The New Sudoku Players' Forum

[DonM]

Hey Jasper32- I have to say thanks for posting this. It turned out to be a far more interesting puzzle than I thought on first glance- a nice challenge. Starting from the posted grid, pretty much, basic groups, ALSs (no nets) and a couple of final loops finished it off.

1. grp(2)r123c7=r7c7-(2=3)r7c3-(3=1)r6c3-r3c3=(1)r3c8 => r3c8<>2
2. (3=2)r7c3-(2=4)r7c7-r9c8=(4)r9c2 => r9c2<>3
3. (1=7)r2c2-(7=4)r9c2-r9c8=(4-6)r3c8=grp(6)r13c8-als(6=27)r3c57-(2)r3c3=(2)r2c1 => r2c1<>1 -> r8c1=1 -> r8c23<>1 -> linebox(6)r45c1: r5c3<>6
4. (7=9)r5c3-(9=6)r4c1-r5c1=r5c7-(6)r3c7=hp(16)r3c38 => r3c3<>7
5. als(7=28)r8c89-(2=4)r7c7-r4c7=(4-6)r4c8=grp(6)r13c8-als(6=27)r3c57-(2)r3c3=(2-7)r2c1=grp(7)r12c2 => r8c2<>7
6. (5=9)r6c5-r6c9=(9-8)r1c9=r8c9-als(8=59)r28c6 => r5c6<>5
7. (5)r1c8=grp(5)r1c45-r2c6=r8c6-(5=6)r8c2-r8c3=r3c3-(6=1)r3c8-als(1=25)r26c8-loop => r3c7<>6, r8c4<>5
-> np(27)r3c57:r3c3<>2 -> r2c1=2 -> xw(6)c17r45: r4c8<>6=4 -> r9c2=4 -> r7c7=4 -> linebox(2)r123c7: r2c7<>2, r1c9<>2
8. (5)r8c6=r2c6-(5=1)r2c8-(1=7)r2c2-(7=6)r1c2-(6=5)r8c2-loop => r2c7<>5 -> r5c7=5, r6c5=5
STTE

(edit: turns out that my previous step 4 for was subsumed by a linebox elim in step 3 -> r5c3<>6 thanks to sharp-eyed Luke. )

[DonM]

Steve, love those 2 AUR moves. I think I may just put them in my pocket and take them home with me! (Would have commented sooner, but only saw them today- I'm sentenced to 'solitary' until I put my own solution up.)

6) (aur27)r13c57=> sis[(2)r9c5,(2)r7c7,(7)r2c7] =>
(2)r9c5.r7c7=(7)r2c7-(7=1)r2c2-(1=5)r2c8-(5=2)r6c8 =>r9c8<>2

7) alt (aur79)r59c13 => sis[(6)r5c1,(23)r7c9] => (6)r5c1=(nt123)r679c3-(1)r3c3=r2c2-(1=5)r2c8-r2c7=(5)r5c7 =>r5c7<>6 => ste[/quote]

[Luke]

I also found this interesting. I've been waiting since January (here) for someone to invoke "sis" so I could see it in action. I think I see how it works now. In this case, it's about "UR avoidance," often discussed around here.

The puzzle is at this stage, (27) UR = * and sis = #:

Code: Select all

 *-----------------------------------------------------------*
 | 3     67    4     | 259  *2579  1     |*279   568   289   |
 | 2     17    8     | 34    6     59    |#759   15    34    |
 | 5     9     167   | 8    *27    34    |*267   16    34    |
 |-------------------+-------------------+-------------------|
 | 69    8     5     | 1     3     2     | 469   46    7     |
 | 679   2     79    | 459   589   489   | 569   3     1     |
 | 4     13    13    | 7     59    6     | 8     25    29    |
 |-------------------+-------------------+-------------------|
 | 8     345   23    | 235   1     7     |#24    9     6     |
 | 1     567   2679  | 259   4     589   | 3     278   28    |
 | 79    47    2379  | 6    #289   389   | 1    -2478  5     | 
 *-----------------------------------------------------------*
Steve K:  6) (aur27)r13c57=> sis[(2)r9c5,(2)r7c7,(7)r2c7] =>
(2)r9c5.r7c7=(7)r2c7-(7=1)r2c2-(1=5)r2c8-(5=2)r6c8 =>r9c8<>2

sis = strong inference set, one candidate must be true.

In this case, at least one of the 2's or the 7 must be true or the UR will exist. This creates a strong link for a chain that ends up pinching off 2r9c8. It's all about finding the next strong link, ain't it?

[DonM]

I also found this interesting. I've been waiting since January (here) for someone to invoke "sis" so I could see it in action. I think I see how it works now. In this case, it's about "UR avoidance," often discussed around here.

It turns out that knowing what the 'sis' is and how to make the best use of it can be two different things. Take Steve's use of the 'sis' below where he exposes and uses a naked triple (sitting right there in front of my eyeballs) to get a strategic elimination.

7) alt (aur79)r59c13 => sis[(6)r5c1,(23)r7c9] => (6)r5c1=(nt123)r679c3-(1)r3c3=r2c2-(1=5)r2c8-r2c7=(5)r5c7 =>r5c7<>6 => ste

Now, take my use of the same 'sis' which resulted in a totally useless elimination (so I didn't post it in my solution) :

AUR(79)r59c13
||
(6-7)r5c1=r5c3
||
(23)r9c3

=> r9c3<>7

[PIsaacson]

I am well aware of ALS’s but have a difficult time locating them and have never heard of an easy way to find them. Even with the ALS’s, that alone would not have solved the puzzle.

You're right, it wouldn't.

I took that as a challenge, albeit for my ALS engine, so I ran it though (illegally) and it spit out the following 7 ALS chains to crack the puzzle:

Code: Select all

do_alschains - reducing r9c2.<347> by <3>
do_alschains - als+[4x6/7] r6c2.<n13> -1- r67c3.<n123> -2- r7c7.<n24> -4- r12678c2.<n134567>

do_alschains - reducing r8c3.<12679> by <1>
do_alschains - als[3x5/6] r6c3.<n13> -3- r7c3.<n23> -2- r4589c1.<n12679>

do_alschains - reducing r5c6.<4589> by <5>
do_alschains - als[4x3/4] r6c5.<n59> -9- r6c9.<n29> -2- r8c9.<n28> -8- r28c6.<n589>

do_alschains - reducing r3c8.<126> by <2>
do_alschains - als+[4x6/8] r3c357.<n1267> -1- r6c3.<n13> -3- r7c37.<n234> -4- r12345c7.<n245679>

do_alschains - reducing r8c1.<12679> by <7>
do_alschains - als[4x6/7] r2459c1.<n12679> -1- b1x259.<n1267> -2- r3c57.<n267> -6- r12368c8.<n125678>

do_alschains - reducing r2c1.<127> by <1>
do_alschains - reducing r8c2.<1567> by <7> base/cover
do_alschains - base/cover {1n28 2n28 6n8 8n8 9n28} {6r1 1r2 4r9 7c2 2578c8}
do_alschains - als+[4x7/7] r29c2.<n147> -4- r123689c8.<n1245678> -6- r3c57.<n267> -2- b1x259.<n1267>

do_pinnings - setting r8c1 = d1 hidden single
do_restrictions - reducing r5c3.<679> by <6> col/box at b4

do_alschains - reducing r2c7.<2579> by <2> dualx
do_alschains - reducing r2c7.<579> by <5> dualx
do_alschains - reducing r3c3.<1267> by <7> dualx
do_alschains - reducing r8c4.<259> by <5> base/cover
do_alschains - base/cover {1n2 2n1268 3n6 5n6 8n26 9n6} {125r2 5r8 6c2 3489c6 7b1}
do_alschains - alsx[4x5/9] r1c2.<n67> -6- r8c2.<n56> -5- r3589c6.<n34589> -9- r2c1268.<n12579>

naked/hidden singles from this point on

The last one is particularly interesting in that the start/end ALSs are linked (the "alsx" tag and "dualx" tags are indicators of this quasi-doubly linked condition) via the direct line-of-sight RCC 7. In essence, it's a continuous ALS loop.

I (my ALS engine that is) also found a solution using 7 shorter ALS chains plus 1 Death Blossom, so there are multiple ways for ALS sets to crack this nut...

[ronk]

I took that as a challenge, albeit for my ALS engine, so I ran it though (illegally) and it spit out the following 7 ALS chains to crack the puzzle:

Code: Select all

[edit: 6th als chain]

do_alschains - reducing r2c1.<127> by <1>
do_alschains - reducing r8c2.<1567> by <7> base/cover
do_alschains - base/cover {1n28 2n28 6n8 8n8 9n28} {6r1 1r2 4r9 7c2 2578c8}
do_alschains - als+[4x7/7] r29c2.<n147> -4- r123689c8.<n1245678> -6- r3c57.<n267> -2- b1x259.<n1267>


The last one is particularly interesting in that the start/end ALSs are linked (the "alsx" tag and "dualx" tags are indicators of this quasi-doubly linked condition) via the direct line-of-sight RCC 7. In essence, it's a continuous ALS loop.

Paul, nice demonstration of the pervasiveness of ALS. Using the same cells (of your 6th chain above), there is a continuous loop of three ALSs. Then the "base/cover" portion is not required.

do_alschains - als+[3x...] r129c2.<n1467> -4- r123689c8.<n1245678> -6- r3c57.<n1267>

[DonM]

Well Paul, if anything, I'd say those results should be an incentive for people to check out the ALS Chain tutorial (you have to start somewhere). I'm actually not totally surprised by your results- I found a number of ALS patterns- unusual for an ER=8.4 puzzle (manually-solving speaking)- but couldn't make any of them result in something useful.

[Luke]

Code: Select all

*-----------------------------------------------------------*
 | 3     67    4     | 259   2579  1     | 279   568   289   |
 | 2     17    8     | 34    6     59    | 579   15    34    |
 | 5     9     167   | 8     27    34    | 267   16    34    |
 |-------------------+-------------------+-------------------|
 | 69    8     5     | 1     3     2     | 469   46    7     |
 | 679   2     79    | 459   589   489   | 569   3     1     |
 | 4     13    13    | 7     59    6     | 8     25    29    |
 |-------------------+-------------------+-------------------|
 | 8     345   23    | 235   1     7     | 24    9     6     |
 | 1     567   2679  | 259   4     589   | 3     278   28    |
 | 79    47    2379  | 6     289   389   | 1     478   5     |
 *-----------------------------------------------------------*
Steve K: 7) (5)r2c6=(5-8)r8c6=(ht238)r9c356-(79)r9c3=(nt123)r679c3-(1)r3c3=r2c2-(1=5)r2c8 loop => r8c6<>9, r9c56<>9,r8c3<>2,r2c7<>5 => ste

Steve (or all,) if you would, I got flummoxed by one of the elims. I’m fine with the cont NL, but can’t get the r8c3<>2. That's a tricky one, with the ht and nt sharing (23.) Any quick explanation would be cool, and thanks.

[Steve K]

In a continuous loop AIC, each weak inference is proved a strong inference. Thus, (nt123)r679c3=(1)r3c3 => (nt123)r679c3=(nt123)r367c3 => r8c3<>2