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Tuesday 20-Aug-2019

... by: Cenoman

Sorry typo in MSLS eliminations: read 7A5 instead of 7A6

Monday 19-Aug-2019

... by: Cenoman

Basics: (5)B9=B45-C6=GJ6-H45=H9 =>-5GJ9
In addition to numpl_npm's MSLS, there an exocet (proof by "no solution") (2357)HJ5, E4, A6 =>-9E4, -14A6 (non-base digits in target cells)
After the 20 MSLS eliminations (48A3, 7A6, 78A7, 8C1, 5C9, 7D7, 49F13, 4F9, 4G3, 45G9, 36B4, 2H4, 3J6) and basics, note E4=5. Therefore 2 is a false base digit (no 2 left in a target cell) =>-2H5, G4=2.
Then puzzle solved by X-wing (3)C46=C9-G9=G6 =>-3A6; singles to the end.

Monday 19-Aug-2019

... by: James Havard

3 minutes and 151 subs to a solution. I couldn't find 2 subs for a basic solution.
With G2=4,J6=4, and J7=3 gives a basic solve.

Monday 19-Aug-2019

... by: Frans Goosens

With trial and error

Combination C2=128 and E7=127

************************************************************
C2=1 E7=1 No solution, Fixed

Combination 2-digits cells

A2=4 Wrong, Undo calculation
A2=8 ( F3=1 ) Solved,


#361--------------------------Solution
500 000 009-------------583 417 269
000 002 010-------------694 832 715
007 090 400-------------217 695 483

305 040 008-------------325 741 698
060 008 030-------------469 528 137
070 000 500-------------871 369 542

700 080 900-------------746 283 951
030 006 000-------------132 956 874
000 100 020-------------958 174 326

Total solving time is : 90 sec.
Number of logical steps is : 8139

Sunday 18-Aug-2019

... by: numpl_npm

u 1236 / c 45789
. _ _ _ . 6 _ _ . _ _ _ .
. _ 2 _ . 3 _ 3 . _ _ _ .
. _ 1 _ . 2 _ 1 . _ 6 _ .
. _ u _ . u _ u . _ u _ .

| 5 o _ | o _ o | _ o 9 | A c 478
| _ _ _ | _ _ 2 | _ 1 _ | B
| _ o 7 | o 9 o | 4 o _ | C c 58

| 3 o 5 | o 4 o | _ o 8 | D c 79
| _ 6 _ | _ _ 8 | _ 3 _ | E
| _ 7 _ | o _ o | 5 o _ | F c 49 (8 in F13)

| 7 o _ | o 8 o | 9 o _ | G c 45
| _ 3 _ | _ _ 6 | _ _ _ | H
| _ _ _ | 1 _ _ | _ 2 _ | J

MSLS found: rACDFG * c2468 --- o 4 * 5 - 1 = u 8 + c 11

Saturday 17-Aug-2019

... by: BobW

With the single cell backdoor H5=5, it's solvable with four inference chains/nets.

With the two cell backdoor A3=3 H5=5, it's solvable with basic methods.

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