first advanced move is hard

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first advanced move is hard

Postby ncantoral » Sun May 18, 2008 10:37 pm

Code: Select all
5 . .|6 . .|. 7 .
. . .|. 4 .|. . 2
9 . .|. . 1|. 5 .
-----+-----+-----
4 . 3|. 7 .|1 . .
. . .|. . .|. . .
. . 6|. 8 .|9 . 3
-----+-----+-----
. 4 .|7 . .|. . 8
1 . .|. 2 .|. . .
. 3 .|. . 4|. . 9


Code: Select all
.---------------------.---------------------.---------------------.
| 5      28     4     | 6      9      28    | 3      7      1     |
| 3      168    18    | 5      4      7     | 68     9      2     |
| 9      2678   278   | 28     3      1     | 468    5      46    |
:---------------------+---------------------+---------------------:
| 4      2589   3     | 29     7      2569  | 1      268    56    |
| 28     12589  12589 | 12349  156    23569 | 24567  2468   4567  |
| 27     1257   6     | 124    8      25    | 9      24     3     |
:---------------------+---------------------+---------------------:
| 26     4      259   | 7      156    3569  | 256    1236   8     |
| 1      5789   5789  | 389    2      35689 | 4567   346    4567  |
| 2678   3      2578  | 18     156    4     | 2567   126    9     |
'---------------------'---------------------'---------------------'


AIC challenge here. can't find a nice solution.
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Postby Steve R » Mon May 19, 2008 12:31 am

Eliminate 2 from r9c8:

r9c8 = 2 => r6c8 = 4 => r6c4 ≠ 4 => r5c4 = 4 => r5c6 = 3 => r7c6 ≠ 3 => r7c8 = 3 => r9c8 = 1

Eliminate 6 from r7c5:

r7c5 = 6 => r7c1 ≠ 6 => r9c1 = 6 => r9c8 = 1 => r7c8 ≠ 1 => r7c5 = 1

Eliminate 1 from r9c4:

XYZ-wing pivoted on r9c5 with pincers r7c5 and r9c8

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Postby submacrolize » Mon May 19, 2008 2:03 am

You can view the solution here...

Click on the gray squares that you want to see. Right click to view the whole solution.
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Postby RW » Mon May 19, 2008 4:56 am

submacrolize, could you please stop posting that link in every thread on the help forum. People here are not interested in the solution as such, but the path that leads to the solution. Clicking a gray square to solve a number is not fun, not challenging and it doesn't bring you a feeling of intellectual accomplishment. It's about as enjoyable as cheating in solitaire.

And for the record, I'm sure ncantoral knew the correct solution long before he posted the puzzle here.

RW
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Postby df » Mon May 19, 2008 3:05 pm

I personally find the links to the solutions very helpful. If I'm stuck on a puzzle and I'm trying to find the solution, having the actual solution is a nice reference. Like you said, it's not about the solution, but the methods employed to arrive at it. Please continue posting these links, submacrolize.
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Postby ronk » Mon May 19, 2008 3:24 pm

df wrote:Please continue posting these links, submacrolize.

Sorry, but I agree with RW. submacrolize's repeated promotion of his website and/or Java software is simply spam.
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Postby RW » Mon May 19, 2008 3:32 pm

df wrote:I personally find the links to the solutions very helpful. If I'm stuck on a puzzle and I'm trying to find the solution, having the actual solution is a nice reference. Like you said, it's not about the solution, but the methods employed to arrive at it. Please continue posting these links, submacrolize.

How nice to see a poster, whose only post so far has been about promoting sudoku100.com, step up to defend another poster whose every post so far has been about promoting the same site. Submacrolize, are you using multiple accounts?
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Postby Draco » Mon May 19, 2008 6:48 pm

One somewhat complex forcing net will reduce the puzzle to singletons. For the #1 path (the network), the nodes are numbred in order of the paths that must be followed. There is a break at segment #4 to keep things from getting too awfully wide. Unfortuantely, what I saw in PREVIEW vs. what posts is different so you will need to copy/paste into Notepad or some other monospaced font editor to see this...
Code: Select all
#1 (net):                 (2)+-[r8c4<>3] r5c4=3 r6c4=4 r6c8=2 [r6c8<>4] => r5c8=4 r4c8=8
                             |                                |                 |      |
r9c4=1 [r9c8<>1] r7c8=1 r7c6=3 r7c3=9                         |                 |      |
                      |                                       |                 |      |
                   (1)+-r8c8=3 [r8c8<>4] + -------------------+                 |      |
                      |                                                         |      |
                      |                    +------------------------------------+      |
                      |                    |                                           |
                      |                    |         +---------------------------------+
                      |                    |         |
                   (3)[r7c8<>6 + r8c8<>6 + r5c8<>6 + r4c8<>6] => r9c8=6

(break for segment #4)
[r8c8<>6] + r9c8=6 [r8c7<>6 r8c9<>6] => r8c6=6 [r8c6<>9] + r7c3=9 [r8c2<>9 r8c3<>9] => r8c4=9
=================
#2 (chain): r9c4=8 r3c4=2

Cancellations: r5c4<>2, r8c4<>8

The network could produce several more cancellations, but that only adds to the complexity... why bother once you have it reduced to singles?

Cheers...

- drac
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Postby daj95376 » Mon May 19, 2008 8:26 pm

Draco, or ...

Code: Select all
                                                      *****************
[r9c4]=1 [r7c8]=1 [r7c6]=3 [r5c4]=3 [r6c4]=4 [r6c8]=2 [r9c8]=6 [r7c1]=6
      |                                            |        ^
      ---------------------------------------------+---------

However, in the original PM:

Code: Select all
( [r9c8]=6 and [r7c1]=6 ) => [r79c5]=15 => [r9c4]<>1
Last edited by daj95376 on Mon May 19, 2008 10:49 pm, edited 1 time in total.
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Postby Draco » Tue May 20, 2008 2:48 am

daj95376 wrote:Draco, or ...

Code: Select all
                                                      *****************
[r9c4]=1 [r7c8]=1 [r7c6]=3 [r5c4]=3 [r6c4]=4 [r6c8]=2 [r9c8]=6 [r7c1]=6

However, in the original PM:

Code: Select all
( [r9c8]=6 and [r7c1]=6 ) => [r79c5]=15 => [r9c4]<>1


Hi Danny,

Nice loop (so to speak).

I like having the explicit "can't be this" bits in there when the chain / net jumps squares like this one does, else I loose track. All a matter of taste I suppose.

And... damn NICE loop to crack it:)

Cheers...

- drac
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Postby daj95376 » Tue May 20, 2008 3:32 am

Draco wrote:Nice loop (so to speak).

I like having the explicit "can't be this" bits in there when the chain / net jumps squares like this one does, else I loose track. All a matter of taste I suppose.

And... damn NICE loop to crack it:)

Thanks for the complement (so to speak).:D

I updated my post to show where it stops being a chain and becomes a network.

I understand your desire for the full interaction being shown. It wasn't until I started writing my chain routine that I drew to dislike tracking eliminations -- multiple elimination paths leading to the same assignment. It doubled the size of my BFS table every time it occurred.

In my first pair, there are two ways the second assignment can occur. Which way is correct? Does it really matter which elimination is selected?

Code: Select all
A)   [r9c4]=1 [r7c5]<>1 [r7c8]=1
B)   [r9c4]=1 [r9c8]<>1 [r7c8]=1

After the first assignment in this sequence, there's only one way the remaining assignments can occur. Does the elimination information really add anything to where the chain is going?

Code: Select all
[r7c8]=1 [r7c6]=3 [r5c4]=3 [r6c4]=4 [r6c8]=2

Cheers back at you:!:
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Postby ncantoral » Tue May 20, 2008 6:03 am

RW wrote:submacrolize, could you please stop posting that link in every thread on the help forum. People here are not interested in the solution as such, but the path that leads to the solution. Clicking a gray square to solve a number is not fun, not challenging and it doesn't bring you a feeling of intellectual accomplishment. It's about as enjoyable as cheating in solitaire.

And for the record, I'm sure ncantoral knew the correct solution long before he posted the puzzle here.

RW


ha! the solution is an afterthought. I like the journey.

r9c8 = 2 => r6c8 = 4 => r6c4 ≠ 4 => r5c4 = 4 => r5c6 = 3 => r7c6 ≠ 3 => r7c8 = 3 => r9c8 = 1


this is a string of strong and weak links, AIC, correct?? and it leads to a contradiction. does it take practice to see this? because I would never know to look for it.
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Postby Glyn » Tue May 20, 2008 9:25 am

Just convinced myself it is an AIC so I thought I'd add it in that form:

(2=4)r6c8-(4)r6c4=(4-3)r5c4=(3)r5c6-(3)r7c6=(3-1)r7c8=(1)r9c8

Either r6c8=2 or r9c8=1 either way r9c8<>2.
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Postby ronk » Tue May 20, 2008 10:32 am

daj95376 wrote:Does the elimination information really add anything to where the chain is going?

Dang tootin'. When posting chains, it's about communicating an inference stream to the reader ... not the size of a table somewhere. That's why we have nice-loop (NL) and alternating-inference-chain (AIC) notation in the first place.
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Postby Carcul » Sun May 25, 2008 11:49 am

After basic moves:

If any of r9c13 is 8 and simultaneously r7c3 is 9 then,

-8,9-[r8c23]-5,7-[r8c79](-6-[r9c8])=5,7|8=[r3c7]-8-[r3c4]-2-[r6c4]-
-{XY-Wing: [r9c4]-1-[r9c8]-2-[r6c8]-4-[r6c4]-1-[r9c4]}=8=[r9c4],

but this cannot be, since one of r9c13 is already 8. Therefore we have here the link =9=[r7c3]-8-[r9c13]. Let's now use it:

[r9c13]-8-[r9c4]-1-[r9c8]=1=[r7c8]=3=[r7c6]=9=[r7c3]-8-[r9c13],

and so [r9c13]<>8 which solves the puzzle.
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