The New Sudoku Players' Forum

by **urhegyi** » Sun Feb 28, 2021 1:55 am

Can anyone find a neat solution not necessary in one step.(perhaps as a bonus)

: 02-03-21-extreme-plus.png (14.8 KiB) Viewed 1029 times

Code: Select all: 39.2.57.......7...4....8.....4.5.2.15.....3..9.23..4...6.....82.......7..4.6.....

by **pjb** » Sun Feb 28, 2021 4:22 am

A mix of different moves, starting with:

Code: Select all: 3 9 8 | 2 14 5 | 7 14 6 6 25 15 | 149 3 7 | 8 124 59 4 257 157 | 19 6 8 | 159 123 359 ------------------------+----------------------+--------------------- 78 3 4 | 78 5 9 | 2 6 1 5 178 6 | 1478 12478 124 | 3 9 78 9 178 2 | 3 178 6 | 4 5 78 ------------------------+----------------------+--------------------- 17 6 3579 | 1457 1479 134 | 159 8 2 128 58 359 | 158 1289 123 | 6 7 4 1278 4 579 | 6 12789 12 | 159 13 359

1. Double ALS at r9c6 and r8c23456, with X-Z values 1 and 2 => -1 r7c456, -1 r9c5, -2 r9c5, -8 r8c1,
2. BUG-lite (type 1) of 178 at r5c259, r6c259 => -1 r5c5, -7 r5c5, -8 r5c5
3. (9=5)r2c9 - (5=9)r3c47 => -9 r3c9
4. (1=7)r6c5 - (7)r45c4 = (7-5)r7c4 = (5)r8c4 - (5=8*)r8c2 - (8)r6c2 = (8)r6c9 - (8=7)r5c9 - (7|8*=1)r5c2 => -1 r5c46, r6c2; stte

Phil

Website · by **denis_berthier** » Sun Feb 28, 2021 5:25 am

.

6 singles to the following

Code: Select all: ;;; Resolution state RS0: 3 9 8 2 14 5 7 14 6 126 125 156 149 13469 7 8 123459 359 4 1257 1567 19 1369 8 159 12359 359 678 3 4 789 5 69 2 69 1 5 178 167 14789 1246789 12469 3 69 789 9 178 2 3 1678 16 4 56 578 17 6 13579 14579 13479 1349 159 8 2 128 1258 1359 1589 12389 1239 6 7 4 1278 4 13579 6 123789 1239 159 1359 359

whip[1]: c6n3{r9 .} ==> r9c5 ≠ 3, r7c5 ≠ 3, r8c5 ≠ 3
whip[1]: c6n6{r6 .} ==> r6c5 ≠ 6, r5c5 ≠ 6
whip[1]: b9n3{r9c9 .} ==> r9c6 ≠ 3, r9c3 ≠ 3

Code: Select all: ;;; Resolution state RS1: 3 9 8 2 14 5 7 14 6 126 125 156 149 13469 7 8 123459 359 4 1257 1567 19 1369 8 159 12359 359 678 3 4 789 5 69 2 69 1 5 178 167 14789 124789 12469 3 69 789 9 178 2 3 178 16 4 56 578 17 6 13579 14579 1479 1349 159 8 2 128 1258 1359 1589 1289 1239 6 7 4 1278 4 1579 6 12789 129 159 1359 359

naked-pairs-in-a-column: c8{r4 r5}{n6 n9} ==> r9c8 ≠ 9, r6c8 ≠ 6, r3c8 ≠ 9, r2c8 ≠ 9
singles ==> r6c8 = 5, r6c6 = 6, r4c6 = 9, r4c8 = 6, r5c8 = 9, r5c3 = 6, r2c1 = 6, r3c5 = 6, r2c5 = 3
whip[1]: c5n9{r9 .} ==> r7c4 ≠ 9, r8c4 ≠ 9
whip[1]: c1n1{r9 .} ==> r7c3 ≠ 1, r8c2 ≠ 1, r8c3 ≠ 1, r9c3 ≠ 1
whip[1]: c3n1{r3 .} ==> r2c2 ≠ 1, r3c2 ≠ 1
whip[1]: c1n2{r9 .} ==> r8c2 ≠ 2

Code: Select all: ;;; Resolution state RS2: 3 9 8 2 14 5 7 14 6 6 25 15 149 3 7 8 124 59 4 257 157 19 6 8 159 123 359 78 3 4 78 5 9 2 6 1 5 178 6 1478 12478 124 3 9 78 9 178 2 3 178 6 4 5 78 17 6 3579 1457 1479 134 159 8 2 128 58 359 158 1289 123 6 7 4 1278 4 579 6 12789 12 159 13 359

From this point on (which some would consider as "no-step"), there are two possibilities:

1) Using SudoRules standard simplest-first strategy, requiring only very simple short reversible chains:
finned-x-wing-in-rows: n8{r9 r4}{c1 c5} ==> r6c5 ≠ 8, r5c5 ≠ 8
whip[1]: c5n8{r9 .} ==> r8c4 ≠ 8
biv-chain[3]: r7c1{n1 n7} - r4n7{c1 c4} - r6c5{n7 n1} ==> r7c5 ≠ 1
biv-chain[3]: r4n7{c1 c4} - b5n8{r4c4 r5c4} - r5c9{n8 n7} ==> r5c2 ≠ 7
biv-chain-rc[3]: r5c2{n1 n8} - r8c2{n8 n5} - r8c4{n5 n1} ==> r5c4 ≠ 1
biv-chain[3]: r9n8{c5 c1} - r8c2{n8 n5} - r8c4{n5 n1} ==> r9c5 ≠ 1
z-chain-rc[3]: r3c4{n9 n1} - r3c7{n1 n5} - r2c9{n5 .} ==> r3c9 ≠ 9
z-chain[3]: r5n2{c5 c6} - b5n1{r5c6 r6c5} - r1c5{n1 .} ==> r5c5 ≠ 4
z-chain-bn[3]: b2n4{r1c5 r2c4} - b5n4{r5c4 r5c6} - b5n1{r5c6 .} ==> r1c5 ≠ 1
Singles+ 1 whip[1] to the end

2) Looking for a 1-step solution
There are:
2 BRT-anti-backdoors: n8r4c4 n7r4c1
3 W1-anti-backdoors: n8r4c4 n7r4c1 n7r3c2
6 S-anti-backdoors: n8r9c1 n5r8c2 n5r7c4 n8r4c4 n7r4c1 n7r3c2 , 4 of which give a 1-step solution:

Code: Select all: whip[8]: r4n8{c1 c4} - c5n8{r6 r8} - r8c2{n8 n5} - r8c4{n5 n1} - c6n1{r9 r5} - r6c5{n1 n7} - c4n7{r5 r7} - c4n5{r7 .} ==> r9c1 ≠ 8 Singles and whips[1] to the end whip[9]: r2c2{n5 n2} - r3c2{n2 n7} - b4n7{r6c2 r4c1} - r4c4{n7 n8} - r8c4{n8 n1} - c6n1{r9 r5} - r6c5{n1 n7} - c4n7{r5 r7} - c4n5{r7 .} ==> r8c2 ≠ 5 stte whip[9]: r4c1{n8 n7} - c2n7{r6 r3} - c2n2{r3 r2} - c2n5{r2 r8} - r8c4{n5 n1} - c6n1{r9 r5} - r6c5{n1 n7} - c4n7{r5 r7} - c4n5{r7 .} ==> r4c4 ≠ 8 stte whip[9]: c2n7{r6 r3} - c2n2{r3 r2} - c2n5{r2 r8} - c4n5{r8 r7} - c4n7{r7 r5} - r4c4{n7 n8} - r6c5{n8 n1} - r1c5{n1 n4} - c4n4{r2 .} ==> r4c1 ≠ 7 stte

Of course, using so long chains for a puzzle that can be solved in S+W3 is questionable.

However, for those who would insist on a 1-step solution starting from the givens, there's one with Forcing-T&E:
FORCING-T&E(BRT) applied to bivalue candidates n7r4c1 and n7r4c4 :
===> 24 values decided in both cases: n1r7c1 n6r2c1 n6r5c3 n9r5c8 n6r4c8 n5r6c8 n9r4c6 n6r6c6 n6r3c5 n3r2c5 n2r2c2 n2r3c8 n3r3c9 n3r9c8 n1r9c7 n2r9c6 n2r8c1 n7r7c5 n1r6c5 n1r5c2 n4r1c5 n2r5c5 n1r1c8 n4r2c8
===> 108 candidates eliminated in both cases: n1r1c5 n4r1c8 n1r2c1 n2r2c1 n1r2c2 n5r2c2 n6r2c3 n1r2c4 n4r2c4 n1r2c5 n4r2c5 n6r2c5 n9r2c5 n1r2c8 n2r2c8 n3r2c8 n5r2c8 n9r2c8 n3r2c9 n1r3c2 n2r3c2 n5r3c3 n6r3c3 n1r3c5 n3r3c5 n9r3c5 n1r3c7 n1r3c8 n3r3c8 n5r3c8 n9r3c8 n5r3c9 n9r3c9 n6r4c1 n9r4c4 n6r4c6 n9r4c8 n7r5c2 n8r5c2 n1r5c3 n7r5c3 n1r5c4 n7r5c4 n9r5c4 n1r5c5 n4r5c5 n6r5c5 n7r5c5 n8r5c5 n9r5c5 n1r5c6 n2r5c6 n6r5c6 n9r5c6 n6r5c8 n9r5c9 n1r6c2 n6r6c5 n7r6c5 n8r6c5 n1r6c6 n6r6c8 n5r6c9 n7r7c1 n1r7c3 n5r7c3 n7r7c3 n1r7c4 n7r7c4 n9r7c4 n1r7c5 n3r7c5 n4r7c5 n9r7c5 n1r7c6 n9r7c6 n1r7c7 n1r8c1 n8r8c1 n1r8c2 n2r8c2 n1r8c3 n5r8c3 n8r8c4 n9r8c4 n1r8c5 n2r8c5 n3r8c5 n2r8c6 n9r8c6 n1r9c1 n2r9c1 n1r9c3 n3r9c3 n9r9c3 n1r9c5 n2r9c5 n3r9c5 n7r9c5 n1r9c6 n3r9c6 n9r9c6 n5r9c7 n9r9c7 n1r9c8 n5r9c8 n9r9c8 n3r9c9

Code: Select all: CURRENT RESOLUTION STATE: 3 9 8 2 4 5 7 1 6 6 2 15 9 3 7 8 4 59 4 57 17 19 6 8 59 2 3 78 3 4 78 5 9 2 6 1 5 1 6 48 2 4 3 9 78 9 78 2 3 1 6 4 5 78 1 6 39 45 7 34 59 8 2 2 58 39 15 89 13 6 7 4 78 4 57 6 89 2 1 3 59

stte

by **Leren** » Sun Feb 28, 2021 8:46 am

Code: Select all: *------------------------------------------------* | 3 9 8 | 2 14 5 | 7 14 6 | | 6 25 15 | 149 3 7 | 8 124 59 | | 4 257 157 | 19 6 8 | 159 123 359 | |---------------+------------------+-------------| |b78 3 4 |a78 5 9 | 2 6 1 | | 5 178 6 | 1478 1247-8 124 | 3 9 78 | | 9 178 2 | 3 17-8 6 | 4 5 78 | |---------------+------------------+-------------| | 17 6 3579 | 1457 1479 134 | 159 8 2 | | 128 58 359 | 15-8 1289 123 | 6 7 4 | |c1278 4 579 | 6 d12789 12 | 159 13 359 | *------------------------------------------------*

Skyscraper : (8) r4c4 = r1c4 - r1c9 = (8) r5c9 => - 8 r56c5, r8c4;

Code: Select all: *--------------------------------------------------* | 3 9 8 | 2 14 5 | 7 14 6 | | 6 25 15 | 149 3 7 | 8 124 59 | | 4 257 157 | 19 6 8 | 159 123 359 | |---------------+--------------------+-------------| | 78 3 4 | e78 5 9 | 2 6 1 | | 5 178 6 | e1478 1247 c124 | 3 9 78 | | 9 178 2 | 3 d17 6 | 4 5 78 | |---------------+--------------------+-------------| | 17 6 3579 | f147-5 1479 b134 | 159 8 2 | | 128 58 359 | a15 1289 b123 | 6 7 4 | | 1278 4 579 | 6 12789 b12 | 159 13 359 | *--------------------------------------------------*

(5=1) r8c4 - r789c6 = r5c6 - (1=7) r6c5 - r45c4 = (7) r7c4; => - 5 r7c4; stte

Leren

by **yzfwsf** » Sun Feb 28, 2021 9:01 am

Same as Leren

Hidden Text: Show

Code: Select all: Hidden Single: 4 in b9 => r8c9=4 Hidden Single: 6 in b9 => r8c7=6 Hidden Single: 3 in b4 => r4c2=3 Hidden Single: 8 in c7 => r2c7=8 Hidden Single: 8 in b1 => r1c3=8 Naked Single: r1c9=6 Locked Candidates 1 (Pointing): 3 in b9 => r9c3<>3,r9c5<>3,r9c6<>3 Locked Candidates 1 (Pointing): 3 in b2 => r7c5<>3,r8c5<>3 Locked Candidates 1 (Pointing): 6 in b2 => r5c5<>6,r6c5<>6 Locked Pair: in r4c8,r5c8 => r5c9<>9,r6c8<>6,r2c8<>9,r3c8<>9,r6c8<>6,r9c8<>9, Hidden Single: 6 in r6 => r6c6=6 Naked Single: r4c6=9 Hidden Single: 9 in b6 => r5c8=9 Hidden Single: 6 in b6 => r4c8=6 Hidden Single: 6 in b4 => r5c3=6 Hidden Single: 6 in b1 => r2c1=6 Hidden Single: 6 in b2 => r3c5=6 Hidden Single: 3 in b2 => r2c5=3 Naked Single: r6c8=5 Locked Candidates 2 (Claiming): 1 in c1 => r8c2<>1,r7c3<>1,r8c3<>1,r9c3<>1 Locked Candidates 2 (Claiming): 2 in c1 => r8c2<>2 Locked Candidates 1 (Pointing): 1 in b4 => r2c2<>1,r3c2<>1 Locked Candidates 1 (Pointing): 9 in b2 => r7c4<>9,r8c4<>9 Skyscraper : 8 in r4c4,r9c5 connected by r49c1 => r56c5,r8c4 <> 8 Grouped AIC Type 2: 7r7c4 = r45c4 - (7=1)r6c5 - r5c6 = r789c6 - (1=5)r8c4 => r7c4<>5 Hidden Single: 5 in b8 => r8c4=5 Naked Single: r8c2=8 Hidden Single: 8 in b8 => r9c5=8 Hidden Single: 8 in b4 => r4c1=8 Full House: r4c4=7 Hidden Single: 7 in b8 => r7c5=7 Hidden Single: 9 in b8 => r8c5=9 Hidden Single: 8 in b5 => r5c4=8 Hidden Single: 8 in b6 => r6c9=8 Full House: r5c9=7 Hidden Single: 7 in b4 => r6c2=7 Full House: r6c5=1 Full House: r5c2=1 Hidden Single: 7 in b1 => r3c3=7 Hidden Single: 7 in b7 => r9c1=7 Hidden Single: 2 in b7 => r8c1=2 Full House: r7c1=1 Hidden Single: 2 in b8 => r9c6=2 Hidden Single: 1 in b8 => r8c6=1 Full House: r8c3=3 Hidden Single: 3 in b8 => r7c6=3 Full House: r5c6=4 Full House: r5c5=2 Full House: r1c5=4 Full House: r1c8=1 Full House: r7c4=4 Hidden Single: 1 in b9 => r9c7=1 Hidden Single: 4 in b3 => r2c8=4 Hidden Single: 2 in b3 => r3c8=2 Full House: r9c8=3 Hidden Single: 3 in b3 => r3c9=3 Hidden Single: 1 in b1 => r2c3=1 Hidden Single: 1 in b2 => r3c4=1 Full House: r2c4=9 Hidden Single: 9 in b3 => r3c7=9 Full House: r3c2=5 Full House: r2c2=2 Full House: r2c9=5 Full House: r7c7=5 Full House: r7c3=9 Full House: r9c3=5 Full House: r9c9=9

by **AnotherLife** » Sun Feb 28, 2021 12:04 pm

After the 'lcls' steps, as Cenoman uses to say, the sudoku state becomes as follows:

Code: Select all: .-----------------.------------------.---------------. | 3 9 8 | 2 14 5 | 7 14 6 | | 6 25 15 | 149 3 7 | 8 124 59 | | 4 257 157 | 19 6 8 | 159 123 359 | :-----------------+------------------+---------------: | 78 3 4 | 78 5 9 | 2 6 1 | | 5 178 6 | 1478 12478 124 | 3 9 78 | | 9 178 2 | 3 178 6 | 4 5 78 | :-----------------+------------------+---------------: | 17 6 3579 | 1457 1479 134 | 159 8 2 | | 128 58 359 | 158 1289 123 | 6 7 4 | | 1278 4 579 | 6 12789 12 | 159 13 359 | '-----------------'------------------'---------------'

Now it is possible to get a one-step solution with a forcing net. Maybe this is not a 'nice' solution, but the logical chains can be easily described graphically. I will try to prove that the premise r4c1=7 leads to r7c4=5 and r7c4=4, which is a contradiction. So r4c1=8, and we get a sequence of singles.
See the picture: https://disk.yandex.ru/i/rBJo3ttnB5sQeg
1. The red chain. r4c1=7 => r4c4<>7 (1.1) => r4c4=8 => r6c5<>8 (1.2)
2. The black chains. r4c1=7 => r56c2<>7 => r3c2=7 =>
2.1. r3c2<>5
2.2. r3c3<>2 =>r3c8=2 => r2c8<>2 => r2c2=2 => r2c2<>5 => (because of 2.1) r8c2=5 => r8c4<>5 => r7c4=5 (2.2.1) => r7c4<>7 => (because of 1.1) r5c4=7 =>
2.2.2. r5c4<>4
2.2.3. r6c5<>7 => (because of 1.2) r6c5=1 => r1c5<>1 => r1c5=4 => r2c4<>4 => (because of 2.2.2) r7c4=4 - contradiction with 2.2.1.
Consequently, r4c1<>7, sste.

Website · by **denis_berthier** » Sun Feb 28, 2021 12:19 pm

AnotherLife wrote:Now it is possible to get a one-step solution with a forcing net. [...]

Starting with the same PM, the same elimination was obtained by my last whip[9] instead of this complicated net.

by **AnotherLife** » Sun Feb 28, 2021 12:53 pm

denis_berthier wrote:whip[9]: c2n7{r6 r3} - c2n2{r3 r2} - c2n5{r2 r8} - c4n5{r8 r7} - c4n7{r7 r5} - r4c4{n7 n8} - r6c5{n8 n1} - r1c5{n1 n4} - c4n4{r2 .} ==> r4c1 ≠ 7

Can you explain your solution graphically?

Website · by **denis_berthier** » Sun Feb 28, 2021 3:43 pm

AnotherLife wrote:
denis_berthier wrote:whip[9]: c2n7{r6 r3} - c2n2{r3 r2} - c2n5{r2 r8} - c4n5{r8 r7} - c4n7{r7 r5} - r4c4{n7 n8} - r6c5{n8 n1} - r1c5{n1 n4} - c4n4{r2 .} ==> r4c1 ≠ 7

Can you explain your solution graphically?

You can find general graphics for whips in the Basic User Manual for CSP-Rules-V2.1: https://github.com/denis-berthier/CSP-Rules-V2.1/blob/master/Docs/2021-BUM-V2.1-r2.pdf, section 1.7

by **yzfwsf** » Sun Feb 28, 2021 3:49 pm

Hi denis:
Can you provide a graphical representation of the specific steps you found in this puzzle? Convenient to compare and see what is the difference with AIC? Thank you .

Website · by **denis_berthier** » Sun Feb 28, 2021 4:02 pm

yzfwsf wrote:Hi denis:
Can you provide a graphical representation of the specific steps you found in this puzzle? Convenient to compare and see what is the difference with AIC? Thank you .

The way whips are written is, in and of itself, a graphical representation. Read the conventions in the BUM and tell me if there is anything in particular that you don't understand.
Graphically, a whip IS the same thing as a bivalue-chain, i.e. a basic AIC and all the elements present in the AIC are present in the whip notation. You need to understand the general nrc notation, the fact that z- and t- candidates are not part of the whip and the reasons why they can be abstracted.
I've met people here who have no sense of abstraction. It's a big problem for them to understand whips.

by **jco** » Sun Feb 28, 2021 4:57 pm

Hello,

After basics

Code: Select all: 1 2 3 4 5 6 7 8 9 .------------------+------------------+---------------. | 3 9 8 | 2 14 5 | 7 14 6 | 1 | 6 25 15 | 149 3 7 | 8 124 59 | 2 | 4 257 157 | 19 6 8 | 159 123 359 | 3 |------------------+------------------+---------------| | d78* 3 4 |e78# 5 9 | 2 6 1 | 4 | 5 178 6 | 1478 1247-8 124 | 3 9 78 | 5 | 9 178 2 | 3 17-8 6 | 4 5 78 | 6 |------------------+------------------+---------------| | 1-7 6 3579 | 1457 1479 134 | 159 8 2 | 7 | 128 58 359 | 15-8 1289 123 | 6 7 4 | 8 |ca1278* 4 a579* | 6 b12789# 12 | 159 13 359 | 9 '------------------+------------------+---------------'

1. (7)r9c13=(7)r9c5-(8)r9c5=(8)r9c1-(8=7)r4c1-(7=8)r4c4 => -7 r7c1, -8 r8c4, -8 r56c5

Code: Select all: 1 2 3 4 5 6 7 8 9 .------------------+------------------+---------------. | 3 9 8 | 2 14 5 | 7 14 6 | 1 | 6 25 15 | 149 3 7 | 8 124 59 | 2 | 4 257 157 | 19 6 8 | 159 123 359 | 3 |------------------+------------------+---------------| | d78 3 4 |e78 5 9 | 2 6 1 | 4 | 5 178 6 | 1478 12478 124 | 3 9 78 | 5 | 9 178 2 | 3 17 6 | 4 5 78 | 6 |------------------+------------------+---------------| | 1 6 3579 |a45-7 479 34 | 59 8 2 | 7 | c28 c58 359 |b15 1289 123 | 6 7 4 | 8 | c278 4 579 | 6 12789 12 | 159 13 359 | 9 '------------------+------------------+---------------'

2. (5)r7c4=r8c4-(5=287)b7p457-(7)r4c1=(7)r4c4 => -7 r7c4; lclste

Regards,
jco

Edit: fixed typos and corrected to lclste.

by **ghfick** » Sun Feb 28, 2021 5:15 pm

Hi Denis,
I think that yzfwsf and Another Life would like to see the example whip[9] explained in detail. I think there are many members of this forum [me included] who would appreciate a visual explanation. If I am correct, the visuals [for this example puzzle] would be many. One will need the grid with candidates displayed in all 4 ways [rc, rn cn, bn] and for each of the 9 parts of the whip[9]. So, I think there might be as many as 36! grids.
I think you may say I am wrong and you will say I should read the BUM. All I can say is that I have tried to read and study the BUM since you first advised us to do so. Indeed, even the notation for whip[1] still eludes me. On pg 16, one reads r3n4{c5 .} => r4c6 ≠ 4. What does the dot . mean? I would appreciate an explanation that is specific to the 9x9 Sudoku and not the 'general' case.
Perhaps you will argue that we all need to 'see' things in 3D. Well, 3D displays are not easy to reconcile. XSudo is amazing but not for the faint of heart.
Regards Gordon

Website · by **denis_berthier** » Sun Feb 28, 2021 6:04 pm

ghfick wrote:Hi Denis,
I think that yzfwsf and Another Life would like to see the example whip[9] explained in detail. I think there are many members of this forum [me included] who would appreciate a visual explanation. If I am correct, the visuals [for this example puzzle] would be many. One will need the grid with candidates displayed in all 4 ways [rc, rn cn, bn] and for each of the 9 parts of the whip[9]. So, I think there might be as many as 36! grids.
I think you may say I am wrong and you will say I should read the BUM. All I can say is that I have tried to read and study the BUM since you first advised us to do so. Indeed, even the notation for whip[1] still eludes me. On pg 16, one reads r3n4{c5 .} => r4c6 ≠ 4. What does the dot . mean? I would appreciate an explanation that is specific to the 9x9 Sudoku and not the 'general' case.
Perhaps you will argue that we all need to 'see' things in 3D. Well, 3D displays are not easy to reconcile. XSudo is amazing but not for the faint of heart.
Regards Gordon

The only grid you need to see to understand whips is the standard one. And the only thing you have to understand is a whip looks like a bivalue-chain, i.e. a basic AIC: z- and t- candidates and the associated links don't belong to the whip and don't have to be displayed.
In

Code: Select all: whip[9]: c2n7{r6 r3} - c2n2{r3 r2} - c2n5{r2 r8} - c4n5{r8 r7} - c4n7{r7 r5} - r4c4{n7 n8} - r6c5{n8 n1} - r1c5{n1 n4} - c4n4{r2 .} ==> r4c1 ≠ 7

,
- is a link, what you would call a "weak link"
{ } is the content of a csp-variable, which would be translated as a "strong link" in AIC notation. Unfortunately, AIC notation is totally inconsistent, as it makes a difference between bivalue and bilocal.

Making AIC notation consistent, you could write the previous whip as:
whip[9]: c2n7(r6=r3) - c2n2(r3=r2) - c2n5(r2=r8) - c4n5(r8=r7) - c4n7(r7=r5) - r4c4(n7=n8) - r6c5(n8=n1) - r1c5(n1=n4) - c4n4(r2=.) ==> r4c1 ≠ 7
(the "." denotes the absence of a candidate)

r3n4{c5 .} => r4c6 ≠ 4 is for Sudoku !!

by **creint** » Sun Feb 28, 2021 7:35 pm

If 7r4c1 is true then it leads to no 4 possible in c4, so it can be removed.
Starting at Blue point (false) and follow the manual added blue line.
Note that blue point was group 7r56c2 = 7r3c2 .., but gets reduced to a single in the end.
Note that 7r4c4 was already excluded at start of chain.
Note that . refers to 4r57c4
Another name could be something like "unit contradiction in net".

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