The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |.3.|81.|4..|
 |...|934|..1|
 |1..|2..|...|
 |---+---+---|
 |.2.|.6.|853|
 |65.|378|.94|
 |348|.9.|.1.|
 |---+---+---|
 |...|..5|..6|
 |7..|189|...|
 |..2|.43|.7.|
 *-----------*


Play/Print this puzzle online

[Leren]

Code: Select all

*--------------------------------------------------------------*
| 25    3    f9-56   | 8     1     7      | 4     26   e59     |
| 258   78   a56     | 9     3     4      |d57    26    1      |
| 1     79    4      | 2     5     6      | 379   38    89     |
|--------------------+--------------------+--------------------|
| 9     2     7      | 4     6     1      | 8     5     3      |
| 6     5     1      | 3     7     8      | 2     9     4      |
| 3     4     8      | 5     9     2      | 6     1     7      |
|--------------------+--------------------+--------------------|
| 4     189   39     | 7     2     5      | 139   38    6      |
| 7     6    b35     | 1     8     9      |c35    4     2      |
| 58    189   2      | 6     4     3      | 159   7     589    |
*--------------------------------------------------------------*

(6=5) r2c3 - (5) r8c3 = r8c7 - r2c7 = (5-9) r1c9 = (9) r1c3 => - 56 r1c3; stte

Leren

[SteveG48]

Code: Select all

.---------------.---------.--------------.
| 25   3   a569 | 8  1  7 | 4    26  59  |
| 258  78   56  | 9  3  4 | 57   26  1   |
| 1    f79  4   | 2  5  6 | g379 h38 I89 |
:---------------+---------+--------------:
| 9    2    7   | 4  6  1 | 8    5   3   |
| 6    5    1   | 3  7  8 | 2    9   4   |
| 3    4    8   | 5  9  2 | 6    1   7   |
:---------------+---------+--------------:
| 4    189  b39 | 7  2  5 | 139  38  6   |
| 7    6    c35 | 1  8  9 | 35   4   2   |
| d58  189  2   | 6  4  3 | 159  7 ej589 |
'---------------'---------'--------------'


Parallel chains:

(9)r1c3 = {(9-3)r7c3 = (3-5)r8c3 = (5)r9c1 - (5)r9c9}|{(9-7)r3c2 = (7-3)r3c7 = (3-8)r3c8 = (8)r3c9 - (8)r9c9} = (9)r9c9 => -9 r1c9 ; stte

[SteveG48]

Code: Select all

*--------------------------------------------------------------*
| 25    3    f9-56   | 8     1     7      | 4     26   e59     |
| 258   78   a56     | 9     3     4      |d57    26    1      |
| 1     79    4      | 2     5     6      | 379   38    89     |
|--------------------+--------------------+--------------------|
| 9     2     7      | 4     6     1      | 8     5     3      |
| 6     5     1      | 3     7     8      | 2     9     4      |
| 3     4     8      | 5     9     2      | 6     1     7      |
|--------------------+--------------------+--------------------|
| 4     189   39     | 7     2     5      | 139   38    6      |
| 7     6    b35     | 1     8     9      |c35    4     2      |
| 58    189   2      | 6     4     3      | 159   7     589    |
*--------------------------------------------------------------*

(6=5) r2c3 - (5) r8c3 = r8c7 - r2c7 = (5-9) r1c9 = (9) r1c3 => - 56 r1c3; stte

Leren

Leren, I guess I'm slow tonight, because I don't follow the logic. I can see where you've shown that if r2c3 is not a 6, then r1c3 must be a 9. But what if r2c3 is a 6? In fact, I see that you've proved that r2c3 is, in fact, a 6, but that leaves you with a 3,5,9 triple on column 3.

[Leren]

SteveG48 wrote: (6=5) r2c3 - (5) r8c3 = r8c7 - r2c7 = (5-9) r1c9 = (9) r1c3 => - 56 r1c3; stte

Leren, I guess I'm slow tonight, because I don't follow the logic. I can see where you've shown that if r2c3 is not a 6, then r1c3 must be a 9. But what if r2c3 is a 6?

Hi Steve, if you look at the chain carefully you'll see that it establishes 2 things:

1. If r2c3 <> 6 then r1c3 = 9 ( in particular <> 6), but also

2. If r8c3 <> 5 then r1c3 = 9 ( in particular <> 5). [Just ignore the first term in the chain and you'll see that it's true].

Obviously if r2c3 = 6 then r1c3 <> 6 => - 6 r1c3 and if r8c3 = 5 then r1c3 <> 5 => - 5 r1c3.

You can state this differently in words and it might be more intuitive: If r8c3 <> 5 then r1c3 = 9 ( in particular <> 5, 6). However if r8c3 = 5 then r1c3 <> 5 but also r2c3 <> 5 (= 6) so r1c3 <> 6 => - 56 r1c3.

Leren

[ronk]

Leren, I guess I'm slow tonight, because I don't follow the logic. I can see where you've shown that if r2c3 is not a 6, then r1c3 must be a 9. But what if r2c3 is a 6? In fact, I see that you've proved that r2c3 is, in fact, a 6, but that leaves you with a 3,5,9 triple on column 3.

SteveG48, it might help to think of it as two different chains:

Code: Select all

             (5) r8c3 = r8c7 - r2c7 = (5-9) r1c9 = (9) r1c3 => - 5 r1c3
(6=5) r2c3 - (5) r8c3 = r8c7 - r2c7 = (5-9) r1c9 = (9) r1c3 => - 6 r1c3; stte

[SteveG48]

Thanks, Leren, Ronk. I wondered about the underlining in the chain. Now I see the significance.

This is a really pretty little puzzle- and a pretty solution.

The more I look at it, the neater it looks. Is that called something in particular?

The way I like to look at it is that 9 in r1c3 is a weak link to the 5,6 pair in r1c3, followed by strong links to the 6 in r2c3 and the 5 in r7c3, setting 5 and 6 in those cells and excluding them in r1c3.

[DonM]

Thanks, Leren, Ronk. I wondered about the underlining in the chain. Now I see the significance.

This is a really pretty little puzzle- and a pretty solution.

The more I look at it, the neater it looks. Is that called something in particular?

The way I like to look at it is that 9 in r1c3 is a weak link to the 5,6 pair in r1c3, followed by strong links to the 6 in r2c3 and the 5 in r7c3, setting 5 and 6 in those cells and excluding them in r1c3.

This was given the name 'pausing chain' some time ago whereby the chain 'pauses' (figuratively speaking) for a moment to declare another elimination using the point at the 2nd underline the same as if it was the beginning of the chain. You can also use a '*' to indicate the point of pausing.

You will also see pausing chains where the chain is continued beyond the point of a first elimination to a point of yet another elimination.

[Leren]

SteveG48 wrote: Is that called something in particular?

"Technically" this is an L2 Wing: (5) r8c3 = r8c7 - r2c7 = (5-9) r1c9 = (9) r1c3 + a bi-value extension Cell r2c3.
The extension cell must see both ends of the L2 Wing and contain one of the end candidates (5 or 9) plus one other candidate (6). The ''other'' candidate can be eliminated from the end of the AIC not containing the
Strong link to the bi-value cell end candidate (5).

Of course this works for any AIC Type 2 - not just L2 wings.

Leren

[pjb]

Code: Select all

 25     3      569    | 8      1      7      | 4      26     59     
 258    78     56     | 9      3      4      | 57     26     1     
 1     b79     4      | 2      5      6      |c379    38     89     
 ---------------------+----------------------+---------------------
 9      2      7      | 4      6      1      | 8      5      3     
 6      5      1      | 3      7      8      | 2      9      4     
 3      4      8      | 5      9      2      | 6      1      7     
 ---------------------+----------------------+---------------------
 4     a189    39     | 7      2      5      |c139    38     6     
 7      6      35     | 1      8      9      |c35     4      2     
a58    a189    2      | 6      4      3      |c159    7      89-5   


(5=9) r7c2, r9c12 - (9=7) r3c2 - (7=5) r3789c7 => -5 r9c9; stte

Phil

[ArkieTech]

Code: Select all

 *--------------------------------------------------*
 | 25   3   b569  | 8    1    7    | 4    26  c59   |
 | 258  78   56   | 9    3    4    | 57   26   1    |
 | 1    79   4    | 2    5    6    | 379  38   89   |
 |----------------+----------------+----------------|
 | 9    2    7    | 4    6    1    | 8    5    3    |
 | 6    5    1    | 3    7    8    | 2    9    4    |
 | 3    4    8    | 5    9    2    | 6    1    7    |
 |----------------+----------------+----------------|
 | 4    189 a39   | 7    2    5    | 139  38   6    |
 | 7    6   a35   | 1    8    9    | 3-5  4    2    |
 | 8-5  189  2    | 6    4    3    | 159  7   d589  |
 *--------------------------------------------------*
als m-wing
(5=9)r78c3-r1c3=(9-5)r1c9=5r9c9 -5r8c7,r9c1; ste

[eleven]

(5=9) r7c2, r9c12 - (9=7) r3c2 - (7=5) r3789c7 => -5 r9c9; stte

(5=9)r78c3-r1c3=(9-5)r1c9=5r9c9 -5r8c7,r9c1; ste

Normally i would write it, as i found it:
Almost pair 56+9 in r12c3:
r1c3=9 -> r1c9=5
r12c3=56 -> r8c7=5 (through r8c3<>5)
or
(5=9)r1c9-r1c3=(9-3)r7c3=(3-5)r8c3=5r8c7

But you seem to like that one more:
(5=9)r1c9-(9=356)r128c3-(3=5)r8c7 => r23c7,r9c9<>5

[SteveG48]

SteveG48 wrote: Is that called something in particular?

"Technically" this is an L2 Wing:

Thanks, Leren. I see you making good use of things I haven't learned about like L2 and H2 wings. Can you point me to a reference on these?

[SteveG48]

The more I look at it, the neater it looks. Is that called something in particular?

This was given the name 'pausing chain' some time ago whereby the chain 'pauses' (figuratively speaking) for a moment to declare another elimination using the point at the 2nd underline the same as if it was the beginning of the chain.

Thanks, Don.

[tlanglet]

I have been gone for a while and now notice a lot of "action" during that time. I expect it will take me a while to catch up on all that activity, but I do have a solution for the puzzle today. It is an interesting AUR(19)r79c27 with extra internal digits in all four cells and a x-wing(1) overlay.

Code: Select all

 *--------------------------------------------------*
 | 25   3    569  | 8    1    7    | 4    26   59   |
 | 258  78   56   | 9    3    4    | 57   26   1    |
 | 1    79   4    | 2    5    6    | 379  38   89   |
 |----------------+----------------+----------------|
 | 9    2    7    | 4    6    1    | 8    5    3    |
 | 6    5    1    | 3    7    8    | 2    9    4    |
 | 3    4    8    | 5    9    2    | 6    1    7    |
 |----------------+----------------+----------------|
 | 4   *189  39   | 7    2    5    |*139  38   6    |
 | 7    6    35   | 1    8    9    | 35   4    2    |
 | 58  *189  2    | 6    4    3    |*159  7    589  |
 *--------------------------------------------------*


The external SIS is simply 9r7c3 & 9r9c9, thus
AUR(19)r79c27[9r9c9=(9-3)r7c3]=3r8c3-(3=5)r8c7 => r9c9<>5 to complete the puzzle.

Also, the four internal SIS, 8r79c2,3r7c7,5r9c7, provide the identical solution.
8r79c2-(8=5)r9c1 => r9c9<>5
||
3r7c7-(3=5)r8c7 => r9c9<>5
||
5r9c7 => r9c9<>5

Ted