## Solution to Ruud's #2/50000

Advanced methods and approaches for solving Sudoku puzzles

### Solution to Ruud's #2/50000

Having just sunk Ruud's flagship #1/50000 on the Discussions Forum, I thought I'd sink his #2 on this.

RUUD'S #2/50000 SOLVED BY FORCING NET, CCW AND AIC

The puzzle:
053070008000000003001300000060800009040090060200007010000004600500000000700050290 or, more legibly,

053 070 008
000 000 003
001 300 000

060 800 009
040 090 060
200 007 010

000 004 600
500 000 000
700 050 290

The solution to Ruud's #2/50000 is very short. Only one forcing net required, including a whiff of CCW and a tiny AIC, to eliminate 4k3, and that's it!
(1) 6c9, 2e9, 1ab7, 1de1, 7de3, 8ef7, 4hk3, 6hk3, 7hk9, 46hk3, 146k349.

(2) ?4k3, 1k9, 6k4, 6h3, 6f5, 45f49, 289bfg3,
( CCW: P: 3f2; Q: 3f7.
p: -3d1; q: -3de7, 3d8, -3d1.) -3d1, 1d1,
(AIC: 3e1=f2 - k2=k6) -3e6, 15e46, 4f4, 5f9, 7g9, 4h9, 3h7, 8h8, 8f7, 9f3, 3f2, 8e1, 7e7, 4d7, 3d8, 5e3, 1e46 ?? -4k3, 6k3, singles to solution.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

To make your "walkthrough" even shorter, you could ditch the commas and just leave one space after each move...

Also, some people like to use "." instead of "0" to represent a blank cell when posting, and include them within [code][/code] tags, such as:
Code: Select all
`.53.7...8........3..13......6.8....9.4..9..6.2....7.1......46..5........7...5.29..53|.7.|..8...|...|..3..1|3..|...---+---+---.6.|8..|..9.4.|.9.|.6.2..|..7|.1.---+---+---...|..4|6..5..|...|...7..|.5.|29.`

But it's probably a personal preference...
udosuk

Posts: 2698
Joined: 17 July 2005

udosuk, gurth comes from another forum where Code blocks aren't available and zero (0) works better than period (.) on proportionally spaced text. He's following a habit!

BTW, I find Forcing Nets more trouble than they're worth in most cases. Especially when you have to resort to an elimination chain anyway.
Code: Select all
`SSTSr6c9    =  5     [r6c9]=4,[r9c9]=1,[r9c4]=6,[r6c4]=5, ...                 [r5c4]=1,[r5c6]=3,[r5c1]=8,[r6c3]=9,[r6c2]=3 => [r9]=INVALIDSSTS`
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

### Another Solution

Code: Select all
` *--------------------------------------------------------------------* | 469     5       3   | 12469   7       1269   | 149     24      8   | | 4689    2789    289 | 124569  12468   125689 | 14579   2457    3   | | 489     2789    1   | 3       248     2589   | 4579    2457    6   | |---------------------+------------------------+---------------------| | 13      6       57  | 8       1234    1235   | 3457    3457    9   | | 138     4       578 | 15      9       135    | 3578    6       2   | | 2       389     589 | 456     346     7      | 3458    1       45  | |---------------------+------------------------+---------------------| | 389     12389   289 | 1279    1238    4      | 6       358     157 | | 5       12389   46  | 12679   12368   123689 | 34      348     147 | | 7       38      46  | 16      5       38     | 2       9       14  | *--------------------------------------------------------------------*`

r5c1=1 or r7c1=9. But:

[r5c1]-1-[r5c4]-5-[r6c4]-{XY-Wing: [r6c9]-4-[r6c4]-6-[r9c4]-1-[r9c9]-4-
[r6c9]}-4-[r6c9]-5-[r7c9]=5=[r7c8]=8=[r8c8]=3=[r4c8]-3-[r4c1]-1-
[r5c1], => r5c1<>1.

So, r7c1=9. After that, the same XY-Wing solves the puzzle.

Carcul
Carcul

Posts: 724
Joined: 04 November 2005

Carcul wrote:r5c1=1 or r7c1=9

It'd have been nice if you had explained that move a little... Some folks might not be able to see it directly...
udosuk

Posts: 2698
Joined: 17 July 2005

I still couldn't understand the above move... I know if we eliminate 1 from r5c1 and 9 from r7c1 the puzzle will run into contradiction eventually, but that's like 50 moves later on Simple Sudoku...

Could some kind soul (Carcul or others) help me out by explaining why r5c1=1 or r7c1=9 must be true in a short logical line? Thanks heaps!
udosuk

Posts: 2698
Joined: 17 July 2005

For all troubled souls like me who didn't understand the above move, here is the explanation provided by our sudoku grandmaster, Carcul :
Carcul wrote:For reference, here is the grid:

Code: Select all
` *--------------------------------------------------------------------*  | 469     5       3   | 12469   7       1269   | 149     24      8   |  | 4689    2789    289 | 124569  12468   125689 | 14579   2457    3   |  | 489     2789    1   | 3       248     2589   | 4579    2457    6   |  |---------------------+------------------------+---------------------|  | 13      6       57  | 8       1234    1235   | 3457    3457    9   |  | 138     4       578 | 15      9       135    | 3578    6       2   |  | 2       389     589 | 456     346     7      | 3458    1       45  |  |---------------------+------------------------+---------------------|  | 389     12389   289 | 1279    1238    4      | 6       358     157 |  | 5       12389   46  | 12679   12368   123689 | 34      348     147 |  | 7       38      46  | 16      5       38     | 2       9       14  |  *--------------------------------------------------------------------* `

(In my private "writtings" regarding Sudoku solutions, I use the symbol "~" with the meaning "is not", which is somewhat different from the symbol "<>" meaning "cannot be").

Now, if r5c1~1 and r7c1~9 then:

{Nice Loop: [r5c6]-3-[r5c1]=3=[r7c1]-3-[r9c2]=3=[r9c6]-3-[r5c6]}-3-
-[r5c46]-5-[r6c4]-{XY-Wing: [r6c9]-4-[r6c4]-6-[r9c4]-1-[r9c9]-4-
[r6c9]}-4-[r6c9]-5-[r7c9]=5=[r7c8]=8=[r8c8](=3=[r4c8]-3-[r4c5|r6c7])
(-8-[r8c56])=3=[r8c7]-3-[r8c56]-{TILA(3): r6c5|r6c2|r5c1|r7c1|r7c5}

and so r5c1=1 or r7c1=9.

In "forcing chain" mode, it means:

r5c1~1 & r7c1~9
=> naked pair {38} in r57c1 => r5c6<>3 (turbot fish of [3] on c1/r9) => naked pair {15} in r5c46 => r6c4<>5 (=4|6) => r6c9<>4 (=5) (xy-wing r69c49) => r7c9<>5 => r7c8=5 => r8c8=8 => ...
... => r4c8=3 => r4c56<>3, r6c7<>3
... => r8c7=3 => r8c56<>3
=> we have a "conjugating loop" of [3] with odd length in (r7c1-r5c1-r6c2-r6c5-r7c5) => contradiction

Therefore r5c1=1 or r7c1=9

Anyone could spot a shorter move?
udosuk

Posts: 2698
Joined: 17 July 2005

In fact, we don't really need that or relationship... SSTS plus a short forcing chain for r5c1<>1 (which Carcul already did) solves the puzzle:

Code: Select all
`SSTS -> *-----------------------------------------------------------------------------* | 469     5       3       | 12469   7       1269    | 149     24      8       | | 4689    2789    289     | 124569  12468   125689  | 14579   2457    3       | | 489     2789    1       | 3       248     2589    | 4579    2457    6       | |-------------------------+-------------------------+-------------------------| |#13      6       57      | 8       1234    1235    | 3457   #3457    9       | |-138     4       578     |#15      9       135     | 3578    6       2       | | 2       389     589     |#456     346     7       | 3458    1      #45      | |-------------------------+-------------------------+-------------------------| | 389     12389   289     | 1279    1238    4       | 6      #358    #157     | | 5       12389   46      | 12679   12368   123689  | 34     #348     147     | | 7       38      46      | 16      5       38      | 2       9       14      | *-----------------------------------------------------------------------------*r5c1=1 => r5c4=5 => r6c4<>5 (=4|6) => r6c9<>4 (=5) (xy-wing r69c49)=> r7c9<>5 => r7c8=5 => r8c8=8 => r4c8=3 => r4c1=1Two 1s on c1!Hence r5c1<>1-> SSTS -> the puzzle is solved`

udosuk

Posts: 2698
Joined: 17 July 2005

Yet another solution:
r8c7=3 => r4c8=3 => r4c1=1
r8c7=4 => r6c9=4 => r4c5=4 => r4c6=2 => r4c1=1
=> r4c1=1 and it solves with turbot fish and xy-wing
ravel

Posts: 998
Joined: 21 February 2006

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