Vanhegan Fiendish July 16, 2013

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Vanhegan Fiendish July 16, 2013

Postby ArkieTech » Tue Jul 16, 2013 11:52 pm

Code: Select all
 *-----------*
 |...|...|327|
 |...|25.|4..|
 |.4.|...|.9.|
 |---+---+---|
 |..1|.72|..6|
 |.2.|...|.8.|
 |3..|54.|1..|
 |---+---+---|
 |.1.|...|.5.|
 |..2|.19|...|
 |983|...|...|
 *-----------*


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dan
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Re: Vanhegan Fiendish July 16, 2013

Postby Leren » Wed Jul 17, 2013 12:26 am

Code: Select all
*--------------------------------------------------------------*
| 1     6     5      | 89    89    4      | 3     2     7      |
|b78    3     9      | 2     5     67     | 4     16   c18     |
| 2     4     78     | 1367  36    1367   | 68    9     5      |
|--------------------+--------------------+--------------------|
|a48    5     1      |g3-8   7     2      | 9    f34    6      |
| 467   2     467    | 1369  369   136    | 5     8    e34     |
| 3     9     68     | 5     4     68     | 1     7     2      |
|--------------------+--------------------+--------------------|
| 46    1     46     | 378   238   378    | 27    5     9      |
| 5     7     2      | 46    1     9      | 68    346  d348    |
| 9     8     3      | 467   26    5      | 27    146   14     |
*--------------------------------------------------------------*

(8) r4c1 = r2c1 - r2c9 = (8-3) r8c9 = r5c9 - r4c8 = r4c4 => - 8 r4c4; stte

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Re: Vanhegan Fiendish July 16, 2013

Postby ArkieTech » Wed Jul 17, 2013 11:22 am

Code: Select all
 *-----------------------------------------------------------*
 | 1     6     5     | 89    89    4     | 3     2     7     |
 | 7-8   3     9     | 2     5     67    | 4     16   a18    |
 | 2     4     78    | 1367  36    1367  | 68    9     5     |
 |-------------------+-------------------+-------------------|
 |c48    5     1     | 38    7     2     | 9    c34    6     |
 | 467   2     467   | 1369  369   136   | 5     8    b34    |
 | 3     9     68    | 5     4     68    | 1     7     2     |
 |-------------------+-------------------+-------------------|
 | 46    1     46    | 378   238   378   | 27    5     9     |
 | 5     7     2     | 46    1     9     | 68    346   348   |
 | 9     8     3     | 467   26    5     | 27    146  b14    |
 *-----------------------------------------------------------*
(8=1)r2c9-(1=3)r59c9-(3=8)r4c18 => -8r2c1; ste
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Re: Vanhegan Fiendish July 16, 2013

Postby tlanglet » Wed Jul 17, 2013 1:32 pm

Another variation on the theme..........

Code: Select all
 *-----------------------------------------------------------*
 | 1     6     5     | 89    89    4     | 3     2     7     |
 | 7-8   3     9     | 2     5     67    | 4     16   d18    |
 | 2     4     78    | 1367  36    1367  | 68    9     5     |
 |-------------------+-------------------+-------------------|
 |a48    5     1     | 38    7     2     | 9     34    6     |
 |b467   2    b467   | 1369  369   136   | 5     8    c34    |
 | 3     9     68    | 5     4     68    | 1     7     2     |
 |-------------------+-------------------+-------------------|
 | 46    1     46    | 378   238   378   | 27    5     9     |
 | 5     7     2     | 46    1     9     | 68    346   348   |
 | 9     8     3     | 467   26    5     | 27    146  d14    |
 *-----------------------------------------------------------*

(8=4)r4c1-r5c13=r5c9-(4=18)r92c9 => r2c1<>8

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Re: Vanhegan Fiendish July 16, 2013

Postby daj95376 » Wed Jul 17, 2013 3:58 pm

Leren wrote:_

(8) r4c1 = r2c1 - r2c9 = (8-3) r8c9 = r5c9 - r4c8 = r4c4 => - 8 r4c4; stte

FWIW, which is probably extremely little, I would tag your chain as:

4-SI L-Wing: (8) r4c1 = r2c1 - r2c9 = (8-3) r8c9 = r5c9 - r4c8 = (3)r4c4 => - 8 r4c4
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Re: Vanhegan Fiendish July 16, 2013

Postby Marty R. » Wed Jul 17, 2013 4:47 pm

Code: Select all
+-----------+---------------+------------+
| 1   6 5   | 89   89  4    | 3  2   7   |
| 78  3 9   | 2    5   67   | 4  16  18  |
| 2   4 78  | 1367 36  1367 | 68 9   5   |
+-----------+---------------+------------+
| 48  5 1   | 38   7   2    | 9  34  6   |
| 467 2 467 | 1369 369 136  | 5  8   34  |
| 3   9 68  | 5    4   68   | 1  7   2   |
+-----------+---------------+------------+
| 46  1 46  | 378  238 378  | 27 5   9   |
| 5   7 2   | 46   1   9    | 68 346 348 |
| 9   8 3   | 467  26  5    | 27 146 14  |
+-----------+---------------+------------+

Play this puzzle online at the Daily Sudoku site

XY-Chain (1=4)r9c9-r89c8=r4c8-(4=8)r4c1-(8=7)r2c1-(7=6)r2c6-(6=1)r2c8=>r2c9<>1
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Re: Vanhegan Fiendish July 16, 2013

Postby Marty R. » Wed Jul 17, 2013 5:07 pm

Leren wrote:
Code: Select all
*--------------------------------------------------------------*
| 1     6     5      | 89    89    4      | 3     2     7      |
|b78    3     9      | 2     5     67     | 4     16   c18     |
| 2     4     78     | 1367  36    1367   | 68    9     5      |
|--------------------+--------------------+--------------------|
|a48    5     1      |g3-8   7     2      | 9    f34    6      |
| 467   2     467    | 1369  369   136    | 5     8    e34     |
| 3     9     68     | 5     4     68     | 1     7     2      |
|--------------------+--------------------+--------------------|
| 46    1     46     | 378   238   378    | 27    5     9      |
| 5     7     2      | 46    1     9      | 68    346  d348    |
| 9     8     3      | 467   26    5      | 27    146   14     |
*--------------------------------------------------------------*

(8) r4c1 = r2c1 - r2c9 = (8-3) r8c9 = r5c9 - r4c8 = r4c4 => - 8 r4c4; stte

Leren


Leren, a technicality, but I just want to see if I understand. I understand your chain as saying if r4c1 is not 8, then r4c4 is 3. That is invalid, so isn't the original premise wrong? Could the conclusion be that r4c1<>not 8, thus =4? I usually write something like that as <>(<>8)=4.
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Re: Vanhegan Fiendish July 16, 2013

Postby Leren » Thu Jul 18, 2013 4:33 am

Marty R wrote: Leren, a technicality, but I just want to see if I understand. I understand your chain as saying if r4c1 is not 8, then r4c4 is 3. That is invalid, so isn't the original premise wrong? Could the conclusion be that r4c1<>not 8, thus =4? I usually write something like that as <>(<>8)=4.

This AIC Type 2 logic often causes confusion because part of it is left out (presumably because it's supposed to be obvious).

The chain shows that if r4c1 is not 8 then r4c4 is 3. The "obvious " part that's left out is that if r4c1 is 8 then r4c4 is not 8 (because it's in the same row).

Now r4c1 can only be (1) not 8 or (2) 8. So r4c4 can only be (1) 3 or (2) not 8. But if r4c4 is 3 then in particular it's not 8 ! So we can say that r4c4 can only be (1) not 8 or (2) not 8. So r4c4 <>8 !!!

It's the underlined bit of the logic that's both obvious when you think about it but at the same time strangely counter-intuitive.

Incidentally, the same chain would also allow the elimination of 3 in r4c1 had it been there, in which case I would have written:

(8) r4c1 = r2c1 - r2c9 = (8-3) r8c9 = r5c9 - (3) r4c8 => r4c1 <>3, r4c4 <> 8. (I've only added (3) in front of the right most term to make the chain easier to read from right to left).

To understand both eliminations you would have to (1) read the chain from left to right and note that if r4c1 = 8 then r4c4 <> 8 and (2) read the chain from right to left and note that if r4c4 = 3 then r4c1 <> 3.

That's why I've adopted the practice of always putting digit numbers in brackets eg (8) r4c1 ... not - 8 r4c1 ... The idea being that you have to either (1) assume or (2) find out the Truth status of the 8 depending on which direction you read the chain. If you read it from left to right assume that the 8 is False. If you read it from right to left assume that the right most digit is False and hopefully you should find out that the 8 ends up as True.

I know this has been a long answer but have a look at the previous days puzzle and the discussion that concerned a similar case - a Type 2 AIC with eliminations at both ends.

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