The New Sudoku Players' Forum

by **ArkieTech** » Tue Oct 03, 2017 2:00 am

Code: Select all: *-----------* |.54|9.3|...| |...|...|...| |6.7|.8.|..3| |---+---+---| |..3|5.8|...| |..6|.3.|8..| |...|4.2|5..| |---+---+---| |2..|.1.|6.8| |...|...|...| |...|8.4|12.| *-----------*

by **Leren** » Tue Oct 03, 2017 3:49 am

Code: Select all: *------------------------------------------------------* | 18 5 4 | 9 27 3 | 27 168 16 | | 189 3 2 | 167 4 1567 | 79 15789 1579 | | 6 19 7 | 12 8 15 |*249 *1459 3 | |----------------+---------------+---------------------| |f1479 *12479 3 | 5 679 8 | 79-4 1679-4 124679 | | 5 *2479 6 | 17 3 179 | 8 479 2479 | | 179 1789 18 | 4 679 2 | 5 3 1679 | |----------------+---------------+---------------------| | 2 *47 59 | 3 1 79 | 6 *4579 8 | | 147 1478 18 | 267 2579 679 | 3 4579 4579 | | 3 6 59 | 8 579 4 | 1 2 579 | *------------------------------------------------------*

Finned Franken Swordfish in 4's r37b4 c278 with a fin Cell r4c1 => - 4 r4c78; lclste

Leren

by **JC Van Hay** » Tue Oct 03, 2017 6:41 am

Or the solutions of {4C19} -> -{4r4c78}

by **Cenoman** » Tue Oct 03, 2017 10:35 am

Code: Select all: +----------------------+----------------------+-------------------------+ | 18 5 4 | 9 27 3 | 27 168 16 | | 189 3 2 | 167 4 1567 | 79 15789 1579 | | 6 19 7 | 12 8 15 | 249 1459 3 | +----------------------+----------------------+-------------------------+ | a1479 12479 3 | 5 679 8 | 79-4 1679-4 d124679 | | 5 2479 6 | 17 3 179 | 8 479 d2479 | | 179 1789 18 | 4 679 2 | 5 3 1679 | +----------------------+----------------------+-------------------------+ | 2 47 59 | 3 1 79 | 6 4579 8 | | b147 1478 18 | 267 2579 679 | 3 4579 c4579 | | 3 6 59 | 8 579 4 | 1 2 579 | +----------------------+----------------------+-------------------------+

X-wing (4)r4c1=r8c1-r8c9=r45c9 => -4 r4c78; lclste (JC's solution)

And now, stte finish:

Code: Select all: 4r7c2 4r7c8 4r3c8 4r3c7 2r3c7 2r3c4 2r1c5 7r1c5 7r9c5 7r9c9 7r7c6 9r7c6 9r5c6 7r5c46 ------------------------------ (ALS r5c46) 4r8c9 4r45c9 4b6p15 7b6p15 ---------------- (ALS b6p15) 7r6c2 7r6c9 7r6c5 7r6c1 7r7c8 7r8c8 7r5c8 7r4c8 7r2c8 8r2c8 8r1c8 6r1c8 6r1c9 6r6c9 6r6c5 7r12c7 7r4c7 6r4c5 7r4c5 9r4c5 7r2c7 9r2c7 9r3c78 9r3c2 9r6c1 9r4c1 9r2c1 TM demonstrating (4)r7c2==(7)r7c6==(7)r6c2 => -7r7c2; stte

Cenoman

by **pjb** » Wed Oct 04, 2017 5:38 am

Using POM analysis, all patterns of 4 include r3c7 so this can be made equal to 4; lclste
Phil

October 3, 2017