## Alternate step for ALS xz-rule in top1465 #13 ??

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

### Alternate step for ALS xz-rule in top1465 #13 ??

#13 of the top1465 is cracked with the techniques of Simple Sudoku and one ALS xz-rule (illustrated below). But if you didn't see that ALS, what would your next step be?
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`top1465_00136.9.....8...7.1...4............6...4.2.....3..3....5...1.5...7.8...9..........2.. After SSTS  6      57     9      | 234    2345   2345   | 137-4 B1245    8 23     58     23     | 7      458    1      |B469   B4569   B569 4      578    1      | 69     2358   69     | 37    B25      237-5----------------------+----------------------+---------------------- 157    9      8      | 123    6      2357   | 17    A12      4 157    2      46     | 1489   1457   45789  | 16789  3       1679 17     3      46     | 12489  1247   2479   | 5      689-12  12679----------------------+----------------------+---------------------- 239    1      23     | 5      34     3468   | 34689  7       369 8      46     57     | 12346  9      23467  | 1346   456-1   1356 39     46     57     | 13468  1347   34678  | 2      45689-1 13569 Sets: A = {r4c8} = {12}       B = {r2c789,r13c8} = {124569}     x,z = 1,2 and x,z = 2,1 (doubly linked, aka dual ALS xz) Elims: r689c8<>1, r6c8<>2, r1c7<>4 and r3c9<>5`

[edit: I'm looking for any step(s), not necessarily ones that duplicate the eliminations above.]
ronk
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ronk, Your [r1c7]<>4 cracks the puzzle. The other eliminations are extraneous.

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`[r1c7]<>4 => N/Triple [c7] => N/Pair [r8] => N/Triple [c8] => [r1c8]=4 => Singles`

There is a short contradiction net for this elimination, but who (in their right mind) would go looking for it.
daj95376
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daj95376 wrote:
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`[r1c7]<>4 => N/Triple [c7] => N/Pair [r8] => N/Triple [c8] => [r1c8]=4 => Singles`

There is a short contradiction net for this elimination ...

Contradiction It looks like an interesting continuous ALS loop -- interesting because of the AALS in r8. I'll try to figure it out later.
ronk
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There is an alternate dual ALS-xz for the elimination r1c7<>4 & r3c9<>5:
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` *-----------------------------------------------------------------------------* | 6       57      9       | 234     2345    2345    |-1347   @1245    8       | | 23      58      23      | 7       458     1       |#469    #4569   #569     | | 4       578     1       | 69      2358    69      | 37     @25     -2357    | |-------------------------+-------------------------+-------------------------| | 157     9       8       | 123     6       2357    | 17     @12      4       | | 157     2       46      | 1489    1457    45789   | 16789   3       1679    | | 17      3       46      | 12489   1247    2479    | 5       12689   12679   | |-------------------------+-------------------------+-------------------------| | 239     1       23      | 5       34      3468    | 34689   7       369     | | 8       46      57      | 12346   9       23467   | 1346    1456    1356    | | 39      46      57      | 13468   1347    34678   | 2       145689  13569   | *-----------------------------------------------------------------------------*`

ALS A (@): r134c8={1245}
ALS B (#): r2c789={4569}
x,z=4|5

Therefore r1c7<>4 & r3c9<>5

But ronk's ALS-xz definitely works too. Actually there is a similar move in the 13-clue Sudoku X thread lately:
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` *-----------------------------------------------------------------------------* |@14      56-789 @3789    | 1245    678     2359    | 56      379     48      | | 2589   @47     @3789    | 45      678     359     | 56      1       239     | | 2589    56-789 @1347    | 145     678     359     | 48      379     239     | |-------------------------\-------------------------/-------------------------| | 589     145789  147-8-9 | 37      2       6       | 48      39      3489    | | 89      4789    47-8-9  | 37      5       1       | 2       6       3489    | | 3       2       6       | 9       4       8       | 1       5       7       | |-------------------------/-------------------------\-------------------------| | 14      14      2       | 6       3       7       | 9       8       5       | | 6       3       5       | 8       9       4       | 7       2       1       | | 7       89     #89      | 25      1       25      | 3       4       6       | *-----------------------------------------------------------------------------*`

Dual ALS-xz rule:
ALS A (@): r1c1+r2c2+r123c3={134789}
ALS B (#): r9c3={89}
x,z=8|9

Therefore r45c3<>{89}
=> r1c1+r2c2+r123c3={13478|13479}, both naked quint in b1
=> {1347} eliminated from elsewhere of b1 => r13c2<>7

To me this smells awfully similar to Sue De Coq or something alike.

I definitely think there should be a particular technique formulated for this pattern.
udosuk

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Joined: 17 July 2005

udosuk wrote:Actually there is a similar move in the 13-clue Sudoku X thread lately:
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` *----------------------------------------------------------*  |@14     56-789 @3789    | 1245  678  2359 | 56  379  48   |  | 2589  @47     @3789    | 45    678  359  | 56  1    239  |  | 2589   56-789 @1347    | 145   678  359  | 48  379  239  |  |------------------------\-----------------/---------------|  | 589    145789  147-8-9 | 37    2    6    | 48  39   3489 |  | 89     4789    47-8-9  | 37    5    1    | 2   6    3489 |  | 3      2       6       | 9     4    8    | 1   5    7    |  |------------------------/-----------------\---------------|  | 14     14      2       | 6     3    7    | 9   8    5    |  | 6      3       5       | 8     9    4    | 7   2    1    |  | 7      89     #89      | 25    1    25   | 3   4    6    |  *----------------------------------------------------------*`

Dual ALS-xz rule:
ALS A (@): r1c1+r2c2+r123c3={134789}
ALS B (#): r9c3={89}
x,z=8|9
[...]
To me this smells awfully similar to Sue De Coq or something alike.

I definitely think there should be a particular technique formulated for this pattern.

All Sue de Coq's can be described in terms of the doubly-linked ALS xz-rule. In this case, split out the AALS r12c3 and you have the Sue de Coq.

P.S. I dislike the adjective "dual" because it's bad English to me.
ronk
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The ALC (1237) in b3\c8 also eliminates 4 from r1c7 and solves the puzzle. However it uses the ALS r457c8 so it may not be considered materially different from the other solutions.

Steve
Steve R

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Joined: 03 April 2006

This is similar to the original ALS, but from a slightly different perspective.

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`Either [r2c789]=469 -or- [r2c8]=5[r2c789]=469 =>                                  [r1c7]<>4[r2c 8 ]=5   => [r3c8]=2 => [r4c8]=1 [r1c8]=4 => [r1c7]<>4`
daj95376
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Steve R wrote:The ALC (1237) in b3\c8 also eliminates 4 from r1c7 and solves the puzzle. However it uses the ALS r457c8 so it may not be considered materially different from the other solutions.

I don't follow, so would you please clarify?
ronk
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In box three the candidates 1, 2, 3 and 7 can be apportioned between column eight and the three cells not in that column open to them. Onr or more must therefore lie in column eight. The rest of column eight means that (a) 3 and 7 cannot be placed in c8b3 and (b) only one of 1 and 2 can be placed there. As a result the three cells in the box but not in the column must contain three of these four candidates. Any other candidate may be eliminated, as, for example, 4 from r1c7.

The method is due to RW, usefully enhanced by ravel.

Steve
Steve R

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Joined: 03 April 2006

A correction to Danny's move:
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`Either [r2c789]=469 -or- [r2c89]=5[r2c789]=469 =>                                  [r1c7]<>4[r2c 89]=5   => [r3c8]=2 => [r4c8]=1 [r1c8]=4 => [r1c7]<>4`
udosuk

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Joined: 17 July 2005

daj95376, udosuk and Steve R, thanks to all for your responses.

All the viewpoints seem to use the same 6 cells in b3 and c8 -- a Sue de Coq or doubly-linked ALS xz-rule. However, daj95376's first post uncovered this very interesting continuous ALS chain yielding [edit: 15] eliminations:
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`top1465_00136.9.....8...7.1...4............6...4.2.....3..3....5...1.5...7.8...9..........2..After SSTS 6      57     9      | 23-4   235-4  235-4  |A1347  C1245    8 23     58     23     | 7      458    1      | 469    469-5   569 4      578    1      | 69     2358   69     |A37    C25      2357----------------------+----------------------+---------------------- 157    9      8      | 123    6      2357   |A17    C12      4 157    2      46     | 1489   1457   45789  | 689-17 3       1679 17     3      46     | 12489  1247   2479   | 5      689-12  12679----------------------+----------------------+---------------------- 239    1      23     | 5      34     3468   | 4689   7       369 8     B46     57     | 123-46 9      237-46 |B1346  B1456    135-6 39     46     57     | 13468  1347   34678  | 2      4689-15 13569Sets:  A = {r134c7} = {1347}       B = {r8c278} = {13456}, an AALS (2 degrees of freedom)       C = {r134c8} = {1245}Elims: r1c456<>4, r5c7<>7, r6c8<>2, r8c46<>4, r8c469<>6,        r5c7<>1, r69c8<>1 and r29c8<>5 (last 5 elims as noted by re'born)A chain attempt:                A                               B                             C -4-(ALS:r1c7=4|7|13=r134c7)-13-(AALS:r8c27=13|46|15=r8c28)-15-(ALS:r134c8=15|2|4=r1c8)-4-`

At least one of r1c7<>4 or r1c8<>4 must be true. If r1c7<>4, then set A becomes a naked triple 137 creating the naked pair 46 within set B. This leaves sets B and C to share digits 1 and 5. Whether set C contains digit 1 or 5, r1c8 will contain digit 4, closing the loop. For the opposite direction, there is a similar result starting with r1c8<>4.

Due to the continuous (closed) loop, the r1c7-4-r1c8 weak link is converted to a conjugate link ... and digits 7, 46, and 2 are locked into sets A, B and C, respectively.

[edit: Although unusual, the "2-digit weak links" (shown in the chain above) yield eliminations as well ... as pointed out by re'born. Consider the digit 1 and 3 weak link between sets A and B. In a left-to-right implication stream, both digits are wholly contained in set A. In a right-to-left implication stream, the digits are shared by sets A and B.]
Last edited by ronk on Tue Aug 07, 2007 10:05 am, edited 1 time in total.
ronk
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ronk wrote:daj95376, udosuk and Steve R, thanks to all for your responses.

All the viewpoints seem to use the same 6 cells in b3 and c8 -- a Sue de Coq or doubly-linked ALS xz-rule. However, daj95376's first post uncovered this very interesting continuous ALS chain yielding 10 eliminations:
Code: Select all
`top1465_00136.9.....8...7.1...4............6...4.2.....3..3....5...1.5...7.8...9..........2..After SSTS 6      57     9      | 23-4   235-4  235-4  |A1347  C1245   8 23     58     23     | 7      458    1      | 469    4569   569 4      578    1      | 69     2358   69     |A37    C25     2357----------------------+----------------------+--------------------- 157    9      8      | 123    6      2357   |A17    C12     4 157    2      46     | 1489   1457   45789  | 1689-7 3      1679 17     3      46     | 12489  1247   2479   | 5      1689-2 12679----------------------+----------------------+--------------------- 239    1      23     | 5      34     3468   | 4689   7      369 8     B46     57     | 123-46 9      237-46 |B1346  B1456   135-6 39     46     57     | 13468  1347   34678  | 2      145689 13569Sets:  A = {r134c7} = {1347}       B = {r8c278} = {13456}, an AALS (2 degrees of freedom)       C = {r134c8} = {1245}Elims: r1c456<>4, r5c7<>7, r6c8<>2, r8c46<>4 and r8c469<>6 A chain form attempt:                A                               B                             C -4-(ALS:r1c7=4|7|13=r134c7)-13-(AALS:r8c27=13|46|15=r8c28)-15-(ALS:r134c8=15|2|4=r1c8)-4-`

At least one of r1c7<>4 or r1c8<>4 must be true. If r1c7<>4, then set A becomes a naked triple 137 creating the naked pair 46 within set B. This leaves sets B and C to share digits 1 and 5. Whether set C contains digit 1 or 5, r1c8 will contain digit 4, closing the loop. For the opposite direction, there is a similar result starting with r1c8<>4.

Due to the continuous (closed) loop, the r1c7-4-r1c8 weak link is converted to a conjugate link ... and digits 7, 46, and 2 are locked into sets A, B and C, respectively.

I think you can also add r29c8<>5, r69c8<>1, r5c7<>1 to your deductions from these cells.
re'born

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re'born wrote:I think you can also add r29c8<>5, r69c8<>1, r5c7<>1 to your deductions from these cells.

Nice catch! I corrected my post to show all 15 elims in one place.
ronk
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### Re: Alternate step for ALS xz-rule in top1465 #13 ??

I've been reading threads on dual-linked ALS. In this thread, Ron asked for an alternate approach to his original DL-ALS. Here's an AIC that's very close to udosuk's dual ALS-xz. It performs the critical elimination.

Code: Select all
`top1465_0013 6.9.....8...7.1...4............6...4.2.....3..3....5...1.5...7.8...9..........2.. after SSTS +--------------------------------------------------------------------------------+ |  6       57      9       |  234     2345    2345    | a137-4  b1245    8       | |  23      58      23      |  7       458     1       | e469    e4569   e569     | |  4       578     1       |  69      2358    69      |  37     d25      2357    | |--------------------------+--------------------------+--------------------------| |  157     9       8       |  123     6       2357    |  17     c12      4       | |  157     2       46      |  1489    1457    45789   |  16789   3       1679    | |  17      3       46      |  12489   1247    2479    |  5       12689   12679   | |--------------------------+--------------------------+--------------------------| |  239     1       23      |  5       34      3468    |  34689   7       369     | |  8       46      57      |  12346   9       23467   |  1346    1456    1356    | |  39      46      57      |  13468   1347    34678   |  2       145689  13569   | +--------------------------------------------------------------------------------+ # 143 eliminations remain (1)r1c7 = r1c8 - (1=2)r4c8 - (2=5)r3c8 - (5=469)r2c789  =>  r1c7<>4`
daj95376
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