als???

Advanced methods and approaches for solving Sudoku puzzles

als???

Postby didier90 » Sun Aug 06, 2006 2:58 pm

hello

At first of all escuser me if my languages is not correct

that think you of this pulze

Code: Select all



 
 *------------------------------------------------------------------------
 | 6       4579    34579 | 2457    57      248   | 1       2478    2378    |
 | 8      *147     147   | 12467   3       9     | 246     5       267     |
 | 145     2       13457 | 14567   16      48    | 346     4678    9       |
 |-------------------------+-------------------------+--------------------
 | 3      *1479    12479 | 169     16      5     | 8       1267    1267   
 |+15     *157     6     | 8       2       3     | 9       17      4       |
 | 129    *19      8     | 169     4       7     | 26      3       5       |
 |-------------------------+-------------------------+--------------------
 | 2459    6       2459  | 2345    8       1     | 7       249     23      |
 | 12459   3       12459 | 2457    579     24    | 2456    124689  1268    |
 | 7       8       12459 | 2345    59      6     | 2345    1249    123     |
 *------------------------------------------------------------------------



als 1 (R2C2, R4C2,R5C2,R6C2)
als 2 (R5C1)
x = 5

I think that I can eliminer the 9 of R1C2

because if 9 in R1C2 => R5C1 = x
it thus remains only (147) for als 1

didier
Last edited by didier90 on Sun Aug 06, 2006 5:00 pm, edited 5 times in total.
didier90
 
Posts: 3
Joined: 06 August 2006

Postby udosuk » Sun Aug 06, 2006 3:23 pm

That seems right... It's a complex xyzw-wing and you could also eliminate 9 from r4c3 & r6c1:

Code: Select all
r2c1/r4c3/r6c1=9 => r6c2=1 => r5c1=5 => r5c2=7 => r2c2=4 => nothing left for r4c2
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby Myth Jellies » Mon Aug 07, 2006 7:11 am

There is no external cell which can see both ALS 1 & 2, so you are going to have problems finding a deduction using just these sets.
Myth Jellies
 
Posts: 593
Joined: 19 September 2005

Postby Steve R » Mon Aug 07, 2006 3:21 pm

True. If working entirely in terms of almost locked sets, a third is needed:

als3 = {r6c2}.

Then als2 and als3 have the restricted common candidate y = 1. Since yx, the three eliminations of udosuk follow.

Steve
Steve R
 
Posts: 74
Joined: 03 April 2006

Postby Myth Jellies » Mon Aug 07, 2006 5:35 pm

Thanks, I see now. Not your typical ALS situation.

From an AIC perspective we have...

9[r6c2]=1[r6c2]-1[r5c1]=5[r5c1]-5[r2456c2]=(1&4&7&9)[r2456c2]
proves that either r6c2 = 9 or r46c2 contains a 9. Kills any other candidate nine that can see both of those in r46c2.

Interesting to see an AIC wrap around on itself like that.
Myth Jellies
 
Posts: 593
Joined: 19 September 2005

Postby ronk » Mon Aug 07, 2006 5:44 pm

I thought this would be a great puzzle for Subset Counting.
Code: Select all

 6     4579 34579 | 2457  57  248 | 1    2478   2378
 8    *147  147   | 12467 3   9   | 246  5      267 
 145   2    13457 | 14567 16  48  | 346  4678   9   
------------------+---------------+------------------
 3    *1479 12479 | 169   16  5   | 8    1267   1267
*15   *157  6     | 8     2   3   | 9    17     4   
 129  *19   8     | 169   4   7   | 26   3      5   
------------------+---------------+------------------
 2459  6    2459  | 2345  8   1   | 7    249    23   
 12459 3    12459 | 2457  579 24  | 2456 124689 1268
 7     8    12459 | 2345  59  6   | 2345 1249   123 

We have a set of five cells (r5c1, r2456) which have five possible digits (14579). The maximum times each digit can occur in this set is 2,1,1,1,1 respectively. IOW the possible digits 14579 have multiplicity 21111, whose sum is 6.

Then I hit a wall and didn't know how to proceed. Is there a way?
ronk
2012 Supporter
 
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Postby re'born » Mon Aug 07, 2006 6:02 pm

ronk wrote:I thought this would be a great puzzle for Subset Counting.
Code: Select all

 6     4579 34579 | 2457  57  248 | 1    2478   2378
 8    *147  147   | 12467 3   9   | 246  5      267 
 145   2    13457 | 14567 16  48  | 346  4678   9   
------------------+---------------+------------------
 3    *1479 12479 | 169   16  5   | 8    1267   1267
*15   *157  6     | 8     2   3   | 9    17     4   
 129  *19   8     | 169   4   7   | 26   3      5   
------------------+---------------+------------------
 2459  6    2459  | 2345  8   1   | 7    249    23   
 12459 3    12459 | 2457  579 24  | 2456 124689 1268
 7     8    12459 | 2345  59  6   | 2345 1249   123 

We have a set of five cells (r5c1, r2456) which have five possible digits (14579). The maximum times each digit can occur in this set is 2,1,1,1,1 respectively. IOW the possible digits 14579 have multiplicity 21111, whose sum is 6.

Then I hit a wall and didn't know how to proceed. Is there a way?


A 9 in r1c2, r4c3 or r6c1 reduces the multiplicity of 9 in your subset to 0 and also forces r6c2 = 1, thus reducing the multiplicity of 1 in your subset to 1. As you now have 5 cells and a total possible multiplicity of 4, one can conclude that r1c2, r4c3, r6c1 are not 9.
re'born
 
Posts: 551
Joined: 31 May 2007

Postby ronk » Mon Aug 07, 2006 6:24 pm

rep'nA wrote:A 9 in r1c2, r4c3 or r6c1 [...] forces r6c2 = 1, thus reducing the multiplicity of 1 in your subset to 1.

"Dynamic" multiplicity!:D
ronk
2012 Supporter
 
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Postby didier90 » Mon Aug 07, 2006 8:28 pm

hi

for me this is that

9 in r1c2 sees all 9 of als1 and also eliminate x (restricted common) of als1 ==> r1c2 <> 9

it is mortal for the pulze

other methode to make

x is in als1 or in als2

if in als1 r5c1 = 1 r6c2 = 9 ==> r1c2 <> 9 (and also (r6c1 r4c3))
if in als2 ==> (r4c2=9 or r6c2=9) ==> r1c2 <> 9 (and also (r6c1 r4c3))

didier
didier90
 
Posts: 3
Joined: 06 August 2006

Postby Havard » Mon Aug 07, 2006 8:42 pm

ronk wrote:Then I hit a wall and didn't know how to proceed. Is there a way?


This is quite a hard puzzle. My solver gives a solution with a lot of very big and long ALS'chains using a lot of cells. Not very user-friendly. I would be interested to see a "simple" solution to this one.

Havard
Havard
 
Posts: 377
Joined: 25 December 2005

Postby ronk » Mon Aug 07, 2006 10:03 pm

Havard wrote:I would be interested to see a "simple" solution to this one.

I doubt there's anything simpler than didier's deduction. I was just trying to do the same with subset counting instead.
ronk
2012 Supporter
 
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Postby daj95376 » Mon Aug 07, 2006 10:39 pm

Everyone seems to have missed the multi-colors on <6> which leads to a simple solution.

Code: Select all
6.....1..8...39.5..2......93....58....68239.4..8.47.35.6..817...3.......78...6...

  c5    -  579   Naked  Triple
    b5  -  9     Locked Candidate (1)

r3c7    <> 6     Multi-Colors
r3c7    =  4     [r3c7]=3 => [c2]=INVALID

[r3c7]=3,[r1c3]=3,[r1c2]=9,[r6c2]=1,[r5c1]=5 => [c2]=INVALID for <47>
Last edited by daj95376 on Wed Sep 13, 2006 5:46 am, edited 1 time in total.
daj95376
2014 Supporter
 
Posts: 2624
Joined: 15 May 2006

Postby ronk » Tue Aug 08, 2006 12:40 am

daj95376 wrote:Everyone seems to have missed the multi-colors on <6> which leads to a simple solution.
[...]
[r3c7]=3 => [r1c3]=3 => [r1c2]=9 => [r6c2]=1 => [r5c1]=5 = [c2]=INVALID for <47>

This INVALID chain uses the same cells as didier's solution plus two more -- not counting the exclusion cell(s). And didier's deduction causes three exclusions rather than just one.
ronk
2012 Supporter
 
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Postby Mike Barker » Tue Aug 08, 2006 1:51 am

I think there's a slightly shorter nice loop which solves the puzzle with an x-wing

Code: Select all
3-element Nice Loop: r1c2=5=r5c2-5-r5c1-1-r6c2~9~r1c2 => r1c2<>9
+---------------------+------------------+---------------------+
|      6 -4579  34579 |   2457   57  248 |     1    2478  2378 |
|      8   147    147 |  12467    3    9 |   246       5   267 |
|    145     2  13457 |  14567   16   48 |   346    4678     9 |
+---------------------+------------------+---------------------+
|      3  1479  12479 |    169   16    5 |     8    1267  1267 |
|    *15  *157      6 |      8    2    3 |     9      17     4 |
|    129   *19      8 |    169    4    7 |    26       3     5 |
+---------------------+------------------+---------------------+
|   2459     6   2459 |   2345    8    1 |     7     249    23 |
|  12459     3  12459 |   2457  579   24 |  2456  124689  1268 |
|      7     8  12459 |   2345   59    6 |  2345    1249   123 |
+---------------------+------------------+---------------------+
Mike Barker
 
Posts: 458
Joined: 22 January 2006

Postby didier90 » Sat Aug 12, 2006 10:15 am

hello

another puzle

Code: Select all



 
 *--------------------------------------------------------------------*
 | 9      1      6      | 238    4      5      | 27     378    78     |
 | 348    47     378    | 1      289    2369   | 26     5      4689   |
 | 3458   2      358    | 389    7      369    | 1      348    4689   |
 |----------------------+----------------------+----------------------|
 | 7      5      4      | 6      3      8      | 9      2      1      |
 | 23     8      39     | 47     1      29     | 57     6      457    |
 | 26     69     1      | 5      29     47     | 8      47     3      |
 |----------------------+----------------------+----------------------|
 | 1      46     2      | 478    5      47     | 3      9      678    |
 | +46     3     *789    | 24789  289    1      |*567   *78    *5678   |
 | 58     79     5789   | 3789   6      39     | 4      1      2      |
 *--------------------------------------------------------------------*





als1 r8c3789 (*)
als2 r8c1 (+)
x = 6


9 in r9c2 sees all 9 of als1 and also x is in als2

if r9c2 = 9 r8c1 = 6


===> r9c2 = 7

it is mortal for this puzle
another methode?

didier
didier90
 
Posts: 3
Joined: 06 August 2006

Next

Return to Advanced solving techniques