The New Sudoku Players' Forum

[didier90]

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

[udosuk]

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

[Myth Jellies]

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.

[Steve R]

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 y ≠ x, the three eliminations of udosuk follow.

Steve

[Myth Jellies]

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.

[ronk]

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?

[re'born]

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.

[ronk]

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

"Dynamic" multiplicity!

[didier90]

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

[Havard]

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

[ronk]

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.

[daj95376]

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>


[ronk]

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.

[Mike Barker]

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 |
+---------------------+------------------+---------------------+

[didier90]

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