## Nice Loop Question

Post the puzzle or solving technique that's causing you trouble and someone will help

### Nice Loop Question

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.
Sped

Posts: 126
Joined: 26 March 2006

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.

GreenLantern

Posts: 26
Joined: 19 August 2005

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.
ravel

Posts: 998
Joined: 21 February 2006

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

GreenLantern

Posts: 26
Joined: 19 August 2005

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?

GreenLantern

Posts: 26
Joined: 19 August 2005

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.
ravel

Posts: 998
Joined: 21 February 2006

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.
Sped

Posts: 126
Joined: 26 March 2006

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!
Sped

Posts: 126
Joined: 26 March 2006

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

GreenLantern

Posts: 26
Joined: 19 August 2005

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 ?)
ravel

Posts: 998
Joined: 21 February 2006

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

GreenLantern

Posts: 26
Joined: 19 August 2005

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.
ronk
2012 Supporter

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

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.
Myth Jellies

Posts: 593
Joined: 19 September 2005