The New Sudoku Players' Forum

by **200e200w** » Sat Feb 17, 2018 11:41 am

Code: Select all: . . .|1 . .|. . . . 2 .|. 3 .|. 4 . 4 . 5|. . .|6 . . -----+-----+----- 7 . .|5 . 6|. 1 . . . .|. 8 .|. . . . 5 .|2 . 3|. . 8 -----+-----+----- . . 4|. . .|9 . 3 . 9 .|. 2 .|. 8 . . . .|. . 7|. . .

200e200w

by **ArkieTech** » Sat Feb 17, 2018 12:04 pm

Code: Select all: *-----------------------------------------------------------------------------* | 3689 3678 36789 | 1 456-79 2458-9 | 23578 23579 2579 | | 1689 2 16789 |*6789 3 58-9 | 1578 4 1579 | | 4 1378 5 |*789 a79 28-9 | 6 2379 1279 | |-------------------------+-------------------------+-------------------------| | 7 348 2389 | 5 49 6 | 234 1 249 | | 12369 1346 12369 | 479 8 149 | 23457 235679 245679 | | 169 5 169 | 2 1479 3 | 47 679 8 | |-------------------------+-------------------------+-------------------------| | 12568 1678 4 |b68 156 158 | 9 2567 3 | | 1356 9 1367 | 34-6 2 145 | 1457 8 14567 | | 123568 1368 12368 | 349-68 14569 7 | 1245 256 12456 | *-----------------------------------------------------------------------------* [(79=68)b2p478-(68=79)r237c4]-79b2p2369,68r89c4; ste

by **SteveG48** » Sat Feb 17, 2018 3:36 pm

ArkieTech wrote:
Code: Select all
*-----------------------------------------------------------------------------* | 3689 3678 36789 | 1 456-79 2458-9 | 23578 23579 2579 | | 1689 2 16789 |*6789 3 58-9 | 1578 4 1579 | | 4 1378 5 |*789 a79 28-9 | 6 2379 1279 | |-------------------------+-------------------------+-------------------------| | 7 348 2389 | 5 49 6 | 234 1 249 | | 12369 1346 12369 | 479 8 149 | 23457 235679 245679 | | 169 5 169 | 2 1479 3 | 47 679 8 | |-------------------------+-------------------------+-------------------------| | 12568 1678 4 |b68 156 158 | 9 2567 3 | | 1356 9 1367 | 34-6 2 145 | 1457 8 14567 | | 123568 1368 12368 | 349-68 14569 7 | 1245 256 12456 | *-----------------------------------------------------------------------------* [(79=68)b2p478-(68=79)r237c4]-79b2p2369,68r89c4; ste

Hi, Dan. I understand the logic of your solution and the sets involved, but I'm confused about the notation. If I expand your chain, I get:

(79)b2p478 = (68)b2p478 - (68)r237c4 = (79)r237c4, or, considering the position of the digits:

(79)b2p478 = (68)r23c4 - (68)r237c4 = (79)r237c4 .

The center terms just don't parse for me, since they seem to be saying that if both 6 and 8 are true in r23c4 then they're not both true in r237c4 .

by **Cenoman** » Sat Feb 17, 2018 5:08 pm

After dan's Sue de Coq, a lot of lclste finish. Let's try:

Code: Select all: +--------------------------+--------------------------+----------------------------+ | 3689 3678 36789 | 1 45679 24589 | 23578 23579 2579 | | 1689 2 16789 | 6789 3 589 | 1578 4 1579 | | 4 1378 5 | 789 c79 289 | 6 2379 1279 | +--------------------------+--------------------------+----------------------------+ | 7 348 2389 | 5 d49 6 | 234 1 249 | | 12369 1346 12369 | aE479 8 149 | 23457 235679 245679 | | 169 5 169 | 2 b1479 3 | 47 679 8 | +--------------------------+--------------------------+----------------------------+ | 12568 1678 4 | 68 156 158 | 9 2567 3 | | 1356 9 1367 | F34-6 2 145 | 1457 8 14567 | | 123568 1368 12368 | F34-689 14569 7 | 1245 256 12456 | +--------------------------+--------------------------+----------------------------+

[(7)r5c4 = r6c5 - (7=9)r3c5 - (9=4)r4c5] - (4)r5c4 = (34)r89c4 => -689 r89c4; stte
Note: -4 r5c4; lclste

by **ArkieTech** » Sat Feb 17, 2018 6:20 pm

SteveG48 wrote:
ArkieTech wrote:
Code: Select all
*-----------------------------------------------------------------------------* | 3689 3678 36789 | 1 456-79 2458-9 | 23578 23579 2579 | | 1689 2 16789 |*6789 3 58-9 | 1578 4 1579 | | 4 1378 5 |*789 a79 28-9 | 6 2379 1279 | |-------------------------+-------------------------+-------------------------| | 7 348 2389 | 5 49 6 | 234 1 249 | | 12369 1346 12369 | 479 8 149 | 23457 235679 245679 | | 169 5 169 | 2 1479 3 | 47 679 8 | |-------------------------+-------------------------+-------------------------| | 12568 1678 4 |b68 156 158 | 9 2567 3 | | 1356 9 1367 | 34-6 2 145 | 1457 8 14567 | | 123568 1368 12368 | 349-68 14569 7 | 1245 256 12456 | *-----------------------------------------------------------------------------* [(79=68)b2p478-(68=79)r237c4]-79b2p2369,68r89c4; ste

Hi, Dan. I understand the logic of your solution and the sets involved, but I'm confused about the notation. If I expand your chain, I get:

(79)b2p478 = (68)b2p478 - (68)r237c4 = (79)r237c4, or, considering the position of the digits:

(79)b2p478 = (68)r23c4 - (68)r237c4 = (79)r237c4 .

The center terms just don't parse for me, since they seem to be saying that if both 6 and 8 are true in r23c4 then they're not both true in r237c4 .

I don't understand your expansion

you have two three cell groupings of 68 and 79

remove the 7 or 9 you have 68 in b2p578 ((79=68)b2p478)
remove the 6 or 8 you have 79 in r237c4 ((68=79)r237c4)

what have I missed?

by **Leren** » Sat Feb 17, 2018 7:38 pm

Code: Select all: *---------------------------------------------------------------* | 3689 3678 36789 | 1 45679 24589 | 23578 23579 2579 | | 1689 2 16789 |a6789c 3 589 | 1578 4 1579 | | 4 1378 5 |a789c b79 289 | 6 2379 1279 | |-------------------+---------------------+---------------------| | 7 348 2389 | 5 b49 6 | 234 1 249 | | 12369 1346 12369 | 479c 8 149 | 23457 235679 245679 | | 169 5 169 | 2 1479 3 | 47 679 8 | |-------------------+---------------------+---------------------| | 12568 1678 4 |a68c 156 158 | 9 2567 3 | | 1356 9 1367 | 346 2 145 | 1457 8 14567 | | 123568 1368 12368 | 3468-9 14569 7 | 1245 256 12456 | *---------------------------------------------------------------*

ALS XY Wing: (9=7) r237c4 - (7=4) r34c5 - (4=9) r2357c4 => - 9 r4c9; stte

Leren

by **SteveG48** » Sat Feb 17, 2018 9:31 pm

ArkieTech wrote:I don't understand your expansion

you have two three cell groupings of 68 and 79

remove the 7 or 9 you have 68 in b2p578 ((79=68)b2p478)
remove the 6 or 8 you have 79 in r237c4 ((68=79)r237c4)

what have I missed?

Your first term says that you have both 6 and 8 in the first cell grouping. How does that show that you don't have both 6 and 8 in the second overlapping grouping? The second grouping contains that same cells in which the first term shows that they are both true.

As I said, I understand the logic. I've never quite figured out how to write a Sue de Coq as a single AIC.

by **pjb** » Sat Feb 17, 2018 9:47 pm

Code: Select all: 3689 3678 36789 | 1 456-79 2458-9 | 23578 23579 2579 1689 2 16789 |b6789 3 58-9 | 1578 4 1579 4 1378 5 |b789 a79 28-9 | 6 2379 1279 ------------------------+----------------------+--------------------- 7 348 2389 | 5 49 6 | 234 1 249 12369 1346 12369 | 479 8 149 | 23457 235679 245679 169 5 169 | 2 1479 3 | 47 679 8 ------------------------+----------------------+--------------------- 12568 1678 4 |b68 156 158 | 9 2567 3 1356 9 1367 | 34-6 2 145 | 1457 8 14567 123568 1368 12368 | 349-68 14569 7 | 1245 256 12456

Another way to view it:
1-3 Double ALS at r3c5 and r237c4, X-Z = 7,9 => -79 r1c5, -9 r123c6, -6 r8c4, -68 r9c4; stte

Phil

The New Sudoku Players' Forum

200e200w's One-Trick Pony #21

200e200w's One-Trick Pony #21

Re: 200e200w's One-Trick Pony #21

Re: 200e200w's One-Trick Pony #21

Re: 200e200w's One-Trick Pony #21

Re: 200e200w's One-Trick Pony #21

Re: 200e200w's One-Trick Pony #21

Re: 200e200w's One-Trick Pony #21

Re: 200e200w's One-Trick Pony #21