No solution, Undo calculation
A7=3 Wrong, Undo calculation
A7=1 No solution, Fixed
A1=2 No solution, Undo calculation
A1=3 ( C7=8 ) Solved,


#365--------------------------Solution
008 005 040-------------368 295 147
090 800 500-------------297 814 536
500 000 002-------------541 367 892

009 600 700-------------419 682 753
650 070 010-------------653 479 218
700 003 000-------------782 153 964

000 040 000-------------125 746 389
800 900 600-------------834 921 675
970 500 400-------------976 538 421

Total solving time is : 55 sec.
Number of logical steps is : 5219

Saturday 14-Sep-2019

... by: James Havard

19 subs and 30 seconds. two subs for basic solution.
C3=1 and G2=2

Saturday 14-Sep-2019

... by: Algo

A4 =2
B3 =7

3.8s

Saturday 14-Sep-2019

... by: BobW

As usual, my solver was not able to find a solution without resorting to T&E.
So, it reverts to the best backdoor search:

With A4=2, it's now solvable with two inference chains:
G4=3? leads to contradiction. Therefore G4≠3
B1=4? leads to contradiction. Therefore B1≠4
Then basics to end.

With backdoor A4=2 and B1=2, it's solvable with singles only.


Experimenting with PM grid formatting (this is where the first inference chain, G4=3? is examined). I hope this works.

1 2 3 4 5 6 7 8 9
+------------------+----------------+--------------------+
A| 13 136 8 | 2 1369 5 | 139 4 7 |
B| 24 9 247 | 8 136 47 | 5 36 136 |
C| 5 1346 13467 | 37 1369 14679 | 1389 3689 2 |
+------------------+----------------+--------------------+
D| 1234 12348 9 | 6 258 28 | 7 2358 3458 |
E| 6 5 23 | 4 7 289 | 2389 1 389 |
F| 7 248 24 | 1 2589 3 | 289 25689 45689 |
+------------------+----------------+--------------------+
G| 123 1236 12356 | 37 4 12678 | 12389 235789 13589 |
H| 8 1234 12345 | 9 123 127 | 6 2357 135 |
J| 9 7 1236 | 5 12368 1268 | 4 238 138 |
+------------------+----------------+--------------------+

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