The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |9.5|3..|7..|
 |..4|..1|...|
 |.3.|9.7|6..|
 |---+---+---|
 |.4.|...|..2|
 |.2.|1.5|.7.|
 |5..|...|.8.|
 |---+---+---|
 |..2|7.9|.4.|
 |...|2..|9..|
 |..7|..8|2.6|
 *-----------*


Play/Print this puzzle online

[SteveG48]

Code: Select all

 *--------------------------------------------------------------------*
 | 9     c168    5      | 3     b2468  b246    | 7      12     148    |
 | 2678  d678    4      |a5-6    2568   1      | 38     9     g358    |
 |e128    3     e18     | 9      2458   7      | 6     f125    1458   |
 *----------------------+----------------------+----------------------|
 | 17     4      19     | 8      79     3      | 5      6      2      |
 | 368    2      3689   | 1      469    5      | 34     7      39     |
 | 5     d67     369    | 46     24679  246    | 134    8      139    |
 *----------------------+----------------------+----------------------|
 | 136    156    2      | 7      1356   9      | 18     4      158    |
 | 13468  1568   1368   | 2      13456  46     | 9      135    7      |
 | 134    9      7      | 45     1345   8      | 2      135    6      |
 *--------------------------------------------------------------------*


(6)r2c4 - r1c56 = r1c2 - (6=78)r62c2 - (8=12)r3c13 - (12=5)r3c8 - (5=38) r2c79 => -6 r2c4 ; stte

[pjb]

Steve, a lovely loop, but I can only make it eliminate the 5 in r2c9. Can you please show how it works?

Phil

[SteveG48]

Phil, the idea is that if r2c4 is a 6, then r2c2 must be an 8 and either r2c7 or r2c9 must also be an 8. Since that can't be, r2c4 can't be a 6.

I could have started with assuming that r2c2 is an 8 and gotten a contradiction that eliminates that 8, but that only gives lclste.

[ArkieTech]

Code: Select all

 *--------------------------------------------------------------------*
 | 9      168    5      | 3      2468   246    | 7     a12     148    |
 | 267-8  67-8   4      | 56     2568   1      |a38     9     a358    |
 |b128    3     b18     | 9      2458   7      | 6     a125    1458   |
 |----------------------+----------------------+----------------------|
 | 17     4      19     | 8      79     3      | 5      6      2      |
 | 368    2      3689   | 1      469    5      | 34     7      39     |
 | 5      67     369    | 46     24679  246    | 134    8      139    |
 |----------------------+----------------------+----------------------|
 | 136    156    2      | 7      1356   9      | 18     4      158    |
 | 13468  1568   1368   | 2      13456  46     | 9      135    7      |
 | 134    9      7      | 45     1345   8      | 2      135    6      |
 *--------------------------------------------------------------------*
(8=12)r2c78,r13c8-(12=8)r3c13 => -8r2c12; lclste

[Luke]

Code: Select all

 *--------------------------------------------------------------------*
 | 9      168    5      | 3      2468   246    | 7     a12     148    |
 | 267-8  67-8   4      | 56     2568   1      |a38     9     a358    |
 |b128    3     b18     | 9      2458   7      | 6     a125    1458   |
 |----------------------+----------------------+----------------------|
 | 17     4      19     | 8      79     3      | 5      6      2      |
 | 368    2      3689   | 1      469    5      | 34     7      39     |
 | 5      67     369    | 46     24679  246    | 134    8      139    |
 |----------------------+----------------------+----------------------|
 | 136    156    2      | 7      1356   9      | 18     4      158    |
 | 13468  1568   1368   | 2      13456  46     | 9      135    7      |
 | 134    9      7      | 45     1345   8      | 2      135    6      |
 *--------------------------------------------------------------------*
(8=12)r2c78,r13c8-(12=8)r3c13 => -8r2c12; lclste

Dan, I don't get Steve's and now I don't get yours, so it must be me.

It looks like you're starting with (8=1235)box 3. Even if I break it down this way ... (8=35)r2c79-(5=12)r13c8 ...I cannot make the leap to your conclusion.

Edit
Looking backwards, I got it. (8=35)r2c79-(5=128)r3c138 (als xz)
Putting r1c8 in there was a red herring

[tlanglet]

Code: Select all

 *--------------------------------------------------------------------*
 | 9     c168    5      | 3     b2468  b246    | 7      12     148    |
 | 2678  d678    4      |a56     2568   1      | 38     9      38-5   |
 |e128    3     e18     | 9      248-5  7      | 6     e125    1458   |
 |----------------------+----------------------+----------------------|
 | 17     4      19     | 8      79     3      | 5      6      2      |
 | 368    2      3689   | 1      469    5      | 34     7      39     |
 | 5     d67     369    | 46     24679  246    | 134    8      139    |
 |----------------------+----------------------+----------------------|
 | 136    156    2      | 7      1356   9      | 18     4      158    |
 | 13468  1568   1368   | 2      13456  46     | 9      135    7      |
 | 134    9      7      | 45     1345   8      | 2      135    6      |
 *--------------------------------------------------------------------*


I initially worked the ANP(38=5)r2c79 which resulted in a contradiction, so I reworked the solution to have a normal deduction.

(5=6)r2c4-6r1c56=6r1c2-(6=78)r62c2-(8=125)r3c318 => r2c9,r3c5<>5

Ted

[tlanglet]

Code: Select all

 *--------------------------------------------------------------------*
 | 9      168    5      | 3      2468   246    | 7     a12     148    |
 | 267-8  67-8   4      | 56     2568   1      |a38     9     a358    |
 |b128    3     b18     | 9      2458   7      | 6     a125    1458   |
 |----------------------+----------------------+----------------------|
 | 17     4      19     | 8      79     3      | 5      6      2      |
 | 368    2      3689   | 1      469    5      | 34     7      39     |
 | 5      67     369    | 46     24679  246    | 134    8      139    |
 |----------------------+----------------------+----------------------|
 | 136    156    2      | 7      1356   9      | 18     4      158    |
 | 13468  1568   1368   | 2      13456  46     | 9      135    7      |
 | 134    9      7      | 45     1345   8      | 2      135    6      |
 *--------------------------------------------------------------------*
(8=12)r2c78,r13c8-(12=8)r3c13 => -8r2c12; lclste

Very nice solution Dan. Thanks also for the explanation Luke.

I mucked around with both the ANP(38=5) and ANP(12=5) in box 3, but failed to consider the als_xz possibility.

Ted

[SteveG48]

Dan, I don't get Steve's and now I don't get yours, so it must be me.

I'd say not, at least with mine, since Phil didn't see it either. Apparently my explanation wasn't clear enough, so I'll try again.

If we assume that r2c4 is a 6, then it follows that r2c2 must be an 8, step d on the grid. I've underlined this to emphasize that it's essential to the final conclusion. It then follows that r2c9 cannot be a 5, step g. We're left with a 3/8 pair in r2c79. (I should have put a g by r2c7 as well as r2c9.) Therefore, either r2c7 or r2c9 must be an 8, contradicting the intermediate conclusion at step d, so we must reject the initial assumption of 6 at r2c4.

This isn't as satisfying as a chain in which the end directly contradicts the beginning, but I think the logic is sound.

[pjb]

Steve, I should have responded earlier, but I am happy with your explanation. A good solution. I'm still looking.

Phil

[SteveG48]

I'm still looking.

Phil

As am I. It's not an easy puzzle for true single steppers. Everything I try brings me back into the same path that I've already been down.

[daj95376]

_

Notation seems to be going to pot around here. Steve's solutions are "unusual" but fun to unravel. Dan's solution is just plain wrong. The <12> cells are r13c8 ... and these cells never directly sees r3c13.

Luke provided an accurate ALS interpretation, but I'm partial to:

Code: Select all

 +-----------------------------------------------------------------------+
 |  9      168    5      |  3      2468   246    |  7      12     148    |
 |  2678   678    4      |  56     2568   1      |  38     9      358    |
 |  128    3      18     |  9      2458   7      |  6      125    1458   |
 |-----------------------+-----------------------+-----------------------|
 |  17     4      19     |  8      79     3      |  5      6      2      |
 |  368    2      3689   |  1      469    5      |  34     7      39     |
 |  5      67     369    |  46     24679  246    |  134    8      139    |
 |-----------------------+-----------------------+-----------------------|
 |  136    156    2      |  7      1356   9      |  18     4      158    |
 |  13468  1568   1368   |  2      13456  46     |  9      135    7      |
 |  134    9      7      |  45     1345   8      |  2      135    6      |
 +-----------------------------------------------------------------------+
 # 94 eliminations remain

 (38=5)r2c79 - (5=128)r3c813  =>  r2c2<>8


This way, the restrictive common to both ALS-XZ sets is distinctly identified.

I don't care if it takes a couple more basic steps as well. If the presence of basic steps at the end of the solution are important, then we should start counting the presence of basic steps before the single-stepper grid. Someone wants their cake and to be able to eat it, too.

[ronk]

Notation seems to be going to pot around here.

Would that be a reference to cannabis or the loo? .

[SteveG48]

_

Notation seems to be going to pot around here. Steve's solutions are "unusual" but fun to unravel.

Thanks. I think. Maybe it's the pot, as Ronk said.

Seriously, on the subject of notation, you never did comment on this: http://forum.enjoysudoku.com/dan-s-donner-december-20-2013-t31536-15.html#p232382 . I'd never seen this sort of thing until a few weeks ago, so I need to figure out what is and isn't considered good form.

I don't care if it takes a couple more basic steps as well. If the presence of basic steps at the end of the solution are important, then we should start counting the presence of basic steps before the single-stepper grid. Someone wants their cake and to be able to eat it, too.

Matter of taste. As long as we're all having fun....

[daj95376]

Seriously, on the subject of notation, you never did comment on this: http://forum.enjoysudoku.com/dan-s-donner-december-20-2013-t31536-15.html#p232382 . I'd never seen this sort of thing until a few weeks ago, so I need to figure out what is and isn't considered good form.

Normally, I disdain some of the ways an ALS is used -- especially when it's to conceal network logic. In the message that you reference, using an ALS to condense a boring sequence through bivalue cells in the same unit/house seemed to be an acceptable alternative. Nothing in that ALS performed anything that wouldn't have been performed through the bivalue sequence! However, there is a BIG caveat that goes along with this approach when a continuous loop is involved. One can easily forget to include eliminations because the weak links between the bivalue cells are now less obvious.