The New Sudoku Players' Forum

[GreenLantern]

ronk mentions 23 eliminations based on Havard's nice loop:

Code: Select all

 3      B168     7       |*59     568(9)  4        |*19     2      *569
 456    B26      2456    | 1      2569   *235679   | 8      3456   *79(3456)
 9      B1268    124568  |*237(5) 2568    23568(7) | 1347   13456   34567
-------------------------+-------------------------+------------------------
 168(7) B12678   1268    |*47(5)  3       15(7)    | 14(7)    9     4568
 167     5       39      | 8      1249    1279     | 12347  1346    3467
 178     4       39      | 6      1259    12579    | 1237   1358    3578
-------------------------+-------------------------+------------------------
 4678    379(68) 468     | 2349   24689   23689    | 5      347(8)  1
 2      B136(8)  156(48) |A345    7       156(38)  |A349   A348    A3489
 14578   379(18) 1458    | 349(5) 14589   13589    | 6      347(8)  2


If you include the eliminations in Box 7 and r8c6<>8 in your list of exclusions, then I think that r56c1/r79c2<>7 should also be included
since r8c2<>3 => r1238c2=1268 => r4c2=7.

[Havard]

Very interesting grouped nice loop! I think that r34c4<>5 can also be added to your list of eliminations that follow from the continuous nice loop.

Great observation, but now I am really confused about the rules for when a loop can "eat itself"... Can anyone make a general statement about this?

ronk: I love your notation!

Havard

[GreenLantern]

Original grid:

Code: Select all

3     168     7       | 59     5689  4        | 19    2     569           
456   26      2456    | 1      2569  235679   | 8     3456  345679
9     1268    14568   | 2357   2568  235678   | 1347  13456 34567         
----------------------+-----------------------+----------------------------
1678  12678   128     | 457    3     15       | 147   9     4568           
167   5       39      | 8      1249  1279     | 12347 1346  3467           
178   4       39      | 6      1259  12579    | 1237  1358  3578           
----------------------+-----------------------+----------------------------
4678  36789   468     | 2349   24689 23689    | 5     3478  1             
2     1368    14568   | 345    7     1568     | 349   348   3489           
14578 13789   1458    | 349    14589 13589    | 6     3478  2             


If I re-write the loop as a multiple implication continuous loop, I don't really see the loop eating itself:

Code: Select all

[r2c6]=7=[r2c9]=9=[r1c79]-9-[r1c4]-5-{Naked Quad (3489): [r8c4789]}
     (-8-[r8c236])(-4-[r8c3])-3-{Naked Quad (1268): [r1238c2]}(-8-[r79c2])
     (-1-[r9c2])=3|7=[r4c2](-7-[r56c1|r79c2])-7-[r4c4]=7=[r3c4]
      -7-[r2c6]

=> r2c9<>3456, r1c5<>9, r34c4<>5, r8c236<>8, r8c3<>4, r9c2<>1, r79c2<>8,
   r56c1/r79c2<>7, r4c17<>7, r3c6<>7


Nice loop experts -- did I write this loop in a valid fashion and are my deductions correct?

[ronk]

... now I am really confused about the rules for when a loop can "eat itself"... Can anyone make a general statement about this?

1. When part of a continuous loop, a weak link ...

cellset1-x-cellset2

... effectively becomes a strong link and digit x may be excluded from any cells that see all digits x of cellset1 and cellset2.

2. When part of a continuous loop, the "extra" digits {st...xy} of an almost-locked-set ...

-r-(ALS:cellset1=r|st...xy|z=cellset2)-z-

... become locked in the ALS. Any digit matching an extra digit, that also sees all instances of that extra digit in the ALS, may be excluded.

For your very nice continuous loop example:

r2c6=7=r2c9=9=r1c79-9-r1c4-5-{A:r8c4=5|489|3=r8c4789} -3- {B(r12348c2):r8c2=3|1268|7=r4c2}-7-r4c4=7=r3c4-7-r2c6

1. Normally, for the (blue) strong link ... r1c4-5-r8c4 ... at least one of r1c4=5 and r8c4=5 is false. Because of the continuous loop, exactly one of r1c4=5 and r8c4=5 is false. Alternatively stated, exactly one of r1c4=5 and r8c4=5 must be true. Therefore, digit 5 may not exist anywhere else in c4 ... whether or not it's in another part of the continuous loop.

2. (I'll try to add a proof here of why digits {489} are locked in ALS set A. Anyone have one?) The (red) extra digits {489} between the vertical bars of A are locked in A because of the continuous loop. They can exist nowhere else in r8.

ronk: I love your notation!

Thanks.

I think that r56c1/r79c2<>7 should also be included since r8c2<>3 ...

I'm quite sure that's incorrect. We know only ... in this continuous loop ... that [edit: exactly one of r8c2=3 or r4c2=7 is true].

[GreenLantern]

I'm quite sure that's incorrect. We know only ... in this continuous loop ... that either r8c2<>3 or r4c2<>7.

Thanks for the clarification, ronk.

[daj95376]

Havard,

What a wonderfully challenging puzzle. Thanks for sharing it.

[Viggo]

Code: Select all

[r2c6]=7=[r2c9]=9=[r1c79]-9-[r1c4]-5-{Naked Quad (3489): [r8c4789]}
     (-8-[r8c236])(-4-[r8c3])-3-{Naked Quad (1268): [r1238c2]}(-8-[79c2])
     (-1-[r9c2])=3|7=[r4c2](-7-[r56c1|r79c2])-7-[r4c4]=7=[r3c4]
      -7-[r2c6]

=> r2c9<>3456, r1c5<>9, r34c4<>5, r8c236<>8, r8c3<>4, r9c2<>1, r79c2<>8,  r56c1/r79c2<>7, r4c17<>7, r3c6<>7

- did I write this loop in a valid fashion and are my deductions correct?

In addition to Ronks comments, I think this part can becomme a problem: {Naked Quad (1268): [r1238c2]} (I have got the same problem in a previous post in this thread). If you consider the ALS in c2 to be a Naked Quad, then it changes position depending on how you read the direction of the loop. In one direction it becomes [r1238c2] and in the other direction [r1234c2]. This makes the statement unclear to me - what becomes the locked part of this ALS?

Therefore I like Ronk's notation better, because it is independent on the direction. Perhaps it is better to consider the ALS in C2 as a naked quint with [r12348c2], because then the ALS do not change position.

/Viggo