The New Sudoku Players' Forum

by **ArkieTech** » Wed Dec 30, 2015 12:38 am

Code: Select all: *-----------* |9..|.25|.6.| |.15|..8|...| |...|...|.89| |---+---+---| |..8|..7|4..| |.2.|.3.|.7.| |..3|5..|2..| |---+---+---| |84.|...|...| |...|6..|84.| |.5.|98.|..1| *-----------*

by **pjb** » Wed Dec 30, 2015 1:36 am

Code: Select all: 9 8 47 | 3 2 5 | 1 6 47 6 1 5 | 47 9 8 | 37 23 247 2347 37 247 | 147 167 16 | 5 8 9 ------------------------+----------------------+--------------------- a15 69 8 | 2 16 7 | 4 b159 3 45-1 2 49-1 | 8 3 19 |d69 7 c56 l17 679 3 | 5 4 169 | 2 19 8 ------------------------+----------------------+--------------------- 8 4 1267 | 17 157 f23 |e3679 59 2567 237-1 379 1279 | 6 157 g23 | 8 4 h257 k237 5 267 | 9 8 4 |i367 j23 1

Getting a bit ridiculous:
(1=5)r4c1 - r4c8 = (5-6)r5c9* = (6-9)r5c7^ = (9-3)r7c7 = r7c6 - (3=2)r8c6 - (25=7)r8c9* - (67=3)r9c7^# - (3=2)r9c8 - (23=7)r9c1# - (7=1)r6c1 => -1 r5c13, r8c1; stte

Phil

by **SteveG48** » Wed Dec 30, 2015 2:03 am

Thanks, Phil. I needed that. I think I'll go get a glass of wine now.

by **Leren** » Wed Dec 30, 2015 3:28 am

Code: Select all: *--------------------------------------------------------------------------------* | 9 8 e47 | 3 2 5 | 1 6 47 | | 6 1 5 |b47 9 8 |a37an 23 247 | |d2347 d37 d247 |c147 c167 16 | 5 8 9 | |--------------------------+--------------------------+--------------------------| | 15 69f 8 | 2 16 7 | 4 159g 3 | | 145 2 149 | 8 3 19 | 69 7 56 | | 17dj 679e 3 | 5 4 169 | 2 19h 8 | |--------------------------+--------------------------+--------------------------| | 8 4 1267 | 17 157 23 | 3679 59 2567 | | 1237 379 1279 | 6 157 23 | 8 4 257 | | 237ck 5 f267 | 9 8 4 | 367bm 23 1 | *--------------------------------------------------------------------------------* 7 r2c7 - r2c4 = r3c45 - r3c123 = r1c3 - r9c3; - r9c3 7 r2c7 - r9c7 = r9c1 - r6c1 = (7-6) r6c2 = (6-9) r4c2 = r4c8 - (9=1) r6c8 - (1=7) r6c1 - r9c1 = r9c7 - 7 r2c7 => - 7 r2c7; stte

Also ridiculous, but effective !

Leren

by **ArkieTech** » Wed Dec 30, 2015 12:49 pm

Code: Select all: *-----------------------------------------------------------* | 9 8 47 | 3 2 5 | 1 6 47 | | 6 1 5 | 47 9 8 | 37 23 247 | | 2347 37 247 | 147 167 16 | 5 8 9 | |-------------------+-------------------+-------------------| | 15 69 8 | 2 16 7 | 4 159 3 | | 145 2 149 | 8 3 19 | 69 7 56 | | 17 679 3 | 5 4 169 | 2 19 8 | |-------------------+-------------------+-------------------| | 8 4 1267 | 17 157 23 | 3679 59 2567 | | 1237 379 1279 | 6 157 23 | 8 4 257 | | 237 5 267 | 9 8 4 | 367 23 1 | *-----------------------------------------------------------* [(7=6)r6c168-(6=9)r356c6-(9=7)b4p1467]-[7r9c1=ss:7r19]-7r2c7|r78c9; ste

I was hoping someone would find something simple. JC where are you?

by **JC Van Hay** » Wed Dec 30, 2015 5:40 pm

I hope there are no typos

10 singles
LC(9B9), NP(23)r29c8; 2 singles
LC(6B6), HP(23)r78c6
*
XYWing(169)R4B5 -> r2c7=3; singles to the end of a unique solution.
Proof of the exclusion of 7r2c7 :

Code: Select all: +---------------------+-----------------+-------------------+ | 9 8 4(7) | 3 2 5 | 1 6 4(7) | | 6 1 5 | 47 9 8 | 3-7 23 247 | | 234-7 37 247 | 147 167 16 | 5 8 9 | +---------------------+-----------------+-------------------+ | 15 69 8 | 2 16 7 | 4 159 3 | | 45(1) 2 49(1) | 8 3 (19) | 69 7 56 | | (+7-1) 69-7 3 | 5 4 16(9) | 2 (19) 8 | +---------------------+-----------------+-------------------+ | 8 4 1267 | 17 157 23 | 3679 59 2567 | | 123-7 379 1279 | 6 157 23 | 8 4 257 | | 23(-7) 5 26(7) | 9 8 4 | 36(7) 23 1 | +---------------------+-----------------+-------------------+

In English :

Therefore, {7R19, 1R5, 9C6, (17)r6c1, (19)r6c8} -> r2c7≠7; stte
However : Wing(1R5, 9C6, (19)r6c8) -> r6c1=7 -> Skyscraper(7R19)-7r2c7; stte
This is a 2 steps solution!
*
To get a single-stepper avoiding r6c1=7, the following combination of constraints is possible :

Code: Select all: +---------------------+---------------+--------------------+ | 9 8 4(7) | 3 2 5 | 1 6 4(7) | | 6 1 5 | 47 9 8 | 3-7 23 247 | | 2347 (37) 247 | 147 167 16 | 5 8 9 | +---------------------+---------------+--------------------+ | 15 6(9) 8 | 2 16 7 | 4 15(9) 3 | | 145 2 149 | 8 3 19 | 69 7 56 | | (17) 679 3 | 5 4 169 | 2 (19) 8 | +---------------------+---------------+--------------------+ | 8 4 1267 | 17 157 23 | 3679 59 2567 | | 1237 (379) 1279 | 6 157 23 | 8 4 257 | | 23(7) 5 26(7) | 9 8 4 | 36(7) 23 1 | +---------------------+---------------+--------------------+

In English :

r2c7=7 -> r1c3=r9c1=7, r3c2=3 and r6c1=1, r6c8=r4c2=9; r8c2={}In exclusion matrix notation :

Code: Select all: 7r2c7 7r1c9 7r1c3 7r9c7 7r9c3 7r9c1 7r6c1 1r6c1 1r6c8 9r6c8 9r4c8 9r4c2 7r3c2 3r3c2 7r8c2 9r8c2 3r8c2

Therefore, {7R19, (17)r6c1, (19)r6c8, 9R4, (37)r3c2, (379)r8c2} -> r2c7≠7; stte
In Eureka notation : Kraken Cell (379)r8c2 -> [7r1c9==7r9c7]-7r2c7; stte

or as an AAIC :

Complement: illustrating solving method of higher level puzzle: before using an eventual derived constraint as a supplement to the B/B-Plot, other Wings are available as a starting point to find the solutions of this puzzle. Let us examine the one where the pivot is 4C1 [refer to Steve Kurzkals blog on the au site] :
4C1 strongly couple the cells r35c1 containing 3 and 2 members of bilocals respectively : 5r5c9=(5-4)r5c1=4r3c1-[2r3c1=2r3c3 and 3r3c1=3r3c2]
4C1 is embedded in a "loop" :

Code: Select all: +----------------------+------------------+----------------------+ | 9 8 (47) | 3 2 5 | 1 6 47 | | 6 1 5 | 47 9 8 | 37 23 247 | | 237(4) 37 (247) | 147 167 16 | 5 8 9 | +----------------------+------------------+----------------------+ | 15 69 8 | 2 16 7 | 4 159 3 | | -1(45) 2 149 | 8 3 19 | (69) 7 (56) | | 17 679 3 | 5 4 169 | 2 19 8 | +----------------------+------------------+----------------------+ | 8 4 (126-7) | (17) (157) 23 | 36-7(9) (59) 26-57 | | 1237 379 19-27 | 6 15-7 23 | 8 4 257 | | 237 5 (267) | 9 8 4 | 37-6 23 1 | +----------------------+------------------+----------------------+

In Eureka notation :

In exclusion matrix notation :

Code: Select all: 4r3c1 4r5c1 5r5c1 5r5c9 6r5c9 6r5c7 9r5c7 9r7c7 9r7c8 5r7c8 5r7c5 1r7c5 7r7c5 1r7c4 7r7c4 1r7c3 7r7c3 2r7c3 6r7c3 2r9c3 6r9c3 7r9c3 2r3c3 7r3c3 4r3c3 7r1c3 4r1c3

The following derived constraints are available :

lclste
PS : typo : [6r5c7==6r9c3]-3r9c7 is to be replaced by [6r5c7==6r9c3]-6r9c7. Thanks eleven !

by **Marty R.** » Wed Dec 30, 2015 7:48 pm

Code: Select all: +---------------+-------------+---------------+ | 9 8 47 | 3 2 5 | 1 6 47 | | 6 1 5 | 47 9 8 | 37 23 247 | | 2347 37 247 | 147 167 16 | 5 8 9 | +---------------+-------------+---------------+ | 15 69 8 | 2 16 7 | 4 159 3 | | 145 2 149 | 8 3 19 | 69 7 56 | | 17 679 3 | 5 4 169 | 2 19 8 | +---------------+-------------+---------------+ | 8 4 1267 | 17 157 23 | 3679 59 2567 | | 1237 379 1279 | 6 157 23 | 8 4 257 | | 237 5 267 | 9 8 4 | 367 23 1 | +---------------+-------------+---------------+

Play this puzzle online at the Daily Sudoku site

Steve: please don't read this :lol:

(6=73)r9c7-(3=7)r2c7-(7=42)r12c9-(2=75)r8c9-(5=6)r5c9=> -6r5c7

by **eleven** » Wed Dec 30, 2015 8:43 pm

Oh, Marty ...

I like JC's loop.
I verified the eliminiations in the grid, while in the "derived constraints" i can't see -7r7c3 (it's from 17r7c45==7r1c3), and -6r9c7 is missing (6r5c7==6r9c3).
[Added:] I made a grid, where the 2 ways are dvided by a '|', so the eliminations can be seen better [edited overlapping cells]:

Code: Select all: +----------------------+------------------+----------------------+ | 9 8 (47|7) | 3 2 5 | 1 6 47 | | 6 1 5 | 47 9 8 | 37 23 247 | |(237|4) 37 (247|2) | 147 167 16 | 5 8 9 | +----------------------+------------------+----------------------+ | 15 69 8 | 2 16 7 | 4 159 3 | |-1(4|5) 2 149 | 8 3 19 |(6|9) 7 (5|6) | | 17 679 3 | 5 4 169 | 2 19 8 | +----------------------+------------------+----------------------+ | 8 4 (26|1)-7 |(17|7)(17|5) 23 |(9|36)-7 (5|9) 26-57 | | 1237 379 19-27 | 6 15-7 23 | 8 4 257 | | 237 5 (267|6) | 9 8 4 | 37-6 23 1 | +----------------------+------------------+----------------------+

This is a uniqueness solution:

Code: Select all: +-------------------+-------------------+-------------------+ | 9 8 47 | 3 2 5 | 1 6 47 | | 6 1 5 | 47 9 8 | 37 23 247 | | 2347 37 247 | 147 #16+7 #16 | 5 8 9 | +-------------------+-------------------+-------------------+ |c15 #69 8 | 2 #16 7 | 4 #19+5 3 | |d145 2 e149 | 8 3 b19 | 69 7 56 | | 17 #69+7 3 | 5 4 a#16+9 | 2 b#19 8 | +-------------------+-------------------+-------------------+ | 8 4 f1267 |g17 157 23 | 3679 59 2567 | | 1237 379 e1279 | 6 157 23 | 8 4 257 | | 237 5 267 | 9 8 4 | 367 23 1 | +-------------------+-------------------+-------------------+

DP169 in marked cells:
7r3c5 or 9r5c3 or 9r6c6 or 9r5c7
9r5c3-r5c6=9r6c6
9r5c7-r5c6=9r6c6
9r6c6-(9=1)r4c6&r5c8-r4c5|r3c8=(1-5)r4c1=(5-4)r5c1=(49-1)r58c3=1r7c3-(1=7)r7c4
=> r23c4<>7, r78c5<>7

by **JC Van Hay** » Wed Dec 30, 2015 9:23 pm

eleven wrote:I verified the eliminiations in the grid, while in the "derived constraints" i can't see -7r7c3 (it's from 17r7c45==7r1c3), and -6r9c7 is missing (6r5c7==6r9c3).

1. Look at the exclusion matrix : r7c45=17 or r7c5=5, r7c8=r5c7=9, r5c9=6, r5c1=5, r3c1=4, r1c3=7, r3c3=2, r9c3=6, r7c3=1 (r1c3=7 -> r7c3≠7 not represented in the exclusion matrix), r7c4=7 => [(17)r7c45==7r7c4]-7r8c5,r7c389
2. Typo : [6r5c7==6r9c3]-3r9c7 is to be replaced by [6r5c7==6r9c3]-6r9c7

Maybe, just one final observation : my favorite analysis of a puzzle is "(multiple) coloring". The exclusions by the "loop" are therefore obvious, but not so (!) when a particular "notation" is used.

by **eleven** » Wed Dec 30, 2015 10:08 pm

Yes, i had to edit my 2-way-grid to reflect that in the other direction r7c4 has to be 7 (not 17).
This other direction is easier to spot in my opinion (just singles in col. 3)
(4r5c1=4r3c1-(4=267)r139c3-(267=1)r7c3-(1=57)r7c45-(5=9)r7c8-r7c7=(9-6)r5c7=(6-5)r5c9=r5c1, loop).

[Added:] What i want is to find the eliminations manually as easy as possible, without the need of writing and analysing any notes. I do it in a multicoloring manner in the grid by marking the one and other direction digits. This should be reflected in my 2way grid above.

by **ArkieTech** » Wed Dec 30, 2015 11:26 pm

Thanks everyone appreciate and learn from this.

by **David P Bird** » Thu Dec 31, 2015 9:08 am

JC Thanks for your detailed explanation of your approach to this puzzle as it give me an appreciation of the way you work.

The goal of finding a one-step solution has provided some interesting almost pattern tricks (especially from you) but it's not my cup of tea as I would only go for a complex structure if it saved a lot of simple ones.

Like Eleven initially I thought some of your eliminations were wrong because I was only considering those from the fantastic closed loop you found, but you were including others from the matrix analysis. I haven't analysed this case but I believe that using AICs these would only be available in follow-on steps which would only be needed if they were significant.

I use Graded Equivalence Marking to solve puzzles which if done to completion shows all the eliminations from chains based stemming from a single strong link. I then select those that appear most significant to implement before moving on to any further strong links that haven't been reached yet. Although GEM can be used to follow branched paths I discipline myself only to use linear chains.

I often find that the number of paths that can be followed from each side of the opening link gives a very good indication of which one is true, the ones from the false side being far more numerous, but of course the challenge is to prove this. I wonder if others also find the same thing?

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