Difference between revisions of "Sixth row template problem"
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Is there any one stone sixth row [[template]] ? | Is there any one stone sixth row [[template]] ? | ||
| − | More generally, it is still unknown whether one stone edge templates can be found for every heights. Such [[Edge templates with one stone|templates]] have been found for sizes up to 5 but none above. Answering with "No" to the former question answers the latter. | + | More generally, it is still unknown whether one stone edge templates (which carrier has not cell upper than the stone) can be found for every heights. Such [[Edge templates with one stone|templates]] have been found for sizes up to 5 but none above. Answering with "No" to the former question answers the latter. |
== Description == | == Description == | ||
Revision as of 19:17, 10 January 2009
As of January 2009 the following problem, initially stated by javerberg and wccanard in the LG forum, is still open:
Is there any one stone sixth row template ?
More generally, it is still unknown whether one stone edge templates (which carrier has not cell upper than the stone) 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.
Contents
Description
Is there a number m such that the game on the board of width m designed as follows, with Blue's turn to play, 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 the n-th row template 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 template IV1a and edge template IV1b, 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!
We can dispose of 3 moves on the left (and, using mirror symmetry, the corresponding 3 moves on the right), as follows:
At this point, we can use the Parallel ladder trick as follows:
...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.
