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Almost remote pairs

 
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Fri Feb 29, 2008 8:41 pm    Post subject: Almost remote pairs Reply with quote

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

>>> play online
There are at least 2 ways to solve it with a remote pair variant.
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Fri Feb 29, 2008 9:27 pm    Post subject: Reply with quote

(I think we are all about to learn something.)

Working through the basics is a challenge on this one. You might get to here:

Code:
+-------------------+-------------------+-------------------+
| 1     78    67    | 5     2     49    | 49    678   3     |
| 23    4     36    | 179   89    1789  | 19    5     26    |
| 5     278   9     | 3     14    6     | 14    278   278   |
+-------------------+-------------------+-------------------+
| 69    123   5     | 124   89    123   | 7     34    689   |
| 4     237   27    | 279   6     23789 | 5     389   1     |
| 679   1367  8     | 1479  5     1379  | 2     3469  69    |
+-------------------+-------------------+-------------------+
| 67    267   14    | 8     14    5     | 3     279   279   |
| 23    9     34    | 6     7     24    | 8     1     5     |
| 8     5     127   | 129   3     129   | 6     27    4     |
+-------------------+-------------------+-------------------+


I will now see if I can take it further.

Keith
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Fri Feb 29, 2008 9:34 pm    Post subject: Reply with quote

There is a UR with strong links on <69> in R46C16. R4C9 cannot be <9>, R6C1 cannot be <6>. Leading to an XY-wing, leading to here:
Code:
+-------------------+-------------------+-------------------+
| 1     78    67    | 5     2     49    | 49    678   3     |
| 23    4     36    | 179   89    1789  | 19    5     26    |
| 5     278   9     | 3     14    6     | 14    278   278   |
+-------------------+-------------------+-------------------+
| 69    123   5     | 124   89    123   | 7     34    68    |
| 4     237   27    | 279   6     23789 | 5     389   1     |
| 79    1367  8     | 147   5     137   | 2     3469  69    |
+-------------------+-------------------+-------------------+
| 67    267   14    | 8     14    5     | 3     279   279   |
| 23    9     34    | 6     7     24    | 8     1     5     |
| 8     5     127   | 129   3     129   | 6     27    4     |
+-------------------+-------------------+-------------------+

HMMM ... Question
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Fri Feb 29, 2008 9:50 pm    Post subject: Reply with quote

I have solved it!

The pattern will be found by Medusa multicoloring on <6> and <7>, but you do not have to resort to that. I am not sure I would have found this one without ravel's hints.

I'll wait a while to post the details.

And, ravel said there are two ways ...

Keith

(ravel, thank you! My kind of puzzle.)
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Marty R.



Joined: 12 Feb 2006
Posts: 5181
Location: Rochester, NY, USA

PostPosted: Fri Feb 29, 2008 10:16 pm    Post subject: Reply with quote

Quote:

The pattern will be found by Medusa multicoloring on <6> and <7>

Not looking to open a can of worms here, but what is Medusa multicoloring? Is that something different from what I know as extended Medusa?
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Johan



Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium

PostPosted: Fri Feb 29, 2008 11:00 pm    Post subject: Reply with quote

I found this one after exploring a Type 3 UR [23] in R28C13, and a Type 6 UR [69] in R46C19 which places <3> in R5C2.
When R2C9=|6| => R4C9≠|6| => R4C1=|6| => R6C1=|9| => R6C9=|6| <=> R2C9≠6 collapsing the puzzle

Code:
+--------------------------+--------------------------+--------------------------+
| 1         78        67   | 5          2       49    | 49        678       3    |
| 23        4         236  | 179        89      1789  | 19        5         2|6| |
| 5         278       9    | 3          14      6     | 14        278       278  |
+--------------------------+--------------------------+--------------------------+
||6|9       123       5    | 124        89      123   | 7         34      -|6|89 |
| 4         237       27   | 279        6       23789 | 5         389       1    |
| 67|9|     1367      8    | 1479       5       1379  | 2         3469     |6|9  |
+--------------------------+--------------------------+--------------------------+
| 67        267       14   | 8          14      5     | 3         279       279  |
| 23        9         234  | 6          7       24    | 8         1         5    |
| 8         5         127  | 129        3       129   | 6         27        4    |
+--------------------------+--------------------------+--------------------------+
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Fri Feb 29, 2008 11:24 pm    Post subject: Reply with quote

Marty R. wrote:
Quote:

The pattern will be found by Medusa multicoloring on <6> and <7>

Not looking to open a can of worms here, but what is Medusa multicoloring? Is that something different from what I know as extended Medusa?

Marty,

Can of worms? Box of snakes?

Yes, it is different. A skyscraper is multi-coloring.

Suppose you have a skyscraper pattern. One skyscraper is green and red. The other is blue and orange. If green and blue are on the ground floor, red and orange are the peaks. Any cell that sees both red and orange cannot be true. Now, instead of just coloring the strong links, do Medusa coloring starting on each tower. Any cell that sees both red and orange cannot be true.

Keith
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Fri Feb 29, 2008 11:34 pm    Post subject: Reply with quote

Code:
+-------------------+-------------------+-------------------+
| 1     78    6g7r  | 5     2     49    | 49    678   3     |
| 23    4     36r   | 179   89    1789  | 19    5     26g   |
| 5     278   9     | 3     14    6     | 14    278   278   |
+-------------------+-------------------+-------------------+
| 6o9   123   5     | 124   89    123   | 7     34    6b8   |
| 4     237   27    | 279   6     23789 | 5     389   1     |
| 79    1367  8     | 147   5     137   | 2     3469  69    |
+-------------------+-------------------+-------------------+
| 6b7o  267   14    | 8     14    5     | 3     279   279   |
| 23    9     34    | 6     7     24    | 8     1     5     |
| 8     5     12-7  | 129   3     129   | 6     27    4     |
+-------------------+-------------------+-------------------+

There is a skyscraper on <6> in R24. One of R2C3 and R4C1 must be <6>. But, note the cells <67>. One of R1C3 and R7C1 must then be <7>.

Which takes out <7> in R9C3, solving the puzzle.

Keith

g - green
b - blue
r - red
o - orange
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Earl



Joined: 30 May 2007
Posts: 546
Location: St Louis MO

PostPosted: Fri Feb 29, 2008 11:42 pm    Post subject: almost Reply with quote

Is Johan's solution a "forcing chain?"

Keith's solution is essentially a w-wing, established by the pincers of the skyscraper (R2C3 and R4C1). But BOTH could be 6, which would not make a strong pair. Does that establish a w-wing? Just wondering.

Earl
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Sat Mar 01, 2008 12:26 am    Post subject: Reply with quote

Quote:
I found this one after exploring a Type 3 UR [23] in R28C13


Johan,

I admire your eye for uniqueness solutions... I just want to make sure of something... isn't that a type 4B in your quote??
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Sat Mar 01, 2008 12:52 am    Post subject: Reply with quote

Marty,

multicoloring medusa chains allows you to make eliminations because you are making the same type of bridge that you can make when you muticolor a single candidate. requires much attention to detail and some of the bridged clusters can cover the entire board.
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Johan



Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium

PostPosted: Sat Mar 01, 2008 1:27 am    Post subject: Reply with quote

Quote:
isn't that a type 4B in your quote??

Norm,

I think you're right about this one, it seems more difficult for me defining those varied UR types than dissecting them.
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Marty R.



Joined: 12 Feb 2006
Posts: 5181
Location: Rochester, NY, USA

PostPosted: Sat Mar 01, 2008 1:58 am    Post subject: Reply with quote

Thanks Keith and Norm.

Quote:
Can of worms? Box of snakes?


Maybe that should be promoted to Pandora's box. Laughing

Let's see if I grasp this. When you have this situation, you start two Medusas, one with r-g and the other with b-o and see where it takes you, making sure that inferences follow the old rules, i.e., an inference from, say, a "g" also must be a "g."

Under this system, does this mean that an r-g or b-o trap is not allowed? That a trap only occurs between one color from each set? Could a trap result from each of r-b, r-o, g-b and g-o?
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Sat Mar 01, 2008 3:03 am    Post subject: Reply with quote

Marty R. wrote:
Thanks Keith and Norm.

Quote:
Can of worms? Box of snakes?


Maybe that should be promoted to Pandora's box. Laughing

Let's see if I grasp this. When you have this situation, you start two Medusas, one with r-g and the other with b-o and see where it takes you, making sure that inferences follow the old rules, i.e., an inference from, say, a "g" also must be a "g."

Under this system, does this mean that an r-g or b-o trap is not allowed? That a trap only occurs between one color from each set? Could a trap result from each of r-b, r-o, g-b and g-o?


Marty,

If I understand the question, the answer is "No". It is not that complicated.

To use my previous notation, suppose you start Medusa coloring on green-red. Any conclusions you make are valid. Suppose you do not get too far.

Suppose you start another Medusa coloring cluster in cells you have not yet used. Any conclusions you make for this cluster (blue - orange) are valid.

How do you connect the two? I suggest it is via the weak link, as in a skyscraper. As in the cells in C9 above. I believe the only conclusion between the two clusters is that a Medusa peer of both red and orange cannot be true.

Keith.

(Damn! Now I have to explain what a "Medusa peer" is.)
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Marty R.



Joined: 12 Feb 2006
Posts: 5181
Location: Rochester, NY, USA

PostPosted: Sat Mar 01, 2008 4:47 am    Post subject: Reply with quote

Quote:
If I understand the question, the answer is "No".

I'm not sure I understand the question. But thanks, I'll try and play around with it.
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Sat Mar 01, 2008 11:56 am    Post subject: Reply with quote

Nice, you have found, what i meant (and more), but in a different way.

My way to get to Keith's elimination of 7 in r9c3 was a w-wing, connected by 2 strong links for 6.
Code:
+-------------------+-------------------+-------------------+
| 1     78   #67    | 5     2     49    | 49   @678   3     |
| 23    4     36    | 179   89    1789  | 19    5     26    |
| 5     278   9     | 3     14    6     | 14    278   278   |
+-------------------+-------------------+-------------------+
| 69    123   5     | 124   89    123   | 7     34    689   |
| 4     237   27    | 279   6     23789 | 5     389   1     |
| 679  @1367  8     | 1479  5     1379  | 2    @3469  69    |
+-------------------+-------------------+-------------------+
|#67   @267   14    | 8     14    5     | 3     279   279   |
| 23    9     34    | 6     7     24    | 8     1     5     |
| 8     5     12-7  | 129   3     129   | 6     27    4     |
+-------------------+-------------------+-------------------+

For those, who are not familiar with our terminology:
The 2 strong links for 6 in colums 2 and 8 form a skyskraper, one of r1c8 and r7c2 must be 6. Then one of r1c3 and r7c1 must be 7.

Johan's chain is exactly describing, what i saw as "half M-wing".
Code:
+-------------------+-------------------+-------------------+
| 1     78    67    | 5     2     49    | 49    678   3     |
| 23    4     36    | 179   89    1789  | 19    5     2-6   |
| 5     278   9     | 3     14    6     | 14    278   278   |
+-------------------+-------------------+-------------------+
|#69    123   5     | 124   89    123   | 7     34   @68    |
| 4     237   27    | 279   6     23789 | 5     389   1     |
|@79   *1367  8     | 147   5     137   | 2    34-69 #69    |
+-------------------+-------------------+-------------------+
| 67    267   14    | 8     14    5     | 3     279   279   |
| 23    9     34    | 6     7     24    | 8     1     5     |
| 8     5     127   | 129   3     129   | 6     27    4     |
+-------------------+-------------------+-------------------+

r6c9=9 => r6c1<>9 => r4c1=9.
Now there is a strong link for 6 from r4c1 to r4c9. So we have either r6c9=6 or (r6c9=9 =>) r4c9=6.
Or you take the strong link to r6c2 to eliminate 6 in r6c8.
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Asellus



Joined: 05 Jun 2007
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PostPosted: Sat Mar 01, 2008 9:34 pm    Post subject: Reply with quote

Some fancy stuff... but, I saw that <6> elimination in r6c8 this way: Take the otherwise useless {679} XY Wing in r67c1|r6c9 and transport the <6> in r6c9 to r6c2 via r4c1.
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Sun Mar 02, 2008 7:58 am    Post subject: Reply with quote

storm_norm wrote:
Marty,

multicoloring medusa chains allows you to make eliminations because you are making the same type of bridge that you can make when you muticolor a single candidate. requires much attention to detail and some of the bridged clusters can cover the entire board.

Norm,

I think I agree, if you just start coloring Medusa clusters, and then try to figure out the bridges (weak links).

However, if you establish the bridge first, as in my link of green - blue in C9 in the topic of this thread, the process is not that complicated, I think.

I simply note the colors in the margin: Eliminations can be made by red - green, blue - orange, AND by red - orange.

You do not have to use colors. I use a circle o and a solid dot first (red - green) then + and x (blue and orange).

Keith

PS: Thank you all. This, and other recent threads, are a great discussion. I've been having a good time!
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Marty R.



Joined: 12 Feb 2006
Posts: 5181
Location: Rochester, NY, USA

PostPosted: Sun Mar 02, 2008 4:21 pm    Post subject: Reply with quote

Marty asked
Quote:
Under this system, does this mean that an r-g or b-o trap is not allowed? That a trap only occurs between one color from each set? Could a trap result from each of r-b, r-o, g-b and g-o?

Keith wrote
Quote:
I simply note the colors in the margin: Eliminations can be made by red - green, blue - orange, AND by red - orange.

Keith, that's the answer I was looking for when I asked the question earlier in the thread.
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