Everything about Sudoku that doesn't fit in one of the other sections

Please let me know if you find these interesting (or not); you can just PM me if you care to reply. If more find them interesting than not, I'll post continually as I find puzzles that look interesting.
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`. 1 . | . . . | . . .. . 3 | 7 . 6 | 2 . .2 7 . | . 3 . | . . 9------+-------+------. . . | . . . | . 8 5. . . | . 1 . | . . .7 9 . | . . . | . . .------+-------+------8 . . | . 9 . | . 3 2. . 6 | 2 . 8 | 7 . .. . . | . . . | . 6 .`

After SSTS:
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`6    1     458  | 4589 2458  2459  | 348  457 37 9    458   3    | 7    458   6     | 2    145 1482    7     458  | 1458 3     145   | 6    45  9  ----------------+------------------+-------------34   346   124  | 349  2467  23479 | 1349 8   5  345  34568 2458 | 3459 1     23459 | 349  247 3677    9     1245 | 3458 24568 2345  | 134  124 36 ----------------+------------------+-------------8    45    7    | 6    9     145   | 145  3   2  1345 345   6    | 2    45    8     | 7    9   14 145  2     9    | 1345 457   13457 | 1458 6   148`

Found a simple net (couldn't quite get it to a chain) that cracks this to SSTS to solve:

r1c9=3 r5c9=7 r6c9=6 [r6c5<>6] r4c5=6 r4c6=7 [r4c5<>2 + r4c6<>2] => r4c3=2 r6c3=1 ... and ...
r1c9=7 [r1c8<>7] r5c8=7 r6c8=2
So: r6c8<>1

Other solutions?

Cheers ...

- drac
Draco

Posts: 143
Joined: 14 March 2008

No other solution, just a shorter net:
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`Forcing Net Contradiction in r6 => [r6c8]<>1  [r6c8]=1=>[r5c8]=2=>[r5c9]=7=>[r6c9]=6=>[r4c5]=6=>([r4c5]<>2=>)[r4c6]=7=>[r4c3]=2=>[r6c3]=1`

Without net I need a triple chain:
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`Forcing Chain Contradiction in r4 => [r6c8]<>1  [r6c8]=1=>[r4c3]=1=>[r4c3]<>2  [r6c8]=1=>[r5c8]=2=>[r5c9]=7=>[r6c9]=6=>[r4c5]=6=>[r4c5]<>2  [r6c8]=1=>[r5c8]=2=>[r5c9]=7=>[r6c9]=6=>[r4c5]=6=>[r4c6]=7=>[r4c6]<>2`
hobiwan
2012 Supporter

Posts: 321
Joined: 16 January 2008
Location: Klagenfurt

Another: Evil from Life.com 5-23-08

Starting with:
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`2 . . | 6 . . | . . 3. . . | . . . | . . .8 7 . | 9 . 5 | . 1 .------+-------+------. . 2 | 5 . 3 | 7 . .. . 5 | . 6 . | 4 . .. . 6 | 2 . 1 | 3 . .------+-------+------. 1 . | 3 . 6 | . 4 5. . . | . . . | . . .5 . . | . . 7 | . . 8`

SSTS takes us to:
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`2   45   19 | 6  17   48  | 58 79 3 346 3456 19 | 78 1237 248 | 58 79 248   7    34 | 9  23   5   | 6  1  24------------+-------------+---------1   49   2  | 5  49   3   | 7  8  6 37  39   5  | 78 6    89  | 4  2  1 47  8    6  | 2  47   1   | 3  5  9 ------------+-------------+---------9   1    7  | 3  8    6   | 2  4  5 346 2346 8  | 14 5    29  | 19 36 7 5   2346 34 | 14 29   7   | 19 36 8 `

I would guess there are many ways to crack it from here; I went after r2c9:

r2c4=7 r5c4=8 r5c6=9 r8c7=9 r8c4=1 r9c4=4 [r9c3<>4] r3c3=4 r2c9=4
and r2c4=8 r5c6=8 r8c6=9 r2c6=2 ==> r2c9<>2.

SSTS to solve.
Draco

Posts: 143
Joined: 14 March 2008

ALS-xz:

ALS A: r3c35={234}
ALS B: r9c457={1249}
restricted common: x=2 (r39c5)
common: z=4 (r3c3+r9c4)

=> r9c3, seeing r3c3+r9c4, can't be 4, must be 3

SSTS to solve.
udosuk

Posts: 2698
Joined: 17 July 2005

ALS and Kraken/mutant fish are two techniques I have net yet learned (maybe I am spending too much time on forcing net / chain variants). Thanks udosuk.

Cheers...

- drac
Draco

Posts: 143
Joined: 14 March 2008

FYI the ALS is just an elegant presentation of logical deductions, and most (if not all) can be represented as a forcing chain. E.g. this move as an XY-chain:

[r9c3] - 4 - [r9c4] - 1 - [r9c7] - 9 - [r9c5] - 2 - [r3c5] - 3 - [r3c3] - 4 - [r9c3]

=> r9c3 <> 4

udosuk

Posts: 2698
Joined: 17 July 2005

Speaking of chain variants:

Almost Locked Set XY-Chain: A=[r3c39] - {234}, B=[r9c3] - {34}, C=[r9c4] - {14}, D=[r9c57] - {129}, RCs=1,3,4, X=2 => [r3c5]<>2
Singles (plus one XY-Wing)

Since two of the four ALS are just bivalue cells, you could probably write it as forcing chain or as nice loop.

A "normal" chain for another cell:

Forcing Chain Contradiction in r9 => [r9c7]=1
[r9c7]<>1=>[r9c7]=9=>[r4c5]=9=>[r4c2]=4=>[r9c2]<>4
[r9c7]<>1=>[r9c7]=9=>[r9c5]=2=>[r3c5]=3=>[r3c3]=4=>[r9c3]<>4
[r9c7]<>1=>[r9c4]=1=>[r9c4]<>4
Singles
hobiwan
2012 Supporter

Posts: 321
Joined: 16 January 2008
Location: Klagenfurt

udosuk wrote:FYI the ALS is just an elegant presentation of logical deductions, and most (if not all) can be represented as a forcing chain. E.g. this move as an XY-chain:

[r9c3] - 4 - [r9c4] - 1 - [r9c7] - 9 - [r9c5] - 2 - [r3c5] - 3 - [r3c3] - 4 - [r9c3] => r9c3 <> 4

I had not thought of that; thanks! Now I'll be looking for chains on ALS's that I see. For instance, a nice short one giving the same results (the same bi-value as my first chain for its start):

r2c4=7 r6c5=7 r4c5=4 r8c7=9 r8c4=1 r9c4=4 +
r2c4=8 r5c6=8 r8c6=9 r2c6=2 r3c5=3 r3c3=4 => r9c3<>4

No doubt there are even more ways through this particular door. Thanks again for the explanation!

Cheers...

- drac
Draco

Posts: 143
Joined: 14 March 2008

Here's a freshly generated puzzle with one path through -- others?
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`......2.7.....914..5.2.....62.......4..576..1.......89.....4.5..846.....2.6......1389 4   139  | 138   6    5    | 2  39  7 378  6   2    | 378   38   9    | 1  4   5 1379 5   1379 | 2     4    137  | 89 6   38--------------+-----------------+----------6    2   13   | 1389  1389 138  | 5  7   4 4    9   8    | 5     7    6    | 3  2   1 137  137 5    | 4     13   2    | 6  8   9 --------------+-----------------+----------1379 137 1379 | 138   2    4    | 78 5   6 5    8   4    | 6     139  137  | 79 139 2 2    137 6    | 13789 5    1378 | 4  139 38`

r2c5=3 r6c5=1
r2c5=3 r4c5=8 r8c5=9 r8c7=7 r3c7=9 r3c9=8 r9c9=3
r4c5=8 r4c4=9
[r6c5<>3 + r4c5<>3 + r4c4<>3] = r4c6=3
[r8c5<>3] + r4c6=3 [r8c6<>3] = r8c8=3
Cannot have two 3’s in box 9, ergo: r2c5<>3

Clear a couple of singles, then:
r3c7=8 r7c4=8 + r3c7=9 r1c3=9 r1c4=1 => r7c4<>1

SSTS to solve

Cheers...

- drac
Draco

Posts: 143
Joined: 14 March 2008

Code: Select all
`......2.7.....914..5.2.....62.......4..576..1.......89.....4.5..846.....2.6...... r3  b3  Locked Candidate 1              <> 8    [r3c16] r7  b7  Locked Candidate 1              <> 9    [r7c47]   c36   X-Wing    f/s                   <> 7    [r7c4] +-----------------------------------------------------------------------+ |  1389   4      139    |  138    6      5      |  2      39     7      | |  378    6      2      |  378    38     9      |  1      4      5      | |  1379   5      1379   |  2      4      137    |  89     6      38     | |-----------------------+-----------------------+-----------------------| |  6      2      13     |  1389   1389   138    |  5      7      4      | |  4      9      8      |  5      7      6      |  3      2      1      | |  137    137    5      |  4      13     2      |  6      8      9      | |-----------------------+-----------------------+-----------------------| |  1379   137    1379   |  138    2      4      |  78     5      6      | |  5      8      4      |  6      139    137    |  79     139    2      | |  2      137    6      |  13789  5      1378   |  4      139    38     | +-----------------------------------------------------------------------+`

It can be solved with: chains, XYZ-Wings, and an XY-Chain; -or-

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`r1c8=3 r89c8=19 r8c7=7 r7c8=8 r7c4,r8c6=13 r8c5=9 r8c8=1 r8c6=3 r7c4=1 ...                r2c5=3       /               \r1c4=8                   contradiction!  =>  [r1c8]<>3       \               /         r9c4=7 r2c4=3`

Singles to complete.
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

Nice network Danny; your new algorithm is really cranking 'em out!

Cheers...

- drac
Draco

Posts: 143
Joined: 14 March 2008

Draco wrote:Nice network Danny; your new algorithm is really cranking 'em out!

Actually, it's not from my new algorithm. In fact, it's from egg on my face in my previous solver. I implemented chains wrong in it and ended up with a great network generator. Unfortunately, it doesn't record the steps, so I have to manually recreate them if I want to use them.

The new algorithm doesn' have n-tuples and locked candidates capability, yet. However, it records the steps!!!

Cheers
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

Draco wrote:Here's a freshly generated puzzle with one path through -- others?

I tried to come up with an ALS-solution, but I still need one chain:

Almost Locked Set XZ-Rule: A=[r7c123] - {1379}, B=[r7c7],[r9c9] - {378}, X=7, Z=3 => [r9c2]<>3
Locked Candidates Type 1 (Pointing): 3 in b7 => [r7c4]<>3
Discontinuous Nice Loop r2c4 =7= r2c1 -7- r3c3 =7= r7c3 -7- r7c7 -8- r7c4 -1- r1c4 =1= r3c6 =7= r2c4 => r2c4<>3, r2c4<>8
Naked Single: [r2c4]=7
Almost Locked Set XY-Wing: A=[r3c679] - {1389}, B=[r7c47] - {178}, C=[r8c7] - {79}, Y,Z=7,9, X=1 => [r1c4],[r89c6]<>1
SSTS
hobiwan
2012 Supporter

Posts: 321
Joined: 16 January 2008
Location: Klagenfurt