## Almost Locked Sets xz-rule (doubly-linked)

Advanced methods and approaches for solving Sudoku puzzles
aran wrote:- any candidate which locks any of 678 in A or any of 128 in B is false (because of the double RC).
This immediately eliminates the extraneous 8s in columns 5 and 6, the extraneous 7 in column 5.
Taking the logic one step further : 8r789c4 must be false since they force 8 in b2c56.
OK we know all that.
But now look at 1r5c4 : true, this forces 1r9c6 which from the above logic is false.
Hence <1>r5c4.
So the count is now up to 13.

I think there is a bit of confusion here. The double-RC, or equivalently the AIC/ALS loops indicate that 6&7&8 are locked in A while 1&2&8 are locked in B. For completeness, 5&9 are locked in the union of A & B. Thus any digit which prevents these placements (not forces them) would be false. Thus you have no reason to eliminate 1r5c4.
Myth Jellies

Posts: 593
Joined: 19 September 2005

Myth Jellies wrote:
aran wrote:- any candidate which locks any of 678 in A or any of 128 in B is false (because of the double RC).
This immediately eliminates the extraneous 8s in columns 5 and 6, the extraneous 7 in column 5.
Taking the logic one step further : 8r789c4 must be false since they force 8 in b2c56.
OK we know all that.
But now look at 1r5c4 : true, this forces 1r9c6 which from the above logic is false.
Hence <1>r5c4.
So the count is now up to 13.

I think there is a bit of confusion here. The double-RC, or equivalently the AIC/ALS loops indicate that 6&7&8 are locked in A while 1&2&8 are locked in B. For completeness, 5&9 are locked in the union of A & B. Thus any digit which prevents these placements (not forces them) would be false. Thus you have no reason to eliminate 1r5c4.

Realised the error walking along a street far from my computer and had hoped to edit before any hawkeye read the post !
Entirely right.
aran

Posts: 334
Joined: 02 March 2007

Subset counting makes the eliminations clear. There are 8 cells and the 7 candidates have max multiplicities 1,1,1,1,1,2,1. Therefore, any placement that would decrease the multiplicity of any candidate by 1 is invalid. Consequently, there are 13 eliminations:
r5c4<>5
r7c5<>7
r12c5<>8
r2c6<>8
r7c456<>8
r8c46<>8
r9c24<>8
r5c4<>9

[Edit: Removed a nonexistent 6 elimination (thanks to aran for pointing it out]
Last edited by re'born on Fri Dec 05, 2008 11:54 am, edited 1 time in total.
re'born

Posts: 551
Joined: 31 May 2007

re'born wrote:Subset counting makes the eliminations clear. There are 8 cells and the 7 candidates have max multiplicities 1,1,1,1,1,2,1. Therefore, any placement that would decrease the multiplicity of any candidate by 1 is invalid. Consequently, there are 14 eliminations:
r5c4<>5
r7c5<>6
r7c5<>7
r12c5<>8
r2c6<>8
r7c456<>8
r8c46<>8
r9c24<>8
r5c4<>9

r7c5<>6
aran

Posts: 334
Joined: 02 March 2007

Please use a pm to notify someone of an obvious typo ... and probably a suspected typo too. Thanks, Ron
ronk
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Location: Southeastern USA

re'born wrote:Subset counting makes the eliminations clear. There are 8 cells and the 7 candidates have max multiplicities 1,1,1,1,1,2,1. Therefore, any placement that would decrease the multiplicity of any candidate by 1 is invalid. Consequently, there are 14 eliminations:
r5c4<>5
r7c5<>6
r7c5<>7
r12c5<>8
r2c6<>8
r7c456<>8
r8c46<>8
r9c24<>8
r5c4<>9

Re'born I see that Ronk thinks I should have sent u a pm (didn't know about that procedure) and "suspects" something. Well I can only say :
1. No offence intended.
2. Didn't know what you meant and...on further investigation...still don't. What did you mean ? (I don't see any placement of 6 that reduces the multiplicity of the 6 subset).
aran

Posts: 334
Joined: 02 March 2007

Hi aran,

No offense taken. I still do most of my solving on paper and occasionally I do something as ridiculous as not removing the 6 from r7c5, despite the presence of 6 in r7c2. Technically, what I wrote is correct...just silly.
re'born

Posts: 551
Joined: 31 May 2007

aran wrote:PS are you considering <8>r9c2 as inherent in the elimination chain, or merely as a by-product of the pointing pair 8r9c56 resulting from earlier eliminations ?

Inherent in the elimination chain, although I see that both ways work. Ronk's cover set group #3 does the work.
Allan Barker

Posts: 266
Joined: 20 February 2008

Allan Barker wrote:
aran wrote:PS are you considering <8>r9c2 as inherent in the elimination chain, or merely as a by-product of the pointing pair 8r9c56 resulting from earlier eliminations ?

Inherent in the elimination chain, although I see that both ways work. Ronk's cover set group #3 does the work.

Allan, on reflection, I don't see that Ronk's cover unit "does the work" in the sense that a cover unit generates eliminations within its own ranks (ie itself) and since Ronk's units are b5,c5,c6, none of those can directly generate : <8>r9c2 nor <8>r789c4.
In fact, I'd consider all of those as by-product and reduce the number of eliminations attributable to the strategy (whether construed as ALS, subset counting, or DDS) to 9.
aran

Posts: 334
Joined: 02 March 2007

Occasionally one ALS in a row and another ALS in a column are doubly-linked, as in the below, where digits 5 and 9 are shared by ALS A in c9 and ALS B in r6.
Code: Select all
`top1465 #1223..92...6....4....82.............7.1..........36......2..3..915.1...75...48....9..After SSTS 58     13457  9      | 2      1358   138    | 3457    6       15-37 6      1357   17     | 4      1359   13     | 2357    2379    8 2      1345   148    | 7      135689 1368   | 345     349     159-3----------------------+----------------------+----------------------- 589    45     248    | 569    2368   7      | 3468-5  1      A359 589    1457   12478  | 1569   12368  123468 | 34678-5 3478-9 A3579 3      6     B1478   | 59-1  B18    B148    |B4578   B4789    2----------------------+----------------------+----------------------- 7      2      3      | 8      4      9      | 1       5       6 1      9      6      | 3      7      5      | 28      28      4 4      8      5      | 16     126    126    | 9       37     A37Sets: A = {r459c9} = {3579}; B = {r6c35678} = {145789}    x,z = 5,9 and x,z = 9,5Elims: r45c7<>5, r5c8<>9, r1c9<>37, r3c9<>3, r6c4<>1`

Two other perspectives:
Code: Select all
`As constraint sets: 8 base sets: r459c9, r6c35678                    8 cover sets: 59b6, 37c9, 1478r6As a continuous nice loop:(r45c9,r9c9 =5|37|9= r45c9) -9- (r6c78,r6c356 =9|1478|5= r6c78) -5- continuous loopEliminations same as above`

A hidden pair, a line\block interaction and singles then complete the puzzle.
ronk
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Location: Southeastern USA

aran wrote:Allan, on reflection, I don't see that Ronk's cover unit "does the work" in the sense that a cover unit generates eliminations within its own ranks (ie itself) and since Ronk's units are b5,c5,c6, none of those can directly generate : <8>r9c2 nor <8>r789c4.

Hmm, I notice that you did not include cover set 8r9 in your list of Ronk's cover units. Only one of Ronk's 4 suggested cover set groups included 8r9, specifically it was group #3:

Base sets: (56789)r4569c5 (12589)r4569c6
Cover sets: #3 = 59b5, 67c5, 12c6, 8b5, 8r9

In this case, <8>r9c2 is within cover set 8r9 and will be eliminated.

aran wrote:In fact, I'd consider all of those as by-product and reduce the number of eliminations attributable to the strategy (whether construed as ALS, subset counting, or DDS) to 9.

I am looking only at base/cover sets, rather than specific stratigies where my understanding is not so good. In this case, 3 different cover set groups can be applied seperately to eliminate all 13 candidates, and any one of the 3 can come first. Thus, none would be a by-product of the other.

Specifically, I'm not sure if 8r9, 8b5, or even 8r4 can be included in a true double linked ALS.
Allan Barker

Posts: 266
Joined: 20 February 2008

Allan Barker wrote:Specifically, I'm not sure if 8r9, 8b5, or even 8r4 can be included in a true double linked ALS.

Maybe not, but there is a triple-link between the ALS in r456c56 and the AALS in r9c56 which results in essentially the same thing.
Myth Jellies

Posts: 593
Joined: 19 September 2005

ronk wrote:Occasionally one ALS in a row and another ALS in a column are doubly-linked, as in the below, where digits 5 and 9 are shared by ALS A in c9 and ALS B in r6.
Code: Select all
`top1465 #1223..92...6....4....82.............7.1..........36......2..3..915.1...75...48....9..After SSTS 58     13457  9      | 2      1358   138    | 3457    6       15-37 6      1357   17     | 4      1359   13     | 2357    2379    8 2      1345   148    | 7      135689 1368   | 345     349     159-3----------------------+----------------------+----------------------- 589    45     248    | 569    2368   7      | 3468-5  1      A359 589    1457   12478  | 1569   12368  123468 | 34678-5 3478-9 A3579 3      6     B1478   | 59-1  B18    B148    |B4578   B4789    2----------------------+----------------------+----------------------- 7      2      3      | 8      4      9      | 1       5       6 1      9      6      | 3      7      5      | 28      28      4 4      8      5      | 16     126    126    | 9       37     A37Sets: A = {r459c9} = {3579}; B = {r6c35678} = {145789}    x,z = 5,9 and x,z = 9,5Elims: r45c7<>5, r5c8<>9, r1c9<>37, r3c9<>3, r6c4<>1`

Two other perspectives:
Code: Select all
`As constraint sets: 8 base sets: r459c9, r6c35678                    8 cover sets: 59b6, 37c9, 1478r6As a continuous nice loop:(r45c9,r9c9 =5|37|9= r45c9) -9- (r6c78,r6c356 =9|1478|5= r6c78) -5- continuous loopEliminations same as above`

A hidden pair, a line\block interaction and singles then complete the puzzle.

Hidden triple nice loop version (in another language) :
379r459c9=5r45c9-5r6c7=14789r6c35678-9r45c9=357r459c9.
aran

Posts: 334
Joined: 02 March 2007

Allan Barker wrote:
aran wrote:Allan, on reflection, I don't see that Ronk's cover unit "does the work" in the sense that a cover unit generates eliminations within its own ranks (ie itself) and since Ronk's units are b5,c5,c6, none of those can directly generate : <8>r9c2 nor <8>r789c4.

Hmm, I notice that you did not include cover set 8r9 in your list of Ronk's cover units. Only one of Ronk's 4 suggested cover set groups included 8r9, specifically it was group #3:

Base sets: (56789)r4569c5 (12589)r4569c6
Cover sets: #3 = 59b5, 67c5, 12c6, 8b5, 8r9

In this case, <8>r9c2 is within cover set 8r9 and will be eliminated

You're absolutely right : those cover sets do all that you say.
I must have misunderstood what Ronk's covers sets were.
aran

Posts: 334
Joined: 02 March 2007

I would like to ask if the following qualifies as a doubly linked ALS? The issue might be that one of the ALSs has only one cell. It can be written (at least) two ways:

Way #1: Sets: A = {r1c156} = {1358}; B = {r2c6} = {13} linked by 13
Way #2: Sets: A = {r1c1} = {58}; B = {r1c56,r2c6} = {1358} linked by 58
Eliminations(8) = r1c279<>5, r2c5<>13, r3c5<>13, r3c6<>13

Code: Select all
`  +--------------------------------------------------------------------------------------+  | (58)     134-57   9        | 2        (1358)   (138)    | 34-57    6        13-57    |  | 6        1357     17       | 4        -1-359   (13)     | 2357     2379     8        |  | 2        1345     148      | 7        -1-35689  -1-368  | 345      349      1359     |  +--------------------------------------------------------------------------------------+  | 589      45       248      | 569      2368     7        | 34568    1        359      |  | 589      1457     12478    | 1569     12368    123468   | 345678   34789    3579     |  | 3        6        1478     | 159      18       148      | 4578     4789     2        |  +--------------------------------------------------------------------------------------+  | 7        2        3        | 8        4        9        | 1        5        6        |  | 1        9        6        | 3        7        5        | 28       28       4        |  | 4        8        5        | 16       126      126      | 9        37       37       |  +--------------------------------------------------------------------------------------+       8r1  5r1  1b2  3b2 (NRC notation)1N1: 811==511                  |    |             1N5: 815==515==115==315      |         |    |   1N6: 816=======116==316                  |    |2N6:           126==326  As cover sets:     4 Base Sets  = {1n156 2n6}     4 Cover Sets = {58r1 13b2}`
Allan Barker

Posts: 266
Joined: 20 February 2008

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