The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |..5|4.3|2..|
 |...|179|...|
 |9..|...|..7|
 |---+---+---|
 |78.|3.1|.95|
 |.5.|...|.2.|
 |19.|8.4|.36|
 |---+---+---|
 |8..|...|..4|
 |...|618|...|
 |..7|9.5|6..|
 *-----------*


Play/Print this puzzle online

[pjb]

Code: Select all

6      7      5      | 4      8      3      | 2      1      9     
234    234    348    | 1      7      9      | 5      6      38     
9      13    c138    | 5      6      2      |b38     4      7     
---------------------+----------------------+---------------------
7      8      6      | 3      2      1      | 4      9      5     
34     5      34     | 7      9      6      |a18     2      18     
1      9      2      | 8      5      4      | 7      3      6     
---------------------+----------------------+---------------------
8      6     d19     | 2      3      7      | 9-1    5      4     
5      234    349    | 6      1      8      | 39     7      23     
23     123    7      | 9      4      5      | 6      8      123   


(1=8)r5c7 - r3c7 = (8-1)r3c3 = r7c3 => -1 r7c7; stte

Phil

[SteveG48]

Code: Select all

 *--------------------------------------------------*
 | 6    7    5    | 4    8    3    | 2    1    9    |
 | 234  234  348  | 1    7    9    | 5    6    3-8  |
 | 9   d13   138  | 5    6    2    |e38   4    7    |
 *----------------+----------------+----------------|
 | 7    8    6    | 3    2    1    | 4    9    5    |
 | 34   5    34   | 7    9    6    | 18   2   a18   |
 | 1    9    2    | 8    5    4    | 7    3    6    |
 *----------------+----------------+----------------|
 | 8    6    19   | 2    3    7    | 19   5    4    |
 | 5    234  349  | 6    1    8    | 39   7    23   |
 | 23  c123  7    | 9    4    5    | 6    8   b123  |
 *--------------------------------------------------*


(8=1)r5c9 - r9c9 = r9c2 - (1=3)r3c2 - (3=8)r3c7 => -8 r2c9 ; stte

[ixsetf]

Code: Select all

+-------------+-------+----------+
| 6   7   5   | 4 8 3 | 2  1 9   |
| 234 234 348 | 1 7 9 | 5  6 38  |
| 9   13  138 | 5 6 2 | 38 4 7   |
+-------------+-------+----------+
| 7   8   6   | 3 2 1 | 4  9 5   |
| 34  5   34  | 7 9 6 | 18 2 18  |
| 1   9   2   | 8 5 4 | 7  3 6   |
+-------------+-------+----------+
| 8   6   19  | 2 3 7 | 19 5 4   |
| 5   234 349 | 6 1 8 | 39 7 23  |
| 23  123 7   | 9 4 5 | 6  8 123 |
+-------------+-------+----------+


I don't really know how the notation works, so I'll try to get the information across as best I can.

r3c3 = 1, r9c2 = 1 (only 1 in box), r7c7 = 1 (only 1 in box), r5c9 = 1 (only 1 in box), r5c7 = 8 (only value in box), r2c9 = 8 (only 8 in box), top left box contains no position for an 8, therefore r3c3 =/= 1.

basics to end.

[Leren]

Code: Select all

*--------------------------------------------------------------*
| 6     7     5      | 4     8     3      | 2     1     9      |
| 234   234   348    | 1     7     9      | 5     6     38     |
| 9    c13   d138    | 5     6     2      |b38    4     7      |
|--------------------+--------------------+--------------------|
| 7     8     6      | 3     2     1      | 4     9     5      |
| 34    5     34     | 7     9     6      |a18    2     18     |
| 1     9     2      | 8     5     4      | 7     3     6      |
|--------------------+--------------------+--------------------|
| 8     6    e19     | 2     3     7      | 9-1   5     4      |
| 5     234   349    | 6     1     8      | 39    7     23     |
| 23    123   7      | 9     4     5      | 6     8     123    |
*--------------------------------------------------------------*

XY Wing with transport (1=8) r5c7 - (8=3) r3c7 - (3=1) r3c2 - r3c3 = (1) r7c3 => - 1 r7c7; stte

Leren

[ArkieTech]

Code: Select all

 *--------------------------------------------------*
 | 6    7    5    | 4    8    3    | 2    1    9    |
 | 234  234 b348  | 1    7    9    | 5    6   a38   |
 | 9    13   138  | 5    6    2    | 8-3  4    7    |
 |----------------+----------------+----------------|
 | 7    8    6    | 3    2    1    | 4    9    5    |
 | 34   5   b34   | 7    9    6    | 18   2    18   |
 | 1    9    2    | 8    5    4    | 7    3    6    |
 |----------------+----------------+----------------|
 | 8    6    19   | 2    3    7    | 19   5    4    |
 | 5    234 b349  | 6    1    8    |c39   7    2-3  |
 | 23   123  7    | 9    4    5    | 6    8    12-3 |
 *--------------------------------------------------*
(3=8)r2c9-(8=9)r258c3-(9=3)r8c7 => -3r3c7,r89c9; ste

[Ngisa]

Code: Select all

+-------------+-------+----------+
| 6   7   5   | 4 8 3 | 2  1 9   |
| 234 234 348 | 1 7 9 | 5  6 38  |
| 9   a13  138 | 5 6 2 | 8-3 4 7   |
+-------------+-------+----------+
| 7   8   6   | 3 2 1 | 4  9 5   |
| 34  5   34  | 7 9 6 | 18 2 18  |
| 1   9   2   | 8 5 4 | 7  3 6   |
+-------------+-------+----------+
| 8   6   c19  | 2 3 7 | d19 5 4   |
| 5   24-3 349 | 6 1 8 | e39 7 23  |
| 23  b123 7   | 9 4 5 | 6  8 123 |
+-------------+-------+----------+


(3=1)r3c2-r9c2-(9=1)r7c3-(1=9)r7c7-(9=3)r8c7 => -3r3c7, r8c2; stte.

[Sudtyro2]

r3c3 = 1, r9c2 = 1 (only 1 in box), r7c7 = 1 (only 1 in box), r5c9 = 1 (only 1 in box), r5c7 = 8 (only value in box), r2c9 = 8 (only 8 in box), top left box contains no position for an 8, therefore r3c3 =/= 1. basics to end.

Code: Select all

+---------------+-------+-----------+
| 6   7    5    | 4 8 3 | 2  1  9   |
| 234 234 f348  | 1 7 9 | 5  6 e38  |
| 9  a13  g38-1 | 5 6 2 | 38 4  7   |
+---------------+-------+-----------+
| 7   8    6    | 3 2 1 | 4  9  5   |
| 34  5    34   | 7 9 6 | 18 2  d18 |
| 1   9    2    | 8 5 4 | 7  3  6   |
+---------------+-------+-----------+
| 8   6    19   | 2 3 7 | 19 5  4   |
| 5   234  349  | 6 1 8 | 39 7  23  |
| 23 b123  7    | 9 4 5 | 6  8 c123 |
+---------------+-------+-----------+

Hi ixsetf,
One alternative to your sequence of assignments is to use an AIC that includes some of your nodes but doesn't require specific parity assignments. Each end-node of the AIC has a strong-inference link, meaning that one or both of those nodes must be true. Hence, any candidate that can “see” (weakly link to) both ends of the chain must be false.

(1)r3c2=(1)r9c2-(1)r9c9=(1-8)r5c9=(8)r2c9-(8)r2c3=(8)r3c3 => r3c3<>1; singles to end.

SteveC

[Marty R.]

Code: Select all

+-------------+-------+----------+
| 6   7   5   | 4 8 3 | 2  1 9   |
| 234 234 348 | 1 7 9 | 5  6 38  |
| 9   13  138 | 5 6 2 | 38 4 7   |
+-------------+-------+----------+
| 7   8   6   | 3 2 1 | 4  9 5   |
| 34  5   34  | 7 9 6 | 18 2 18  |
| 1   9   2   | 8 5 4 | 7  3 6   |
+-------------+-------+----------+
| 8   6   19  | 2 3 7 | 19 5 4   |
| 5   234 349 | 6 1 8 | 39 7 23  |
| 23  123 7   | 9 4 5 | 6  8 123 |
+-------------+-------+----------+


Play this puzzle online at the Daily Sudoku site

I wasted too much time and tried too hard just to fail in an attempt to play the BUG+8

(1=3)r3c2-(3=8)r3c7-(8=1)r5c7-r5c9=r9c9=>r9c2<>1

[Marty R.]

ixsetf,

top left box

Boxes are numbered as seen below. A box is referred to, for example, as box 5 or b5

Code: Select all

+-------+-------+-------+
| . . . | . . . | . . . |
| . 1 . | . 2 . | . 3 . |
| . . . | . . . | . . . |
+-------+-------+-------+
| . . . | . . . | . . . |
| . 4 . | . 5 . | . 6 . |
| . . . | . . . | . . . |
+-------+-------+-------+
| . . . | . . . | . . . |
| . 7 . | . 8 . | . 9 . |
| . . . | . . . | . . . |
+-------+-------+-------+


Play this puzzle online at the Daily Sudoku site

[Sudtyro2]

Code: Select all

+-------------+-------+----------+
| 6   7   5   | 4 8 3 | 2  1 9   |
| 234 234 348 | 1 7 9 | 5  6 38  |
| 9   13  138 | 5 6 2 | 38 4 7   |
+-------------+-------+----------+
| 7   8   6   | 3 2 1 | 4  9 5   |
| 34  5   34  | 7 9 6 | 18 2 18  |
| 1   9   2   | 8 5 4 | 7  3 6   |
+-------------+-------+----------+
| 8   6   19  | 2 3 7 | 19 5 4   |
| 5   234 349 | 6 1 8 | 39 7 23  |
| 23  123 7   | 9 4 5 | 6  8 123 |
+-------------+-------+----------+


I wasted too much time and tried too hard just to fail in an attempt to play the BUG+8
(1=3)r3c2-(3=8)r3c7-(8=1)r5c7-r5c9=r9c9=>r9c2<>1

Hi Marty,
I can't handle BUG+8, but there is an AUR(34)[r25c13] => (8)r2c3=(2)r2c1, via internals.
So, I can form (8)r2c3=(2-4)r2c1=(4-3)r5c1=(3)r5c3 => r2c3<>3, but that's not enough for basics to the end. Maybe externals?

I think there's also a 6-cell ADP(34)[r2c12,r5c13,r8c23], but that's getting way beyond my pay grade. Any ideas?

SteveC

[Marty R.]

Code: Select all

+-------------+-------+----------+
| 6   7   5   | 4 8 3 | 2  1 9   |
| 234 234 348 | 1 7 9 | 5  6 38  |
| 9   13  138 | 5 6 2 | 38 4 7   |
+-------------+-------+----------+
| 7   8   6   | 3 2 1 | 4  9 5   |
| 34  5   34  | 7 9 6 | 18 2 18  |
| 1   9   2   | 8 5 4 | 7  3 6   |
+-------------+-------+----------+
| 8   6   19  | 2 3 7 | 19 5 4   |
| 5   234 349 | 6 1 8 | 39 7 23  |
| 23  123 7   | 9 4 5 | 6  8 123 |
+-------------+-------+----------+


I wasted too much time and tried too hard just to fail in an attempt to play the BUG+8
(1=3)r3c2-(3=8)r3c7-(8=1)r5c7-r5c9=r9c9=>r9c2<>1

Hi Marty,
I can't handle BUG+8, but there is an AUR(34)[r25c13] => (8)r2c3=(2)r2c1, via internals.
So, I can form (8)r2c3=(2-4)r2c1=(4-3)r5c1=(3)r5c3 => r2c3<>3, but that's not enough for basics to the end. Maybe externals?

I think there's also a 6-cell ADP(34)[r2c12,r5c13,r8c23], but that's getting way beyond my pay grade. Any ideas?

SteveC

Steve,

The notation is a little hard to follow with all those b's in there, but I don't see how you conclude that r2c3<>3 from that notation. You've shown that if r2c1=2, then r5c3=3. What if r2c3=8? That'll also show r2c3<>3, but isn't notated. I tried playing the 2 in r2c1 as far as I could, then doing the same with the 8 in r2c3. I found five common results, that is, five cells under each scenario could not be equal to 3, but that didn't get me very far. I couldn't do anything trying to use the 2 and 8 as pincers.

As to the ADP, which I call a potential DP and others call a BUG Lite, I didn't try that but I don't see much just from eyeballing it.

Cheers,
Marty

[Sudtyro2]

Code: Select all

+--------------+-------+----------+
| 6   7    5   | 4 8 3 | 2  1 9   |
|*234 234 *348 | 1 7 9 | 5  6 38  |
| 9   13   138 | 5 6 2 | 38 4 7   |
+--------------+-------+----------+
| 7   8    6   | 3 2 1 | 4  9 5   |
|*34  5   *34  | 7 9 6 | 18 2 18  |
| 1   9    2   | 8 5 4 | 7  3 6   |
+--------------+-------+----------+
| 8   6    19  | 2 3 7 | 19 5 4   |
| 5   234  349 | 6 1 8 | 39 7 23  |
| 23  123  7   | 9 4 5 | 6  8 123 |
+--------------+-------+----------+

So, I can form (8)r2c3=(2-4)r2c1=(4-3)r5c1=(3)r5c3 => r2c3<>3, but that's not enough for basics to the end. Maybe externals?

The notation is a little hard to follow with all those b's in there, but I don't see how you conclude that r2c3<>3 from that notation. You've shown that if r2c1=2, then r5c3=3. What if r2c3=8? That'll also show r2c3<>3, but isn't notated...

Thx, Marty, for the feedback! I may have been a little too brief in the chain notation, so let me expand on that a bit.

To prevent the UR(34)r25c13, one must place (8)r2c3 and/or (2)r2c1. Hence, there exists a strong-inference link between the two, which I write simply as (8)r2c3=(2)r2c1. I can then use that link to form a bidirectional AIC:
(8)r2c3=(2)r2c1-(4)r2c1=(4-3)r5c1=(3)r5c3 => r2c3<>3

The AIC shows a derived strong inference between (8)r2c3 and (3)r5c3, which become the “pincers” (if I understand that term correctly) and therefore provide for the exclusion without having to assign any specific parities.

It seemed easier for me to use the AIC directly rather than having to run separate implication streams on (8)r2c3 and (2)r2c1 to look for common eliminations.

I think technically one should probably write the strong-inference link between (8)r2c3 and (2)r2c1 as something like (8=34)r25c13-UR-(34=2)r25c13, but I chose the simpler form.

So, the AIC => r2c3<>3, but that's not very useful for a “one-stepper” solution.

SteveC

[Marty R.]

Code: Select all

+--------------+-------+----------+
| 6   7    5   | 4 8 3 | 2  1 9   |
|*234 234 *348 | 1 7 9 | 5  6 38  |
| 9   13   138 | 5 6 2 | 38 4 7   |
+--------------+-------+----------+
| 7   8    6   | 3 2 1 | 4  9 5   |
|*34  5   *34  | 7 9 6 | 18 2 18  |
| 1   9    2   | 8 5 4 | 7  3 6   |
+--------------+-------+----------+
| 8   6    19  | 2 3 7 | 19 5 4   |
| 5   234  349 | 6 1 8 | 39 7 23  |
| 23  123  7   | 9 4 5 | 6  8 123 |
+--------------+-------+----------+

So, I can form (8)r2c3=(2-4)r2c1=(4-3)r5c1=(3)r5c3 => r2c3<>3, but that's not enough for basics to the end. Maybe externals?

The notation is a little hard to follow with all those b's in there, but I don't see how you conclude that r2c3<>3 from that notation. You've shown that if r2c1=2, then r5c3=3. What if r2c3=8? That'll also show r2c3<>3, but isn't notated...

Thx, Marty, for the feedback! I may have been a little too brief in the chain notation, so let me expand on that a bit.

To prevent the UR(34)r25c13, one must place (8)r2c3 and/or (2)r2c1. Hence, there exists a strong-inference link between the two, which I write simply as (8)r2c3=(2)r2c1. I can then use that link to form a bidirectional AIC:
(8)r2c3=(2)r2c1-(4)r2c1=(4-3)r5c1=(3)r5c3 => r2c3<>3

The AIC shows a derived strong inference between (8)r2c3 and (3)r5c3, which become the “pincers” (if I understand that term correctly) and therefore provide for the exclusion without having to assign any specific parities.

It seemed easier for me to use the AIC directly rather than having to run separate implication streams on (8)r2c3 and (2)r2c1 to look for common eliminations.

I think technically one should probably write the strong-inference link between (8)r2c3 and (2)r2c1 as something like (8=34)r25c13-UR-(34=2)r25c13, but I chose the simpler form.

So, the AIC => r2c3<>3, but that's not very useful for a “one-stepper” solution.

SteveC

Steve,

Keep in mind that I'm not the sharpest pencil in the box. We agree that r2c3<>3, but I still don't see it from the notation. Yes, the 2 proves that r5c3=3 and that r2c3 has to be<>3. Then what? The reader has to look and notice that if r2c3=8 that creates a 13 pair in box 1 which also says r2c3<>3. But I really don't know if that's the way it should be.

The strong inference is correctly written as 8r2c3=2r2c1 (or 2=8 depending on where things are going). There is no need for the more complicated notation to express the inference.

I'm in full agreement that an AIC is preferable to separate implication streams which look for common outcomes, although I believe the latter is necessary at times. In this case the separate streams proved five cells to be <>3, even though that didn't get us very far.

As to your understanding of the term "pincers", my definition is that a pincer situation exists when it's been proven that at least one of two cells must be of a certain value such that any cell seeing both can have the number removed. And once you have pincers, however they came to be, the door is open for pincer transport, a very powerful technique. In you initial strong inference of 8=2, you'd have a classic pincer situation if the 2 proved an 8 somewhere, then any 8s could be removed from cells seeing both pincer cells.

Take care,
Marty

[Sudtyro2]

Code: Select all

+--------------+-------+----------+
| 6   7    5   | 4 8 3 | 2  1 9   |
|*234 234 *348 | 1 7 9 | 5  6 38  |
| 9   13   138 | 5 6 2 | 38 4 7   |
+--------------+-------+----------+
| 7   8    6   | 3 2 1 | 4  9 5   |
|*34  5   *34  | 7 9 6 | 18 2 18  |
| 1   9    2   | 8 5 4 | 7  3 6   |
+--------------+-------+----------+
| 8   6    19  | 2 3 7 | 19 5 4   |
| 5   234  349 | 6 1 8 | 39 7 23  |
| 23  123  7   | 9 4 5 | 6  8 123 |
+--------------+-------+----------+

... We agree that r2c3<>3, but I still don't see it from the notation. Yes, the 2 proves that r5c3=3 and that r2c3 has to be<>3. Then what? The reader has to look and notice that if r2c3=8 that creates a 13 pair in box 1 which also says r2c3<>3. But I really don't know if that's the way it should be.

Marty,
As I mentioned to ixsetf earlier in this thread, the real beauty (and power) of an AIC is that the nodes don't require specific parity assignments. By definition, each end-node of the AIC has a strong-inference link, meaning that one or both of those nodes must be true. Hence, any candidate that can “see” (weakly link to) both ends of the chain must be false.

Let's take the AIC in question: (8)r2c3=(2)r2c1-(4)r2c1=(4-3)r5c1=(3)r5c3 => r2c3<>3
Now, one or both of the two end-nodes must be true. We don't know which is what, but we do know that (3)r2c3 can definitely “see” both of those nodes. Hence, it can be excluded. That's really all there is to it!

One can inspect the alternating inferences in the AIC to see what actually happens. Written as above, the chain reads from left to right:
If (8)r2c3 is false, then (2)r2c1 must be true, so (4)r2c1 must be false, and so on, until you reach the other end where (3)r5c3 must be true. (3)r2c3 “sees” that last node, so (3)r2c3 must be false.

Being bidirectional, one can also read (or write) the AIC from right to left:
(3)r5c3=(3-4)r5c1=(4)r2c1-(2)r2c1=(8)r2c3.
If (3)r5c3 is false, then (3)r5c1 must be true, so (4)r5c1 must be false, and so on, until you reach the other end where (8)r2c3 must be true. (3)r2c3 “sees” that last node, so again (3)r2c3 must be false.

Note that the bidirectional property separately accounts for both possible parities of the UAR “internals” (8)r2c3 and (2)r2c1, depending on which way you read the chain.

OK, now I've got a question: How would you apply Ted's UAR “externals” approach to show r2c3<>3? I'm still struggling to understand his five External Principles!

SteveC