Difference between revisions of "Sixth row template problem"

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(Possible paths to answer: Starting the solution of the 6th row template)
(...answering "Yes": Fixing the number of rows (silly mistake))
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Here is a start.  Just from edge [[Edge template IV1a]], Blue's first move must be one of the following:
 
Here is a start.  Just from edge [[Edge template IV1a]], Blue's first move must be one of the following:
 
<hex>
 
<hex>
R6 C19 Q0
+
R7 C19 Q0
 
1:BBBBBBBBBRBBBBBBBBB
 
1:BBBBBBBBBRBBBBBBBBB
Si2 Sj2
+
Rj2
Si3
+
Si3 Sj3
Sg4 Sh4 Si4 Sj4
+
Si4
Sf5 Sg5 Sh5 Si5 Sj5
+
Sg5 Sh5 Si5 Sj5  
Se6 Sf6 Sg6 Sh6 Si6 Sj6
+
Sf6 Sg6 Sh6 Si6 Sj6
 +
Se7 Sf7 Sg7 Sh7 Si7 Sj7
 
</hex>
 
</hex>
 
Many of these moves will be easy to dismiss.  Others will benefit from the [[Parallel ladder]] trick.  Of course, symmetry will cut our work in half!
 
Many of these moves will be easy to dismiss.  Others will benefit from the [[Parallel ladder]] trick.  Of course, symmetry will cut our work in half!

Revision as of 04:24, 10 January 2009

As of January 2009 the following problem is still open. Javerberg-Wccanard Problem is simply put as follow.

Is there any one stone sixth row template ?

It is still unknown whether one stone edge templates can be found for every heights. Such templates have been found for sizes up to 5 but none above. Answering with "No" to the former question answers the latter.

Description

Is there a width n such that the game on the board of width n designed as follow with turn to Blue is won by Red ?

Generalisation

The general problem of knowing if there is n such that there is no one stone edge template on the n^th row<math>n^th</math> is also referred to as Javerberg-Wccanard Problem.

Possible paths to answer

By "hand"...

...answering "Yes"

This would involve placing a stone on the 6th row of a sufficiently wide board, and showing how to always connect to the bottom. (Note this does not necessarily identify the minimal template needed.)

Here is a start. Just from edge Edge template IV1a, Blue's first move must be one of the following:

Many of these moves will be easy to dismiss. Others will benefit from the Parallel ladder trick. Of course, symmetry will cut our work in half!

...answering "No"

This would involve showing how to connect (in the diagram above) the Blue stones to the right (plus Blue stones on the far right edge) to Blue stones on the left (plus Blue stones on the far left edge), no matter how wide the board is.

Computer Aided demonstration ...

... answering "Yes"

Such a proof would use the computer to find the template and it's carrier. Afterwards it should be easy to manually check that every Blue intrusion does not prevent Red from connecting to bottom.

... answering "No"

TODO

See Also

External link

  • The thread were the names were associated.