Nice Loop Question

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Nice Loop Question

Postby Sped » Sun May 07, 2006 5:39 pm

Code: Select all
 *-----------*
 |2..|8..|..7|
 |..8|9.1|2..|
 |...|.4.|.8.|
 |---+---+---|
 |64.|...|.7.|
 |193|...|64.|
 |87.|...|.31|
 |---+---+---|
 |...|.2.|...|
 |..1|3.9|8..|
 |3..|..6|..5|
 *-----------*


 *-----------*
 |2..|8..|..7|
 |..8|9.1|2..|
 |...|.4.|.8.|
 |---+---+---|
 |64.|...|.7.|
 |193|...|64.|
 |87.|...|.31|
 |---+---+---|
 |...|.2.|...|
 |..1|3.9|8..|
 |3..|..6|..5|
 *-----------*

 
 *-----------------------------------------------------------*
 | 2     1356  469   | 8     356   35    | 134   1569  7     |
 | 47    356   8     | 9     3567  1     | 2     56    346   |
 | 59    1356  67    | 2567  4     2357  | 13    8     369   |
 |-------------------+-------------------+-------------------|
 | 6     4     25    | 12    1389  238   | 59    7     28    |
 | 1     9     3     | 257   58    2578  | 6     4     28    |
 | 8     7     25    | 246   69    24    | 59    3     1     |
 |-------------------+-------------------+-------------------|
 | 59    568   467   | 457   2     4578  | 1347  16    369   |
 | 47    256   1     | 3     57    9     | 8     26    46    |
 | 3     28    479   | 147   18    6     | 47    29    5     |
 *-----------------------------------------------------------*


I'm just learning nice loops and am not real sure about things yet. Are the following nice loop exclusions valid?

[r8c2]-5-[r7c1]=9=[r7c9]=3=[r7c7]=1=[r7c8]-6-[r8c9]-4-[r8c1]-7-[r8c5]-5-[r8c2], =>r8c2<>5
[r7c4]-5-[r7c1]=9=[r7c9]=3=[r7c7]=1=[r7c8]-6-[r8c9]-4-[r8c1]-7-[r8c5]-5-[r7c4], =>r7c4<>5
[r7c6]-5-[r7c1]=9=[r7c9]=3=[r7c7]=1=[r7c8]-6-[r8c9]-4-[r8c1]-7-[r8c5]-5-[r7c6], =>r7c6<>5

After thinking about it, the above loop is not valid. Jumping from a strong link to a weak link must be done on the same candidate.

Back to the drawing board.

I've been informed that the grid can be solved with using Almost Locked Sets. That's another technique I don't understand.
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Postby GreenLantern » Mon May 08, 2006 6:18 am

I am also a Nice Loop Novice. I thought this puzzle looked familiar.
Michael Mepham's Daily Sudoku Diabolical #294 (April 30, 2006). After
three X-wings, I got almost the same starting candidates grid as you:
Code: Select all
+-------------------+-------------------+-------------------+
| 2     1356  469   | 8     356   35    | 134   1569  7     |
| 47    356   8     | 9     3567  1     | 2     56    346   |
| 59    1356  67    | 2567  4     2357  | 13    8     369   |
+-------------------+-------------------+-------------------+
| 6     4     25    | 12    1389  238   | 59    7     28    |
| 1     9     3     | 257   58    2578  | 6     4     28    |
| 8     7     25    | 246   69    24    | 59    3     1     |
+-------------------+-------------------+-------------------+
| 59    568   467   | 1457  2     4578  | 1347  16    369   |
| 47    256   1     | 3     57    9     | 8     26    46    |
| 3     28    479   | 147   18    6     | 147   129   5     |
+-------------------+-------------------+-------------------+

How did you determine that r7c4<>1?

I managed to solve this puzzle after finding 4 simple nice loops that
did not make use of Almost Locked Sets. Afterwards, I discovered that
I had overlooked an XYZ-wing which would have eliminated the same
candidate I was able to eliminate with one of my loops.
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Postby ravel » Mon May 08, 2006 9:20 am

Sped wrote:Are the following nice loop exclusions valid?

I dont know, but the logic is correct. [Edit: not my day:( I just overlooked that the beginning was wrong: [r8c2]-5-[r7c1]=9=..]
I would prefer to write the 3 loops as one xy-chain.
GreenLantern wrote:How did you determine that r7c4<>1?

I only found this:
[r7c4]-1-[r4c4]=1=[r4c5]=3=[r4c6]-3-[r1c6]-5-[r7c6]=5=[r8c5]-5-[r5c5]-8-[r9c5]-1-[r7c4]

PS: this forcing chain is easier and gives more:
r7c3=6 => r7c8=1
r3c3=6 => r2c1=7 => r8c1=4 => r8c9=6 => r7c8=1
[Edit: Thanks to GreenLantern - here a chain is missing, showing r1c3=6 => r7c8=1, and i found only a very long one for that (see below)]
Last edited by ravel on Mon May 08, 2006 10:31 am, edited 2 times in total.
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Postby GreenLantern » Mon May 08, 2006 11:29 am

Sped wrote:I'm just learning nice loops and am not real sure about things yet. Are the following nice loop exclusions valid?

[r8c2]-5-[r7c1]=9=[r7c9]=3=[r7c7]=1=[r7c8]-6-[r8c9]-4-[r8c1]-7-[r8c5]-5-[r8c2], =>r8c2<>5
[r7c4]-5-[r7c1]=9=[r7c9]=3=[r7c7]=1=[r7c8]-6-[r8c9]-4-[r8c1]-7-[r8c5]-5-[r7c4], =>r7c4<>5
[r7c6]-5-[r7c1]=9=[r7c9]=3=[r7c7]=1=[r7c8]-6-[r8c9]-4-[r8c1]-7-[r8c5]-5-[r7c6], =>r7c6<>5


These loops can be correctly written as:

[r8c2]-5-[r7c1]-9-[r3c1]=9=[r1c3]=4=[r2c1]-4-[r8c1]-7-[r8c5]-5-[r8c2] => r8c2<>5
[r7c4]-5-[r7c1]-9-[r3c1]=9=[r1c3]=4=[r2c1]-4-[r8c1]-7-[r8c5]-5-[r7c4] => r7c4<>5
[r7c6]-5-[r7c1]-9-[r3c1]=9=[r1c3]=4=[r2c1]-4-[r8c1]-7-[r8c5]-5-[r7c6] => r7c6<>5
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Postby GreenLantern » Mon May 08, 2006 12:02 pm

ravel wrote:I only found this:
[r7c4]-1-[r4c4]=1=[r4c5]=3=[r4c6]-3-[r1c6]-5-[r7c6]=5=[r8c5]-5-[r5c5]-8-[r9c5]-1-[r7c4]

Thanks. I was just asking because I thought there was some simple elimination
I was overlooking in setting up the initial candidates grid.
ravel wrote:PS: this forcing chain is easier and gives more:
r7c3=6 => r7c8=1
r3c3=6 => r2c1=7 => r8c1=4 => r8c9=6 => r7c8=1

I don't quite understand this forcing chain. Wouldn't it also have to
include a chain for r1c3=6?
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Postby ravel » Mon May 08, 2006 2:11 pm

GreenLantern wrote:I don't quite understand this forcing chain. Wouldn't it also have to
include a chain for r1c3=6?

Yes, of course, thanks, i just missed the hardest part.
r1c3=6 => r9c3=9 => r7c1=5 => r8c5=5 => r5c5=8 => r9c5=1 => r9c2=8 => r9c8=2 => r8c8=6 => r7c8=1
So its not easy after all.
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Postby Sped » Mon May 08, 2006 6:10 pm

GreenLantern wrote:I am also a Nice Loop Novice. I thought this puzzle looked familiar.
Michael Mepham's Daily Sudoku Diabolical #294 (April 30, 2006). After
three X-wings, I got almost the same starting candidates grid as you

How did you determine that r7c4<>1?


Yes, this is the April 30th Diabolical.

Simple Sudoku gets this far and is stuck:
Code: Select all
 *-----------------------------------------------------------*
 | 2     1356  469   | 8     356   35    | 134   1569  7     |
 | 47    356   8     | 9     3567  1     | 2     56    346   |
 | 59    1356  67    | 2567  4     2357  | 13    8     369   |
 |-------------------+-------------------+-------------------|
 | 6     4     25    | 12    1389  238   | 59    7     28    |
 | 1     9     3     | 257   58    2578  | 6     4     28    |
 | 8     7     25    | 246   69    24    | 59    3     1     |
 |-------------------+-------------------+-------------------|
 | 59    568   467   | 1457  2     4578  | 1347  16    369   |
 | 47    256   1     | 3     57    9     | 8     26    46    |
 | 3     28    479   | 147   18    6     | 147   129   5     |
 *-----------------------------------------------------------*


Then the XY chain

1-(r7c8)-6-(r8c9)-4-(r8c1)-7-(r8c5)-5-(r5c5)-8-(r9c5)-1

eliminates the 1s in r7c4, r9c7 and r9c8.

In nice loop notation:

[r7c4]-1-[r7c8]-6-[r8c9]-4-[r8c1]-7-[r8c5]-5-[r5c5]-8-[r9c5]-1-[r7c4], => r7c4<>1
[r9c7]-1-[r7c8]-6-[r8c9]-4-[r8c1]-7-[r8c5]-5-[r5c5]-8-[r9c5]-1-[r9c7], => r9c7<>1
[r9c8]-1-[r7c8]-6-[r8c9]-4-[r8c1]-7-[r8c5]-5-[r5c5]-8-[r9c5]-1-[r9c8], => r9c8<>1

That doesn't solve it though.

Tha ALS solution involves 5 candidates in 4 cells in box 1 and 4 candidates in 3 cells in row 8. The two sets both have 5s restricted to column 2. It shows that of the 47 pair in column 1, the 4 must be in r2c1 and the 7 in r8c1, otherwise a quad in box 1 and a triple in row 8 would each put a 5 in column 2. I'm sure I'm not using ALS terminology correctly, but I'm new to ALS. The sets are marked A and B below;


Code: Select all
 *-----------------------------------------------------------*
 | 2     1356A 469   | 8     356   35    | 134   1569  7     |
 | 47    356A  8     | 9     3567  1     | 2     56    346   |
 | 59    1356A 67A   | 2567  4     2357  | 13    8     369   |
 |-------------------+-------------------+-------------------|
 | 6     4     25    | 12    1389  238   | 59    7     28    |
 | 1     9     3     | 257   58    2578  | 6     4     28    |
 | 8     7     25    | 246   69    24    | 59    3     1     |
 |-------------------+-------------------+-------------------|
 | 59    568   467   | 1457  2     4578  | 1347  16    369   |
 | 47    256B  1     | 3     57    9     | 8     26B   46B   |
 | 3     28    479   | 147   18    6     | 147   129   5     |
 *-----------------------------------------------------------*


r2c1<>7 cracks the puzzle wide open.

I did not come up with this solution, but thought I'd pass it along.
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Postby Sped » Mon May 08, 2006 6:29 pm

GreenLantern wrote:
These loops can be correctly written as:

[r8c2]-5-[r7c1]-9-[r3c1]=9=[r1c3]=4=[r2c1]-4-[r8c1]-7-[r8c5]-5-[r8c2] => r8c2<>5
[r7c4]-5-[r7c1]-9-[r3c1]=9=[r1c3]=4=[r2c1]-4-[r8c1]-7-[r8c5]-5-[r7c4] => r7c4<>5
[r7c6]-5-[r7c1]-9-[r3c1]=9=[r1c3]=4=[r2c1]-4-[r8c1]-7-[r8c5]-5-[r7c6] => r7c6<>5


Thanks! I see it!
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Postby GreenLantern » Mon May 08, 2006 8:21 pm

Sped wrote:Tha ALS solution involves 5 candidates in 4 cells in box 1 and 4 candidates in 3 cells in row 8. The two sets both have 5s restricted to column 2. It shows that of the 47 pair in column 1, the 4 must be in r2c1 and the 7 in r8c1, otherwise a quad in box 1 and a triple in row 8 would each put a 5 in column 2. I'm sure I'm not using ALS terminology correctly, but I'm new to ALS. The sets are marked A and B below;

r2c1<>7 cracks the puzzle wide open.

Code: Select all
 *-----------------------------------------------------------*
 | 2     1356A 469   | 8     356   35    | 134   1569  7     |
 | 47    356A  8     | 9     3567  1     | 2     56    346   |
 | 59    1356A 67A   | 2567  4     2357  | 13    8     369   |
 |-------------------+-------------------+-------------------|
 | 6     4     25    | 12    1389  238   | 59    7     28    |
 | 1     9     3     | 257   58    2578  | 6     4     28    |
 | 8     7     25    | 246   69    24    | 59    3     1     |
 |-------------------+-------------------+-------------------|
 | 59    568   467   | 1457  2     4578  | 1347  16    369   |
 | 47    256B  1     | 3     57    9     | 8     26B   46B   |
 | 3     28    479   | 147   18    6     | 147   129   5     |
 *-----------------------------------------------------------*

That's an interesting solution which makes perfect sense now that you've explained it all. Working backwards from your solution, I came up with the following nice loop using an ALS link that leads to the same elimination:

[r2c1]=4=[r8c1]=7=[r8c5]=5=[r7c46]-5-[r7c1]-9-[r3c1]=9=[r3c1]-9-[r9c3]=9|7=[r3c3]-7-[r2c1] => r2c1<>7
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Postby ravel » Mon May 08, 2006 10:32 pm

Sped,
this is no standard ALS, because you need the 47-pair to connect the almost locked sets. Maybe the name is ALS-chain (?)

GreenLantern,
i understand, that r2c1=7 => (some steps) r3c1=5 => (ALS) r3c3=7 => r2c1=4, but not your notation (are there typos ?)
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Postby GreenLantern » Mon May 08, 2006 11:07 pm

ravel wrote:i understand, that r2c1=7 => (some steps) r3c1=5 => (ALS) r3c3=7 => r2c1=4, but not your notation (are there typos ?)


Oops. Yes, thanks. Meant to write:

[r2c1]=4=[r8c1]=7=[r8c5]=5=[r7c46]-5-[r7c1]-9-[r3c1]=9=[r1c3]-9-[r9c3]=9|7=[r3c3]-7-[r2c1] => r2c1<>7
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Postby ronk » Tue May 09, 2006 12:18 am

ravel wrote:this is no standard ALS, because you need the 47-pair to connect the almost locked sets. Maybe the name is ALS-chain (?)

If one of the 47 cells was labeled set C, then it would be the ALS xy-rule.

My preference would be to attach one of the '47' cells to set A or set B -- r1c2 to set A or r1c8 to set B -- and then it would be the ALS xz-rule.
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Postby Myth Jellies » Tue May 09, 2006 6:41 am

Code: Select all
 *-----------------------------------------------------------*
 | 2     1356A 469   | 8     356   35    | 134   1569  7     |
 |-47    356A  8     | 9     3567  1     | 2     56    346   |
 | 59    1356A 67A   | 2567  4     2357  | 13    8     369   |
 |-------------------+-------------------+-------------------|
 | 6     4     25    | 12    1389  238   | 59    7     28    |
 | 1     9     3     | 257   58    2578  | 6     4     28    |
 | 8     7     25    | 246   69    24    | 59    3     1     |
 |-------------------+-------------------+-------------------|
 | 59    568  -467   | 1457  2     4578  | 1347  16    369   |
 | 47B   256B  1     | 3     57    9     | 8     26B   46B   |
 | 3     28   -479   | 147   18    6     | 147   129   5     |
 *-----------------------------------------------------------*

Using an AIC/ALS chain
A(1&3&6&7) = A5 - B5 = B(2&4&6&7). This implies that either (A = 1 & 3 & 6 & 7) or (B = 2 & 4 & 6 & 7) or both must be true. Looking at the sevens in those groupings, you have r3c3 (A) = 7 and/or r8c1 (B) = 7, therefore any cells seeing those two (conveniently marked with a '-') cannot be a 7.
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