## Need help with this...

Post the puzzle or solving technique that's causing you trouble and someone will help
RW wrote:I found a short trail to reduce candidate 2 from r2c7, but couldn't find any shorter than 13 steps for candidate 8. So here's the complete solution, only the neccessary steps included:

r1c2=9

RW,

Quick typo. Your first line says r1c2 = 9, when you mean r2c2=9. I went through the rest of your deduction just now. Bravo. Very well done.
re'born

Posts: 551
Joined: 31 May 2007

Thanks rep'nA, typo corrected.

RW
RW
2010 Supporter

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For those, who still seek a big challenge:
AFAIK only 3 of the 7 hardest puzzles of my list here are solved so far
ravel

Posts: 998
Joined: 21 February 2006

### Another solution

RW, thank you for your great solution on this puzzle. It has inspired me (as a beginner) to make another try on this puzzle and with Carsuls way of nice loop notation. I hope this second solution is still welcome. At the same time I include some illustrations by into Sudoku.

In RW's solution, the puzzle is solved in only 15 steps. However some of the steps are quite complex.

RW wrote:4. If r2c7=8 => r1c4=8 => r4c4=7 => r2c6=4 => r2c9=2 => r1c9=1 => r7c9=4 => r9c5=1 => r9c7=2 => r5c7=4 => r5c4=6 => r9c4=9 => r7c5=8 => r5c5=empty cell => r2c7<>8

I think the NLN of this step is like this:

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`[r2c7](-8-[r2c9])      (-8-[r5c7])      (-8-[r1c9]=8=[r1c4](-8-[r5c4])                         -8-[r4c4]-7-[r9c4])      -8-[r2c6](-4-[r2c9](-2-[r1c9])                         -2-[r9c9]=2=[r9c7]-2-[r5c7](-4-[r5c5])                                                    -4-[r5c4]-6-[r9c4]-9-[r7c5])               -4-[r3c4]=4=[r3c78]-4-[r1c9](-1-[r7c9]-4-[r7c5])                                           -1-[r9c9]=1=[r9c5](-1-[r5c5])                                                             -1-[r7c5]-8-[r5c5]=> r5c5 has no candidates => r2c7<>8.`

I have made a little addition in the notation by including a line feed when a different chain is progressed, because it makes it simpler to read by me. The into sudoku graphics is like this:

The following solution has 20 steps, but I think each step seems simpler for me. I do not claim, that it is better...

The start is:

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` 56    26    246   | 1458  7     9     | 3     1246  1248  1     9     2467  | 458   3     48    | 24578 2467  248   357   37    8     | 145   2     6     | 457   147   9    -------------------+-------------------+------------------- 4     2378  1     | 78    5     2378  | 289   239   6     368   5     9     | 468   1468  12348 | 248   234   7     3678  23678 2367  | 46789 4689  23478 | 1     5     2348 -------------------+-------------------+------------------- 2     178   5     | 3     1489  1478  | 6     1479  14    36789 13678 367   | 2     14689 1478  | 479   13479 5     3679  4     367   | 679   169   5     | 279   8     123  `

1. X-wing, [r9c1]-3-[r89c3]=3=[r6c3]-3-[r6c9]=3=[r9c9]-3-[r9c13] => r9c1 <> 3 (as RW)

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`2. [r5c8]-3-[r6c9]=3=[r9c9](=1=[r9c5])                           =2=[r9c7]-2-[r5c7]=2=[r5c6]=1=[r5c5]   => r9c5=1 and r5c5=1 - contradiction  => r5c8<>3`

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`3. [r6c123](-3-[r6c6])           -3-[r6c9](=3=[r9c9]=1=[r9c5])                    =3=[r4c8]-3-[r4c6]=3=[r5c6]=1=[r5c5]   => r9c5=1 and r5c5=1 - contradiction  => r6c123<>3`

=> c12 box line reduction => r8c12<>3

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`4. [r7c5](-1-[r5c5])         (-1-[r9c5]=1=[r9c9]=3=[r6c9]-3-[r4c8])         =9=[r7c8]-9-[r4c8](-2-[r5c7])                           -2-[r5c8](-4-[r5c45])                                    -4-[r5c7](-8-[r5c5])                                             -8-[r5c4]-6-[r5c5]    => r5c5 has no candidates => r7c5<>1`

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`5. [r5c1](-6-[r6c123])         (=3=[r4c2]-3-[r4c8])         =3=[r5c6]=1=[r5c5]-1-[r9c5]=1=[r9c9]=2=[r9c7]-2-[r5c7]=2=[r5c8]           -2-[r4c8](-9-[r8c8])                    -9-[r7c8](=9=[r8c7]-9-[r8c1]=9=[r9c1]-9-[r9c5])                             =9=[r7c5]-9-[r6c5]=9=[r6c4]=6=[r6c5]-6-[r9c5]   => r9c5 has no candidates => r5c1<>6`

=> r6c123 box line reduction => r6c45<>6

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`6. [r5c1](=3=[r4c2]-3-[r4c8]=3=[r6c9])         (=3=[r5c6]=1=[r5c5]=6=[r5c4]-6-[r9c4])         -8-[r5c7]=8=[r4c7](-8-[r4c4]-7-[r9c4])                           =9=[r4c8]-9-[r7c8]=9=[r7c5]-9-[r6c5]=9=[r6c4]-9-[r9c4]   => r9c4 has no candidates => r5c1=3   => single r3c2=3`

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`7. [r8c5](-1-[r9c5]=1=[r9c9](=2=[r9c7]-2-[r45c7])                            =3=[r6c9]-3-[r4c8])         -1-[r5c5]=1=[r5c6]=2=[r5c8]-2-[r4c8](-9-[r4c7]-8-[r4c4]-7-[r9c4])                                             -9-[r78c8]=9=[r8c7]-9-[r8c1]                                               =9=[r9c1](-9-[r9c4]-6-[r9c5])                                                        -9-[r9c5]   => r9c5 has no candidates => r8c5<>1`

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`8.  [r9c5](-9-[r7c5]=9=[r7c8](-9-[r4c8])-9-[r8c7])          =1=[r9c9](-1-[r7c9]-4-[r8c7]-7-[r8c3])                   =3=[r8c8](-3-[r8c3])                            -3-[r4c8](-2-[r5c7])                                     -2-[r5c8](-4-[r5c4])                                              -4-[r5c7]-8-[r5c4]-6-[r9c4]                                                =6=[r8c5]-6-[r8c3]    => r8c3 has no candidates => r9c5<>9`

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`9.  [r8c8](-9-[r8c7])          (-9-[r7c8]=9=[r7c5](-9-[r8c5])(-9-[r9c4])-9-[r6c5]=9=[r6c4])          =3=[r9c9](=1=[r9c5](-1-[r8c6])                             -1-[r5c5]=1=[r5c6])                   -3-[r6c9]=3=[r4c8]-3-[r4c6]=3=[r6c6]=2=[r4c6](-2-[r4c2])                                                                =7=[r4c4]                (-7-[r9c4]-6-[r8c5])                -7-[r4c2]-8-[r78c2]=8=[r8c1](-8-[r8c6])                                            -8-[r8c5](-4-[r8c7])                                                     -4-[r8c6]-7-[r8c7]    => r8c7 has no candidates => r8c8<>9`

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`10. [r9c7]=2=[r9c9](=1=[r9c5]-1-[r5c5]=1=[r5c6])                   (=3=[r9c3]-3-[r8c3])                   =3=[r6c9]-3-[r4c8]=3=[r4c6]=2=[r6c6]-2-[r6c23]=2=[r4c2]                     =7=[r4c4]-7-[r9c4]=7=[r9c1](-7-[r3c1]=7=[r2c3]-7-[r6c3]                                                -7-[r8c3]-6-[r6c3]    => r6c3 has no candidates => r9c7<>9`

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`11. [r4c7](-2-[r5c7])          (-2-[r5c8](-4-[r5c4])                    (-4-[r5c5])                    -4-[r5c7](-8-[r5c4])                             -8-[r5c5])          -2-[r9c7]=2=[r9c9]=1=[r9c5]-1-[r5c5]-6-[r5c4]    => r5c4 has no candidates => r4c7<>2`

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`12. [r2c8](-2-[r45c8])          =6=[r1c8](-6-[r1c2]-2-[r46c2]=2=[r6c3]-2-[r6c9]=2=[r5c7]-2-[r9c7])                   -6-[r1c1]-5-[r3c1]-7-[r3c78]=7=[r2c7]-7-[r9c7]    => r9c7 has no candidates => r2c8<>2`

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`13. [r2c7](-2-[r2c9])          (-2-[r2c3])          -2-[r9c7]=2=[r9c9](=1=[r9c5]-1-[r5c5]=1=[r5c6])                            =3=[r6c9]-3-[r4c8]=3=[r4c6]=2=[r6c6]-2-[r6c23]                              =2=[r4c2]-2-[r1c2]=2=[r1c3]=4=[r2c3](-4-[r2c9])                                                               -4-[r2c6]-8-[r2c9]    => r2c9 has no candidates => r2c7<>2`

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`14. [r5c7](-2-[r9c7]-7-[r23c7])          -2-[r45c8]=2=[r1c8](=6=[r2c8])                             -2-[r1c23]=2=[r2c3]=7=[r1c3]-7-[r3c78]    => Box 3 has no candidates for no. 7. => r5c7<>2    => single r9c7=2`

15. [r7c8]=9=[r4c8]=3=[r6c9]-3-[r9c9]-1-[r7c8] => r7c8<>1

16. [r7c8]=9=[r4c8]=3=[r6c9]-3-[r9c9]-1-[r7c9]-4-[r7c8] => r7c8<>4

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`17. [r7c8](=9=[r8c7]-9-[r8c1]=9=[r9c1]-9-[r9c4])          =9=[r4c8](=3=[r6c9]-3-[r9c9]-1-[r9c5]-6-[r9c4])                   -9-[r4c7]-8-[r4c4]-7-[r9c4]    => r9c4 has no candidates => r7c8=9`

=> single r4c7=9
=> pointing pair with 7 in r8
=> empty rectangle box 4 => r6c6<>7
=> empty rectangle box 2 => r5c4<>8

18. [r6c6]=3=[r4c6]-3-[r4c8]-2-[r5c8]=2=[r5c6]-2-[r6c6] => r6c6<>2

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`19. [r7c6](-4-[r8c6])          (-4-[r7c5]-8-[r8c6])          -4-[r7c9]-1-[r9c9]=1=[r9c5]-1-[r8c6]    => r8c6 has no candidates => r7c6<>4`

=> empty rectangle box 6 => r5c5<>4

20. [r5c6]=2=[r5c8]-2-[r4c8]-3-[r6c9]=3=[r9c9]=1=[r9c5]-1-[r5c5]=1=[r5c6]
=> r5c6<>4, r5c6<>8, r6c9<>2

=> pointing pair c8 => r1c8<>2
=> box line reduction => r4c2<>2
=> Naked pair 78 r4
=> single r7c6=7
=> box-line-reduction with 7 in c13
=> empty rectangle box 6 => r5c5<>4
=> empty rectangle box 4 => r6c5<>8
=> XYwing box 5 => r6c4<>9
=> just singles to solution

/Viggo
Viggo

Posts: 60
Joined: 21 April 2006

Yup, definitely works. I agree some of my steps were complex, but I suppose you also found out that it wasn't very easy to find simple reductions in this grid. Thank you for supplying the graphics, makes it a lot easier to follow, especially in the more complex steps like step 9. The line feed when moving to a different chain is a vey good idea, I've never been very good at reading NLN, but this made it a lot clearer.

One typo (?):
Viggo wrote:=> empty rectangle box 6 => r5c5<>4
=> empty rectangle box 4 => r6c5<>8
=> XYwing box 5 => r6c4<>9
=> just singles to solution

You already eliminated 4 from r5c5 with an empty rectangle earlier. But there is a skyscraper to eliminate 8 from r5c5 at this point...

Then I have some suggestions to make steps 4 and 5 shorter:

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`4. [r7c5]-1-[r5c5]=1=[r5c6]      =9=[r7c8]-9-[r4c8]      -1-[r9c5]=1=[r9c9]=3=[r6c9]-3-[r4c8]-2-[r5c78]   => no 2 in row 5`

(This is the first time i attempt NLN, what I'm trying to say is: If r7c4=1 => r5c6=1 => r7c8=9 => r9c9=1 => r6c9=3 => r4c8=2 => no 2 in row 5, tell me if the notation is wrong.)

My suggestion for step 5 is an AUR, I'm not going to attempt NLN on this one:

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`5. If r5c1=6 => r5c6=3 => r5c5=1      => r6c45=69 => AUR in r69c45: r9c4<>69 => r6c4=9 => r6c5=6    => no 6 in box 8`

Of course, no solution is better than another, but new interesting solutions are always welcome. An interesting feature of your solution is that only 4/20 steps make use of rows 1-3. The first 9 steps only used rows 4-9. Is there any particular reason to this?

RW
RW
2010 Supporter

Posts: 1000
Joined: 16 March 2006

Viggo wrote:=> empty rectangle box 6 => r5c5<>4
=> empty rectangle box 4 => r6c5<>8
=> XYwing box 5 => r6c4<>9
=> just singles to solution

These are btw exactly the same as my steps 13-15, so with your way of counting my solution was only 12 steps. But that was after I excluded all unneccessary steps. Have you checked if all your long trails are neccessary? At least the finned x-wing could be left out as the box line reduction after step 3 would make the same elimination.

RW
RW
2010 Supporter

Posts: 1000
Joined: 16 March 2006

RW wrote:I agree some of my steps were complex, but I suppose you also found out that it wasn't very easy to find simple reductions in this grid.

Yes, I agree - and some of my steps are complex too.

RW wrote:You already eliminated 4 from r5c5 with an empty rectangle earlier. But there is a skyscraper to eliminate 8 from r5c5 at this point...

You are right. 4 was eliminated in step 19. However, I'am not able to find the skyscraper. I can find a discontinuos nice loop for this elimination with three strong links:

[r5c5]-8-[r7c5]=8=[r7c2]-8-[r8c1]=8=[r6c1]-8-[r6c9]=8=[r5c7]-8-[r5c5]

RW wrote:Then I have some suggestions to make steps 4 and 5 shorter:

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`4. [r7c5]-1-[r5c5]=1=[r5c6]      =9=[r7c8]-9-[r4c8]      -1-[r9c5]=1=[r9c9]=3=[r6c9]-3-[r4c8]-2-[r5c78]   => no 2 in row 5`

(This is the first time i attempt NLN, what I'm trying to say is: If r7c4=1 => r5c6=1 => r7c8=9 => r9c9=1 => r6c9=3 => r4c8=2 => no 2 in row 5, tell me if the notation is wrong.)

I think the notation is all right, but I'am no specialist in NLN. And you are right, this step is simpler. Here is the graphics:

RW wrote:My suggestion for step 5 is an AUR...

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`5. If r5c1=6 => r5c6=3 => r5c5=1      => r6c45=69 => AUR in r69c45: r9c4<>69 => r6c4=9 => r6c5=6    => no 6 in box 8`

Unique rechtangles are still somewhat sofisticated for me. I shall try to do the graphics here:

I do not understand how you can eliminate 69 in r9c4. Please explain or make a link to some explanation on AUR.

RW wrote:An interesting feature of your solution is that only 4/20 steps make use of rows 1-3. The first 9 steps only used rows 4-9. Is there any particular reason to this?

Possibly I got familiar with the strong links in row 9 and the strong links with candidate 9, and they are placed in the lower part of the puzzle. So I choose candidates able to activate these links.

RW wrote:These are btw exactly the same as my steps 13-15, so with your way of counting my solution was only 12 steps. But that was after I excluded all unneccessary steps. Have you checked if all your long trails are neccessary? At least the finned x-wing could be left out as the box line reduction after step 3 would make the same elimination.

Yes you are right - I did not discover that. I have not made such a check. I think you mean, that some steps may not depend on previous steps, and they might be able to cause simpler way of elimination than the previous steps - am I right?

/Viggo
Viggo

Posts: 60
Joined: 21 April 2006

Viggo wrote:I think you mean, that some steps may not depend on previous steps, and they might be able to cause simpler way of elimination than the previous steps - am I right?

I mean that some eliminations might be totally unneccessary. If you read the earlier post on this thread, you'll see that I started with a few simple steps that solved r1c3=4. When I had solved the entire puzzle I noticed that this had no effect on any later (advanced) steps so I could leave out these steps and solve r1c3 as a single after the advanced steps.

The AUR step was unneccessarily long, I just noticed that it's enough with

If r5c1=6 => r6c45=69 => r69c4=69 => AUR: r9c5<>69 => r9c5=1 => r5c6=1 => no 3 in row 5

The elimination can be made because if this

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`|69    69||--------||69    69|`

was part of the solution, then the solution wouldn't be unique, and as we know all valid puzzles have unique solutions. You can read Keiths introduction to Unique rectangles here and for more information, check out the uniqueness related threads in Mike Barker's list.

RW
RW
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RW, thank you for the interesting links. It has helped med understand some more of the AUR's and your new step 5.

I'am currently looking into possible elimination of steps, as you suggested.

/Viggo
Viggo

Posts: 60
Joined: 21 April 2006

As RW suggested, some steps may be eliminated. I have tried to do that, and ends up with 14 steps as given below. I'am not sure, that I could have selected these steps beforehand. Some of the eliminated steps just turned out to be not esential for solving the puzzle.

As RW commented, I choose not to count empty rechtangles, XY-wing etc. as a step.

Step 1, 2, and 3 is as before:

1. X-wing, [r9c1]-3-[r89c3]=3=[r6c3]-3-[r6c9]=3=[r9c9]-3-[r9c13] => r9c1 <> 3 (as RW)

Code: Select all
`2. [r5c8]-3-[r6c9]=3=[r9c9](=1=[r9c5])                           =2=[r9c7]-2-[r5c7]=2=[r5c6]=1=[r5c5]   => r9c5=1 and r5c5=1 - contradiction  => r5c8<>33. [r6c123](-3-[r6c6])           -3-[r6c9](=3=[r9c9]=1=[r9c5])                    =3=[r4c8]-3-[r4c6]=3=[r5c6]=1=[r5c5]   => r9c5=1 and r5c5=1 - contradiction  => r6c123<>3    => c12 box line reduction => r8c12<>3(previous step 5)4b. [r5c1](-6-[r6c123])          (=3=[r4c2]-3-[r4c8])          =3=[r5c6]=1=[r5c5]-1-[r9c5]=1=[r9c9]=2=[r9c7]-2-[r5c7]=2=[r5c8]            -2-[r4c8](-9-[r8c8])                     -9-[r7c8](=9=[r8c7]-9-[r8c1]=9=[r9c1]-9-[r9c5])                              =9=[r7c5]-9-[r6c5]=9=[r6c4]=6=[r6c5]-6-[r9c5]    => r9c5 has no candidates => r5c1<>6    => r6 box line reduction => r6c45<>6(previous step 6)5b. [r5c1](=3=[r4c2]-3-[r4c8]=3=[r6c9])          (=3=[r5c6]=1=[r5c5]=6=[r5c4]-6-[r9c4])          -8-[r5c7]=8=[r4c7](-8-[r4c4]-7-[r9c4])                            =9=[r4c8]-9-[r7c8]=9=[r7c5]-9-[r6c5]=9=[r6c4]-9-[r9c4]    => r9c4 has no candidates => r5c1=3    => single r3c2=3(previous step 10)6b. [r9c7]=2=[r9c9](=1=[r9c5]-1-[r5c5]=1=[r5c6])                   (=3=[r9c3]-3-[r8c3])                   =3=[r6c9]-3-[r4c8]=3=[r4c6]=2=[r6c6]-2-[r6c23]=2=[r4c2]                     =7=[r4c4]-7-[r9c4]=7=[r9c1](-7-[r3c1]=7=[r2c3]-7-[r6c3]                                                -7-[r8c3]-6-[r6c3]    => r6c3 has no candidates => r9c7<>9(previous step 11)7b. [r4c7](-2-[r5c7])          (-2-[r5c8](-4-[r5c4])                    (-4-[r5c5])                    -4-[r5c7](-8-[r5c4])                             -8-[r5c5])          -2-[r9c7]=2=[r9c9]=1=[r9c5]-1-[r5c5]-6-[r5c4]    => r5c4 has no candidates => r4c7<>2(previous step 12)8b. [r2c8](-2-[r45c8])          =6=[r1c8](-6-[r1c2]-2-[r46c2]=2=[r6c3]-2-[r6c9]=2=[r5c7]-2-[r9c7])                   -6-[r1c1]-5-[r3c1]-7-[r3c78]=7=[r2c7]-7-[r9c7]    => r9c7 has no candidates => r2c8<>2(previous step 13)9b. [r2c7](-2-[r2c9])          (-2-[r2c3])          -2-[r9c7]=2=[r9c9](=1=[r9c5]-1-[r5c5]=1=[r5c6])                            =3=[r6c9]-3-[r4c8]=3=[r4c6]=2=[r6c6]-2-[r6c23]                              =2=[r4c2]-2-[r1c2]=2=[r1c3]=4=[r2c3](-4-[r2c9])                                                               -4-[r2c6]-8-[r2c9]    => r2c9 has no candidates => r2c7<>2(previous step 14)10b. [r5c7](-2-[r9c7]-7-[r23c7])           -2-[r45c8]=2=[r1c8](=6=[r2c8])                              -2-[r1c23]=2=[r2c3]=7=[r1c3]-7-[r3c78]     => Box 3 has no candidates for no. 7. => r5c7<>2     => single r9c7=2(almost as previous step 9)11b. [r8c8](-9-[r8c7])           (-9-[r7c8]=9=[r7c5](-9-[r8c5])(-9-[r9c4])-9-[r6c5]=9=[r6c4])           =3=[r9c9](=1=[r9c5](-1-[r8c6])                              (-1-[r7c5])                              -1-[r5c5]=1=[r5c6])                    -3-[r6c9]=3=[r4c8]-3-[r4c6]=3=[r6c6]=2=[r4c6](-2-[r4c2])                                                                 =7=[r4c4]                 (-7-[r9c4]-6-[r8c5])                 -7-[r4c2]-8-[r78c2]=8=[r8c1](-8-[r8c6])                                             -8-[r8c5](-4-[r8c7])                                                      -4-[r8c6]-7-[r8c7]     => r8c7 has no candidates => r8c8<>9(previous step 17 with ekstra eliminations)12b. [r7c8](=9=[r8c7]-9-[r8c1]=9=[r9c1]-9-[r9c4])           =9=[r4c8](=3=[r6c9]-3-[r9c9]-1-[r9c5]-6-[r9c4])                    -9-[r4c7]-8-[r4c4]-7-[r9c4]     => r9c4 has no candidates => r7c8=9`

=> single r4c7=9
=> pointing pair with 7 in r8 => r8c1236<>7
=> empty rectangle box 4 => r6c6<>7
=> empty rectangle box 2 => r5c4<>8

(previous step 20 with ekstra eliminations)
13b. [r5c6]=2=[r5c8]-2-[r4c8]-3-[r6c9]=3=[r9c9]=1=[r9c5]-1-[r5c5]=1=[r5c6]
(continuous nice loop)
=> r5c6<>4, r5c6<>8, r6c9<>2, r78c5<>1
=> pointing pair c8 => r1c8<>2

(previous step 18)
14b. [r6c6]=3=[r4c6]-3-[r4c8]-2-[r5c8]=2=[r5c6]-2-[r6c6] => r6c6<>2

=> box line reduction => r4c2<>2
=> Naked pair 78 r4
=> single r7c6=7
=> box-line-reduction with 7 in c13
=> empty rectangle box 6 => r5c5<>4
=> empty rectangle box 4 => r6c5<>8
=> XY-wing box 5 => r6c4<>9
=> just singles to solution

I have compared the solution with RW's:

Step 1 is the same.
Step 7b eliminates the same candidate as RW's step 6, but in another way.
Step 10b eliminates 9 in r9c7 as RW's step 7, but in another way.
Step 2 eliminates the same candidate as RW's step 8 almost the same way.
Step 12b eliminates 8 in r4c7 as RW's step 9, but in another way.
Step 10b eliminates 2 in r5c7 as RW's step 9, but in another way.
Step 5b eliminates 7 in r3c2 as RW's step 10, but in another way.
Step 14b eliminates 7 in r7c2 as RW's step 11, but in another way.
Step 12b eliminates 7 in r7c8 as RW's step 12, but in another way.

The following graphics illustrate the elimination until the later simpler reductions:

The blue dots represent my "hard" eliminations, and the brown dots represent the implied easier removed candidates. The pink dots represent RW's "hard" eliminations, and the yellow dots are on candidates, where RW's and my eliminations are the same.

In general I'think the two ways, the puzzle is solved, are very different. Except for the first finned X-wing and step 2, all the error nets are different. When other equal candidates are eliminated, the premis are different too. However some of the later simpler reductions are the same.

/Viggo
Viggo

Posts: 60
Joined: 21 April 2006

Viggo wrote:2. [r5c8]-3-[r6c9]=3=[r9c9](=1=[r9c5])
=2=[r9c7]-2-[r5c7]=2=[r5c6]=1=[r5c5]
=> r9c5=1 and r5c5=1 - contradiction => r5c8<>3

Starting the error net with r9c9=3 would be a shorter net, allow elimination at a strong link, and make "hard" step 3 unnecessary.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

ronk wrote:Starting the error net with r9c9=3 would be a shorter net, allow elimination at a strong link, and make "hard" step 3 unnecessary.

I'have just tried that, but it seems complicated to me. Can you supply the error net starting with r9c9=3?

/Viggo
Viggo

Posts: 60
Joined: 21 April 2006

Viggo wrote:Can you supply the error net starting with r9c9=3

Nope. I made my statement based solely on your error net expression. Were it correct, you could drop off the first two "dangling" links of ...

[r5c8]-3-[r6c9]=3=[r9c9](=1=[r9c5])=2=[r9c7]-2-[r5c7]=2=[r5c6]=1=r5c5]

But looking at the actual puzzle, I see a link ... [r5c8]=2=[r5c67] ... was omitted. I'm not quite sure how to include that, but maybe this works:

[r5c8](=2=[r5c67])-3-[r6c9]=3=[r9c9](=1=[r9c5])=2=[r9c7]-2-[r5c7]=2=[r5c6]=1=r5c5]
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Ronk, thank you for your comment!

Unfortunately, I'am still confused regarding your statement. Do you think, that you are not able to exclude the candidate 3 in r5c8 in this way?

When you start the error net by r5c8=3, then r5c8<>2. In this way you at the same time prepare the strong link [r5c6]=2=[r5c7]. Do you think, that you need to make a specific statement about that in the notation - like in your red comment?

In the notation I'have used with this puzzle, I'have at more places prepared strong links by previously excluding candidates in the error net and with no comment about that.

/Viggo
Viggo

Posts: 60
Joined: 21 April 2006

Viggo wrote:Do you think, that you are not able to exclude the candidate 3 in r5c8 in this way?

To clarify, the deduction is AOK. I only took issue with the correctness -- and perhaps I should say completeness -- of the error net expression.

Viggo wrote:When you start the error net by r5c8=3, then r5c8<>2. In this way you at the same time prepare the strong link [r5c6]=2=[r5c7]. Do you think, that you need to make a specific statement about that in the notation - like in your red comment?

Let's say we have two almost identical cases. In the first case, digit 2 is a candidate in cell r5c8. In the second case, it is not. I certainly don't think the error net expression for the deduction r5c8<>3 should be identical for the two cases.

That said, I don't claim to know the "right way" to show that "preparation of a strong link." Perhaps Carcul will weight in on this point.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

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