May 8, 2014

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May 8, 2014

Postby ArkieTech » Wed May 07, 2014 11:01 pm

Code: Select all
 *-----------*
 |.17|.64|...|
 |6..|.2.|.1.|
 |.5.|.3.|2..|
 |---+---+---|
 |...|..2|..7|
 |92.|...|.53|
 |5..|3..|...|
 |---+---+---|
 |..5|.8.|.6.|
 |.9.|.7.|..1|
 |...|14.|89.|
 *-----------*


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Re: May 8, 2014

Postby SteveG48 » Wed May 07, 2014 11:49 pm

I'll get this one out of the way:

W-wing: 4/8 r4c8,r5c3 SL B5 => -8 r4c2 ; stte
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Re: May 8, 2014

Postby SteveG48 » Thu May 08, 2014 12:18 am

Code: Select all
 *-----------------------------------------------------------*
 | 2     1     7     | 589   6     4     | 35    38    589   |
 | 6     38    389   | 5789  2     5789  | 45    1     4589  |
 | 48    5     489   | 89    3     1     | 2     7     6     |
 *-------------------+-------------------+-------------------|
 | 3     468   1     | 468   5     2     | 9     48    7     |
 | 9     2     48    | 478   1     78    | 6     5     3     |
 | 5     7     468   | 3     9     68    | 1     248   248   |
 *-------------------+-------------------+-------------------|
 | 1    b34    5     | 29    8     39    | 7     6    c24    |
 | 48    9     2468  | 256   7     56    | 345   234   1     |
 | 7    a36   a236   | 1     4     356   | 8     9     5-2   |
 *-----------------------------------------------------------*


(2=36)r9c23 - (3=4)r7c2 - (4=2)r7c9 => -2 r9c9 ; stte
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Re: May 8, 2014

Postby Leren » Thu May 08, 2014 1:19 am

Code: Select all
*--------------------------------------------------------------*
| 2     1     7      | 589   6     4      | 35    38    589    |
| 6     38    389    | 5789  2     5789   | 45    1     4589   |
| 48    5     489    | 89    3     1      | 2     7     6      |
|--------------------+--------------------+--------------------|
| 3    d46-8  1      |c468   5     2      | 9     48    7      |
| 9     2    a48     |b478   1     78     | 6     5     3      |
| 5     7     468    | 3     9     68     | 1     248   248    |
|--------------------+--------------------+--------------------|
| 1     34    5      | 29    8     39     | 7     6     24     |
| 48    9     2468   | 256   7     56     | 345   234   1      |
| 7     36    236    | 1     4     356    | 8     9     25     |
*--------------------------------------------------------------*

H2 Wing: (8=4) r5c3 - r5c4 = (4-6) r4c4 = (6) r4c2 => - 8 r4c2; stte

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Re: May 8, 2014

Postby tlanglet » Thu May 08, 2014 2:51 am

Code: Select all
 *-----------------------------------------------------------*
 | 2     1     7     | 589   6     4     | 35    38    589   |
 | 6     38    389   | 5789  2     5789  | 45    1     4589  |
 | 48    5     489   | 89    3     1     | 2     7     6     |
 |-------------------+-------------------+-------------------|
 | 3     468   1     | 468   5     2     | 9     48    7     |
 | 9     2     48    | 478   1     78    | 6     5     3     |
 | 5     7     468   | 3     9     68    | 1     248   248   |
 |-------------------+-------------------+-------------------|
 | 1     34    5     |c29    8     39    | 7     6    b24    |
 | 48    9     2468  |d256   7    d56    | 34-5  234   1     |
 | 7     36    236   | 1     4     356   | 8     9    a25    |
 *-----------------------------------------------------------*

(5=2)r9c9-r7c9=r7c4-(2=56)r8c46 => r8c7<>5

Ted
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Re: May 8, 2014

Postby Marty R. » Thu May 08, 2014 4:20 pm

SteveG48 wrote:I'll get this one out of the way:

W-wing: 4/8 r4c8,r5c3 SL B5 => -8 r4c2 ; stte


Same. There's also an SL 4 in r6.
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Re: May 8, 2014

Postby Sudtyro2 » Thu May 08, 2014 5:40 pm

Code: Select all
*--------------------------------------------------------------*
| 2     1     7      | 589   6     4      | 35    38    589    |
| 6     38    389    |*5789  2    *5789   | 45    1     4589   |
| 48    5     489    | 89    3     1      | 2     7     6      |
|--------------------+--------------------+--------------------|
| 3     468   1      | 468   5     2      | 9     48    7      |
| 9     2     48     |*478   1    *78     | 6     5     3      |
| 5     7     468    | 3     9     68     | 1     248   248    |
|--------------------+--------------------+--------------------|
| 1     34    5      | 29    8     39     | 7     6     24     |
| 48    9     2468   | 256   7     56     | 345   234   1      |
| 7     36    236    | 1     4     356    | 8     9     25     |
*--------------------------------------------------------------*

Ted,
I'm slowly working through your extensive posting on ADP External Inferences, but some confusion (and laziness) always abounds for me. I look at the current puzzle and readily see ADP(78)r25c46. Next, r5 has only one external ADP digit at (8)r5c3, so this choice seems promising. But, then the remaining ADP cells in r2, say, have no external 7s, but there are three external 8s to be considered. A strong inference between (8)r2c2 and (8)r5c3 would reproduce Leren's (8)r4c2 elimination (and subsequent stte), but how does one proceed from there to properly consider the remaining two external 8s in r2?

TIA for any help! And comments from others are most welcome!

SteveC
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Re: May 8, 2014

Postby Marty R. » Thu May 08, 2014 5:58 pm

Sudtyro2 wrote:
Code: Select all
*--------------------------------------------------------------*
| 2     1     7      | 589   6     4      | 35    38    589    |
| 6     38    389    |*5789  2    *5789   | 45    1     4589   |
| 48    5     489    | 89    3     1      | 2     7     6      |
|--------------------+--------------------+--------------------|
| 3     468   1      | 468   5     2      | 9     48    7      |
| 9     2     48     |*478   1    *78     | 6     5     3      |
| 5     7     468    | 3     9     68     | 1     248   248    |
|--------------------+--------------------+--------------------|
| 1     34    5      | 29    8     39     | 7     6     24     |
| 48    9     2468   | 256   7     56     | 345   234   1      |
| 7     36    236    | 1     4     356    | 8     9     25     |
*--------------------------------------------------------------*

Ted,
I'm slowly working through your extensive posting on ADP External References, but some confusion (and laziness) always abounds for me. I look at the current puzzle and readily see ADP(78)r25c46. Next, r5 has only one external ADP digit at (8)r5c3, so this choice seems promising. But, then the remaining ADP cells in r2, say, have no external 7s, but there are three external 8s to be considered. A strong inference between (8)r2c2 and (8)r5c3 would reproduce Leren's (8)r4c2 elimination (and subsequent stte), but how does one proceed from there to properly consider the remaining two external 8s in r2?

TIA for any help! And comments from others are most welcome!

SteveC


Steve,

You are not obligated to use the externals in the row only, you have choices. For example, for the cells in row 2 you don't have to use the externals in row 2, you can use them in the box. Both are in c4, so combined they work with the 8 in r5c3 to eliminate the 8 from r4c4. I don't know that that helps to solve the puzzle, but it's just an example of the flexibility.
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Re: May 8, 2014

Postby Sudtyro2 » Thu May 08, 2014 6:57 pm

Marty R. wrote:
You are not obligated to use the externals in the row only, you have choices. For example, for the cells in row 2 you don't have to use the externals in row 2, you can use them in the box. Both are in c4, so combined they work with the 8 in r5c3 to eliminate the 8 from r4c4. I don't know that that helps to solve the puzzle, but it's just an example of the flexibility.

Thx, Marty...I think I did notice that b2 only had the two external 8s, vice the three in row2, but I couldn't make a useful chain. Also, I'm probably being slow again, but if one does use the two (assumed grouped) 8s in b2, how does (8)r5c3 "see" (8)r4c4?

SteveC
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Re: May 8, 2014

Postby Marty R. » Thu May 08, 2014 7:57 pm

Sudtyro2 wrote:
Marty R. wrote:
You are not obligated to use the externals in the row only, you have choices. For example, for the cells in row 2 you don't have to use the externals in row 2, you can use them in the box. Both are in c4, so combined they work with the 8 in r5c3 to eliminate the 8 from r4c4. I don't know that that helps to solve the puzzle, but it's just an example of the flexibility.

Thx, Marty...I think I did notice that b2 only had the two external 8s, vice the three in row2, but I couldn't make a useful chain. Also, I'm probably being slow again, but if one does use the two (assumed grouped) 8s in b2, how does (8)r5c3 "see" (8)r4c4?

SteveC


Steve, sorry, just a stupid typo on my part. The 8s in c4 act as pincers with the 8 in r5c3 to eliminate the 8 from r5c4.
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Location: Rochester, New York, USA

Re: May 8, 2014

Postby tlanglet » Fri May 09, 2014 1:36 pm

Sudtyro2 wrote:
Code: Select all
*--------------------------------------------------------------*
| 2     1     7      | 589   6     4      | 35    38    589    |
| 6     38    389    |*5789  2    *5789   | 45    1     4589   |
| 48    5     489    | 89    3     1      | 2     7     6      |
|--------------------+--------------------+--------------------|
| 3     468   1      | 468   5     2      | 9     48    7      |
| 9     2     48     |*478   1    *78     | 6     5     3      |
| 5     7     468    | 3     9     68     | 1     248   248    |
|--------------------+--------------------+--------------------|
| 1     34    5      | 29    8     39     | 7     6     24     |
| 48    9     2468   | 256   7     56     | 345   234   1      |
| 7     36    236    | 1     4     356    | 8     9     25     |
*--------------------------------------------------------------*

Ted,
I'm slowly working through your extensive posting on ADP External Inferences, but some confusion (and laziness) always abounds for me. I look at the current puzzle and readily see ADP(78)r25c46. Next, r5 has only one external ADP digit at (8)r5c3, so this choice seems promising. But, then the remaining ADP cells in r2, say, have no external 7s, but there are three external 8s to be considered. A strong inference between (8)r2c2 and (8)r5c3 would reproduce Leren's (8)r4c2 elimination (and subsequent stte), but how does one proceed from there to properly consider the remaining two external 8s in r2?

TIA for any help! And comments from others are most welcome!

SteveC


Steve,

Marty has responded to your specific question but I would like to emphasize a couple of points.

1) As Marty noted, you are not confined to using a single house, such as a row, but may use the house which seems to offer the most practical opportunity for each ADP digit.

2) Remember that all external occurrences of the ADP digit in a house must be included in the inference analysis. In your example you have a total of four inference to resolve that will hopefully lead to a common result. You considered two of the inferences but need to pursue the two remaining realizing that every situation that we consider will not be fruitful.

Consider another ADP in the same puzzle: AUR(48)r38c13.

Code: Select all
 *-----------------------------------------------------------*
 | 2     1     7     | 589   6     4     | 35    38    589   |
 | 6     38    389   | 5789  2     5789  | 45    1     4589  |
 |*48    5    *489   |b89    3     1     | 2     7     6     |
 |-------------------+-------------------+-------------------|
 | 3     468   1     | 468   5     2     | 9     48    7     |
 | 9     2     48    | 478   1     78    | 6     5     3     |
 | 5     7     468   | 3     9     68    | 1     248   248   |
 |-------------------+-------------------+-------------------|
 | 1    a34    5     | 29    8     39    | 7     6     24    |
 |*48    9    *2468  | 256   7     56    | 345   234   1     |
 | 7     36    236   | 1     4     356   | 8     9     25    |
 *-----------------------------------------------------------*
 

First, look at the ADP digit 4. For the two ADP cells r3c13, box 1 (and row3) does not have an external digit (4) which is the ideal situation, and for ADP cells r8c13 I would suggest the external 4r7c2 in box 7 since it is a single digit whereas both r8 and c3 have two external occurrences.

Now, consider ADP digit 8. For the two ADP cells in r3c13, my first choice would be r3 which contains only 8r3c4. To our delight, we note that no external 8s exist in the box or row that "sees" r8c13.

So the result of our first inference analysis is the simple inference 4r7c2=8r3c4 which results in

AUR(48)r38c13[4r7c2=8r3c4]-(8=4)r3c1 => r8c1<>4 which is not particularly useful except as an example.

Ted
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Re: May 8, 2014

Postby Sudtyro2 » Sat May 10, 2014 3:13 pm

Marty and Ted...many thx for the helpful and encouraging feedback.
This area is just a tad more complicated than I first thought.
Much additional study ahead!

SteveC
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